Solution Manual For Use but dont count Fundamentals of Aerodynamics, 7th Edition John Anderson

Solution Manual For Use but dont count SM Fundamentals of Aerodynamics, 7th Edition John Anderson Chapters 1-15

Chapter 1 1.1

(a)

(b)



p 1.9  104   0.326 kg/m3 RT (287)(203)

T

p 1058   501 R R 1.23  103 (1716)





1.2

N    

TE LE

 pu cos   u sin   ds u

TE LE

 p cos    sin   ds

(1.7)

ds cos   dx ds cos   dy

Hence,

 pu  p  dx   LE  u    dy LE

N   

TE

N   

TE

TE

 p  p     p  p    dx    u    dy LE  u LE TE

Divide by q S  q c(1)

  N 1 TE  p  p   p  p   1 TE   u     u     dx   LE   dy LE qc c c  q q   q    q    cn 









1 c 1 TE cpe  cpu dx   cf  cfe dy  c 0 c LE n

This is Eq. (1.15).

A  

 p u sin  u cos  ds u

TE LE



TE LE

 p sin   cos  ds

(1.8)

 pu  p  dy   LE  u    dx LE

A  

TE

A  

TE

TE

 p  p    p  p    dy    u    dx LE  u 0 c

  A 1 TE  p  p   p  p   1 c     u  dy    u      dx q  c c LE  q   q   c 0  q q 

ca 









1 TE 1 c cpu  cp dy   cfu  cf dx  c LE c 0

This is Eq. (1.16).

MLE  

TE

 p u cos  u sin   x   p u sin   u cos  y  ds u

TE

  p cos   sin   x   p sin    cos   y  ds

LE 

 MLE  

 pu  p  x dx   LE  u    x dy LE

TE

 MLE   

LE 

TE

 pu  p  y dy   LE  u    y dx LE

TE

TE

 p  p    p  p   x dx    u    x dy LE  u LE

TE

TE

 pu  p    p  p  y dy   LE  u    y dx LE

TE

TE

Divide by q  c2 : M LE   1 TE  p  p   p  p    1 TE    2   u  x dx  2   u      x dy 2 qc c LE  q    q    c LE  q q  



  1 TE  p u  p   p  p   1 TE  u     y dy  2  LE   y dx 2  LE  q q q q c c            

cm c 



1 c  Cpu  Cp c2  0 

TE LE

 x dx    C  C  x dy TE

fu

LE

f

 C  C  y dy    C  C  y dx c

pu

p

0

fu

f

This is Eq. (1.17).

1.3

MLE     p  p u  (dx) (1) x   p  p u   x dx c

c

0

0

MLE    p  p u 

c2 2

N    p  pu  dx   p  p u  c c

0

 c2   p  p     u 2 MLE  x cp    N  p  pu  c

x cp  c/2 1.4

For a flat plate,   0 in Eqs. (1.7)  (1.11). Hence,

N    p  p u  dx    2  104 (x  1) 2  1.19  105  dx c

1

0

0

4 x

1

1  N  2  10   x  x   1.19  105 x   1.12  105 N 0 3 0 3

2

A      u  dx   c

1

0

0

 731 x

0.2



 288 x 0.2 dx

1

A  1274 x 0.8   1274 N 0

L  N cos  A sin  1.12  105 cos 10  1274 sin 10  1.105  105 N

D  N sin  A cos  1.12  105 sin 10  1274 cos  2.07  10 N 4

MLE    p u  p  x dx    2  104 (x  1) 2  1.19  105  x dx c

1

0

0

4 x

1

2x 3 x 2  5 2 1 4  2  10      0.595  10 x  0  5.78  10 Nm 4 3 2  0 4

Mc/ 4  MLE  L(c/4)  5.78  104  1.105  105 (0.25)  3.02  104 N/m



5.78  10 M x cp   LE   N 1.12  105

4

  0.516 m

1.5

c  cn cos  ca sin

 (1.2) cos 12  (0.3) sin   1.18 cd  cn sin   ca cos  (1.2) sin 12  (0.3) cos  0.279

1.6

cn  c cos  cd sin

Also, using the more accurate Nʹ rather than Lʹ in Eq. (1.22), we have

x cp 

 cm  c Mc/4 c    c  c/ 4  4 N 4  cn 

Hence:

  

cn

x cp /c

2.0 0 2.0 4.0 6.0 8.0 10.0 12.0 14.0

0.0498 0.25 0.44 0.639 0.846 1.07 1.243 1.402 1.52

1.09 0.41 0.336 0.306 0.293 0.284 0.277 0.271 0.266

Note that x cp moves forward as  is increased, and that it closely approaches the quarterchord point in the range of  of 10° to 14°. At higher angles-of-attack, beyond the stall (  > 16°), x cp will reverse its movement and move rearward as  continues to increase. Compare the above variation with the center-of-pressure measurements of the Wright Brothers on one of their airfoils, shown in Fig. 1.28.

1.7

K = 3 (mass, length, and time)

f1  D,  , V , c, g   0

Hence N  5

We can write this expression in terms of N − K = 5 − 3 = 2 dimensionless Pi products:

f 2  Π1 , Π 2  where

Π1  f3   , V , c, D  Π 2  f 4   , V , c, g  Let

Π1   a V b cd D



1  m 3

  t  m t   0 a

1

b

2

c

mass: a  1  0

a  1

length: 3a  b  c  1  0

b  2

b  2  0

c  2

time:

Hence:

Let

Π1 

D D , or Π1  2 2 1  V c  V 2c2 2

Π1 

D q  c2

Π 2  2 V cb g d



1  m 3

  t   t  0 a

1

b

2

d

mass: a  0

a0

length: 3a  1  b  d  0

d  1/2

time:

1  2d  0

b  1/2

Π2 

V cg

Hence,

Thus:  D V  f 2  Π1 , Π 2   f 2  , 0  q c2 cg   

or:

CD  f  Fr  1.8



Dw  f1  , V , c, a  , cp , cv



K  4 (mass, length, time, degrees)





f 2 Dw ,  , V , c, a  , cp , c v  0 Hence, N = 7. This can be written as a function of N − K = 7 − 4 = 3 pi products:

f3   Π1 , Π 2 , Π3   0 where:













Π1  f 4  , V , c, cp , D

Π2  f5  , V , c, cp , a  Π3  f6  , V , c, cp , c v

The Dimensions of cp and cv are





m t 2 ( ) energy (force)(distance)  cp     mass    mass    m 

cp  

2

1

t 2    where    degrees.

For П1:

1 V j ck cp n D  Π1

 m   t     t     m t   1 3

mass:

i

1

j

k

2

2

n

i 1  0

n

2

i  1

length: 3i  j  k  2n  1  0

n0

 j  2n  2  0

j  2

time:

degrees: n  0

Hence:

Π1 

D D , or Π1  2 2 qc  V c

For Π 2 :

Π2  i V c j cp k a  n

k  2



1  m 3 mass:

  t    i

1

i

2

t 2

   t  k

k

1

n

i0

i0

length: 3i  1  j  2k  n  0

k0

1  2k  n  0

n  1

time:

j0

degrees: k  0

Hence: Π2 

V a

For Π3:

Π3  i V j ck cp n cv



1  m 3 mass:

  t   t     t   i

1

j

k

2

2

n

0

n

2

i0

0

1

i0

length: 3i  j  k  2n  2  0

n  1

 j  2n  2  0

j0

time:

k0

degrees: n  1  0

Hence:

Π3 

cv . We can take the reciprocal, and still have a dimensionless product. cp

Π3 

cv  cp

Hence,

Thus,

 D V cp  f3  , ,   q S a  c v  or,

CD  f  M  ,   1.9

M1 V a V  1 2  1 M2 V2 a1 V2

T2 100 800  1 T1 200 200

Hence, the Mach numbers of the two flows are the same.

Re1  Vc     Vc  1 1 1 2 1 1 1 Re2  V2c2  1   V2c2

T2  1.23  100  1  800   0.354    T1  1.739  200  2  200

The Reynold‟s numbers are different. Hence, the two flows are not dynamically similar.

1.10 Denote free flight by subscript 1, and the wind tunnel by subscript 2. For the lift and drag coefficients to be the same in both cases, the flows must be dynamically similar. Hence M1  M2

and Re1  Re2

For Mach number: V1 V2  a1 a2

Since a  T, we have

V2 V 250  1   16.7 T2 T1 223 For Reynolds number:

1 V1c1  2 V2c  1 2

Assume, as before, that   T. Hence

2 V2c2 T2



1 V1c1 T1

(1)

or,

2 V2 T1



1 V1  c1 

  T1  c2 

(0.414)(250)  5    223 1

or,

2 V2 T2

 34.65

(2)

Finally, from the equation of state: 2T2 

p2 1.01  105   351.9 R 287

(3)

Eqs. (1)  (3) represent three equations for the three unknowns, 2, V2 , and T2. They are summarized below:

V2  16.7 T2

2 V2

(1)

 34.65

(2)

2T2  351.9

(3)

T2

From Eq. (3):

2  351.9/T2

(4)

Subst. (4) into (2):

351.9  V2     34.65 T2  T2  Subst. (1) into (5):

Hence,

(5)

351.9 (16.7)  34.65 T2

T2 

(351.9)(16.7)  169.6K (34.65)

From Eq. (1): V2  16.7 T2  16.7 169.6  217.5 From Eq. (3): 2 

1.11

m sec

351.9 351.9 kg   2.07 3 T2 169.6 m

p b  pa   g Δ h





 1.01  105  1.36  104 (9.8)(0.2) pb  7.43  104 N/m2

1.12

Weight = Buoyancy force + lift W 

L

B

B  (15, 000)

(1.1117)

(9.8)

volume (m3)

air density at 1000m (kg/m3)

acceleration of gravity (m/sec)

q 



1.634  105 N

1 1  V 2  (1.1117)(30)2  500 N/m 2 2 2

S  d2 /4  (14)2 /4  153.9 m2 L  qS CL  (500)(153.9)(0.05)  3487 N

Hence: W  1.634  105  3847  1.67  105 N

1.13 Let us use the formalism surrounding Eq. (1.16) in the text. In this case, cd = ca , and from Eq. (1.16), neglecting skin friction cd 





1 TE Cpu  Cp dy c  LE

(1)

From Eq. (1.13) in the text, Eq. (1) above can be written as

cd 





1 TE Cpu  Cp ( sin  ds) c  LE

(2)

Draw a picture: Following our sign convention, note that  is drawn counterclockwise in this sketch, hence it is a negative angle, .

From the geometry:   

Hence, sin ()   sin   sin (  )  cos  Substitute this into Eq. (2), noting also that ds  rd and the chord c is twice the radius, c  2r.

From Eq. (2),





cd 

1 TE Cpu  Cp cos  r d  2r  LE

cd 

1 TE Cpu  Cp cos  d  2  LE

cd 

1 TE 1 TE Cpu cos  d    Cp cos  d   2 LE 2 LE





(3)

Consider the limits of integration for the above integrals. The first integral is evaluated from the leading edge to the trailing edge along the upper surface. Hence,   0 at LE and π at TE.

The second integral is evaluated from the leading edge to the trailing edge along the bottom surface. Hence,   2 at LE and at the TE. Thus, Eq. (3) becomes cd 

1  1  Cpu cos  d    Cp cos  d   2 0 2 2

(4)

In Eq. (4), Cpu  2 cos 2 

for 0    /2

Cpu  0

for

Cp  2 cos2

for

Cp  0

for    

 2



3    2 2 3 2

Thus, Eq. (4) becomes

cd  

 /2 0

cos3  d   

3 / 2 2

cos3  d 





1  3 2 Since cos  d    sin   cos   2 , Eq. (5) becomes 3  /2

 1   cd   sin   cos 2   2    3 0





3 / 2

 1     sin   cos 2   2    3  2





1 1 cd    (1)(2)    (1)(2) 3  3 cd  4/3

15 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

1.14

Consider the arbitrary body sketched above. Consider also the vertical cylinder element inside the body which intercepts the surface area dA1 near the top of the body, and dA2 near the bottom of the body. The pressures on dA1 and dA2 are p1 and p2 respectively, and makes angles 1 and 2 respectively with respect to the vertical line through the middle of dA1 and dA2. The net pressure force in the y-direction on this cylinder is: dFy  p1 cos 1 dA1  p2 cos 2 dA2

(1)

Let dAy be the projection of dA1 and dA2 on a plane perpendicular to the y axis. dA y  cos 1 dA1  cos 2 dA2

Thus, Eq. (1) becomes

dFy   p 2  p1  dA y

(2)

From the hydrostatic equation

p2  p1  

h2 h1

gd y

(3)

Combining Eqs. (2) and (3),

dFy  

h2 h1

 g dy dA y

(4)

However, dy dAy = dV = element of volume of the body. Thus, the total force in the y direction, Fy, is given by Eq. (4) integrated over the volume of the body 16 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Force on body 1.15

 v  g dV



Fy

Weight of fluid displaced by body.

From Eq. (1.45)

CL 

L 2W 2(2950)   2 q S  V S (0.002377)V 2 (174)

CL 

14265 V 2

(1)

Also,

CD  0.025  0.054 CL2

(2)

Tabulate Eqs. (1) and (2) versus velocity.

V (ft/sec)

CL

CD

L CL  D CD

70 90 110 130 150 170 190 210 230 250

2.911 1.761 1.179 0.844 0.634 0.494 0.395 0.323 0.270 0.228

0.483 0.192 0.100 0.063 0.047 0.038 0.033 0.031 0.029 0.028

6.03 9.17 11.79 13.40 13.49 13.0 11.97 10.42 9.31 8.14

These results are plotted on the next page.

17 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Examining this graph, we note, for steady, level flight: 1. The lift coefficient decreases as V increases.

2. At lower velocity range, the drag coefficients decreases even faster than the lift coefficient with velocity. (Note that on the graph the scale for CD is one-tenth that for CL.) 3. As a result, the lift-to-drag ratio first increases, goes through a maximum, and then gradually decreases as velocity increases.

It can be shown that the maximum velocity for this airplane is about 265 ft/sec at sea level.

As seen in the graph, the maximum value of L/D occurs around V = 140 ft/sec, which is much lower than the maximum velocity. However, at higher velocity the value of L/D decreases only gradually as V increases. This has the practical implication that at higher speeds, even though the value of L/D is less than its maximum, it is still a reasonably high value. The range of the aircraft is proportional to L/D (see for example, Anderson, Aircraft Performance and Design, McGraw-Hill, 1999, or Anderson, Introduction to Flight, 4th ed., McGraw-Hill, 2000). To obtain maximum range, the airplane should fly at the velocity for maximum L/D, which for this case is 140 ft/sec. However, one reason to fly in an airplane is 18 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

to get from one place to another in a reasonably short time. By flying at the low velocity of V = 140 ft/sec, the flight time may be unacceptably long. By cruising at a higher speed, say 200 ft/sec, the flight time will be cut by 30%, with only an 18% decrease in L/D. 1.16

From Eq. (1.59), in the text,

 dV    dy  y  0

W    or,

 dV    w     dy  y  0 As given in Section 1.11, μ at standard sea level temperature is

  1.7894  105 kg/(m)(s) Thus,

 dV   282  w   1.576  107 sec 1   5 dy  1.7894  10   y 0

1.17

From Eq. (1.60)

 dT  q  w   k  dy  y  0 The heat transfer into the surface is 0.03 MW/m2, and from Section 1.11, the thermal conductivity at standard sea level temperature is k  2.53  102 J/(m)(s)(K)

19 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

qw  0.03 MW/m2  0.03  106 W/m 2  dT  0.03  106 K   1.186  106   2 m  dy  y  0 2.53  10

1.18 (a) At standard sea level, from Appendix E, ρ∞ = 0.002377 slug/(ft3). Also, from Section 1.11, μ∞ = 3.7373 ×10−7 slug/(ft)(s). Putting the velocity in consistent units, V  (200)

88  293.3 ft/sec 60

Thus, Re 

 V c (0.002377)(293.3)(14.25)   3.7373  107

Re  2.66  107

(b)

Re 

V c (0.002377)(1340)(21.5)   3.7373  107

Re  1.83  108

1.19 From Figure 1.65, we have the normal and tangential force coefficients at 3-degres angle of attack given as 0.546 and 0 respectively. From Eqs. (1.1) and (1.2), written in terms of force coefficients, and using Lilienthal‟s nomenclature: CL   cos    sin  CD   sin    cos 

The ratio of lift-to-drag is L CL  cos    sin    D CD  sin    cos 

20 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

L 0.546 cos 3  0   cot 3  19.08 D 0.546 sin 3  0

At α = 2°, from the Lilienthal Table, η = 0.489 and θ = 0.008 L 0.489 cos 2  0.008 sin 2 0.4887  2.79  104   D 0.489 sin 2  0.008 cos 2 0.0171  8  103 L 0.488   19.44 D 0.0251

At α = 1°, η = 0.434 and θ = 0.016 L 0.434 cos 1  0.016 sin 1  D 0.434 sin 1  0.016 cos 1

L 0.4339  2.792  104 0.4336    18.56 3 D 7.574  10  0.01599 0.023356 At α = 4°, η = 0.6 and θ = 0.007

L 0.6 cos 4  0.007 sin 4 0.5985  4.883  104   D 0.6 sin 4  0.007 cos 4 0.04185  6.9829  103 L 0.59899   17.18 D 0.03487

So we have the following tabulation: α-degrees

L/D

1°

18.56

2°

19.44

3°

19.08

4°

17.18

The Wright brothers chose a three-degree angle of attack for their design point because, from the Lilienthal Table, it corresponded very nearly to the maximum, L/D.

21 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

1.20

Consider the existence of a forward-facing aerodynamic force on an airfoil, as discussed in Section 1.16. Can a forward-facing axial force exist on a flat plate at an angle of attack in a flow? Thoroughly explain your answer.

Solution: The net aerodynamic force acting on a cambered airfoil is the combined effect of many small deflections as the air moves around the airfoil‟s surface. This means that in some places on the airfoil‟s surface the flow is being deflected at an angle that is greater than the nominal angle of attack (the angle of the airfoiled cord to the free stream) producing a larger aerodynamic force locally whereas at other places, the flow is being deflected less than the nominal angle of attack. Thus, in certain situations, the combined effect of these deflections causes the ratio of total lift to total drag to exceed the cotangent of the angle of attack leading to a forward-facing axial force. Such a force is not possible on a flat plate because the angle of attack is the same everywhere along its surface.

1.21

Derive P = ρRT from PV  nRuT where n is the number of moles.

Solution: Start with the „chemistry‟ version of the ideal gas law: pV  nRuT Here, p is the pressure, V is the volume, n is the number of moles, T is the temperature, and Ru is the universal gas constant. Now, rearrange: n p  RuT V Then, multiply and divide by the molecular weight M as follows: nM Ru p T V M Recognize that nM/V=ρ (the density) and R_u/M=R (the ideal gas constant) so p   RT

1.22

Calculate R for a 20% mixture of H2 in N2.

Solution: When one speaks about mixtures of gases, the units are assumed to be moles unless other instruction is provided. Thus, when the problem refers to a 20% mixture of H2 and N2 , it means that there are 0.2 moles of H2 and 0.8 moles of N2 for every mole of gas mixture. Since the gas constant is the universal gas constant divided by the molecular weight, our first task is to calculate the molecular weight of the mixture: M mix  0.2M H2  0.8M N2 kg  kg  kg   M mix  0.2  2   0.8  28   23.2 kmol  kmol   kmol  Once we have the molecular weight of the mixture, we can calculate the gas constant for the mixture: J 8314 Ru kmolK  358.4 J R  kg M mix kgK 23.2 kmol

22 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Finally, there are some situations where it may be more useful to describe composition on a mass basis. In these situations, one would add the words “on a mass basis” to the description of the mixture. In the above example, one would say: “a 20% mixture of H2 and N2 on a mass basis” 1.23

Calculate the lift force generated by a NACA 1408 airfoil with a chord of 1.5 meters and a span of 15 meters operating at 50 m/s and an angle of attack of 8 degrees in air at a temperature of 20C and a pressure of 1 atm. The kinematic viscosity of air at 20 C is 15.11x10-6 m2/s.

Solution: The first step is to recognize that you can calculate the lift a wing produces if you know the lift coefficient ( cl ), the dynamic pressure ( q ), and its area (S).

Rearranging the definition of cl gives: L  cl q S

Incorporating the definitions of dynamic pressure ( q 

U 2 2

) and area (span x

chord) gives:

  U2  L  cl      c   2  We calculate the air density using the ideal gas law  Mair =288.7 J/kgK  :

p 1atm    RT  J   288.7 kgK   273  20  K  

N m2  1.20 kg atm m3

101324

At this point we know everything except cl so now we go to the lift coefficient data for an 8 degree angle of attack. However, when we look at the data closely, we see that the lift coefficient can depend on the Reynolds number if it is high enough. So, we need to calculate the Reynolds number. In order to do that, we need to choose a length scale. The appropriate length scale here is the chord because whether or not transition occurs depends, among other things, on the length of the interface between the air and the airfoil‟s surface which, to first order, is the chord. So,  m 50 1.5m  U  c  s  Re    4.96 x106 2 m  15.11x106 s This Reynolds number puts us somewhere between the round and square symbols on the plot of cl vs angle of attack (α). However, you will notice that lines put through these symbols are virtually on top of each-other. This means that the value of cl can be read directly off the chart without any need to interpolate to account for the effect of Re. Doing so gives cl  0.95 . Substituting everything back into the equation for lift gives: 23 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

2

1.20 L  0.95

1.24

kg  m   50  m3  s  15m 1.5m   32.1kN  7, 210lbf 2

Starting with equations 1.1 and 1.2, use the definitions of the force coefficients to show that: CL  CN cos   CA sin  and CD  CN sin   C Acos  Solution: From equations 1.1 and 1.2:

L  N cos  Asin

(1.1)

D  N sin  Acos

(1.2)

Start by noting that the force coefficients are defined as follows L D N A CL  ; CD  ; CN  ; CA  q S q S q S q S where the dynamic pressure is given by: 1  U 2 2 and the area is: S  lc . Thus, our goal is to replace Dividing equations 1.1 and 1.2 by q S gives: q 

24 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

L N A  cos  sin (a) q S q S q S D N A  sin  cos (b) q S q S q S

Incorporating the definitions of the force coefficients into (a) and (b) leads to:

 a   CL  CN cos  C A sin  b   CD  CN sin  C A cos 1.25

Derive the expression for the moment coefficient about the leading edge (equation 1.17) from the expression for the moment about the leading edge (equation 1.11) Solution: From equation 1.11: ' M LE   [ pu cos   u sin  x   pu sin  u cos  yu ]dsu TE

LE

    pl cos   l sin  x   pl sin   l cos  yl  dsl LE TE

(a)

From geometry:

dx  ds cos dy  ds sin

(b) (c)

S  c 1 Substituting (b) and (c) into (a) gives: ' M LE   [ pu cosu  u sinu  x   pu sinu  u cosu  yu ] TE

LE



(d) dx cosu

dx   pl cosl )  l sinl  x   pl sinl  l cosl  yl  LE cos

TE

l

Rearranging gives: ' M LE   [ pu   u tanu  x   pu tanu  u  yu ] dx TE

LE

    pl   l tanl  x   pl tanl   l  yl  dx (e) LE The angles of the upper and lower surfaces are related to the slopes of the upper and lower surfaces. Thus we can use equations (b) and (c) to write: dy dy tanu   u ; tanl   l dx dx Substituting these into equation (e) gives: TE  dy  dy   ' M LE   [ pu  u u  x    pu u  u  yu ] dx LE dx  dx    TE

TE  dy   dy       pl  l l  x    pl l  l  yl  dx LE dx   dx   

25 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Rearranging gives: TE  dy   dy  ' M LE   [ pu  u u  x    pl  l l  x] dx LE dx   dx  

TE  dy dy        pu u  u  yu    pl l  l  yl  dx LE dx dx      The pressure coefficient is defined as: p  p Cp  q

(f)

Thus, pu  C p,u q  p

(g)

pl  C p,l q  p

(h)

Similarly, the friction coefficient is defined as:  cf  q So,

 u  c f ,u q  l  c f ,l q

Substituting (g), (h), (i), and (j) into (f) and dividing by q gives: ' TE  dy   dy  M LE   [ C p ,u  c f ,u u  x   C p ,l  c f ,l l  x] dx LE q dx   dx   TE  dy dy        C p ,u u  c f ,u  yu   C p ,l l  c f ,l  yl  dx LE dx dx      Note that the p terms cancel. Combining the pressure and stress terms, changing the limit from 0 to c (chord length) and diving by c2 gives: ' c  dy dy  M LE 1  c  2    Cpu  Cpl  x dx    cf u u  cf l l  x dx  2 0 q c c  0 dx dx   



c  dy dy 1  c   Cpu u  cfu  yu dx    Cpl l  cfl  yl dx  2  0 0 c   dx dx    

The left hand side is the moment coefficient so, cmLE 

c  dy dy  1  c Cpu  Cpl  x dx    cfu u  cf l l  x dx  2  0  0 c  dx dx   



c  dy dy 1  c   Cpu u  cfu  yu dx    Cpl l  cfl  yl dx  2  0 0 c   dx dx    

26 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

(i) (j)

1.26

You work for a company that is developing a new family of propeller for fixed wing UAVs. Your task is to design and test the propellers in a wind tunnel. The key performance parameters are the propeller‟s thrust T and its power consumption P. Assume that these will be determined by the free stream density  , velocity U  , propeller diameter D, and propeller speed ω, use dimensional analysis to derive the main non-dimensional parameters associated with the performance of the propeller.

How should you use these parameters in the conduct of your job?

Solution: Apply the Buckingham-Pi theorem: 1. List all parameters involved: ρ, U, D, ω, T, P. Thus, N=6.

2. Select a fundamental set of dimensions: m, , t Thus, k=3 3. List dimensions of all parameters in terms of the primary dimensions:  D T P  U 1 m m 2 3 t t2 t t3 4. Select k repeating parameters: Any can be chosen but experience tells us that choosing ρ, U, and D will lead to „familiar looking‟ dimensionless groups. Later, we will check ourselves by seeing what happens if we choose different repeating parameters.

5. Set up N-k dimensional equations. Use the 3 repeating parameters and each of the 3 remaining parameters in succession to determine the Π groups by requiring that each Π group be dimensionless. m

a

b

m   1 Π1   U D    3    c t   t m: a0 a0  D  : 3a  b  c  0   b  1 so, Π1  U t: b  1  0  c 1  a

b

c

a

b

m   m  Π2   U D T   3    c  2    t  t  m: a 1  0  a  1 T  : 3a  b  c  1  0   b  2 so, Π 2  U 2 D 2  t: b  2  0  c  2 a

b

c

2 m   m  Π3   U D T   3    c  3    t  t  m: a 1  0  a  1 P  : 3a  b  c  2  0   b  3 so, Π 3  U 3 D 2  c  2 t: b  3  0  The result is a dimensionless expression that relates the power consumption of the propeller to its speed, diameter, and thrust: a

a

b

b

c

27 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

 D  P T  f , 3 2 2 2  U D  U U D 

(a)

Equation (a) helps with your job because it tells you several important things. First, it tells you what is important in the relationship between power, speed and thrust so that you can devise a set of experiments that yields the most information for the fewest experimental runs. Second, it helps you understand how changing the design (propeller diameter and speed) will change its performance so that you don‟t have to do a whole new set of tests any time you make a design change.

Finally, let‟s check ourselves by trying a different set of repeating parameters: ρ, ω, D 5a. Set up the dimensional equations again: a

b

 m  1   Π 4   a b DcU   3    c     t   t  m: a0 a0  U  Same as Π1. : 3a  c  1  0   b  1 so, Π 4   D t: b  1  0   c  1 a

b

 m  1  m  Π5   a b DcT   3    c  2    t   t  m: a  1  0  a  1  T : 3a  c  1  0   b  2 so, Π 5   2 D 4 t: b  2  0  c   4  2  m  1 c  m  Π6    D P   3     3    t  t  m: a  1  0  a  1  P : 3a  c  2  0   b  3 so, Π 6   3 D 5 t: b  3  0  c   5  This gives the following dimensionless expression that relates the power consumption of the propeller to its speed, diameter, and thrust: a

a

b

b

c

 U  P T  g , 3 5 3 4   D   D  D 

(b)

Uh oh, (a) and (b) aren‟t the same! Was our previous analysis wrong? The answer is no because Π4, Π5, and Π are not „linearly independent‟ from Π1 , Π2 and Π3 ie. we can form Π4, Π5, and Π from Π1 , Π2 and Π3: Π4 

Π Π 1 , Π 5  22 , and Π 6  33 Π1 Π1 Π1

28 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

This is a fancy way of saying that both sets of parameters are equivalent on a „physical‟ basis but because they are different mathematically, different functions (f vs. g) describe the relationship between them. The second set of parameters is what is usually used to report propeller performance because they don‟t „blow up‟ when the forward speed U goes to zero. Finally, remember that the functions f and/or g are not known a-priori and are what we seek to find out from experiments.

1.27

Consider two airfoils tested in two different environments where Airfoil 2 is a three times scale replica of Airfoil 1.

Factors

Airfoil 1

Airfoil 2

ρ

V

1.28kg/m3 100m/s

1.01kg/ m3 300m/s

T

200 K

1800 K

a. Determine if these flows are dynamically similar assuming that both μ and a 1 2

are proportional to T . b. If the flows are similar, explain why. If the flows are not similar, what one parameter would you change in Test 2 to make them similar and what would its new value be? c. Now assume that Airfoil 2 is no longer a 3x scale replica of Airfoil 1. Airfoil 2 still has a chord that is 3x larger than that of Airfoil 1, but now Airfoil 1 is symmetric whereas Airfoil 2 has positive camber. Are these two flows 1

dynamically similar (assuming both μ and a are proportional to T 2 ) or could they be made similar by changing one experimental parameter as in b?

Solution: a)

The flows are dynamically similar if the bodies are geometrically similar and the values of the dimensionless parameters that describe the flow (in this case the Reynolds and Mach numbers) are the same. The problem states that Airfoil 2 is a 3x scale replica of Airfoil 1 so the two airfoils are geometrically similar by definition. Thus the first criteria is fulfilled. We will need to calculate the Mach and Reynolds numbers to check the second criteria. More specifically, we need to check that M 1 Re1  1 M 2 Re2

Starting with the Mach numbers:

29 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

V1 V1 m  100  T1 V1 M 1 a1 T2  1800 K s    *  1  V2 M 2 V2 V2 T1  300 m  200 K s  a2  T2

So, the values of the first dimensionless parameter – the Mach number – are the same. Now let‟s check the second dimensionless parameters (Reynolds numbers):

kg  m  1.28 3 100  1m   Re1 1V1c12 1V1c1 T2 1800 K m  s      0.07  1 kg  m Re2  2V2c2 1 2V2 c2 T1  200 K 1.01 3 1800   3m  m  s  The Reynolds numbers are not the same. Since dynamic similarity requires that the values of all dimensionless parameters be the same, the flows are not dynamically similar. b) It has been shown above that the Mach numbers are the same so we will focus on the Reynolds number. We will hold the temperature and velocity the same because we don‟t want to change the Mach number. The ratio of the chord lengths is fixed (one airfoil is 3x the size of the other) so we can‟t change the airfoil geometrically. Thus, the only other parameter left to change is the density. Since the Reynolds numbers must be identical, we write:

kg  m  1.28 3 100  1m  1800 K  Re1 Vc  m  s 1 1 1 1 2   m Re2 2V2c2 1   (  2 )  300   3m  200 K s  Solving for the density gives: 2  0.4267 kg / m3 . c) Since the airfoils are no longer geometrically similar, the flows cannot be dynamically similar.

1.28

Consider an airplane in level flight at 578 MPH at an altitude of 38,000 ft. It has a rectangular wing 25 ft long and 4 ft wide with a drag coefficient of 0.012. Find the weight of the airplane if its lift to drag ratio is 15.

Solution: Since we are given L/D and CD, we can solve for CL: CL  CD

L   0.012 15  0.18 D

In level flight, the lift force equals the weight:

L W Dividing by the dynamic pressure gives:

30 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

L W  q S q S

Inserting the definition of lift coefficient on the LHS and the definition of dynamic pressure on the RHS gives: CL 

W 1 2   V S 2

Rearranging gives: W

CL V 2 S 2

Appendix E tells us that the density at 38,000 ft is 6.463x10-4 slug/ft3. Substituting into the equation gives:  slug  ft .18  6.4629 x104 3   847.73   25 ft x 4 ft  2 ft   s  C  V S  W L     4,180 lbf 2 2 2

Note that we converted 578 MPH to 847.7 ft/s so that the set of units were consistent.

A good way to avoid errors with units is to carry them through the problem so you can see that they cancel in way that produces the correct unit for the quantity of interest – in this case the weight. A way to avoid confusion when dealing with mass units in the Imperial system (ie. slugs etc.) is to convert everything to SI units, do the calculation, and then convert back to Imperial units when reporting your final answer.

1.29

Derive an expression for dynamic pressure in terms of pressure and the Mach number.

Solution: Start with the definition of dynamic pressure: q 

1 2  u 2

(a)

From the definition of the Mach number: u  Ma

(b)

From the definition of the speed of sound:

a   RT

(c)

Substituting (c) into (b) and then into (a) gives: 1 q   RTM 2 2

From the ideal gas law: p   RT So, 31 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

(d)

1 q   pM 2 2

CHAPTER 2 2.1

However, the integral of the surface vector over a closed surface is zero, i.e., 

 d S  0 S

Hence, combining Eqs. (1) and (2), we have 

F0

2.2

32 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Denote the pressure distributions on the upper and lower walls by pu(x) and p (x). respectively.

The walls are close enough to the model such that pu and p are not necessarily equal to p.

Assume that faces ai and bh are far enough upstream and downstream of the model such that p  p

and v  0

and ai and bh.

Take the y-component of Eq. (2.66)  



L    ( V dS) v   (p dS)y abhi

S

 

The first integral = 0 over all surfaces, either because V ds  0 or because v  0. Hence b



h

L    (p dS)y     p u dx   p dx   a  i abhi

Minus sign because y-component is in downward Direction.

Note: In the above, the integrals over ia and bh cancel because p = p on both faces. Hence h

b

i

a

L   p dx   p u dx

2.3

 

 

2 2 dy v cy/ x  y y    2 2 dx u cx/ x  y x

dy dx  y x

n y  n x  c1  n (c2 x) y  c2 x

The streamlines are straight lines emanating from the origin. (This is the velocity field and streamline pattern for a source, to be discussed in Chapter 3.)

2.4

dy v x   dx u y

y dy   x dx

33 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

y2  x 2  const x 2  y2  const.

The streamlines are concentric with their centers at the origin. (This is the velocity field and streamline pattern for a vortex, to be discussed in Chapter 3.)

2.5 From inspection, since there is no radial component of velocity, the streamlines must be circular, with centers at the origin. To show this more precisely, u  V sin   cr v  V cos   cr

y  cy r

x  cx r

dy v x   dx u y

y2  x 2  const.

This is the equation of a circle with the center at the origin. (This velocity field corresponds to solid body rotation.)

2.6

dy v y   dx u x dy dx  y x

n y  x n x  c1 y  c2 /x

The streamlines are hyperbolas.

34 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

2.7

(a)

 1 D( v)  V  v Dt 

In polar coordinates:   V  1   r Vr  

r r

Transformation:

1 V r 

x  r cos  y  r sin  Vr  u cos   v sin  V  u sin   v cos 

35 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

  c cos  cr cos 2 r r

cx

u

x  y 

v

x  y 

2

2

cy

2

Vr 

2

  c sin  cr sin 2 r r

c c c cos 2  sin 2  r r r

c c V   cos  sin   cos  sin   0 r r 

V 

(b)

1 1  (0) (c)  0 r r r 

From Eq. (2.23)  V 1  Vr   V   V  ez      r r    r

  V  ez [0  0  0]  0

The flowfield is irrotational.

2.8

u

v

cy

x  y  2

2

cx

x  y  2

Vr 

2

  c sin  cr sin 2 r r

   c cos  cr cos 2 r r

c c cos  sin   cos sin   0 r r

c c c V   sin 2  cos 2   r r r 

(a)

V 

1 1 (0)   (c/r)  0  0  0 r r r 

36 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

(b)

  c 1  (0)     V  ez   (c/r)  2  r   r  r   c  c  ez  2   0 r2 r 



  V  0 except at the origin, where r = 0. The flowfield is singular at the origin.

2.9

Vr  0.

V  c r

  cr 1  (0)     V  ez   (c/r)   r r    r 



 e z (c  c  0)  2c e z

The vorticity is finite. The flow is not irrotational; it is rotational.

2.10

37 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Mass flow between streamlines      ρ V  n    ρ Vθ  r  ρ Vr (rθ)

Let cd approach ab dψ   ρ Vθ dr  ρ r Vr dθ

(1)

Also, since    (r, ), from calculus d     dr    d r 

(2)

Comparing Eqs. (1) and (2) ρ V 

 r

and  r Vr   

or:  Vr 

2.11

1  r 

 V  

 

u  cx 

 :   cxy  f (x) y

v  cy  

 :   cxy  f (y) x

(1) (2)

38 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Comparing Eqs. (1) and (2), f(x) and f(y) = constant   c x y  const.

u  cx 

(3)

 :   cx 2  f (y) x

(4)

 :   cy 2  f (x) y

(5)

v  cy 

Comparing Eqs. (4) and (5), f (y)  cy2 and f (x)  cx 2



  c x 2  y2



(6)

Differentiating Eq. (3) with respect to x, holding   const.

0  cx

dy  cy dx

or,

 dy    y/x    dx   const

(7)

Differentiating Eq. (6) with respect to x, holding   const.

0  2cx 2cy

dy dx

or,

 dy   x/y    dx   const

(8)

Comparing Eqs. (7) and (8), we see that 1  dy      dy   dx   const    dx   const

Hence, lines of constant  are perpendicular to lines of constant .

39 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

2.12. The geometry of the pipe is shown below.

As the flow goes through the U-shape bend and is turned, it exerts a net force R on the internal surface of the pipe. From the symmetric geometry, R is in the horizontal direction, as shown, acting to the right. The equal and opposite force, R, exerted by the pipe on the flow is the mechanism that reverses the flow velocity. The cross-sectional area of the pipe inlet is d 2 /4 where d is the inside pipe diameter. Hence, A  d2/4  (0.5)2/4  0.196m2. The mass flow entering the pipe is 

m  1 A V1  (1.23)(0.196)(100)  24.11 kg/sec.

Applying the momentum equation, Eq. (2.64) to this geometry, we obtain a result similar to Eq. (2.75), namely

R    ( V  dS) V

(1)

Where the pressure term in Eq. (2.75) is zero because the pressure at the inlet and exit are the same values. In Eq. (1), the product ( V  dS) is negative at the inlet (V and dS are in opposite directions), and is positive at the exit (V and dS) are in the same direction). The magnitude of  V  dS is simply the mass flow, m. Finally, at the inlet V1 is to the right, hence it is in the positive x-direction. At the exit, V2 is to the left, hence it is in the negative xdirection. Thus, V2  V1. With this, Eq. (1) is written as 





R  [ m V1  m V2 ]  m  V1  V2  



 m  V1   V1   m  2V1 

R  (24.11)(2)(100)  4822 N

40 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

2.13

From Example 2.1, we have  h 2  2 x  2 y/  u  V 1   cos e      

(2.35)

and v  V h

2  2 x  2y/  sin e  

(2.36)

Thus,  V h   2    2 x  2y/  u  V        cos e x     

(2.35a)

Integrating (2.35a) with respect to x, we have  V h   2   2x  1 2 y/  f (y)    sin  2 e          

  V x  

  V x 

V h  2x  2y/  f (y)  sin e   

(2.35b)

From (2.36)  2  2 x  2y/  v  V h  sin e y  

(2.36a)

Integrating (2.36a) with respect to y, we have 2x  2y/  2     sin  e   



  V h 



V h  2 x  2y/  sin  e   



1  f(x)   2 

  f(x)

 

 

(2.36b)

41 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Comparing (2.35b) and (2.36b), which represent the same function for  , we see in (2.36b) that f (x)  V x. So the velocity potential for the compressible subsonic flow over a wavy well is:   V x 

2.14

V h  2 x  2y/  sin e   

The equation of a streamline can be found from Eq. (2.118) dy v  dx u

For the flow over the wavy wall in Example 2.1, 2  2 x  2 y/ sin  e dy    dx  h 2  2 x  2 y/  V 1   cos e       V h

As y  , then e2y/  0. Thus, dy 0  0 dx V  0

The slope is zero. Hence, the streamline at y   is straight.

2.15

A pressure field is defined by P  x, y   x 2  y 2  2 . Here, x and y are in units of meters, and the pressure P has units of Pa. Calculate the resulting force that acts on the top of a circle of radius 2 meters centered at the origin.

Solution: -

Convert from Cartesian to polar coordinates: P  r ,   r  2 Since the surface lies on the x-y plane, the force will act in the z-direction only Since we are considering the pressure on the top surface, the force will be directed downward.

Now recall that force is obtained by integrating pressure over the area.

Also recall how to perform a surface integral in polar coordinates: dA  rdrd

42 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

F   ∬ PdA kˆ 2 2  r3 2 8       r  2  rdrd k  2   r 2  kˆ  2   4  kˆ 3  3 0 0 0 40   πN kˆ 3

2.16

2.17

A jet engine undergoing ground testing vents its exhaust through a smokestack.

Investigators set up two probes to measure the dispersion of the exhaust into the atmosphere. Probe A is set up at a fixed location near the exit of the smokestack.

Probe B is attached to a balloon that is carried along by the local air current which, for the purposes of this analysis, will assumed to be steady. Define what type of derivatives these probes represent and identify which probe corresponds to which type of derivative.

Solution: These probes represent the local and convective derivative respectively. The local  derivative is the time rate of change of a fluid property measured at a fixed point. t The convective derivative ( V ) is the time rate of change of a fluid property caused by the movement of the fluid. These two components make up the substantial derivative. Since Probe A is at a fixed location, it corresponds to the local derivative.

Since Probe B is moving with the fluid, it corresponds to the convective derivative.

If V   3x  2 y  i  4  x  y  2z  j   x  3z  k please: a. Compute the curl of V and explain what the curl signifies physically. b. Compute the divergence of V and explain what the divergence signifies physically.

Solution: a) The ‘curl’ of a vector is defined as the cross product of the gradient operator with the vector. One way to compute a cross product is to take the determinant:

43 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

iˆ  V 

ˆj

  x y 3 x  2 y 4  x  y  2z 

kˆ  z  x  3z 

 ˆ   y  x  3z   z 4  x  y  2z  i     ˆ   z  3 x  2 y   x  x  3z   j    ˆ   x 4  x  y  2z   y  3 x  2 y   k     V  8i  1 j  2k   V is a vector quantity whose magnitude is proportional to the angular rate of rotation of a fluid element (if you take a more advanced fluid mechanics class, you will show that it is exactly two times the ‘vorticity’) and points along the axis about which the fluid is rotating. If   V  0 , then the flow is termed ‘irrotational’ because the angular rotation rate is zero ie. there is no angular rotation. b) The divergence of a vector is defined as the dot product of the gradient operator with the vector:

 V

    3x  2 y   4  x  y  2z    x  3z   4 x y z

 V is a scalar quantity that represents the time rate of change of a volume element. If  V  0 , the flow is neither converging or ‘running away’ from itself. If the flow is also steady,  V  0 implies that the density is constant.

2.18

Write the expression for the substantial derivative and explain its physical significance.

Solution: The substantial derivative is given by: D        u   Dt t It is the sum of the time rate of change due to the passage of time (first term on RHS) and the time rate of change due to motion through the property field. As such, it represents the total ‘net’ rate of change due to the passage of time and fluid motion.

It may also be referred to as the ‘convective derivative’.

44 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

2.19

Show how the differential form of the x-component momentum equation (eq. 2.70a) can be expressed using the substantial derivative:   u  p     uV      f x   Fx viscous t x Next, show how your result above can be generalized for the full momentum conservation equation (in differential form):   V     VV   p   f  Fviscous t Solution: Use the chain rule to expand the left hand side of eq. 2.70a noting that u is a scalar that represents the x-component of the velocity and V is the velocity vector.  u p u   u  V   V u     f x   Fx viscous t t x Now, rearrange the LHS to give: p     u  u     V      V u      f x   Fx viscous x  t   t  Now, notice that the first term in brackets on the LHS is the continuity equation in the form of 2.52 and thus equals zero. Meanwhile the second term in brackets on the LHS is the substantial derivative of the velocity. Therefore: Du p      f x   Fx viscous a Dt x Now, how can we generalize this for momentum conservation in all three directions?

Replacing u with v and w for the y and z-directions respectively gives: Dv p      f y   Fy  (b) viscous Dt y Dw p      f z   Fz viscous (c) Dt z Now, note that: V  uiˆ  vjˆ  wkˆ f  f iˆ  f ˆj  f kˆ x

y

z

 F viscous   Fx viscous iˆ   Fz viscous ˆj   Fz viscous kˆ Summing equations (a), (b) and (c) gives: DV   p   f   F viscous Dt

45 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

2.20

Consider steady flow through the duct with circular cross section illustrated below.

Use mass conservation to find U 2 in terms of U1 assuming D2 / D1  2 and: a) The flow is steady, incompressible, and uniform across the entrance and exit of the duct. b) The flow is steady, incompressible, with a parabolic distribution across the duct exit. c) The flow is steady, compressible, and uniform across the entrance and exit of the duct with T2 / T1  5 and P2 / P1  2 Solution: Start by drawing a control volume. Next, write the integral form of mass conservation for the control volume:   dV   u dA  0 t V A Then, look at the problem statement which says that the flow is steady. If the flow is steady, the left hand integral is zero so this is the form of the mass conservation equation that we will work with here:

 u dA  0 A

Now, answer the questions. a) Since the velocities are constant along the entrance and exit, the surface integral simplifies to: 1U1 A1  2U 2 A2 because the velocity profile is uniform. Since the flow is steady and incompressible, the density is constant. Thus: U1 A1 U 2 A2 Since A  

D2 4

D  D2 U  1 U  2 2

1

2

4

4

2

D  2 U 2 U1  1   U1  2   4U1  D2 

b) The exit velocity distribution is parabolic so, the velocity now depends on the radius:  4  u2  r   U 2 1  2 r 2   D2  46 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Incorporating the velocity distribution into the integral form of mass conservation gives: D /2  D2 4   1U1 1    2U 2 1  2 r 2  2 rdr  0 4  D2  0

 1U1

D2 2

r D r   22U 2   2   0 4  2 D2  0 2 1

2

4

2

D  2 U 2  2U1  1   2U1  2   8U1  D2  So, the parabolic distribution doubles the peak velocity exiting the nozzle over what one would expect for a flat distribution. c) Since the flow is uniform, the integral form of mass conservation reduces as in part a) to 1U1 A1  2U 2 A2 However, we are also told that the flow is compressible which means that we cannot assume that the densities upstream and downstream of the duct are constant. Therefore we write: 2

 D  U 2  U1 1  1   2  D2  Using the ideal gas law to re-write the densities in terms of the quantities that are given (temperature and presssure) gives: 2

p T D  2 1 U 2 U1 1 2  1   U1    5  2   10U1 p2 T1  D2  2

2.21

RP-1 flows steadily through the 90 reducing elbow shown in the illustration. The absolute pressure at the inlet to the elbow is 107.3 kPa and the cross-sectional area is 0.01 m2. The cross-sectional area at the outlet is 0.0025 m2 and the velocity there is 16m/s. The elbow discharges to the atmosphere. Find the velocity of the propellant entering the elbow and the force required to hold the elbow in place. Assume that the outside pressure is 101kPa, the density of RP-1 is constant at 810 kg/m3, that the volume of the elbow is 1 l, and that gravity points in the -z direction. Neglect the weight of the elbow itself.

Solution: Problems like these where conditions are known at the entrance and exit are best analyzed by enclosing the system in a control volume and enforcing the integral forms of the conservation equations along the control volume’s boundaries. While there are many control volumes that could be used to solve the problem, all must be drawn 47 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

so that all information needed to evaluate the conservation integrals is known along the entire surface of the control volume. Below is a sketch of the control volume that we will use to solve this problem.

The illustration on the left is a cross-sectional view showing the flows of fluid in and out of the elbow along with the pressures and velocities along the various surfaces that make up the elbow. Flow enters the pipe at pressure p1 and at velocity u1 which is perpendicular to area A1 and exits the pipe at pressure p2 (equal to atmospheric pressure) and at velocity v2 which is perpendicular to the area A2. There is no fluid flow across any other boundaries. The only boundary along which the pressure is not equal to p2 (atmospheric pressure) is the flow entrance. We will solve this problem using the integral forms of the conservation laws. We do this by drawing a control volume around the elbow in such a way that all fluid properties needed to evaluate the integrals are known along the entire control surface. If the goal is to find a force, as it is here, then the force vector must touch the control volume. Note that any control volume can be chosen as long as the values of the fluid properties needed to evaluate the integrals are known along the entire control surface. One also tries to choose volumes with surfaces that make evaluating the integrals as easy as possible.

We start by using the integral form of mass conservation in steady flows to find the velocity of the flow entering the elbow. It is given by:

   u nˆ  dA  0

(a)

CS

It is helpful to break the control volume into a series of surfaces as illustrated below when evaluating (a).

First, recognize that:

   u nˆ  dA     u nˆ  dA    u nˆ  dA   u nˆ  dA CS

1a

1b

2

    u nˆ  dA     u nˆ  dA     u nˆ  dA 3a

3b

3c

    u nˆ  dA    u nˆ  dA 3d

4

Next, recognize that the only places that the velocity is non-zero are along surfaces 1a and 2. Thus only two terms on the RHS of equation (a) contribute to the integral:

   u nˆ  dA     u nˆ  dA   u nˆ  dA  0 CS

1a

2

Simplifying the integrals leads to:   u1 A1   u2 A2  0 Solving for u1 gives: 48 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

u1  u2

A2  m   0.0025m 2  m  16    4 2  A1  s   0.01m  s

Now we are ready to solve the momentum equation. Note, however, that it is a vector equation which means that we will have to keep track of the net force in each of the coordinate directions. The integral expression of momentum conservation in steady flows is given by: ˆ   gdV  F  u  u nˆ  dA    pndA CS

CS

V

where F is the force required to hold the control volume in place. Solving for F gives: ˆ   gdV F    u  u nˆ  dA   pndA CS

CS

V

The first integral on the RHS is called the ‘flux’ term because it arises from the passage of fluid through the control surface. The second term is called the ‘pressure’ term because it arises from pressure forces acting on the control surface. The third term is the body force term because it arises from body forces (ie. gravity) acting on the material enclosed in the control volume. We will evaluate each integral separately and place our results into a table. This will make it easier to keep track of things and to see the various contributors to the net force acting on the elbow. Beginning with the flux term, only two surfaces have non-zero velocities so only two contribute to the integral:

 u  u nˆ  dA   u  u nˆ  dA  u u nˆ  dA CS

1a

2

Evaluating the flux term gives: ∫ ⃗ ( ⃗ ̂)

⃗ ( ̇[ ⃗ ⃗ ]

)

⃗ (

)

)*

̂

(

̂)+

( ∫

⃗ (⃗

̂)

̂

̂

Evaluating the pressure term is more complex because there are more surfaces to contend with. Since the control volume is cylindrical, A1 is circular, A3 is annular, and A4 is circumferential. The subscripts a and b denote the front and back ends of the control volume. Note that A2 is the area through which the flow exits the elbow and that it is included in A4. Note also that the pressure in the fluid along the exit plane also equals atmospheric pressure. Breaking up the pressure integral into the contributions from each surface makes it easier to evaluate: ∫

̂

∫

̂

∫

̂

∫ ̂

∫

̂

∫

̂

49 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

While this is a lot of terms, a look back at the control volume shows that the pressure forces on surfaces 3a and 3b will cancel each-other and that the net pressure force acting on surface 4 is zero. This leaves us with: ∫

̂

∫

̂

∫

̂

which simplifies to: ∫

̂

̂

(

̂ ∫

)̂

(

̂

)(

) ̂

̂

The body force term is given by: ∫

̂)

(

)(

(

̂

∫

Putting it all together gives: ( ̂

)̂

)(

̂)

(

̂)

(

̂)

̂ ̂ ̂ This is the force exerted on the pipe elbow to hold it stationary. ⃗

50 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

2.22

The space shuttle main engine consumes approximately 140.3kg/s of liquid hydrogen and 374.2 kg/s of liquid oxygen at full power. The overall ratio of hydrogen to oxygen on a mole-basis is approximately 6:1 so the combustion products are a gaseous mixture of water and excess hydrogen. The latter is used to cool the engine. The nozzle expands the hot, high pressure, gaseous combustion products to produce thrust – a process we will learn a lot more about later. If the temperature and pressure at the nozzle’s exit are 12.7 kPa and 390K respectively, and the effective diameter of the nozzle exit is 2.16m, find the speed of the gas mixture at the nozzle exit.

Solution: Begin by making a simple sketch of the system that includes representations of what you are given in the problem and what you are trying to find. Here you are given the mass flow rates of H2 and O2 and want to find the exit velocity so we put those on our sketch (see Figure a). (a)

(b)

While you are making your sketch, ask yourself two questions: ‘what am I asked to find’ and ‘what physical principles could help me do it?’ We know that mass conservation involves velocity so perhaps that can help us. The integral form of mass conservation (Equation 2.48) is:   d V  ∬V dS  0 t  v S

While the principle of mass conservation applies to any volume, we must define the volume (V) and its associated surface (S) so that it passes through places where we know enough about the flow to evaluate the integrals. We choose the dashed rectangle in Figure 2.20 because it is simple and we know the values of V S everywhere along it except along the exit plane (which is what we are trying to find).

Now we can evaluate the integrals. The first integral is zero because the flow is steady. The second integral is evaluated by recognizing that it can be separated into pieces associated with different parts of the control surface. Also important to recognize is that the surface area vector dS can be written as the product of a unit normal vector to the surface and the scalar area dS ie. dS  nˆ dS . Thus, the second integral becomes:

∬V dS  ∬ V nˆ  dS  ∬ V nˆ  dS  ∬ V nˆ  dS  ∬ V nˆ  dS S

1

2

3

4

Now, let’s think about what each integral means physically. Only O2 and H2 pass through surface 1 so the integral over surface 1 can be written in terms of the velocities of H2 and O2 as well as the areas through which each flow:

51 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

∬ V nˆ  dS  ∬ V H2

1

H2







nˆH 2 dS  ∬O2 VO2 nˆO1 dS

AH 2

AO 2

Expressing the velocities as vectors leads to VH2  VH2 j , VO2  VO2 j , nˆH2  j , and

nˆO2  j where n̂ is the unit normal to the surface and j is a unit vector pointing in the +y direction. The dot products evaluate to scalars as follows: VH2 nˆH2  VH2 and , VO2 nˆO2  VO2 leading to

∬ V H2

H2











nˆH 2 dS  ∬ H 2 VH 2 dS  mH 2

AH 2

AH 2

∬ V O2

O2



nˆO2 dS  ∬O2 VO2 dS  mO2

AH 2

AH 2

and finally:

∬ V nˆ  dS    m  m  H2

O2

1

Recall that the minus sign arises because the flows of H2 and O2 are into the control volume.

Next, let’s consider surface 3. Combustion products (a mixture of H2O and unused H2) flow through it along its entire length. Following the same line of reasoning as above leads to:

∬ V nˆ  dS  ∬ V exit

3

exit

nˆ3  dS  exitVexit Aexit

3

where exit is the density of the H2O-H2 mixture passing across surface 3 and Vexit is the average velocity of this mixture across the surface. This is what we are asked to find. Note that the value of the integral ( exitVexit Aexit ) is positive because it is leaving the control volume.

Finally, let’s consider surfaces 2 and 4. Since nothing flows across them, V2  V4  0 ,

 V nˆ   0 , and thus the integrals of  V nˆ  along surfaces 2 and 4 must be zero.

Having evaluated all of the integrals, we substitute back into the integral expression of mass conservation giving:   mH  mO   exitVexit Aexit  0 2

2

Solving for the exit velocity gives: Vexit 

mH 2  mO2

exit Aexit

However, we aren’t finished because we don’t know exit . Finding this takes a little more work and physical reasoning. The density of the exhaust gas can be calculated using the ideal gas law: pexit exit  RexitTexit 52 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

We know everything in this expression except the gas constant. Recall that the gas constant is defined as the universal gas constant divided by the molecular weight.

Since the gas exiting the nozzle is a mixture of H2 and O2, we perform a massweighted average to find the average gas constant. This gives: R R Rexit  YH 2O u  YH 2 u WH 2O WH 2 where YH 2O and YH2 , and WH2O and WH 2 are the mass fractions and molecular weights of water and hydrogen respectively in the exhaust gas. The mass fractions are determined from the stoichiometry of the reaction: 6 moles of fuel (H 2) for every 1 mole of oxidizer (O2). This leads to the following overall balanced chemical reaction occurring in the engine: 6 H 2  O2  2H 2O  4H 2 So, kg   2 18  2WH 2O kmol   YH 2O    0.82 kg   kg  2WH 2O  4WH 2  2 18   4 2   kmol   kmol   kg  4 2  2WH 2  kmol  YH 2    0.18 kg   kg  2WH 2O  4WH 2  2 18   4 2   kmol   kmol  The average gas constant of the gas exiting the nozzle is: kJ  kJ     8.314 kmolK   8.314 kmolK  kJ Rexit  0.82   0.18   1.127   kmolK  18 kg   2 kg  kmol  kmol    The density of the gas flow at the exit is: pexit 12.6kPa kg exit    0.019 3 kJ  RexitTexit  m 1.127   590 K  kmolK   The effective area of the exit is:

 2.16m   3.66m2 D2 Aexit   e   4 4 Finally, we can calculate the exit velocity: kg kg 140.3  374.2 mH 2  mO2 s s  7389 m Vexit   kg  exit Aexit s  2  0.019 3  3.66m m   The speed of sound at the exit is about 938 m/s so this corresponds to a Mach number of about 7.9! However, the actual Mach number will be substantially lower because of frictional and thermal losses in the nozzle.

53 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. 2





CHAPTER 3 4.

Consider steady, inviscid flow. x-momentum:

u

u u u p  v  w  x y z x

(1)

y-momentum:

u

v v v p  v  w  x y z y

(2)

z-momentum:

u

w w w p  v  w  x y z z

(3)

Multiply (1), (2), and (3) by dx, dy, and dz respectively: u

u u u 1 p dx  v dx  w dx   dx x y z  x

(4)

u

v v v 1 p dy  v dy  w dy   dy x y z  y

(5)

u

w w w 1 p dz  v dz  w dz   dz x y z  y

(6)

Add (4) + (5) + (6):

u v w  v w  u u dx  dy  dz   v  dx  dy  dz x x  y  y  x  y

v w  1 p p p  u  w dx  dy  dz     dx  dy  dz z z   x y  z  z

(7)

For irrotational flow (see Eq. (2.119)):  V  0 Hence: w v u w v u  ;  ;  y z z x x y

(8)

Subt. Eqs. (8) into (7):

54 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

u v u u  v v  u dx  dy  dz   v  dx  dy  dz y z  y  z  x x

w w  1 p p p  w  w dx  dy  dz     dx  dy  dz z z   z y  z   z u du  v dv  w dw  



1





dp

 

1 1 1 d u 2  v 2  w 2  d V 2  VdV   dp 2 2 

dp  V dV which integrates to

p

1  V 2  const.

2

For incompressible flow.

5.

V1 

(2)  p1  p 2   A  2    1   1  A 2  

p1  2116 lb/ft 2 ; p2  2100 lb/ft 2, A2 /A1  0.8

V1 

2(2116  2100)  154.7 ft/sec  1  2  (0.002377)    1  0.8  

6.

 A  2  p1  p2   V  1   1  12 (1.23)(90)2 (1/0.85)2  1  1913 N/m2  A 2  

7.

V2 

1 2

2 1

2wh 2  1   A 2 /A1    

55 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

m   w  m g  1.36 104  9.8 2   1.33 105 N/m 2 sec  





h  10 cm  0.1 m;   1.23 kg/m3,

V2 

8.





2 1.33 105 (0.1)   1 2  (1.23) 1       12  



A2 1  A1 12

 147 m/sec



p1  p 2  w h  1.33 105 (0.1)  1.33 104 N/m 2 p2  p1  1.33 104  1.01105  1.33 104  8.77 104 N/m2

p0  p 2 

1 1  V22  8.77 104  (1.23)(147) 2  1.01105 N/m 2 2 2

Note: It makes sense that the total pressure in the test section would equal one atmosphere, because the flow in the tunnel is drawn directly from the open ambient surroundings, and for an inviscid flow, we have no losses between the inlet and the test section.

9.

p0  p 

V 

1  V2 2

2  p0  p 

 2

10.

11.



2(1.07  1.01) 105 m  98.8 1.23 sec

 V   130  Cp  1     1    0.73  98.8   V  2





V  V i

V  u  constant 0 0 0  u v w V     0 x y z

It is a physically possible incompressible flow.

56 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

V 







i

j

k

 x

 y

 z

u

v

0

0 0   u u   i (0  0)  j  0    k 0 x  y    



0 w



V  0

12.

The flow is irrotational.

For a source flow, 



V  Vr e r 

  er 2 r

In polar coordinates: 

1 1  V  r Vr   r r r 



1   Λ  1  (0) r  r  r  2 r  r 

V  V 

0 1  Λ V   0 0 r  r  2  

Hence, the flow is a physical possible incompressible flow, except at the origin where r  0. What happens at the origin?

Visualize a cylinder of radius r wrapped around the line source per unit depth perpendicular to the page. The volume flow across this cylindrical surface is

36 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.





 V  dS

(1)

S

Since we are considering a unit depth, then we have the volume flow per unit depth. This is precisely the definition of source strength, Λ. Hence, from (1), 



Λ  constant   V  dS

(2)

S

From the divergence theorem: 





 V  dS   (  V) dV s

(3)

v

Combining Eqs. (2) and (3) 

 (  V) dV  Λ  constant

(4)

v

Shrink the volume to an infinitesimal value, V , around the origin. Eq. (4) becomes 

(  V) V  Λ Taking the limit as V  0 

(  V)  lim



Λ  . V

Hence   V   at origin 

To show that the flow is irrotational, calculate   V.



V 



1 r



  V  r e 













er

re

ez

er

re

ez

 r

 

 z

 r

 

 z

Vr

rV

Vz

Λ 2 r

0

0



1 r

0 0 0 0   0 Λ / 2 r   0 Λ / 2 r       ez   0 z     r  r

15 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Hence, 

  V  0 everywhere.

13.

  V ; x

  V x;

 2 0 y 2

  2  0; 0 y y 2

Hence, Laplaces equation:  2  2   0  0  0 is identically satisfied. x 2 y 2

Similarly, for   V y;

  2  0, 0 x x 2

  2  V, 0 y y 2

Hence, Laplaces equation:  2ψ  2ψ   0  0  0 is identically satisfied. x 2 y 2

14.



Λ 2

n r;

 Λ 1  2 Λ 1  ,  r 2 r r 2 2 r 2   2  0, 0   2

Hence, Laplace‟s equation

0 1     1   1      00 r  r r  r  r 2  2 r r    is identically satisfied.

2

16 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

 

Λ  ; 2

  0. r

 2 0 r 2

 Λ  2  , 0  2  2

Hence, Laplaces equation 1     1  2 1  1  (0)  2 (0)  0 r  2 2 r r  r  r  r r r

is identically satisfied.

15.

The stagnation point is a distance Λ/2V upstream of the source. Hence, Λ  1, or 2 V

Λ  2V

The shape of the body is given by

  V r sin  

Λ Λ  2 2

or, r sin  

Λ Λ  2 V 2V

r sin  

2 V 2 V  2 V 2V

or,

or, r sin      r

Equation of the semi-infinite body.

  sin 

17 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

(rad)

r

x  r cos 

y  r sin 

π

1

−1

0

3

1.0033

−0.990

0.1416

2.8

1.02

−0.961

0.3416

2.5

1.072

−0.859

0.6416

2.0

1.255

−0.522

1.142

π/2

1.57

0

1.57

1.3

1.91

0.511

1.84

1.0

2.54

1.372

2.14

0.75

3.509

2.57

2.39

0.5

5.51

4.84

2.64

Cartesian Coordinates of Body To plot the pressure coefficient:

Vr  V cos  

2πV V Λ  V cos    V cos    2 r 2 r r

V  V sin  or, Vr 1  cos   V r V   sin  V 2

2

2

 V   Vr   V  2 1 2 1 2        cos   cos   2  sin   1  cos   2 r r r r  V   V   V 

18 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

2

 V  2 1 Cp  1      cos   2 r r  V  (rad)

r

Cp

π

1

1.0

3

1.00

0.98

2.8

1.02

0.886

2.5

1.072

0.624

2.0

1.255

0.0283

π/2

1.57

−0.4057

1.3

1.91

−0.554

1.0

2.54

−0.580

0.75

3.509

−0.4982

0.5

5.51

−0.352

19 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

16.

20 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

At point A: Velocity due to freestream  V Velocity due to source 

Λ 2 (r  b)

(note that it is in the negative x-direction) Velocity due to sink 

( Λ) 2 (r  b)

(note that it is in the positive x-direction) Total velocity at Point A: VA  V 

Λ 1 Λ 1  2π (r  b) 2π (r  b)

From point A to be a stagnation point, VA  0.

0  V 

Λ  1 1    2  (r  b) (r  b) 

0  V 

Λ  r  b  (r  b)  Λ ( 2b)  V    2  (r  b)(r  b)  2 r 2  b 2



 2Λ (2b)  Λb 

V r 2  b 2 

17.

r2 

Λb  b2  V

r

Λb  b2  V

Vr 

 1   r r 

For a doublet:   

(1)

k sin  2 r

  cos    2 r

(2)

21 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Substitute (2) into (1)  1   cos    cos     r r  2 r  2 r 2

Integrating with respect to r    1     cos      2  r 

or,



18.

 cos  2 r

 R2     V r sin   1  2   r  Vr 

 R2  1    V cos   1  2  r   r 

V  

 R2     1  2  V sin  r r   2

2

 R2   R2  V  V  V  1  2  V2 cos 2   1  2  V sin 2  r  r    2

2 r

2 

2

2

 R2   R2  V2 2 Cp  1  2  1  1  2  cos θ  1  2  sin 2  V r  r    At the surface, r  R

Cp  1  4 sin 2 

19.

From Eq. (3.93):

Vr  R 2   1   cos  V  r 2 

22 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

 R2  V   1  2  sin  V  r 

From Eq. (3.94):

At any given point (r, ), Vr and V are both directly proportional to V . Hence, the 

direction of the resultant, V, is the same, no matter what the value of V may be. Thus, the shape of the streamlines remains the same.

20.

From Eq. (3.119):

Vr  R 2   1   cos V  r 2 

From Eq. (3.94):

 R2  V Γ   1  2  sin   V 2 rV  r 

Note that V /V is itself a function of V via the second term. Hence, as V changes, the direction of the resultant velocity at a given point will also change. The shape of the streamlines changes when V changes.

21.

L   V Γ Γ

L 6   0.163 m2 /sec  V (1.23)(30)

23 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

22.

At standard sea level conditions,

  0.002377

slug ft 3

   3.737  10 7

slug (ft)(sec)

Also: ft  88  V  120 mph  120   ft/sec  176 sec  60 

q 

1 1  V 2  (0.002377)(176) 2  36.8 lb/ft 2 2 2

For the struts: D  2 in  0.167 ft.

Re 

 VD (0.002377)(187.7)(0.167)   199,382  3.737 107

From Fig. 3.39, C D =1. The total frontal surface area of the struts is (25)(0.167)  4.175 ft 2 .

Hence, Drag due to struts:

DS  q  S CD  (36.8)(4.175)(1)  153 lb For the bracing wires: D 

3 in  0.0078 ft 32

 0.0078  Re  199382    9312  0.167 

From Fig. 3.39, CD  1. The total frontal surface area of the wires is (80)(0.0078)  0.624 ft 2 .

Hence, Drag due to wires:

Dw  q S CD  (36.8)(0.624)(1)  23 lb Total drag due to struts and wires  DS  D W  153  23  176

24 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

The total zero-lift drag for the airplane is (including struts and wires) CD0  q S CD0  (36.8)(230)(0.036)  304.8

176  0.58 of the total 304.8 drag – i.e., 58 percent of the total drag. This clearly points out the drag reduction that was achieved in the early 1930‟s when airplane designers started using internally braced wings with one or more central spars, thus eliminating struts and wires completely.

Note that, for this example, the drag due to the struts and wires is

23.

The flow over the airfoil in Figure 3.37 can be syntheized by a proper distribution of singularities, i.e., point sources and vortices. The strength of the vortices, added together, gives the total circulation, Γ , around the airfoil. This value of Γ is the same along all closed curves around the airfoil, even if the closed curve is drawn a very large distance away from the airfoil. In this case, the airfoil becomes a speck on the page, and the distributed point vortices appear as one stronger point vortex with strength Γ .

This is exactly equivalent to the single point vortex in Figure 3.27 for the circular cylinder, and the lift on the airfoil where the circulation is taken as the total Γ is the same as for a circular cylinder, namely Eq. (3.140),

L   V Γ

25 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

24.

From Eq. (3.92), the stream function for the flow over a circular cylinder is:

 R2     V r sin θ  1  2   r  At    /2,

 R2    V r 1  2  r   Streamline  2 goes through point (1.2R,  /2).

 R2   2  V (1.2R) 1  2  (1.2R)   1   1.2R V    R V  1.2 

 (1.2  0.833) R V  0.3667 R V Hence, the volume flow rate between  2 and  1 is

 2  1  0.3667 R V  0

 2  1  0.3667 R V The volume flow rate between  3 and  1 is twice this amount:

 3  1   3  0  2  0.3667 R V  Thus,

 3  0.7333 R V.

Since the general expression for the stream function is given by Eq. (3.92), the coordinate directly above the top of the cylinder through which  3 flows is obtained from

 R2  0.7333 R V  V r 1  2   r 

26 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

 R2  0.7333 R  r 1  2   r  r  R2  r R 0.7333  1  2    R r  R r Multiply both sides by r/R: 2

r r 0.7333     1 R R

or: 2

r r    0.7333  1  0 R R

From the quadratic formula: r 0.7333  0.5377  4  R 2

Only the plus sign makes physical sense here. Thus, r 0.7333  4.5377   1.4317 R 2

The point directly above the top of the cylinder through which  3 flows is (1.4317R,p /2).

Note that the spacing between  3 and  2 over the top of the cylinder is

(1.4317R - 1.2R )= 0.2317R, whereas the spacing between  2 and  1 is smaller, (1.2R - R )= 0.2R .Therefore, for equal volume flow between consecutive streamlines, the streamlines become farther apart the farther away they are from the surface of the cylinder.

25.

From Section 3.13, the maximum velocity in the flowfield over a circular cylinder occurs at the top and bottom of the cylinder, and is equal to 2 V .

2 V  250 mph V  125 mph

27 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

26.

In Section 3.6, the condition for incompressible flow is  V  0. If this condition does not hold for the flowfield specified in Example 2.1, then that flowfield is not incompressible, i.e., it is a compressible flow.

Indeed, in the solution provided in Equation 2.1, the expression for  V is obtained as

 1  2   2x  2 y/   V      V h    sin e       2

There is no way that this expression is zero for all values of x and y. Hence the flow is not incompressible, i.e., it is compressible.

27.

Consider the velocity given by: V   9 x 2  5 y  iˆ   5 x  ˆj  kˆ Does a velocity potential exist and if so, what is it?

Solution: The velocity potential  is defined as follows:   V where V is the velocity vector. From the definition of the gradient:  

 ˆ  ˆ  ˆ i j k x y z

Therefore:     x  2   9x  5 y     5 x  y    1     z  This yields three differential equations each of which may be integrated.     9 x 2  5 y  dx  3x3  5 yx  C1  y, z 

    5 x  dy  5 yx  C2  x, z 

   1 dz  z  C3  x, y  Combining the results gives:

  3x3  5 xy  z  C  x, y, z  where C  x, y, z   C1  y, z   C2  x, z   C3  x, y  .

28 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

28.

Prove that flows that can be represented using a velocity potential are irrotational.

Solution: The velocity potential  is defined as follows:   V An irrotational flow is one in which V  0 . Therefore, our task is to whether the following is true:     0 ?

Working out the cross product gives:   2  2  ˆ   i  y  z  z  y ˆ ˆi ˆj   k   2   2  ˆ   j z  zx xz     2  2  ˆ z  xy  yx  k   In order for the cross product to be zero, all of its components must be zero. Is this the case? Yes because:  2  2  0 yz zy  2  2  V 

 x  x

 y  y

zx



xz

0

 2  2  0 xy yx

29.

Calculate the stream function and angular velocity associated with the following velocity field: V   y3iˆ  x4 ˆj Can this flow have a velocity potential?

Solution: The stream function  is a single function that can be used to represent the x and y components of velocity. It satisfies the following  (a) u y

 (b) x where u and v are the components of the velocity in the x and y directions respectively. We find the stream function by integrating equations (a) and (b) and applying boundary conditions. From (a): v

29 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

u

 1     y 3dy     y 4  c1  x  y 4

and from (b): v 

ψ 1    x 4 dx     x5  c2  y  x 5

Summing the particular solutions gives the full solution.

1  

1  

     y 4    x5  C  x, y  4 5 where

C  x, y   c1  x   c2  y  To calculate the angular velocity (ω), we recognize that the vorticity (ξ ) is two times the angular rotation rate and that the vorticity is the curl of the velocity field. This leads to:

2ω  ξ  2  V  where

 V 

iˆ

ˆj

kˆ

 x  y3

 y x4

  4 x 3  3 y 2 kˆ z 0





Substituting back into the expression for the angular rotation rate gives: ω  0iˆ  0 ˆj   4 x3  3 y 2  kˆ The fact that ω is not zero means that the flow is rotational. However, a condition for the velocity potential to exist is that the flow is irrotational. Therefore, this flow cannot have a velocity potential.

CHAPTER 4 q 

1 1  V2  (0.002377)(50) 2  2.97 lb/ft 2 2 2 c  0.64 and cmc/ 4  0.036 L  q  s c  (2.97)(2)(1)(0.64)  3.80 lb per unit span

Mc/ 4  q S c cm c/ 4  (2.97)(2)(1)(2)(0.36)  0.428 ft/ lb per unit span

30 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

q 

1 1  V2  (1.23)(50) 2  1538 N/m 2 2 2 c 

L 1353   0.44 q  S (1538)(2)

From Fig. 4.5,

  2 



Γ   V  ds c 

  DΓ DV     ds   V  ds Dt Dt c c



 D ds  dV Dt

Hence, the second term in Eq. (1) becomes    V2  V  d V  d    2   0 c c

From the momentum equation,

31 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.



DV 1   p (neglecting body forces) Dt 

Hence, the first term in Eq. (1) becomes 

 DV  1 dp  ds    p  ds   Dt    c c c

When   const, or   (p), then  c

dp



 0. Hence, from Eq. (3)



DV   Dt  ds  0 c

Substituting Eqs. (2) and (4) into (1), we obtain

DΓ  0 Note: See Karamcheti, Ideal-Fluid Aerodynamics, for more details (pp. 239–242).

Dt

c

MLE   V    ( ) d 0

  V    V



c c (1  cos )  ( ) sin  d 0 2 2





 c2 2  V  1  cos 2  d 0 4

c2    1     V        V2  c2 2 2 2 2 

 q  c 2

 2

This is Eq.  4.36 .

c  2   where  is in radians. Hence

32 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

 1.5  c  2    0.164  57.3  cm \ c  c /r  0.041

(a) For 0 

x  dz  x  0.4:    0.2  0.5   c  dx 1 c

For 0.4 

x  dz  x  1:    0.0888  0.2222   c  dx  2 c

Since x 

c (1  cos ), then 2

 dz     0.05  0.25 cos , for 0    1.3694  dx 1  dz     0.0223  0.1111 cos , for 1.3694      dx 2

L 0   

 dz

1

(cos  1) d   0 dx

1

1.3694

 0

(0.05  0.25 cos)(cos  1) d 

1



 1.3694

 (0.0223  0.1111 cos)(cos  1) d



1

1.3694

 0

 0.05  0.3 cos  0.25 cos  d 2



 0.0223  0.13334 cos  0.1111 cos  d  1.3694 1



2

1.3694

1   1    0.05  0.3 sin   0.25   sin2     2 4  0

33 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.





1   1  0.0223   0.1334 sin   0.111   sin 2      2 4   1.3694



1



[0.06847  0.2939  0.1712  0.0245] 

1







0.2281



1



[0.0701  0.1745]

[0.0305  0.1307  0.0761  0.0109]

 0.0726 rad  4.16

(b)

c  2      L  0  where  is in radians

c 

A1 

2



2 [3  (4.16)]  0.782 57.3

dz

cos d   0 dx 

2

1.3694

 0

(0.05  0.25 cos) cos d

 

2

1.3694

π 0



2

 1.3694

(0.0223  0.1111 cos) cos d

 0.05 cos  0.25 cos  d  2

 0.0223 cos  0.1111 cos  d  1.3694 2



2

1.3694

2  1    0.05 sin   0.25   sin2    2 4  0







2



2  1  (0.0233) sin   0.1111   sin2      2 4   1.3694

[0.04899  0.25 (0.6847  0.09800)  0.1745

34 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

 0.02185  0.1111 (0.6847  0.09800)]

A1  (0.2561) A2 



2



2

1.3694

2



 0.1630

dz

cos2 d   o dx

 0

(0.05  0.25 cos) cos 2d  

2



(0.0223  1.3694

 0.1111 cos) cos d 1.3694

2 1  sin  sin 3     (0.05) sin 2  0.25     2 6   0  2







2



2 1  sin sin 3   (0.0223) sin 2  0.1111      2 6   1.3694  2

[0.009800  0.25 (0.4899  0.1372)  0.004371  0.1111 (0.4899  0.1372)]

 (0.0436) cmc/ 4  x cp c



 4

2



 0.0277

 A 2  A1  

 4

(0.0277  0.1630)  0.1063

 1 1 π   (0.1630  0.0277)   0.386 1   A1  A 2    1  4 c   4  0.782

35 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Experiment (Ref. 11)

Theory

% Difference

L  0

−3.9

−4.16

6.25%

c

0.76

0.782

2.8%

c mc/ 4

−0.095

−0.1063

10.6%

c

MLE   V    ( ) d 0

MLE

cmc/4 



1  V2 c2 2



c 2   ( ) d 2 0 V c

(1)

c (1  cos) 2

d 

c sin  d 2

 (1  cos  )    ()  2 V  A 0   A n sin n  sin  n 1   With the above, Eq. (1) becomes











n 1

0

cmc /4    A0 1  cos 2  d    A n (1  cos) sin  sin n d 0

(2)

Note the following definite integrals: 

2



2



 0 cos  d  2 

 0 sin  d  2 

 0 cos sin  d  0 2

36 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.



 0 sin  sin n d  0

for n  2, 3, 





 0 cos sin  sin 2 d  4 

 0 cos sin  sin n d  0

for n  3, 4, 

Hence, Eq. (2) becomes:

     cm c    A0  A0  A1  A 2  2 2 4   cm c  



A2   A 0  A1   2 2 

The slope of the lift curve is a0 

0.65  (0.39)  0.104 per degree 4  (6)

The slope of the moment coefficient curve is m0 

0.037  (0.045)  8  104 per degree 4  (6)

From Eq. (4.71),

x ac  

mo 8  104  0.25    0.25  0.242 ao 0.104

L  1 2  V2 c c From Appendix D at 3 km,   0.90926 kg/m . From the data given in problem 10, at   4, c  0.65.

3

37 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Hence: L  1 2 (0.90926)(60)3 (2)(0.65)  2128 N per meter

From the Kutta-Joukowski theorem (Equation (3.140) in the text), L   V Γ

Thus,

m2 L 2128 Γ   39 sec  V (0.90926)(60)

The circulation induced in the flow by the rotating cylinder is equal to its circumference times the velocity at the surface, Γ  (2R) (R)  2R 2

or,



Γ 2 R 2

From Problem 2, Γ  39 m2 /sec, Thus, 

39  6.21 radians per second 2 (1)2



6.21 (60)  59.3 rpm 2

or,

38 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Note: This result compares closely with the experimental data in Hoerner, Fluid Dynamic Lift, 1975, p 21-4, Fig. 5. In this figure, c is plotted versus U/V , where U is the tangential velocity of the circumference of the cylinder. For this problem, U  R  (1)(6.21)  6.21 m/sec. Thus, U/V  0.1035. From Hoerner‟s Fig. 5, the corresponding experimental value of lift coefficient is about 0.65, which is the value of the lift coefficient of the cylinder in this problem, i.e, c  0.65 as stipulated in Problem 11.

(a)

c 

dc     L  0 d

 (0.1) [5  (3)]  0.8 (b) When the airfoil is turned upside-down, its zero-lift angle of attack becomes plus 3°. It produces negative (downward lift) at angles of attack smaller than 3°. The lift slope (based on thin airfoil theory) remains the same.

c 

dc   L  0  d

 0.1(5  3)  0.2 (c) In order to produce a lift coefficient equal to that for the right-side-up airfoil at 5° angle-of-attack, namely c  0.8 , the upside-down airfoil must be pitched to a higher angle of attack, determined by:

c 

dc   L  0  d



c  L 0 dc /d

or,

where  L  0 is plus 3 .

0.8    3  11 0.1

39 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

(a)

Thus,

Re 

 V c 

V 

 Re  c

At 18,000 ft, from Appendix E,   1.3553  103 slug/ft 3. From Section 1.11,

  3.7373  107 slug/(ft)(s). Assuming that  varies as the square root of the temperature, and noting that at standard sea level T = 518.69°R, and at 18,000 ft.

T = 454.44°R, we have at an altitude of 18,000 ft.



  3.7373  10

7



1/ 2

 454.55     518.69 

  3.5  107 slug/(ft)(s)

Hence,

3.5  10 9  10   279 ft/sec V  1.3553  10  (8.33) 7



6

3

(b) Modeling the airfoil with a flat plate for the estimation of skin-friction drag, we have from Eq. (4.88), Cf 

0.074 0.074 0.074    0.003 1/5 0.2 24.6 Re 9  106





This is for one side of the plate. Taking into account both the top and bottom surfaces,

Cf  2 (0.003)  0.006

40 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

The total profile drag coefficient for the airfoil section is given experimentally as 0.006. This implies that almost all the drag is due skin friction drag, and that pressure drag due to flow separation is negligible for the given conditions.

Using Figure 4.44 as a guide, first calculate x cr  x1

Rexcr Rec

x 106   0.111  1 6 c 9  10

Since c  8.33 ft, then x1  0.111 (8.33)  0.9256 ft . Over the laminar portion of the flow, from Eq. (4.86)

1.328 1.328    1.328  103  Cf,1 lam  1.328 3 6 10 Re x1

10

For a turbulent flow over the length x1, 0.074 0.074 0.074   1.2   Cf,1 turb  0.074 0.2 15.85 Re (10)0.2 10

 4.669  103

Referring to Eq. (4.92), the net skin friction coefficient over one side of the plate is,

Cf 

x1 x Cf,1    Cf, c   1  Cf,1   lam turb turb c c







Cf  (0.111) 1.328  103  0.003  (0.111) 4.669  10 3



 1.474  104  0.003  5.183  10 4  0.00263

41 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Taking into account both sides of the plate

Cf  2(0.00263)  0.00526 The percentage of profile drag coefficient due to pressure drag is

(0.006  0.00526)100  12.3% 0.006

CHAPTER 5

d  r  (R d ) e

where e is a unit vector perpendicular to the plane of the loop, directed into the page.

V=

 d r  2 R  R d  e     (2 R) e  e   3 2 0 4 r3 4 2R R 4 R

\

Since d and r are always perpendicular (by inspection of the figure),

dV 

 d r  d  3 4 r 4 r 2

By symmetry, the resultant velocity due to the entire loop must be along the x-axis. Hence,

42 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

  2 R d  V   d V cos    2  cos   4 0 r   1 (2 R) cos  2 4 (A  R 2 )

63 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

 R R  R2  2 (A2  R 2 ) R 2  A2 2(A2  R 2 )3/2

a

a , a 1 (1   )  AR

where a  0.1080 per degree  6.188 per radian

From Fig. 5.18:     0.054.

a

6.188  4.91 per rad.

6.188 1 (1  0.054)  (8)  0.0857 per degree

CL  a (   L  0 )  0.0857 [7  (1.3)]  0.712 CD1 

AR 

C2L (0.712)2 (1  )  (1.054)  0.0212  aR  (8)

b 2 (32) 2   6.02 S 170

At standard sea level, p  0.002377 slug/ft

3

 88 ft/sec  V  120 mph    176 ft/sec  60 mph  1 1 q    V2  (0.002377)(176) 2  36.8 lb/ft 2 2 2 a  0.1033 per degree  5.92 per rad CL =

L W 2450    0.3916 q S q S (36.8)(170)

14 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

a

a 1

a (1   )  AR



5.92  4.38 per rad 5.92 1 (1  0.12)  (6.02)  0.0764 per deg

C L  a (   L  0 ) 

C Di 

CL 0.3916  L0   3  2.12 a 0.0764

C2L (0.3916)2   0.01267  eAR  (.64)(6.02) Di  q S CDi  (36.8)(170)(0.01267)  79.3 lb

To be consistent, we will use Helmbold’s equations for both the straight and swept wings. (a) a  0.1 per degree  0.1 (57.3)  5.73 per radian

a 5.73   0.304  AR  (6) From helmbold‟s equation for a straight wing, Eq. (5.81), a 

a 1  [a /( AR)]2  a /( AR) 5.73 1  (0.304)  0.304 2



5.73  4.247 per radian 1.349

(b) From Helmbold‟s equation for a swept wing, Eq. (5.82), where

a cos   5.73cos 45  4.05 per radian and

a cos  4.05   0.215  AR  (6) we have

15 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

a cos 

a

1  [a cos  /( AR)]2  a cos  /( AR) 4.05





1  (0.215) 2  0.215

4.05  3.27 per radian 1.23785

Comparing the results of parts (a) and (b), we readily conclude that the effect of wing sweep is to reduce the lift slope. Moreover, the reduction is substantial.

Again, we use Helmbold’s equations.

a  5.73 per radian

(a)

a 5.73   0.608  AR  (3) a 

a 1  [a /( AR)]2  a /( AR) 5.73 1  (0.608)  0.608 2



5.73  3.222 per radian 1.778

a cos   4.05

(b)

a cos  4.05   0.43  AR  (3) a 

a cos  1  [a cos  /( AR)]2  a cos  /( AR) 4.05 1  (0.43) 2  0.43



4.05  2.667 1.5185

In Problem 5.6, with an aspect ratio of 6, we had

a swept a straight



3.27  0.77 4.247

16 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

The lift slope for the swept wing is only 77% of that for the straight wing when the aspect ratio of both wings is 6.

In Problem 5.7, with aspect ratio 3, we have

a swept a straight



2.667  0.83 3.222

The lift slope for the swept wing is 83% of that for the straight wing.

Conclusion: Wing sweep decreases the lift slope. Moreover, wing sweep affects the lift slope to a greater degree for higher aspect ratio wings than for lower aspect ratio wings. This makes some sense, because the lift slope for low aspect ratio wings is already considerably reduced just due to the aspect ratio effect.

17 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

For the Wrights’ 1900 glider, AR  3.5. For the Lilienthal data, AR  6.48.

1 2 Eq. (5.70), written for the Wrights’ 1900 glider, is a1 

a0  a  1   0  (1  1 )  AR1 

and for the Lilienthal data. a2 

a0  a0  1   (1  2 )  AR 2 

Hence,

 a  1   0  (1  2 ) a1  AR 2   a2  a  1   0  (1  1 )  AR1  We set a 0  2 . Also, for the rectangular wing planform (taper ratio = 1.0) of the Wrights, from Eq. 5.20 1  0.06. For the ogival planform with pointed tips (taper ratio = 0) for the Lilienthal data,  2  0.12. Thus,

 2  1   (1.12) 1.3457 a1 6.48      0.83 a2 1.6057  2  1   (1.06)  3.5 

18 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Hence, to apply the Lilienthal data the Wrights‟ 1900 glider taking into account the difference in aspect ratio, the value of lift coefficient of 0.546 from the table should be reduced to

  (0.546)(0.83)  0.45 The Wrights should have used   0.45 instead of 0.546.

(Note: We continue here Lilienthal‟s use of  for the normal force coefficient, as seen in Figure 1.65. It is defined in conjunction with Eq. (1.63).)

(a)

b2 (36.1)2 AR    5.385 S 242  88  V  362 mph  362    530.9 ft/sec  60 

From Appendix E, at 18,500 ft,   0.0013329 slug/ft . In steady, level flight, L = W. Thus 3

L  W  1  V 2 S C L 2

or,

CL 

2W 2(5820)  2  VS (0.0013329)(530.9) 2 (242)

19 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

CL  0.128 From Eq. (5.62) C2L eAR

CD,i 

For the elliptical wing of the Spitfire, e = 1. Thus (0.128) 2 CD,i   0.000968 (5.385)

(b) D  T  P/V 

550  HP 550(0.9)(1050)   979 lb.

V 530.9

D  1  V 2 S C D 2

CD 

2D 2(979)   0.0215 2  VS (0.0013329)(530.9) 2 (242)

CD,i



Thus,

CD

0.000968  0.045 0.0215

The induced drag is only 4.5% of the total drag. This is no surprise because at Vmax, the lift coefficient is very small and therefore so is the induced drag coefficient.

20 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Since only the wing planform shape is different from that in Problem 5.9, everything else being the same, the lift coefficient and aspect ratio are the same as in Problem 5.9.

Only the span efficiency factor, e, will be different. From Fig. 5.20, for a taper ratio of 0.4 and AR  5.385, Thus   0.005. e

1 1   0.995 1   1.005

C2L (0.128)2 CDi    0.000973 eAR (0.995)(5.385)

The value of C Di obtained from Problem 5.9 is 0.000968. By changing the planfrom from an elliptical shape to a tapered wing with taper ratio 0.4, the increase in the induced drag coefficient is only 0.000973  0.000968  5  106 , or only a 0.5 percent increase. This underscores that a tapered wing with a taper ratio around 0.4 is just about as aerodynamically efficient at the optimum elliptical wing shape.  88  V  70    102.67 ft/sec  60 

At standard sea level,   0.002377 slug/ft . Thus, 3

CL 

2W 2(5820)  2  VS (0.002377)(102.67) 2 (242)

21 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

CL  1.92 C2L (1.92) 2 C Di    0.218 eAR (1)(5.385)

For this low-speed case, the induced drag coefficient is much larger than in the high-speed case, 0.218 compared to 0.000968. Indeed, for an airplane flying at low speeds near stalling speed, induced drag is by far the dominant source of drag, whereas at high speeds it is only a small fraction of the total drag.

CHAPTER 6 Vr 

6.1

V 

c , V  0, V  0 r2

1 r sin 2

er

r e

(r sin) e

 r c r2

 

 

0

0

 c / r 2   1  c / r 2    2  0   r sin  e  0   er (0  0)  r e     r sin       

1 {0  0  0}  0 r sin  2

Flow is irrotational.

c , V  0, V  0 r2

6.2

Vr 

V 

1   2  c  1  1 (0) r  2   (0)   2 r sin   r r   r   r sin  

V 

1 c 00  000  0 r 2 r

The flow is a physical possible incompressible flow.

22 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

6.3 For the sphere:

(Cp )  1 

9 2 sin  4

23 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

For the cylinder:

(Cp )cyl.  1  4 sin 2 

At the top of the sphere:

  /2, hence

(Cp )sphere  5/4  1.25

For no manometer deflection, (Cp )sphere  (Cp )cyl.

1.25  1  4 sin 2  sin 2   0.5625 sin   0.75 Hence:   48.6

The pressure tap on the cylinder must be located at an angular position 48.6° above or below the stagnation point.

CHAPTER 7 7.1

p  RT 

7.2

p (7.8)(2116)   0.0103 slug/ft 3 RT (1716)(934)

(a)

cp 

R (1.4)(1716)   6006 ft lb  1 0.4 slug  R

cv 

R 1716   4290 ft lb   1 0.4 slug  R

e  c v T  4290 (934)  4.007  106 ft lb slug 6 h  cp T  6006 (934)  5.610  10 ft lb slug

24 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

(b) For a calorically perfect gas, cp and cv are constants, independent of temperature.

Hence, we have again cp  6006 ft lb slug R c v  4290 ft lb slug R

Also, at standard sea level, R  519R . Hence E  4290 (519)  2.227  106 ft lb slug

h  6006 (519)  3.117  106 ft lb slug

25 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

7.3

cp 

R (1.4)(287) joule   1004.5  1 0.4 kg K

cv 

R 287 joule   717.5   1 0.4 kg K

h 2  h1  cp  T2  T1   (1004.5)(690  288)  4.038  105

e2  e1  c v  T2  T1   (718.5)(690  288)  2.884  105

s2  s1  cp n

7.4

 

 RT

joule kg

joule kg

T2 p 690 joule  R n 2  (1004.5) n  (287) n 8.656  258.2 T1 p1 288 kg K

4.35  104  0.6186 kg/m3 (287)(245)



1/

 p       p 

1/

 p        p 

7.5

1/1/4

 3.6  104   0.6186  4   4.35  10 

kg m3

   T  1

p   po  To 



 p   1 1 T  To    500    10   po  

7.6

 0.5404

 RT



0.2857

 259K

1.01  105  1.359 kg/m3 (287)(259)

pv  RT, hence v 

RT p

26 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

 v  RT v    2  p p  p T 1  v  1 T      v  p T p Note: 1 atm  1.01  105 N/m2 2 1 1 5 m T    4.95  10 p (0.2) 1.01  105 N









  v  p For an isentropic process: 1   1    2  p2  2   v1 

i.e., p1v1  p2 v2 or pv  constant  c1 1/

c  v  1 p

1/  v  1 1 1 1/  (1/ ) 1   pv (p)( 1 )/   v p1     c1  (p)    p s 

 

1  v  1 v  1 s           v  p s v p p

s 

7.7

cp 

1



(1.4)(0.2) 1.01  105

m2 N

R (1.4)(1716) ft lb   6006  1 (0.4) slug R

ho  h 

7.8



 3.536  105

V2 V2 (1300)2 6 ft lb  cp T   (6006)(480)   3.728  10 slug 2 2 2

Let  h o res  total enthalpy of the reservoir  cp  To res

27 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

 h o e  total enthalpy at the exit  cp Te 

Ve2 2

For an adiabatic flow, h o = constant. Hence

 ho res   ho e Ve2 cp  To res  cp Te  2 Ve  2 cp  To res  Te   2(1004.5)(1000  600)  896.4 m/sec

7.9



(0.61) 1.01  10 p T     R (0.819)(287) T  p    T  p 

( 1)/

5

  262.1 K

 p  ; T  T    p 

( 1)/

 0.5   262.1    0.61 

0.2857

 247.6 K

Since the flow is isentropic, it is also adiabatic. Hence, h o = constant

h  V

V2 V2 h 2 2

2  h   h   V2  2 cp  T  T   V2  2(1004.5)(262.1  247.6)  (300) 2

 345 m/sec

7.10

p  

V

V2 V2  p 2 2

2 p  p 



 V  2





2 1.01  105 (0.61  0.5) 0.819

 (300)2  342.2 m/sec

 345  342.2  % error     100  0.81% 345  

28 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

7.11

 p  T  T    p 

( 1)/

 0.3   262.1    0.61 

0.2857

 214 K

V  2(1004.5)(262.1  214)  (300)2  432 m/sec

7.12

V





2 1.01  105 (0.61  0.3) 0.819

 (300) 2  408 m/sec

 432  408.7  % error     100  5.55% 432   7.13

From Eq. (7.53)

h

V2  constant 2

From Eqs. (7.6b) and (7.9), h  cpT 

 RT  1

(1)

From the equation of state,

RT  p/ 

(2)

Combining Eqs. (1) and (2),

h



p   1   

(3)

Hence, Eq. (7.53) can be written as



 p  V2   constant   1    2

(4)

In the limit of   , Eq. (4) becomes

29 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

p





V2  constant 2

or,

p  1 2  V 2  constant which is Bernoulli‟s equation. Hence, the energy equation for compressible flow can be reduced to Bernoulli‟s equation for the case of   . Hence, the ratio of specific heats for incompressible flow is infinite, which of course does not exist in nature. This is just another example of the special inconsistencies associated with the assumption of incompressible flow, i.e., constant density flow, which of course does not exist in nature. This is why we have stated earlier in this book that incompressible flow is a myth.

As to the question whether Bernoulli‟s equation is a statement of Newton‟s second law or an energy equation, we now see that it is both. For an incompressible flow, the application of the fundamental principles of Newton‟s second law and the conservation of energy are redundant, both leading to the same equation, namely Bernoulli‟s equation. However, philosophically this author feels strongly that Bernoulli‟s equation is fundamentally a statement of Newton‟s second law – it is a mechanical equation. This is how we derived Bernoulli‟s equation in a very straightforward manner in Chapter 3. For the study of inviscid incompressible flow, we need only to apply the fundamental principles of mass conservation and Newton‟s second law. The principle of conservation of energy is redundant and is not needed.

7.14

Derive the 1-D energy equation for adiabatic compressible flow staring from the first law of 2 thermodynamics and Euler‟s equation i.e. derive c pT  u  C from de   q   w and

1 2

udu  

dP



where C is a constant. Hint: assume you have a simple compressible system with

constant specific heats undergoing an internally reversible (i.e. isentropic) process.

Solution: Starting with the first law (where  w is the work done on the control volume):

de   q   w Recognizing that  w   pdv gives:

de   q  pdv Incorporating the definition of enthalpy ( h  e  pv ) leads to:

dh  pdv  vdp   q  pdv

30 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Incorporating the second law for a reversible process ( ds   q ) and rearranging leads T

to: dh  Tds 

dp



Incorporating the additional assumption that the flow is isentropic (ie. ds  0) gives: dh 

dp



Substituting this into Euler‟s equation gives: udu  dh

Integrating both sides gives: 1 2 u  h  C 2

Recall that for a calorically perfect gas, h  c pT . So: 1 2 u  c pT  C 2

Finally, recognize that the constant is related to the total temperature (ie. the temperature that the gas reaches if it were adiabatically decelerated to zero velocity).

This gives:

1 2 u  c pT  c pTo 2 7.15

An airplane flies at Mach 0.8 at an altitude of 10 km. Its engine draws in air from the free stream and raises its pressure by a factor of 80 by the time it exits the engine‟s compressor.

Assume that the flow is steady, inviscid, and that air is a calorically perfect gas with c p  1004.5J /kgK and   1.4 . Also assume that the velocity of the flow exiting the engine‟s compressor is much smaller than the speed of sound (ie. M 2  0) . Please: a. Find the total pressure and temperature of the air exiting the compressor and the work done on the air per unit mass (ie. J/kg) assuming that the entire compression process from the free stream to the compressor‟s exit is isentropic. Note that the work input to the compression process equals the increase in enthalpy. You will learn why in the next problem (7.16). b. Repeat part a assuming that 100J/kg of entropy is generated during the compression process. c. The efficiency of the compression process is the ratio of the work required by the ideal (ie. isentropic) process divided by the work required by the real process. Calculate the efficiency of this compression process.

31 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

d. Sketch the ideal and real compression processes on a T-s diagram. Be sure that your diagram includes isobars corresponding to conditions in the free stream, compressor entrance, and compressor exit. e. How does the T-s diagram change if the Mach number downstream of the compressor is not negligible? Where will the conditions at the exit of the engine‟s inlet – which sits between the free stream and the compressor‟s entrance appear in the T-s diagram? Sketch a new T-s diagram that shows these features and explain.

Solution: Begin by sketching the problem. Air flows along a streamline that originates ahead of the engine inlet and terminates at the compressor face:

𝑀 ≈

𝑀𝑎 0.8 𝑇𝑎 , 𝑝𝑎 a 1

2

Figure 1. Schematic illustration of compression process where Mach number at the compressor exit is assumed to be small and thus approximately equal to zero.

For the purposes of this problem, we will think about the air at the beginning of the streamline as moving at the speed of the aircraft at the temperature and pressure associated with an altitude of 10 km in the standard atmosphere. Consulting a table of the standard atmosphere gives T1   49C and p1  26.5 kPa . We will think of the air at the end of the streamline as nearly at rest so that the temperature and pressure are nearly equal to the total conditions there. a) Let‟s start by finding the total conditions of the free stream. From the isentropic flow tables (A.2): T0a Ta

 1.128 and

p0 a

pa

 1.524

So, T0a  Ta

T0a   224K 1.128  252.7 K Ta

and P0a  Pa

P0a   26.5kPa 1.524   40.4kPa Pa

The pressure ratio of the compression process is given so: P  p02  pa  02   26.5kPa  80   2,120 Pa  Pa 

Since the compression process is assumed to be isentropic,

32 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

 1

T02, s Ta

p     02   pa 

So,  1

1.4 1 p   T02, s  Ta  02    224 K  80  1.4  783.4 K  pa 

The shaft work/mass is equal to the increase in total enthalpy across the compressor so: Wisent J kJ  h02, s  ha  c p T02, s  Ta   1004.5  783.4K  224K   561.9 m kgK kg

b) Now the process is not isentropic but we are given the overall entropy change so we use Gibbs‟s equation (7.25) to find the temperature change: T  p  s02  s1  c p ln  02   Rln  02   T1   p1 

Rearranging and solving for T02 gives:  s R  p02     ln    c p c p  p1   T02  T1e

Note that we can use Gibbs equation with either static or total conditions as long as we are consistent (ie. don‟t mix static and total conditions at the same point in the flow). We are not given the gas constant so we must calculate it from: cp 

R  1

Solving for R gives: R  cp

 1 J 1.4  1 J  1004.5  287  kg 1.4 kg

Solving for the temperature:

T02   224 K  e

 0.100 kJ / kg 0.288 kJ / kgK   ln  80    1.0045 kJ / kg 1.0045 kJ / kgK 

 865.4 K

Finally Wreal J kJ  h02  ha  c p T02  Ta   1004.5 865.4K  224K   644.3 m kgK kg

c) From the definition of compressor efficiency given in the problem:

c 

Wisent 633.9kJ / kg   0.872 Wreal 644.3kJ / kg

33 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

d) Below is the T-s diagram for the compression process from the free stream to the compressor. No information about the inlet is provided so the compression process is assumed to occur in a single step from a to 02. Static and total conditions are equal at the exit because M 2  0 .

T (K) 1000 02

865.4K 800

600

02,s

Ideal Process

783.4K

P02= 2,120kPa

p0a Pa= 26.5kPa

400 0a

224 K

200

a

0 0

100

200

S (J/ kgK)

Figure 2. Sketch of compression process of Figure 1 in T-s space. e) If the Mach number exiting the compressor is not negligible (see Figure 3), the pressure ratio of 80:1 will now be associated with p2 /pa instead of p02 /pa and the total temperature and pressure exiting the compressor will be higher. How much higher depends on the exit Mach number. The function of the engine‟s inlet is to slow down the air before it enters the compressor. This causes the temperature and pressure to rise. How much depends on the losses which could be different (and are usually much lower) than the losses in the compressor itself. Therefore, the thermodynamic path breaks into two parts where the entropy change associated with going from a to 1 is much smaller than the entropy change associated with going from 1 to 2. There are total conditions associated with state 1 which will be higher than the static conditions for both real and isentropic processes. However, T01 and p01 cannot exceed T0a and p0a because the inlet is a passive component (ie. no energy is added to the

34 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

flow). The diagram here is drawn assuming that the inlet is adiabatic (ie.T01  T02 ) which is usually a good assumption. The p01 isobar is not shown because it is so close to but slightly below the p0a isobar. You will learn more about thermodynamic processes occurring in engines when you take courses in aerospace propulsion.

Figure 3. Schematic illustration of compression process where the Mach number exiting the compressor is not zero (but still small). This is a more realistic scenario than what is shown in Figure 1.

T (K) 1000

p02 P2= 2120kPa

02

783.4K

800

600

2

02,s 2,s

Ideal Process

865.4K

p0a p1 Pa= 26.5kPa

400 0a

224 K

200

01

a 1

0 0

100

200

S (J/ kgK)

Figure 4. Sketch of the more realistic compression process of Figure 3 in T-s space.

35 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

7.16

Starting with the integral form of energy conservation (2.95 and 7.43), derive an expression for energy conservation in a steady, inviscid, non-chemically reacting flow without body forces along a streamline that passes through a compressor or turbine. Use Wshaft to represent the work done on the system by the compressor or turbine per unit mass of flow in J/kg. Your result should be analogous to equations 7.53 or 7.54 except that the total enthalpy will not be constant.

Solution: Begin by writing the energy equation (2.95) and thinking about what each of its terms mean:

 q d V  Qviscous   pV  dS     f  V  d V  Wviscous v

S



 t



   e  

v

v 2

V  dV  2 



   e  

S

V2  V  dS 2 

The right hand side (RHS) represents the rate of change in energy of the control volume due to the passage of time (first term on the RHS) and convection of material in and out of the control volume (second term on the RHS). The left hand side (LHS) represents rate at which energy is added to the control volume by various means. For this reason they are often called „source terms‟. The first term on the LHS represents the rate at which energy is added by a chemical reaction or other heat source/sink inside the control volume. The second term represents the rate at which energy is added due to viscous shearing within the control volume (ie. frictional heating due to turbulence within the CV). The third term ( Qviscous ) represents the rate at which work is being done by normal forces (ie. pressure forces) acting at the boundaries or, equivalently, due to flow into and out of the control volume at different pressures. The fifth term ( Wviscous ) represents the rate at which work is done on the control volume by shear forces acting on the control volume surface. The fourth term represents the rate at which the potential energy of the system is changing as it moves up or down in a potential field. If the potential field is due to gravity, then f   g ˆj where g is the acceleration due to gravity and ĵ is a unit vector pointing „up‟ against gravity. Both Qviscous and Wviscous are written as „lumped‟ terms because the processes that generate them are complex and do not need to be understood until later (chapter 15). Thinking about the compressor or turbine as another complex process that adds energy to the flow, we represent its effects by add Wshaft to the RHS recognizing that

Wshaft will be positive in the case of a machine like a compressor that does work on the flow and negative in the case of a machine like a turbine that does work on the environment.

Making this modification leads to:

 q d V  Qviscous   pV  dS     f  V  d V  Wviscous  Wshaft v

S



 t



   e  v



v

V2   dV  2 



   e  S



V2  V  dS 2 

36 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

The energy equation simplifies to the following in this problem where the flow is steady, inviscid, non-chemically reacting flow and lacks body forces: 



pV  dS  Wshaft 

S



   e  S



V2  V  dS 2 

Recognizing that doing work on the fluid in the control volume also changes its internal energy, we express this in a mathematically convenient way by writing Wshaft 

  eref V  dS S

where eref is a reference energy that defined such that Wshaft  eref ,2  eref ,1 . Converting surface integrals to volume integrals using the divergence theorem (2.26) gives: 



   pV  d V 

v



    eref V  d V 

  V2       e   V  d V  2      v



v

Rearranging gives:    p V2     e    eref   Vd V  0    2     

 or

 p V2   e     eref  constant  2   or

 V2   h    eref  constant 2   where h  e 

p



Applying this to two points along a streamline upstream and downstream of the compressor/ turbine leads to: h1 

V12 V2  eref ,1  h2  2  eref ,2 2 2

or h1 

V12 V2  Wshaft  h2  2 2 2

or

h01  Wshaft  h02

37 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

7.17

In a thermally perfect, but not calorically perfect gas, the specific heat at constant pressure can no longer be considered constant. How would this effect the prediction of the stagnation temperature of air flowing at 300K and 1500 m/s (about Mach 4.3)? a. Begin by showing that the differential form of the energy equation for steady, onedimensional, constant area, adiabatic flow is: dh  udu  0 (1) b. Integrate equation 1 assuming cp is constant to recover the simple 1-D expression for energy conservation along a streamline. c. Now, incorporate the effect of temperature variation by representing the specific heat of air as a power law in temperature (T is the temperature and the units are J/KgK)

c p  996.96   0.0357  T   0.0002 T 2  8E  8 T 3 d. Use your result to determine the total temperature of air having a static temperature of 300K and a velocity of 1500 m/s assuming variable specific heat. Note that you will have to use an iterative approach to get the temperature. e. How does this differ from the result assuming constant specific heat (pick the value corresponding to 300K)? f. Based on what you have seen here, under what conditions would it be important to consider the temperature variation of specific heat when predicting the total temperature?

Solution: a. Start with the steady 1-D energy equation and consider variations in the x-direction: h

u2  h0  constant 2

so,

  u2  h   0 x  2  h u u 0 x x  dh  udu  0

(1)

b. Since h  c pT

dh  c p dT The 1-D energy equation becomes:

c p dT  udu  0 or c p dT  udu

38 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

(2)

Integrating (2) for constant c p gives: T2

u2

1

1

T c p dT  u udu

(3)

 u2 u2  c pT2  c pT1    2  1  2   2 c pT1 

u12 u2  c pT2  2 2 2

c. Now, assume that c p is a function of temperature:

c p T   A  BT  CT 2  DT 3 Then (3) becomes: T2



u2





A  BT  CT 2  DT 3 dT   udu

T1

u1

Integrating gives:











u12 u22 B 2 C 3 D 4   A T2  T1   T2  T12  T2  T13  T2  T14 2 2 2 3 4

where A  996.96



J J J J , B  0.0357 , C  0.0002 and D  8E  8 .

2 3 kgK kgK kgK kgK 3

d. We are interested in the stagnation temperature so u2  0 and T2  T0 in the above expression. Thus, we want to find the value of T0 that satisfies:













u2  B 2 C 3 D 4  0   1   A T0  T1   T0  T12  T0  T13  T0  T14  2  2 3 4  Use MATLAB‟s „fzero‟, Excel‟s „goal seek‟ function, or some other software that finds zeros (roots) of a user-supplied function. The result is that T0  1309.1K

e. If c p is constant then: c pTo  c pT1 

u12 2

Solving for T0 gives: To  T1 

u12 2c p

We choose the value of c p corresponding to T1 so: c p  996.96

  J J  J  2   0.0357   300 K    0.0002   300K  2 3 kgK  kgK  kgK    J  3   8E  8 300 K  4  kgK  

39 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

c p  300 K   1023.8

J kgK

and 2

m  1500  s To  300 K    1399 K  J  2 1023.8  kgK  

This is about 6% higher than the value we predicted assuming that cp varies with temperature. This is because air‟s specific heat increases with temperature enabling it to hold more energy per unit mass. f.

As the speed of the flow increases, the total temperature increases and thus the difference between the specific heat at the static temperature and at the total temperature becomes more important. The plot below shows how the „error‟ changes with Mach number. % Error  100 

To, const cp  To To

To Error Caused By Assuming Constant Cp Percent Overprediction

12 10 8 6 4 2 0 0

1

2

3

4

5

6

Mach Number Figure 1. Percent error in total temperature caused by neglecting the temperature dependence of C p as a function of Mach number The error is relatively small until the Mach number gets large. The shape of the curve reflects the temperature dependence of c p with temperature.

CHAPTER 8 40 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

1.1

a   RT  (1.4)(287)(230)  304 m/sec

2.1

Ve2 cp To  cp Te  2

Ve2 (1385)2 Te  To   519   359.3 °R 2cp 2(6006)

a e   RTe  (1.4)(1716)(359.3)  929.1 °R Me 

3.1

Ve 1385   1.49 ae 929.1

a   RTe  (1.4)(287)(300)  347.2 m/sec M

From Tables:

V 250   0.72 a 347.2 To p  1.104 and o  1.412 T p

To  1.104 T  1.104(300)  331.2 °K

po  1.412 p  1.412(1.2)  1.694 atm p* p* po   (0.528)(1.412)  0.7455 p po p

p*  0.7455 p  0.455(1.2)  0.8946 atm T* T* To   (0.8333)(1.104)  0.92 T To T

T*  0.92(300)  276 °K

a*   RTe  (1.4)(287)(276)  333 m/sec

41 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

M* 

4.1

V 250   0.75 a* 333

a   RT  (1.4)(1716)(700)  1297 ft/sec M

v 2983   2.3 a 1297

From Tables:

To p  2.058 and o  12.5 T p

To  2.058 T  2.058(700)  1441 °R po  12.5 p  12.5(1.6)  20 atm T* T* To   (0.8333)(2.058)  1.715 T To T

T*  1.715 T  1.715(700)  1200 °R p* p* po   (0.528)(12.5)  6.6 p po p

p*  6.6 p  6.6(1.6)  10.56 atm

a*   RT*  (1.4)(1716)(1200)  1698 ft/sec M* 

V 2983   1.757 a* 1698

po T  7.824 and o  1.8 p T Hence, for the test section flow,

5.1

From Tables:

po  7.824 p  7.824(1)  7.824 atm To  1.8 T  1.8(230)  414 °K

42 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Since the flow is isentropic, both p o and To are constant throughout the flow. Also, in the reservoir, M  0. Hence, the reservoir pressure and temperature are po  7.824 atm To  414 °K

6.1

From the Standard Altitude Tables, at 10,000 ft.,

p  1455.6 lb/ft 2 and T  483.04 °R From Table A.1:

For M  0.82; For M  1;

po T  1.555, o  1.134 p T

po T  1.893, o  1.2 p T

Since the flow is isentropic, po  constant and To  constant.

7.1

p

p po 1 p  (1.555)(1455.6)  1196 lb/ft 2 po p 1.893

T

T To 1 T  (1.134)(483.04)  456.5 °R To T 1.2

From Table A.2:

p2  T  7.72, 2  3.449, 2  2.238, p1 1 T1

po2 p1

 9.181, M 2  0.5039 ,

po2 po1

 0.4601

Hence, p2 

p2 p1  7.72(1)  7.72 atm p1

T2 

T2 T1  2.238(288)  644.5 °K T1

43 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

1 

p1 (1)(1.01  105 )   1.222 kg/m3 RT1 (287)(288)

2 

2 1  3.449(1.222)  4.21 kg/m3 1

po2 

po2 p1

p1  9.181(1)  9.181 atm

To2  To1 

To1 T1

T1  (2.352)(288)  677.4 °K (using Table A.1)

s 2  s1  R n

8.1

po2 po1

 (287) n 0.4601  222.8

joule kg °K

po T 2  10.33. From Table A.2, M1  3.0 , 2  2.679, 2  12.06 T1 p1 1

Thus, T1 1 T2  (1390)  518.9 °R T2 2.679

T1 

From Table A.1, for M1  3.0,

To2  To1  po2 

9.1

po2 po1

po2 p1

To1 T1

To1 T1

 2.8

T1  2.8 (518.9)  1453 °R

p1  (12.06)(1)  12.06 atm

 e (s2 s)/R  e (199.5)/287  0.499 1

From Table A.2: M1  2.5

44 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

10.1

From Table A.2:

T2  2.799 and M 2  0.4695 T1

Hence, T2 

T2 T1  2.799(480)  1343.5°R T1

a 2  (1.4)(1716)(1343.5)  1796.6 ft/sec V2  M 2 a 2  (0.4695)(1796.6)  843.5 ft/sec From Table A.1, for M 2  0.4695,

T2 * 

To2 T1

 1.044

T2 * To 2 T2  (0.8333)(1.044)(1343.5)  1169°R To 2 T2

A 2 *   RT2*  (1.4)(1716)(1169)  1676 ft/sec M*2 

11.1

V2 843.5   0.503 1676 a *2

po 1   1.893, which is p 0.528 higher than 1.555. Hence, the flow is subsonic. From Table A.1, for

Is the flow subsonic or supersonic? For sonic flow,

po  1.555, M  0.82. p

a   RT  (1.4)(287)(288)  340.2 m/sec V  Ma  (0.82)(340.2)  278.9 m/sec

12.1

7712.8  3.645 is larger than 1.893. Hence, the flow is supersonic. This 2116 means that a normal shock wave exists in front of the nose of the Pitot tube. From Table A.2, for

The ratio

45 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

po2 p1



7712.8  3.645, M1  1.56 2116

a1   RT1  (1.4)(1716)(519)  1116.6 ft/sec

V1  M1 a1  (1.56)(1116.6)  1742 ft/sec 13.1

p 1.01  105   1.22 kg/m3 RT (287)(288)

(a)  

V

2(po  p)



% error 

(b)   V



303  278.9  8.69% 278.9

p 2116   0.002376 slug/ft 3 RT (1716)(519)

2(po  p)



2(7712.8  2116)  2170.5 ft/sec INCORRECT 0.002376



2170.5  1742  24.6% 1742

% error 

14.1

2(1.555  1.0)(1.01  105 )  303 m/sec INCORRECT 1.22

p2   1  2 M12  2 1    2 M12 2 1 M12  1   p1  1  1  1

  1  p2 

po2







 1

  2  1

2

M2  



1  (  1)/2 M12 2 M2   M12    1)/2





Working with the expression inside the parenthesis of Eq. (2):

46 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

(1)

(2)

     1 2     1 2  1  M1  1    M1     1 2  1 2  2       1  (  1)   1 M2  1  2 2 2   M12  (  1)/2   2 M1  (  1)       2  (  1) M12  4 M12  2(  1)  2(  1)  (  1) 2 M12 = 1  (  1)   2 4 M12  2 (  1)  4 M1  2(  1) 

4 M12  ( 2  2  1) M12 ( 2  2  1) M12   4 M12  2 (  1) 4 M12  2 (  1)



(  1) 2 M12 4 M12  2 (  1)

(4)

Combining Eqs. (4), (2), and (1), we have: 

 (  1) 2 M12   1 1    2 M12       which is Eq. (8.80) p1 p 2 p1  4 M12  2 (  1)    1 

po2

15.1

po2 p 2

At 80,000 ft., T  389.99°R  88  V  2112    3097.6 ft/sec  60 

a    RT  (1.4)(1716)(389.99)  967.9 ft/sec M 

3097.6  3.2 967.9

From Appendix A: For M   3.2,

To  3.048 T

To  3.048 T  3.048 (389.99)  1188.7 °R

47 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Since, 0F  460°R, the To  728.7°F

16.1

po2 p1



1.13  11.3.

0.1

From Appendix B, M   2.9 17.1 The temperature at the stagnation point is the total temperature in the freestream, because the total temperature is constant across the normal shock. From Eq. (8.40), To  1 2 1.4  1 1 M  1  (36)2  260.2 T 2 2

Since, T  300 K, we have

To  (260.2)(300)  78,060K This is an ungodly high temperature. It is also incorrect, because long before the air would reach this temperature, it would chemically dissociate and ionize. In such a chemically reacting gas, the specific heats are not constant, which means that Eq. (8.40) is not valid for such a chemically reacting flow. In reality, the temperature at the stagnation point on the Apollo was close to 11,000 K, much lower than our estimate above, but still plenty high. Air at 11,000 K is a partially ionized plasma. For the analysis of high temperature, chemically reacting flows, techniques much different than those discussed in this book must be used.

See for example Anderson, Modern Compressible Flow, 2nd ed., Mc-Graw-Hill, 1990, or Anderson, Hypersonic and High Temperature Gas Dynamics, Mc-Graw-Hill, 1989, reprinted by the American Institute of Aeronautics and Astronautics, 2000.

18.1 Use Eq. (8.40) To  1 2 1 M T 2

For To  11,000 K, T  300 K, and M   36, this equation becomes: 11,000  1 1 (36) 2 300 2 35.67  648   648

48 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

or,

 

683.7  1.055 648

In order to use Eq. (8.40) to estimate a reasonably valid stagnation temperature for the Apollo, we have to use an “effective gamma” of 1.055. To double check this, return to Eq. (8.40), insert   1.055, and calculate To .

To  1 2 1.055  1 1 M  1  (36) 2  36.64 T 2 2

or,

To  36.64 T  36.64 (300)  11,000 K

49 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

19.1 Consider two points in an isentropic flow, points 1 and 2. From the definition of isentropic flow, the entropy at these two points, s1 and s2, is the same. From equation (7.25), s 2  s1  0  cp n

T2 p R n 2 T1 p1

Furthermore, assume at each point 1 and 2, we slow the flow to zero velocity isentropically.

The entropy at each point remains the same, but the temperature is now, by definition, the total temperature, and pressure is now the total pressure. Thus, 0  cp n

T0,2 T0,1

p0,2

R n

p0,1

From Equation (7.56), To is constant throughout the flow, i.e., T0,1  T0,2 . Thus 0  cp n(1)  R n

p0,2 p0,1

or, n

p0,2 p0,1

 0.

Hence,

p0,2  p0,1 20.1

See Figure 1. The plots were produced using the normal shock relations: p2 2  1 M12  1 p1  1



M 22 

1



 1 2

(8.65)

M12

(8.59)

  1 M  2 1

2

2   2     1 M1  T2  2  1  M12  1    2 T1    1      1 M1 





(8.67)

50 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

and the definitions of total pressure and temperature: T0  1 2  1 M T 2

(8.40) 

p0    1 2   1  1  M  p  2 

(8.42)

Figure 1 Changes in properties across normal shocks in air (  1.4) and Krypton (  1.68)

Decreasing the specific heat ratio decreases the static temperature and pressure ratio, increases M2, and decreases the total pressure ratio across the normal shock. The specific heats are functions of temperature. As the Mach number increases, the temperature downstream of the shock increases nonlinearly meaning that temperature changes across the shock become large. Since the specific heat at constant pressure increases about 20% between 300 and 1500K, it will affect the temperature downstream of the shock and thus all of the other properties but this variation is not accounted for in the expressions used to calculate the property changes. The data used to generate these plots are presented in Table 1 and Table 2:

51 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Table 1 Property ratios across a normal shock occurring in air (  1.4)

M1

P02 /P01

T02 /T01

M2

T2 /T1

P2 /P1

1.000 1.250 1.500 1.750 2.000 2.250 2.500 2.750 3.000 3.250 3.500 3.750 4.000 4.250 4.500 4.750 5.000 5.250 5.500 5.750 6.000 6.250 6.500 6.750 7.000 7.250 7.500 7.750 8.000 8.250 8.500 8.750 9.000 9.250 9.500 9.750 10.000

1.000 0.987 0.930 0.835 0.721 0.606 0.499 0.406 0.328 0.265 0.213 0.172 0.139 0.113 0.092 0.075 0.062 0.051 0.042 0.035 0.030 0.025 0.021 0.018 0.015 0.013 0.011 0.010 0.008 0.007 0.006 0.006 0.005 0.004 0.004 0.003 0.003

1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000

1.000 0.813 0.701 0.628 0.577 0.541 0.513 0.492 0.475 0.462 0.451 0.442 0.435 0.429 0.424 0.419 0.415 0.412 0.409 0.406 0.404 0.402 0.400 0.399 0.397 0.396 0.395 0.394 0.393 0.392 0.391 0.390 0.390 0.389 0.389 0.388 0.388

1.000 1.159 1.320 1.495 1.688 1.901 2.138 2.397 2.679 2.985 3.315 3.669 4.047 4.449 4.875 5.325 5.800 6.299 6.822 7.369 7.941 8.536 9.156 9.801 10.469 11.162 11.879 12.621 13.387 14.177 14.991 15.830 16.693 17.580 18.492 19.427 20.388

1.000 1.656 2.458 3.406 4.500 5.740 7.125 8.656 10.333 12.156 14.125 16.240 18.500 20.906 23.458 26.156 29.000 31.990 35.125 38.406 41.833 45.406 49.125 52.990 57.000 61.156 65.458 69.906 74.500 79.240 84.125 89.156 94.333 99.656 105.125 110.740 116.500

Table 2 Property ratios across a normal shock in Kr (  1.68)

M1

P02 /P01

T02 /T01

M2

T2 /T1

1.000 1.250 1.500 1.750 2.000 2.250 2.500 2.750

1.000 0.988 0.938 0.858 0.765 0.670 0.581 0.501

1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000

1.000 0.819 0.716 0.652 0.608 0.577 0.555 0.537

1.000 1.247 1.503 1.784 2.096 2.444 2.829 3.252

P2 /P1 1.000 1.705 2.567 3.586 4.761 6.093 7.582 9.228

52 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

3.000 3.250 3.500 3.750 4.000 4.250 4.500 4.750 5.000 5.250 5.500 5.750 6.000 6.250 6.500 6.750 7.000 7.250 7.500 7.750 8.000 8.250 8.500 8.750 9.000 9.250 9.500 9.750 10.000

0.431 0.372 0.321 0.278 0.242 0.211 0.185 0.163 0.144 0.127 0.113 0.101 0.091 0.082 0.074 0.067 0.060 0.055 0.050 0.046 0.042 0.039 0.036 0.033 0.030 0.028 0.026 0.024 0.023

1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000

0.524 0.514 0.505 0.498 0.493 0.488 0.484 0.480 0.478 0.475 0.473 0.471 0.469 0.468 0.466 0.465 0.464 0.463 0.462 0.462 0.461 0.460 0.460 0.459 0.459 0.458 0.458 0.457 0.457

3.713 4.213 4.753 5.331 5.949 6.607 7.304 8.040 8.816 9.632 10.488 11.383 12.318 13.293 14.307 15.361 16.455 17.588 18.762 19.975 21.227 22.520 23.852 25.224 26.636 28.087 29.579 31.110 32.680

11.030 12.989 15.104 17.377 19.806 22.392 25.134 28.034 31.090 34.302 37.672 41.198 44.881 48.720 52.716 56.869 61.179 65.646 70.269 75.049 79.985 85.078 90.328 95.735 101.299 107.019 112.896 118.929 125.119

53 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

21.1

When 0  M1  1 there is no normal shock at the probe tip so p02  p01 . Therefore: 

p 02 p01    1 2   1   1 M1  p1 p1  2 

When M1  1 one can either use the Rayleigh pitot formula or compute the pressure ratio from the Mach numbers upstream and downstream of the shock. We will use the latter approach: 

    1 2   1 p02 p 2 p02  2   1  M12  1  1  M2  . p1 p1 p 2    1 2  





The first term on the RHS is the static pressure rise across a normal shock and the second term in the RHS is the total to static pressure ratio downstream of the normal shock where M 2 is the Mach number downstream of the normal shock:

M  2 2

1

 1 2

 M12 

M12

  1 2

Figure 1 Pressure ratio for a wind tunnel pitot-static tube where the static pressure is measured upstream of the shock (air,   1.4 ). Data used to produce figure are presented in Table 1.

54 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Table 1 Data used to produce Figure .

M1

p01 /p

M2

p2 /p1

p02 /p2

p02 /p1

0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3 3.25 3.5 3.75 4 4.25 4.5 4.75 5

1.000 1.044 1.186 1.452 1.893 2.590 3.671 5.324 7.824 11.563 17.086 25.140 36.733 53.196 76.272 108.202 151.835 210.755 289.414 393.297 529.090

1.000 0.813 0.701 0.628 0.577 0.541 0.513 0.492 0.475 0.462 0.451 0.442 0.435 0.429 0.424 0.419 0.415

1.000 1.656 2.458 3.406 4.500 5.740 7.125 8.656 10.333 12.156 14.125 16.240 18.500 20.906 23.458 26.156 29.000

1.893 1.544 1.388 1.304 1.253 1.220 1.197 1.180 1.167 1.157 1.150 1.144 1.139 1.135 1.131 1.128 1.126

1.893 2.557 3.413 4.443 5.640 7.002 8.526 10.213 12.061 14.071 16.242 18.574 21.068 23.723 26.539 29.516 32.653

CHAPTER 9  1    Sin 1    41.8  1.5 

h  559 Tan   559 Tan 41.8 h  500 ft

Mn1  M1 sin   (4.0) sin 30  2 From Table A.2, for M n1  2:



po p2 T  4.5; 2  1.687, 2  0.7209, M n 2  0.5774 p1 T1 po1



p2 

p2 p1  (4.5) 2.65 104  1.193 105 N/m 2 p1

T2 

T2 T1  (1.687)(223.3)  376.7K T1

55 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

From the --M diagram: θ  17.7 M2 

M n2 sin(    )



0.5774  2.71 sin(30  17.7)

From Table A.1, for M1  4: po 2 

po 2 po1 po1 p1

To 2  To 1 

To 1 T1

po 1 p1

 151.8,

To1 T1

 4.2





p1  (0.7209)(151.8) 2.65  104  2.9  106 N/m 2

T1  (4.2)(223.3)  937.9K

106 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

s 2  s1  R n

po2 po1

 (287) n 0.7209  93.9

joule kgm K

Consider an oblique shock. For such a case, po 2 po 1

 po 2     p2 



 p2     p1 



Depends on actual Mach number behind the shock M2, not Mn2

(1)

Depends on normal Mach number upstream of the shock, Mn1.

In the derivation of Eq. (8.80), We related M 2 Directly to M1 trough Eq. (8.78). This holds only for a normal shock. If we wish to use Eq. (8.78) for an oblique shock, then both M 2 and M1 in Eq. (8.78) are replaced by M n 2 and M n1. However, in Eq. (1) above, po2 /p2 Depends on M 2 , not

Mn2. Because Eq. (8.78) does not relate M 2 to M1 for an oblique shock (it relates M n2 to M n1), then Eq. (8.78) cannot be used for the derivation of po2 /p1 for an oblique shock. Therefore, the derivation of Eq. (8.80) holds only for a normal shock. It can not be used for an oblique shock, even with M1 replaced by M n1. On the other hand, s 2  s1  cp n

p2 T R n 2 p1 T1

where p 2 /p1 and T2 /T1 for an oblique shock depend only on M n1. Since clearly

po 2 po1

po 2 p1

po1

 e s2 s1  /R then

depends only on M n1. For these reasons, when using Table A.2 to determine changes

across an oblique shock, using Mn1, the total pressure ratio giving

p o2

po 2 po1

is a valid column, but the column

is not valid.

To CORRECTLY calculate po 2: Mn1  M1 sin β = 3 sin 36.87  1.8

107 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

From Table A.2, for Mn1  1.8: From Table A.1, for M1  3:

p o2

po2 po1 po1 p1

p o2 po1

p o2 p1

 0.8127

 36.73

p I  (0.8127)(36.73)(1)  29.85 atm

(b) The INCORRECT calculation of po2 would be as follows:

From Table A.2, for Mn1  1.8:

po2

p o2 p1

p o2 p1

 4.67

p1  4.67 (1 atm)  4.67 atm. Totally WRONG

% error 

29.85  4.67 100  539%  a terribly large error.

4.67

From the --M diagram:

  46

Mn1  M1 sin   2.5 sin 46  1.8 From Table A.2, for M n1  1.8, p2 

p2 T  3.613, 2  1.532, M n 2  0.6165 p1 T1

p2 p1  3.613 (1 atm)  3.613 atm p1

108 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

T2 =

T2 T1  (1.532)(300)  459.6K T1

M2 

M n2 sin(    )



0.6165  1.546 sin(46  22.5)

From the --M diagram, shock detachment occurs when   28.7. When     28.7,   64.5.

Mn1  M1 sin   2.4 sin 64.5  2.17 From Table A.2, for M n1  2.17: pmax 

p2  5.327 p1

p2 pI  5.327 (1 atm)  5.327 atm p1

and the maximum pressure occurs when   28.7

From the --M diagram:   48 Mn1  M1 sin   3.5 sin 48  2.60 From Table A.2:

po 2 p o

 0.4601, M n 2  0.5039,

109 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

M2 

M n2 Sin(    )



0.5039  1.648 Sin(48  30.2)

From Table A.2, for M 2  1.648;

From Table A.1, for M  3.5: p o3 

p o3 p o 2 p o  p o 2 p o p 

a)

po 1 p1

p

po2

 0.876

 76.27

p  (0.876)(0.4601)(76.27)(0.5)  15.37 atm

From Table A.1, for M1  4, Hence, po 1 

p o

p o3

po1 p1

 151.8

p1  151.8 (1 atm)  151.8 atm. f)

From Table A.2, for M1  4: po 2  g)

po 2 po1

p o2 po1

 0.1388

po1  0.1388 (151.8)  21.07 atm

Loss in total pressure  po3  po2  151.8  21.07  130.7 atm b)

110 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

From the --M diagram,   38.7 Mn1  M1 sin   4 sin 38.7  2.5 From Table A.2, for M n1  2.5: M2 

M n2 sin(    )



p o2 po1

0.513  2.21 sin(38.7  25.3)

From Table A.2, for M 2  2.21:

p o3 

po3 po2 po1 po1 po1 p1

 0.499, M n 2  0.513

p o3 p o2

 0.6236

p1  (0.6236)(0.499)(151.8)(1 atm)  47.24 atm

Loss in total pressure  po1  po3  151.8  47.24  104.6 atm c)

From part (b) above, M 2  2.21,

po2 po1

 0.499.

From the --M diagram: 2  47.3

111 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

For the second shock: Mn2  M2 sin 2  2.21sin 47.3  1.624

From Table A.2, for M n 2  1.624: M3 

M n3

sin   2  2 



po4 po3 po2 po1 po3 po2 po1 p1

po

 0.8877, M n3  0.6625

0.6625  1.444 sin(47.3  20)

From Table A.2, for M 3  1.444:

p o4 

p o3

p o4 p o3

 0.947

p1  (0.947)(0.8877)(0.499)(151.8)

po4  63.68 atm

Loss in total pressure  po1  po4  151.8  63.68  88.1 atm CONCLUSION: To decrease a supersonic flow to subsonic speeds via a shock system, a series of oblique shocks followed by a normal shock yields a smaller total pressure loss than a normal shock by itself. Hence, a system of oblique shocks, followed by a normal shock is a more efficient means of slowing a supersonic flow to subsonic speeds than a single normal shock itself.

112 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

From the --M diagram, 1  34.2

Mn1  M1 sin 1

 (3.2) sin 34.2  1.8 From Table A.2, for M n1  1.8:

p2 T  3.613, 2  1.532, p1 T1

Mn2  0.6165

M2 

Mn2

sin  1  1 



0.6165  2.24 sin(34.2  8.2)

For the Reflected Shock: From the --M diagram, for M 2  2.24 and   18.2: 2  44

Mn2  M2 sin 2  2.24 sin 44  1.56

From Table A.2, for M n 2  1.56:

M3 

M n3

sin   2   



p3 T  2.673, 3  1.361, M n3  0.6809 p2 T2

0.6809  1.56 sin(44  18.2)

Note: The fact that M 3 and M n 2 are equal is just a coincidence.

Φ  2    44  18.2  25.8

p3 

p3 p 2 p1  (2.673)(3.613)(1 atm)  9.66 atm p2 p1

T3 

T3 T2 T1  (1.361)(1.532)(520)  1084R T2 T1

113 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

For M1  2, 1  

From Table A.3:

 2    1      

Hence,

M 2  3.0

From Table A.1, for M1  2:

po1

For M 2  3:

po2

p1

p2

 7.824,

To1

 36.73,

To2

T1

T2

 1.8

 2.8

However: po1  po 2 and To1  To2. Thus p2 

p 2 po1  1  p1    (7.824)(0.7)  0.149 atm po2 p1  36.73 

T2 

T2 To1  1  T1    (1.8)(630)  405R To2 T1  2.8 

2 

p2 (0.149)(2116)   4.537 104 slug/ft 3 RT2 (1716)(405)

po2  po1 

po1 p1

p1  (7.824)(0.7)  5.477 atm

15 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

To 2  To1 

To1 T1

T1  (1.8)(630)  1134R

for M1  2, 1  30

From Table A.3:

For M 2  3, 2  19.47

Referenced to the upstream direction: Angle of forward Mach line  1  30

Angle of rearward Mach line   2    19.47  23.38  3.91

Note: The rearward Mach line is below the upstream direction for this problem.

From Table A.1, for M1  1.58: po 2 p2



po1 p2



po1 p1

 4.127

 1   (4.127)    31.6 p1 p 2  0.1306 

po1 p1

From Table A.1, for

po 2 p2

 31.6, M 2  2.9

From Table A.3, for M1  1.58; 1  14.27, for M2  2.9:  2  47.79    2  1  47.79  14.27  33.52

16 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

From the --M diagram: For M1  3 and   30.6,   53.1

Mn1  M1 sin   3 sin 53.1  2.4 From Table A.2, for M n1  2.4:

M2 

M n1 sin(    )



po p2 T  6.553, 2  2.04, 2  0.541, M n 2  0.531 p1 T1 po1

0.5231  1.37 sin(531  30.6)

From Table A.3: For M 2  1.37,  2   3  8.128  30.6  38.73 From Table A.3: For 3  38.73, M3  2.48

From Table A.1: For M1  3,

For M3 = 2.48,:

p o3 p3

po1 p1

 36.73,

 16.56,

To3 T3

To1 T1

 2.8

 2.23

p3 

p3 po3 po2 po1  1  p1    (1)(0.5401)(36.73)(1 atm)  120 atm po3 po2 po1 p1  16.56 

T3 

 1  T1    (1)(1)(2.8)(285)  357.8K To3 To2 To1 T1  2.23  T3 To3 To2 To1

Clearly, p3  p1 , T3  T1 , and M3  M. Why? Because there is an entropy increase across the shock wave, which permanently alter the thermodynamic state of the original flow, even after it is brought back to its original direction.

17 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

(a)

For M1  2.6 and θ  5,   26.5

Mn1  M1 sin   2.6 sin 26.5  1.16 From Table A.2:

p3  1.403 p1

From Table A.1, for M1  2.6:

p o1 p1

 19.95

From Table A. 3, for M1  2.6: 1  41.41  2  1    41.41  5  46.41  M 2  2.83

From Table A.1, for M 2  2.83:

p o2 p2

 28.4

p 2 p 2 po2 po1   (0.0352)(1)(19.95)  0.7022 p1 po2 po1 p1

c 

L  p3  p 2  c cos   p3  p 2    cos  q S q  c (1) q

 p1 1V12  p1V12  p1M12 q   q1  2 1 V    2  p1 2 a12 2 1

2 1

18 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

c 

(b)

2  p3  p 2 

 p1 M

2 1

cos  

2  p3 p 2     cos   M12  p1 p1 

c 

2 (1.403  0.7022) cos 5  0.148 (1.4)(2.6) 2

cd 

2  p3 p 2  sin  sin5   sin   c  0.148  0.0129 2   M1  p 2 p1  cos  cos5

For M1  2.6 and   15,   35.9

Mn1  M1 sin   2.6 sin 35.9  1.525 From Table A.2:

p3  2.529 p1

From Table A.1, for M1  2.6:

po1 p1

 19.95

From Table A.3, for M1  2.6: 1  41.41  2  1    41.41  15  56.41  M 2  3.37

From Table A.1, for M 2  3.37:

p o2 p2

 63.33

p 2 p 2 po 2 po1  1     (1)(19.95)  0.315 p1 po 2 po1 p1  63.33  c 

2  p3 p 2  2   cos   (2.529  0.315) cos 15  0.452 2   M1  p1 p1  (1.4)(2.6) 2

cd  c

(c)

sin  sin15  0.452  0.121 cos  cos15

For M1  2.6 and   30,   59.3

Mn1  M1 sin   2.6 sin 59.3  2.24

19 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

po p3  5.687, 1  19.95, 1  41.41 p1 p1  2  1    41.41  30  71.41  M2  4.46

po2 p2

 275.25

p 2 p 2 po2 po1  1     (1)(19.95)  0.0725 p1 po2 Po1 p1  275.25  c 

2  p3 p 2  2   cos   (5.687  0.0725)  1.19 2   M1  p1 p1  (1.4)(2.6) 2

cd  1.19

sin 30  0.687 cos 30

20 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

For region 2: 1  49.76

 2  1    49.76  5  54.76  M 2  3.27

po1

For M1  3:

 36.73:

p1

For M 2  3.27,

p o2

 54.76

p2

For region 3: 3   2    54.76  20  74.76  M3  4.78

p o3

For M3  4.78:

p3

 407.83

For region 4: M1  3 and   25    44

Mn1  M1 sin   3 sin 44  2.08 po p4  4.881, M n 4  0.5643, and 4  0.6835 p1 po1 M4 

M n4 sin(    )



  1.733. sin(44  25)

Thus,

5  18.69,

po4 p4

 5.165

For region 5: 5   4    18.69  20  38.69  M5  2.48

21 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

p o5 p5

 16.56

Pressure ratios p 2 p 2 po1 po1  1     (1)(36.73)  0.6707 p1 po2 po1 p1  54.76 

p 3 p 2 p 3 p 2 p 3 p o3 p o 2  1     (0.6707)   (1)(54.76)  0.09 p1 p1 p 2 p1 po3 po2 p 2  407.83  p4  4.881 p1

p5 p5 po5 po4 po1  1     (1)(0.6835)(36.73)  1.516 p1 po5 po4 po1 p1  16.56 

Let  length of each face of the diamond wedge.

L  p4 cos 25  p5 cos 5  p 2 cos 5  p3 cos 25 L   p 4  p3  cos 25   p5  p 2  cos 5

c 

c 

 p 4 p 3   p p  L L 2     cos 25   5  2  cos 5    2 q S  p M 2c  M1 c  p1 p1   p1 p1   1 1 2

2 (4.881  0.09) cos 25  (1.516  0.6707) cos 5 (1.4)(3) 2 c

c  0.823 . c However, c/2  cos 10

c



1  0.5077 2 cos 10

c  (0.823)(0.5077)  0.418

22 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

D  p4 sin 25  p5 sin 5  p2 sin 5  p3 sin 25 D   p 4  p3  sin 25   p5  p 2  sin 5

cd 

cd 

 p 4 p 3   p p  D D 2     sin 25   5  2  sin 5    2 q S  p M 2 c  M1 c  p1 p1   p1 p1   1 1 2

2 (4.881  0.09) sin 25  (1.516  0.6707) sin 5 (1.4)(3)2 c

cd  0.333

c

 0.333 (0.5077)  0.169

The maximum expansion would corresponds to M 2  . From Eq. (9.42) in the text,

lim  2

   1   1 lim  tan 1 M 22  1  tan 1 M 2  1    11    1 





M2  



M2  



 1     1      1  2.277 rad  130.45   1 2 2    1  2

Since, for M1  1, 1  0, then    2  1  130.45  0  130.45

max

23 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

For the cylinder, with cd based on frontal area,

 Dcyl  q  S cd  q  d(1)/(4/3) 

4 (d) q  3

For the dimensional wedge airfoil, referring to Figure 9.27.

 Dw   p2  p3  t Hence, 4

 Dcyl 3 (d) q   Dw  p2  p3  t However, t  d and q  

 2

p1 M12

Thus,

4  2 2  M12   M1 3 2     3   D  w  p 2  p3   p 2  p3       p1 p1   p1 p1 

 Dcyl

To calculate p2 /p1 , we have, for M1  5 and   5,   15.1 M n,1  M1 sin   5 sin 15.1   1.303

From Appendix B, for M n,l  1.302, M2 

M n,2 sin(    )



p2  1.805. Also, p1

0.786  4.48 sin(15.1  5)

p3 , the flow is expanded through an angle of 10. From Table C, for M 2  p1 4.48,  2  71.83 (nearest entry)

To calculate

3   2    71.83  10  81.38

Hence, M3  5.6 (nearest entry)

24 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

From Appendix A: For M1  5,

po1 p3

For M3  5.6,

 529.1

p o3 p3

From Appendix B: For M n1  1.303,

 1037 p o2 po1

 0.9794

Thus, p3 p3 po3 po2 po1  1     (1)(0.9794)(529.1)  0.5 p1 po3 po2 po1 p1  1037 

Hence, 2

2

  p1

 p1 

2

(1.4)(5)  D cyl 3  M1 3    17.9  D w  p2  p3  (1.805  0.5) 2

Note: This is why we try to avoid blunt leading edges on supersonic vehicles. (However, at hypersonic speeds, blunt leading edges are necessary to reduce the aerodynamic heating.)

The supersonic flow over a flat plate at a given angle of attack in a freestream with a given Mach number, M , is sketched below.

25 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

The flow direction downstream of the leading edge is given by line ab. The flow direction is below the horizontal (below the direction of M  ) because lift is produced on the flat plate, and due to overall momentum considerations, the downstream flow must be inclined slightly downward. Also, line ab is a slip line; the entropy in region 4 is different than in region 5 because the flows over the top and bottom of the plate have gone through shock waves of different strengths. The boundary condition that must hold across the slip line is constant pressure, i.e., p 4  p5. It is this boundary condition that fixes the strengths of the expansion wave and the shock wave at the trailing edge.

To calculate the trailing edge shock and expansion waves, and the flow direction downstream, use the following iterative approach: 1. Assume a value for .

2. Calculate the strength of the trailing edge shock for the local deflection angle - This gives, among other quantities, a value of p 4.

3. Calculate the strength of the trailing edge expansion wave for a local expansion angle of - This gives a value for p5.

4. Compare p 4 and p 5 from steps 3 and 4. If they are different, assume a new value of .

5. Repeat steps 2-4 until p4  p5. When this condition is satisfied, the iteration has converged, and the trailing edge flow is now determined.

26 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

1  sin 1 1/M1  sin 1 ( 1 2)  30 h1  Tan 1  Tan 30  0.5774 xA x a  h1 /0.5774  1.732 h1 For M1  2, 1   from Appendix C

 2    1  3  26.38  29.38 From Appendix C, for ν 2  29.38, M 2  

 1   1   2  sin 1  2   sin 1    28.29 M   2.11    2    28.29  3  25.29 y2  x A Tan   1.732 h1 Tan  25.29   0.8183 h1 y3  x A Tan   1.732 h1 Tan  3   0.0908 h1

y1  h1  y2  h1  0.8183 h1  0.1817 h1

h 2  y1  y2  y3  0.1817 h1  0.8183 h1  0.0908 h1  1.0908 h1 Assuming a linear variation of M over the distance y1 , the average value of M over y1 is

(2  2.11)/2  2.055. The average value of M over the whole vertical distance h 2 is:

 y   y  y3  M ave  2.055  1   2.11  2   h2   h2   0.1817 h1  (0.8183  0.908 h1 )  2.055    (2.11) 1.0908 h1  1.0908 h1 

 0.3423  1.7586  2.10

27 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Refer to the diagram in the solution for problem 18. 1 is the same value, 30. Also, the following same values carry over: h1 /x A  0.5774, x A  1.732 h1 , and 1  26.38  2    1  30  26.38  56.38 From Appendix C, for  2  56.38, M2  3.368

 1  1  1   2  sin 1    sin    17.27  3.368   M2     2    17.27  30  12.73

Note: The downstream trailing edge of the expansion wave is above the horizontal.

y 2  x A Tan   1.732 h1 Tan 12.73   0.391 h1 y3  x A Tan   1.732 h1 Tan  30   1.0 h1

h 2  h1  y3  h1  h1  2h1

The vertical distance through the expansion wave  h1  y2  h1  0.391 h1  1.391 h1. Again assuming a linear variation of M through the wave, the average value of M over the distance h1  y2 is (2  3.368)/2  2.684.

28 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

The average value of M over the whole vertical distance h 2 is:

 h  y2   y3  y 2  M ave  2.684  1   3.368    h2   h2   h  0.391 h1   h1  0.391 h1   2.684  1   3.368   2h1 2h1    

 2.89

From Appendix A: For M1  3,

po1 p1

 36.73

From Appendix C: At M1  3,  1  49.76

 2   1    49.76  20  69.76

2  4.3 Hence:

p o2 p2

 224.7 from Appendix A.

From the --M diagram, Fig. 9.9, with   20 and M 2  4.3, we have   31.5

Mn2  M2 sin   4.3 sin 31.5  2.246, From Appendix B, for Mn1  2.246, p3  5.74 p2 p3 

p3 p 2 p o  1  p1  (5.74)   (36.73)(1)  0.938 atm p 2 p o p1  224.7 

The pressure downstream of the shock wave is lower than the pressure upstream of the expansion wave, even though both flows are horizontal, because there is an entropy increase across the shock wave, hence the thermodynamic state of the downstream flow is different from that of the upstream flow.

29 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Step 1: Find  given  and M1 .

Since one cannot use a chart for this problem, it is necessary to solve the     M relation (Eq. 1) for  .

 M sin    1 tan    M   cos    2 2 1

2 1

2

(1)

2

This is accomplished by re-writing (1) as (2) and adjusting the value of  until the following is true:









tan    M12   cos 2   2   M12sin 2   1  0  

(2)

Whether or not one converges on the strong or weak solution depends on the initial guess for  . Choosing  slightly larger than sin 1  M1  (the Mach angle) will result in convergence on the weak solution whereas choosing  slightly smaller than 90 will result in convergence on the strong solution. Alternatively, one can specify M1 and  and adjust theta until (2) is true. In this case, the only sensible solutions are ones for which   0 .

There are several ways to perform the iteration. If you are using MATLAB you can use fzero and if you are using Excel you can use „goal seek‟. Both functions use Newton iteration to find  . If you are using another programming language you can write your own Newton solver (See Problem 9.33).

Step 2: Calculate the properties across the shock.

We learned in class that the change in pressure across the wave depends entirely on the Mach number of the flow that is perpendicular to the wave ( M n1 ). At this point, the problem becomes just like all of the other oblique shock problems. So, start by calculating M n1 .

Mn1  M1sin

Then, treat the problem like a normal shock:

p2 2 1 M 2n1  1 p1  1





Step 3: Repeat steps 1 and 2 for 0    max.

Note that you will need to record both strong and  weak each time you do step 1 and then calculate the pressure ratio for each in order to get both „halves‟ of (1).

Step 4: The solution for max    0 is the same as for 0     max so you can just copy your results from steps 1-3 to get the solution for negative turning angles.

30 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

The results of iterating on  are presented in Table .

Table 1 Results of iterative solution to BTM.

M1

β θ (deg) (deg)

θ BTM (eq. 3.2) (deg)

2 0.1 ‒89.42 2.18E-06 2 5 ‒62.50 1.10E-05 2 10 ‒41.29 ‒9.14E-06 2 15 ‒26.28 3.72E-05 2 20 ‒15.33 3.40E-05 2 25 ‒6.88 ‒2.40E-04 2 30 0.00 2.09E-06 2 35 5.75 7.19E-05 2 40 10.62 3.10E-05 2 45 14.76 3.62E-04 2 50 18.13 1.82E-05 2 55 20.74 2.26E-04 2 60 22.43 2.90E-04 2 65 22.97 −7.50E-05 2 70 22.15 7.81E-04 2 75 19.49 ‒4.05E-05 2 80 14.81 −2.36E-05 2 85 8.10 2.93E-04 2 90 0.00 3.47E-07 Finally, we plot the results in Figure :

0.00 5.75 10.62 14.76 18.13 20.74 22.43 22.97 22.15 19.49 14.81 8.10 0.00

θ (deg)

Mn1

0.00 ‒5.75 ‒10.62 ‒14.76 ‒18.13 ‒20.74 ‒22.43 ‒22.97 ‒22.15 ‒19.49 ‒14.81 ‒8.10 0.00

0.00 ‒0.17 0.17 ‒0.13 0.35 ‒0.03 0.52 0.15 0.68 0.38 0.85 0.67 1.00 1.00 1.15 1.37 1.29 1.76 1.41 2.17 1.53 2.57 1.64 2.96 1.73 3.33 1.81 3.67 1.88 3.95 1.93 4.19 1.97 4.36 1.99 4.46 2.00 4.50

P2/P1

Figure 1 Pressure-deflection diagram for air (= 1.4) at an initial Mach number of 2.

31 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Simply put, this is two oblique shock problems and one normal shock problem put together.

All you have to do to solve it is to work through things from left to right. The basic picture looks like this:

Figure 1 Basic ramjet inlet flow structure used to formulate solution to problem.

1. Find conditions in the free stream (region 1): g. You are given M1 , p1 , and T1. h. Compute the total temperature and pressure using the isentropic flow relations: 

   1 2   1 p01  p1 1  M1  2  

  1 2 T01  T1 1  M1  2   2. Find conditions downstream of the first oblique shock (region 2) Find 1 from  2 and M1 . This can be done using the charts or by iteratively finding  that satisfies:









tan  2   M12   cos 2   2   M12sin 2   1  0   i.

Next, find the normal component of the Mach number entering the shock ( M n1 ) using:

Mn1  M1sin j.

Compute the Mach number normal to the wave downstream of the shock

M n2 

1

 1 2

 M 2n1 

M 2n1

  1 2

32 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

k. Now, use M n1 in the normal shock relations to find the conditions downstream of the shock:

  2 2 p 2  p1 1  M n1 1    1 





2    2     1 M n1  2 2 T2  T1 1  M n1  1    2   1      1 M n1 





T02  T01

 p  p  p  p02  p01  1  2  02   p01   p1  p 2  





    1 2   1 2    1 2   1   p01 1  M1  1 M 2n1  1  1  M n2   2 2      1  l.





Finally, compute M 2 from M n1 using:

M2 

Mn2 sin     

3. Find conditions downstream of the second oblique shock (region 3) Use same procedure as for first oblique shock but with   30 and upstream conditions in region 2.

4. Find conditions downstream of the normal shock In this case we can use the normal shock relations directly to find conditions downstream:

  2 p 4  p3 1  M32  1    1 





2    2     1 M 3  2 2 T4  T3 1  M3  1    2   1      1 M 3 





m. compute M 4 from M 3 using:

M4 

1

 1 2

 M 32 

M 32

  1 2

33 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

5. Compute the entropy associated with each region in the ramjet inlet.

Use equation (7.25) for constant specific heats to compute the changes in entropy across each shock:

T  p  s12  cp ln  2   Rln  2   T1   p1 

T  p  s 23  c p ln  3   Rln  3   T2   p2  T  p  s34  c p ln  4   Rln  4   T3   p3 

Note: Assuming that R  287 J/kgK and   1.4 (for air) leads to cp  1004.5 J/kgK. n. Get entropy in a region by adding up entropy changes across all preceding shocks. Assume that the entropy in region 1 is zero

s1  0

s 2  s1  s12

s3  s 2  s 23 s 4  s3  s34

6. Use the expressions for a thermally and calorically perfect gas undergoing a reversible process (Eq. 7.25) to find the equations of lines of constant pressure in T-s space for each region. From 7.25, the change in entropy from some initial state (1) to some new state is given by: p T s  s1  c v ln    Rln    p1   T1 

If this process occurs at constant pressure, the first term on the RHS is zero (because p  p1 ) leading to: T s  s1  Rln    T1 

or s1  s T  T1e R

where s1 is the entropy at the particular flow region. Therefore, the equations for lines of constant pressure in T-s space are: 34 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

s1  s

Region 1: T  T1e R Region 2:

s2  s T  T2e R

Region 3:

s3  s T  T3e R

Region 4:

s4  s T  T4e R

7. Compute conditions at the exit of the inlet if recovery occurs via a single normal shock. In this case:

  2 p4  p1 1  M12  1    1 





2    2     1 M1  2 2 T4  T1 1  M1  1    2   1      1 M1 





T04  T01

M4 

1

 1 2

 M12 

M12

  1 2 





    1 2   1 2    1 2   1  p4  p01 1  M1  1 M12  1  1  M4   2 2      1 





8. Finally, compute everything, put in a table, and plot. The results are summarized in 9. Table and plotted in Figure . Entries in bold are the data given in the problem statement.

Entries in red are adjusted iteratively until  it (green entries) is zero.  it is the value of the left hand side of Eq. 2a above. I used Excel‟s goal seek to do the iteration (ie. adjust  until it  0 but you could do the same thing using fzero() in MATLAB. The  it entries appear in the lower „utility‟ table because they aren‟t really a property of the regions. The second row of the „utility‟ table gives the component of the Mach number normal to the shock and downstream of it. It is put here so as not to be confused with normal Mach numbers upstream of the shock waves. Mn2 is used to compute the Mach number of the flow downstream of the oblique shock as well as the total pressure downstream of the shock.

35 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Table 1 Summary of conditions in the ramjet inlet.

Region Param 1

2

3

4

4NS

M

3.5

2.91

1.35

0.76

p T

26500 223.3 2021215.

9 770.385 0 24.37721 1.444596 4 0

60,102.05 286.64 1,912,806.4 4 770.39 10 53.14

369,132.85 564.73 1,094,474.8 6 770.39 30 90

726,076.20 691.30 1,060,806.2 6 770.39 0 N/A

0.45 374,312.5 0 740.25 430,413.0 2 770.39 0 N/A

2.33 15.82

1.35 176.05

N/A 185.02

N/A 443.91

po To ∆θ β Mn s

θit Mn2

– 0.000311 2 0.721654 9

Units

N/m2 K

K deg deg

J/kgK

–0.0003778 0.531429

36 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Figure 2 T-S diagram for Ramjet Inlet. Solid blue line shows the path of the compression process through the inlet. Dashed blue line shows path of compression process via a single normal shock. Dotted red line shows the total temperature of the flow which is constant because the compression process is adiabatic. The dashed green lines are isobars (lines of constant pressure) associated with each step of the compression process. The pressures associated with each isobar are shown at the ends of the lines. Static pressure increases through the inlet whereas total pressure decreases. The general relationship of the pressures is given by: p01  p04  p4  p04,NS  p3  p2  p1. Note that p4,NS  p3 but p4,NS and p3 are close enough (within 1% of each-other) that the two isobars are basically on top of each other in the chart. The difference between the static and total temperature at each station equals u2 / 2cp . The purpose of an inlet is to slow the flow with as little pressure loss as possible.

Therefore, recovery through a system of oblique shocks is preferred to recovery through a single normal shock. The system of oblique shocks is more efficient because it produces less entropy than a single normal shock.

T2 2 T1 We use Appendix B to look up the Mach number associated with a static temperature ratio of 2. It isn‟t exact but M = 2.4 is pretty close (within < 2%). Since property changes are determined by the normal component of the upstream Mach number, this Mach number is the normal component of the Mach number. Therefore,

Mn,1  2.4 37 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Since

M n,1  M1 sin    Solving for M1 gives: M1 

Mn,1

sin   



2.4  4.95 sin  29 

Note that choosing Mn,1  2.4 is an approximation. Choosing Mn,1  2.357 gives a static temperature ratio that is closer (within <.02%) to the target of 2. This leads to M1  4.86 which is closer to the exact answer. The difference is relatively small (<2%) in this case so choosing Mn,1  2.4 is good enough. However, it illustrates an important aspect of problems like these which is that because the governing equations are non-linear.

Step through the shocks sequentially from left to right. Beginning with the first oblique shock:

M  3   10   27.38 M n,  M sin     1.38    1 1  M 2n,    2   0.748 M n,1     1  M 2n,     2  M1 

M n,1

sin     

 2.505

We use the same procedure to find M 2 noting that while the angle of the second ramp is 20  with respect to the horizontal, the flow is only deflected by 10 as it passes from the first to the 

second part of the ramp. Thus  2  10 for the purpose of evaluating the     M relation.

M1  2.505 2  10  2  31.80 Mn,12  M1 sin  2   1.319

38 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

M n,2 

M2 

   1 1  M 2n,1    2   0.776    1  M 2n,1     2  M n,2

sin   2  2 

 2.09

The last shock is a normal shock so we can just use the expression for the downstream Mach number directly:

M3 

   1 1  M 22    2   0.563    1  M 22     2 

Figures 9.23 and 9.24 in the text show us that the two-dimensional bow shock encapsulates all possible oblique shock solutions for a given free stream Mach number. In the case of the strut illustrated here, the flow along the plane of symmetry experiences a normal shock as it decelerates to meet the tip of the projectile whereas the flow far away from the strut experiences a Mach wave.

The latter is the key insight needed to solve this problem. The angle of the bow shock far away from the strut is the Mach angle. Therefore:  1  26  sin 1    M1 

or M1 

1

 

sin 26

 2.28

The speed of the flow is the Mach number times the local speed of sound or:

 J  m u1  M1  RT1   2.28  1.4   288   300 K   793 kgK  s  As a final note, what if this had been a body of revolution (like a bullet) instead of a strut? The flow around the bullet would be three-dimensional and not planar like the flow around the strut. Since 3-D shocks are weaker than 2-D shocks at the same Mach number, the speed of the flow around the bullet would be higher than the speed of the flow around the strut.

The first step is to sketch a representative streamline in order to develop a physical understanding of what the flow is doing. The first oblique shock turns the flow toward itself 39 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

(ie. downward). The reflected oblique shock forms in order to turn the flow back to the horizontal reflecting the fact that the flow cannot pass through a solid boundary. A representative streamline is sketched in Figure 1.

Figure 1. A representative streamline through the reflected shock system.

Next, we add the normal components of the Mach number because we know we will need these to compute the pressure and temperature changes across the waves. We also add r which is the angle between the streamline in region 2 and the reflected wave. It is very important to recognize that r does not necessarily equal Φr and that the angle between the incident and reflected waves is not necessarily 90 even though it may look like it in the sketch.

Figure 2. Reflected shock system with normal components of Mach number.

Now that the geometry is established, we can go about solving the problem which just boils down to two oblique shocks in series. This begins by finding the upstream Mach number.

Since we are given the shock and turning angles, we use the     M relation to find that M1  2.0.

Now we begin work on the first (or incident) shock. The normal component of the Mach number is: M n.1  M1sini  1.286

40 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

The normal component of the Mach number downstream of the incident shock is: M 2n 2,i 

1

 1 2

M 2n,1

 1  M 2n,1 

 M n 2i  0.793

2

From this, we calculate the Mach number in region 2 downstream of the incident shock: M n,2

M2 



sin  i  2 

0.7934  1.614 sin  40  10.62 

Knowing the Mach number in region 2 and the amount of turning that is needed to straighten the flow out ( 10.62 ) enables us to use the     M relation again to find  r  51.385.

Note that while it was fairly straightforward to use the     M plot to find M1 , finding  r graphically is harder because one must interpolate between contours of Mach number.

Thus, we solved the     M relation numerically.

We compute the normal component of the Mach number with respect to the reflected wave:

Mn2,r  M2sinr  1.260 and use it to compute the normal component of the Mach downstream of the reflected wave as well as the Mach number in region 3: M 2n3 

1

 1 2

M 2n,2

 1  M 2n,2 

 M n3  0.806

2

M3 

Mn,3

sin  r  r 

 1.236

Our last task is to compute the temperature and pressure changes across the shock system so that we can find the conditions in region 3. To do this, we use the expressions for the static temperature and pressure ratios across a normal shock with the normal components of the Mach number. Therefore: 2   2     1 M n1  T2  2  1  M 2n1  1     1.182 2 T1    1      1 M n1 





2 T3    2     1 M n2,r  2  1  M 2n2,r  1     1.167 2 T2    1      1 M n2,r 





41 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

and

p2 2 1 M 2n1  1  1.761 p1  1





p3 2 1 M 2n2,r  1  1.693 p2  1





Alternatively, the values of these ratios can also be found by interpolating in Appendix B. Finally, putting it all together gives:  T  T  T3  T1  2   3    300 K 1.182 1.167   413.8 K  T1   T2   P  P  P3  P1  2   3   1 atm 1.7611.693  2.981 atm  P1   P2 

Begin the solution by sketching the flow field along with two streamlines: one passing through both shocks formed by the ramps and one passing through the single shock that begins where the two lower shocks intersect.

Figure 1. Shock system associated with the sequential ramps.

Next, calculate the pressures in regions 3 and 5. Starting with the flow along the lower streamline, we find the pressure in region 2 downstream of the first oblique shock wave. The normal component of the upstream Mach number is: M n.1l  M1lsin1  1.379

42 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

where 1  27.4 from the     M relation ( 1l  10 and component of the Mach number downstream of the first shock is:

M 2n



1

 1 2

M 2n,1l

 1  M 2n,1l 

M1l  2 ).

The normal

 M n  0.749

2

From this, we calculate the Mach number in region 2 downstream of the incident shock:

M2 

Mn 0.749   2.508 sin  1  1l  sin  27.4  10 

We repeat the procedure for the second shock where  2  10 and M 2  2.508 leads to  2  31.8.

M n.2  M 2sin 2  1.321

The normal component of the Mach number downstream of the second shock is:

M 2n



1

 1 2

M 2n,2

 1  M 2n,2 

 M n  0.775

2

The Mach number in region 3 downstream of the second shock is:

M3 

Mn 0.775   2.090 sin   2   2l  sin  31.8  10 

The static pressure ratios across each shock are:

p2 2 1 M 2n1  1  2.052 p1  1





p3 2 1 M 2n2  1  1.869 p2  1





The pressure in region 3 is given by:

p3  p1

p 2 p3  1 atm  2.052 1.869   3.836 atm p1 p2

43 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Now we turn our attention to the flow above the intersection point where 1u  20 and M1u  3 leads to 1u  37.8 assuming the weak solution. The normal component of the Mach number with respect to the upper wave is: M n.1u  M1u sin1u  1.837

The normal component of the Mach number downstream of the upper shock is:

M 2n



1

 1 2

M 2n,1u

 M nu  0.775

 1  M 2n,1  2

The static pressure ratio across the upper shock is:

p5 2 1 M 2nu  1  3.771 p1  1





So, the pressure in region 5 is:

p5  p1

p5  1 atm  3.771  3.771 atm p1

Since p 5 is slightly lower than p 3 , the boundary conditions across the slip line, which are that the pressures and flow directions in regions 4 and 5 are equal, are not satisfied. Two things happen to satisfy the boundary conditions. First, the angle of the upper oblique shock increases a little in order to raise the pressure a little. This also has the effect of increasing the angle of the flow in region 5 ( 5 ) a little. Second, an expansion wave forms at the intersection of the two lower oblique shocks to lower the pressure of the flow in region 5 a little and make  4  5 . Therefore, the reflected wave referred to in the problem statement is a weak expansion wave. Note however, that this is because we chose the weak solution for the upper oblique shock wave. Choosing the strong solution, would have resulted in a situation where p5  p3 and a shock wave emanating from the intersection point would be needed to further raise the pressure of the lower stream.

The Mach number downstream of the expansion is related to the ratio of the total to static pressures as follows: 

pO2    1 2   1  1  M2  p2 2  

44 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Solving for M 2 gives:

M2 

 1   2  pO2    1      1  p 2   

1.4 1   2  151.835 atm  1.4  1  8.00   1.4  1  0.01555 atm   

From the Prandtl-Meyer function:

    M 2     M1  So,

  M1    M 2     95.625  30  65.625 M1  4.00 The total pressure is constant across the expansion wave so:

p1  p 2

pO2 pO1 p1  1   0.0155 atm  9763 1    1.00 atm p 2 pO2 pO1  151.8 

In an expansion around a sharp corner, all of the Mach waves associated with the turning process originate from the corner. However, in an expansion around a smooth corner, the Mach waves originate from different points along the smooth contour as illustrated below.

Figure 1. Sketch of flow structures along a smooth expansion corner.

Even though the Mach waves are now distributed along the contour, the overall physics of the problem are the same and thus the process of finding the conditions downstream is the same.

So,

    M 2     M1  So,

  M 2    M1     40.28  15  55.28 M 2  3.30

45 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

The static pressure downstream of the expansion is given by:

p 2  p1

pO1 pO2 p 2  1   1 atm 18.47 1    0.321 atm p1 pO1 pO2  57.50 

The Mach angles of the leading and triling Mach lines are:  1  1  1    sin    23.1  2.55   M1 

1  sin 1 

 1  1  1    sin    17.6  3.30   M2 

2  sin 1 

Figure 1. Flow structure downstream of an isentropic compression region.

The isentropic compression wave can also be thought of as the opposite of an expansion corner so the angle through which the flow is turned can be thought of as negative. Therefore:

  M 2     M1   Δ So,

  M 2     4.1  10  67.08  10  57.08 M 2  3.4 The compression process is isentropic so the total pressure remains constant through the compression wave. Therefore, the static pressure downstream of the wave is:

p 2  p1

pO1 pO2 p 2  1   1 atm 173.31    2.621 atm p1 pO1 pO2  66.12 

The Mach angles upstream and downstream of the compression region are:  1  1  1    sin    14.1  4.1   M1 

1  sin 1 

 1  1  1    sin    17.1  3.4   M2 

2  sin 1 

46 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

The coalescence point breaks the flow into at least two regions downstream: one associated with flow that passes through the system of Mach waves (this is the part of the problem addressed in problem 9.29) and another associated with flow that passes through the oblique shock. Since the two flows experience different processes thermodynamically (the entropy of the upper flow increases because it passes through the shock whereas the entropy of the lower flow does not), a slip line must exist between these regions. Thus, Figure 1 provides a more complete representation of the flow structure.

Figure 1. Flow structures associated with turning through a smooth compression corner.

However, it is not clear whether other types of structures also emanate from the coalescence point to satisfy the conditions needed across a slip line – namely that the pressures and flow directions on each side of the line must be the same. We must calculate the conditions downstream of the isentropic compression wave and oblique shock in order to find out.

From the solution to Problem 9.29, we know that: M2  3.4 and p2  2.621 atm . The total pressure downstream of the compression wave is given by: p  pO2  pO1  p1  O1   1 atm 173.3  173.3 atm  p1 

Now let‟s calculate the conditions downstream of the oblique shock that forms above the coalescence point. The normal component of the upstream Mach number is: M n.1  M1sin1  1.536

where 1  22 from the     M relation ( 1  10 and component of the Mach number downstream of the shock is: M 2n3 

1

 1 2

M 2n,1

 1  M 2n,1 

M1  4.1 ).

The normal

 M n3  0.689

2

47 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

From this, we calculate the Mach number in region 2 downstream of the incident shock:

M3 

M n3 0.689   3.31 sin  1  1  sin  22  10 

The static pressure downstream of the oblique shock is given by: p  p3  p1  3   1 atm  2.585   2.585  p1 

where p3 /p2 is the static pressure ratio across a normal shock with M n.1  1.536 .

The total pressure downstream of the oblique shock is given by: p  pO3  pO1  O3   173.3 atm  0.940   162.6 atm  pO1 

where pO3 /pO2 is the static pressure ratio across a normal shock with M n.1  1.536 . Now let‟s summarize our findings in a table.

Table 1. Differences in flow properties across the slip line.

M p pO



2 3.4 2.621 173.3 10

> > > =

3 3.31 2.585 162.6 10

atm atm °

As you might have suspected, the pressures in region 2 downstream of the isentropic compression wave are higher than the pressures behind the oblique shock. This is because the isentropic process recovers pressure more efficiently than the non-isentropic one. However, this poses a problem for the existence of a slip line because even though the flow directions are the same, the pressures on either side of the slip line are different. Therefore, another set of structures must form to equilibrate the pressures while preserving the flow direction. This can be accomplished by an expansion wave emanating from the coalescence point into the lower part of the flow followed by an oblique shock that turns the flow back to 10 as illustrated in Figure 2. Finding this combination requires a numerical solution.

Newton‟s method is an iterative process for finding the roots (or zeros) of a function. In this case the function is:

f     tan    2cot   

M12sin 2     1

M12   cos  2    2

48 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

(1)

and we are looking for values of  that make f     0 . Figure 1 shows how f    varies with  for a turning angle of 15 and two different Mach numbers (1.8 and 4).

Figure 1. Plot of f    vs  for   15 and two different Mach numbers (1.8 and 4) showing how the solutions to the     M relation manifest as zeros on the plot.

The basic idea behind Newton‟s method is to evaluate the function at two points and then try to predict the value of  that makes f     0 by drawing a line between the two points that intersects the horizontal axis. The function is then evaluated again, a new line is drawn, and the estimate is updated until the estimate converges to the solution. The process is illustrated graphically in Figure 2.

Figure 2. Plot of f    vs  for   15 and M  4 illustrating the use of Newton‟s method to find the zeros of the function. In this case, we are finding the solution on the left or 49 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

the „weak‟ solution. The procedure for finding the strong solution is the same but begins on the right hand side of the plot.

The first step of the process is to choose (somewhat arbitrarily) two points values of  for which to evaluate f    . The slope of the straight line between them is:

mo 

f   2   f  1 

 2  1

(2)

where the subscript „o‟ indicates that this is the starting point or „zeroth‟ iteration. The equation of the straight line between these points is: f o  mo   b o

(3)

where bo is the y-intercept. Note that it can be calculated using either the first or second point ie:

bo  f  1   mo 1  f   2   mo  2

(4)

Since we are looking for the value of  that makes f     0 , we set the LHS of Equation 3 to zero and rearrange to make a new „guess‟ for what value of  will make f     0. Thus:

3  

bo mo

(5)

At this point, we check to see if f  3   Tol where Tol is the convergence tolerance. If it is, we are done. If not, then we repeat the process again:

m1 

f  3   f   2 

3   2

(6)

Where the subscript „o‟ indicates that this is the starting point or „zeroth‟ iteration. The equation of the straight line between these points is: f1  m1  b1

(7)

where bo is the y-intercept. Note that it can be calculated using either the first or second point ie:

b1  f   2   m12  f  3   m13

(8)

Since we are looking for the value of  that makes f     0 , we set the LHS of Equation 7 to zero and rearrange to make a new „guess‟ for what value of  will make f     0. Thus: 50 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

4  

b1 m1

(9)

Again, we check to see if f   4   Tol . If the answer is „yes‟ then we are done. If the answer is „no‟ we repeat the process is until f   n   Tol . The value of n depends on Tol (ie. how close you want to get to zero) but convergence is quick and n is usually < 10 for Tol  10 7 .

Below is the code used to implement the above algorithm in Excel using Visual Basic. You may do this yourself by going to the Developer tab1 and clicking „Visual Basic‟. This creates a new window with a new set of commands. Clicking Insert->Module creates a window in which you can create your own functions that can be called from a cell in a spreadsheet. Two functions were created to solve this problem. BTM has arguments beta, theta, and M and is used to evaluate Equation 1. Note that gamma = 1.4 is assumed. Find Beta_BTM solves the     M relation for  and has arguments theta, M, and S. Theta and M are the flow turning angle and Mach number respectively. S is a flag that determines which solution is provided. If S = 1, the strong solution is provided. Otherwise, the weak solution is provided.

Finally, this example is implemented in Excel because it is probably available to the largest number of students. However, there are many other ways that this algorithm can be implemented. MATLAB is particularly convenient because it has a built-in Newton solver called fzero.

Public Function BTM(Beta, theta, M) 'computes the value of the theta-beta-Mach number relation 'used in iterative procedure where a solution to the beta-theta-M 'relation corresponds to BTM = 0 'Beta and theta are assumed to be in DEGREES ' 'convert beta and theta to radians B = Beta * 3.14159 / 180 T = theta * 3.14159 / 180 ' gamma = 1.4 A1 = (M * M * Sin(B) * Sin(B)) - 1 A2 = M * M * (gamma + Cos(2 * B)) A3 = 2 / Tan(B) BTM = Tan(T) - (A3 * A1 / (A2 + 2)) End Function Public Function FindBeta_BTM(Th, M, S) 'Uses Newton iteration to find the value of beta associated 'particular Mach number (M)and turning angle (Th).

'Works with function BTM. Solution <=> BTM=0.

'If S=1, find strong solution; otherwise find weak solution Db = 1 'Initial perturbation of theta to calculate derivative Etol = 0.0000001 'Convergence tolerance

with

1

If the Developer tab doesn‟t show on your version of Excel, go to File->Options->Customize Ribbon and add „Developer‟.

51 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

a

If S = 1 Then B1 = 89 V1 = BTM(B1, Th, M) B2 = B1 − Db V2 = BTM(B2, Th, M) While Abs(V2) > Etol Slope = (V2 − V1) / (B2 − B1) Intercept = V2 − (Slope * B2) B1 = B2 V1 = V2 B2 = − Intercept / Slope V2 = BTM(B2, Th, M) Wend Else B1 = 1 V1 = BTM(B1, Th, M) B2 = B1 + Db V2 = BTM(B2, Th, M) While Abs(V2) > Etol Slope = (V2 − V1) / (B2 − B1) Intercept = V2 − (Slope * B2) B1 = B2 V1 = V2 B2 = −Intercept / Slope V2 = BTM(B2, Th, M) Wend End If FindBeta_BTM = B2 End Function

Newton‟s method is an iterative process for finding the roots (or zeros) of a function. In this case the function is:

Err  M       M 

(1)

where   M  is the Prandtl-Meyer function:

 M 

  1 1   1 2 tan M  1  tan 1 M 2  1  1  1





(2)

Note that Equation 2 gives the value of the Prandtl-Meyer function in radians so we will need to convert it to degrees (multiply Equation 2 by 180/ ) if we wish to work with angles in degrees in Equation 1. We are looking for values of M that make Err  M   0 . Figure 1 shows how Err  M  varies with M for values of the Prandtl-Meyer function of 40 and 76 . The plot shows that the zeros are at about M  5 for   76 and M  2.5 for   40 .

Unlike the     M relation, there is only one solution.

52 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Figure 1. Plot of Error vs M for   76 and   40 showing how the solutions to Equation 1 manifest as zeros on the plot.

The basic idea behind Newton‟s method is to evaluate the function at two points and then try to predict the value of M that makes Error  0 by drawing a line between the two points that intersects the horizontal axis. The function is then evaluated again, a new line is drawn, and the estimate is updated until the estimate converges to the solution. The process is illustrated graphically in Figure 2.

Figure 2. Plot of Err  M  vs M for   76 illustrating the use of Newton‟s method to find the Mach number associated with Err  0.

The first step of the process is to choose (somewhat arbitrarily) two points values of M for which to evaluate Err . The slope of the straight line between them is:

mo 

Err  M 2   Err  M1  M 2  M1

53 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

(3)

where the subscript „o‟ indicates that this is the starting point or „zeroth‟ iteration. The equation of the straight line between these points is: Erro  mo M  bo

(4)

where b o is the y-intercept. Note that it can be calculated using either the first or second point ie:

bo  Err  M1   mo M1  Err  M 2   mo M 2

(5)

Since we are looking for the value of M that makes Err  M   0, we set the LHS of Equation 4 to zero and rearrange to make a new „guess‟ for what value of M will make Err  M   0.

Thus:

M3  

bo mo

(6)

At this point, we check to see if Err  M3   Tol where Tol is the convergence tolerance. If it is, we are done. If not, then we repeat the process again:

m1 

Err  M3   Err  M 2 

(7)

M3  M 2

Where the subscript „o‟ indicates that this is the starting point or „zeroth‟ iteration. The equation of the straight line between these points is: Err1  m1M  b1

(8)

where b o is the y-intercept. Note that it can be calculated using either the first or second point ie:

b1  Err  M 2   m1M 2  Err  M3   m1M3

(9)

Since we are looking for the value of M that makes Err  M   0 , we set the LHS of Equation 8 to zero and rearrange to make a new „guess‟ for what value of M will make Err  M   0. Thus:

M4  

b1 m1

(10)

Again, we check to see if Err  M 4   Tol . If the answer is „yes‟ then we are done. If the answer is „no‟ we repeat the process is until Err  M n   Tol . The value of n depends on Tol 54 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

(ie. how close you want to get to zero) but convergence is quick and n is usually < 10 for Tol  10 7.

Below is the code used to implement the above algorithm in Excel using Visual Basic. You may do this yourself by going to the Developer tab2 and clicking „Visual Basic‟. This creates a new window with a new set of commands. Clicking Insert->Module creates a window in which you can create your own functions that can be called from a cell in a spreadsheet. Two functions were created to solve this problem. PMF has arguments gamma (specific heat ratio) and M (Mach number) and returns the value of the Prandtl-Meyer function in radians.

FindM_PMF has arguments theta (value of the Prandtl-Meyer function in radians) and gamma (specific heat ratio). It finds the Mach number associated with theta.

Finally, this example is implemented in Excel because it is probably available to the largest number of students. However, there are many other ways that this algorithm can be implemented. MATLAB is particularly convenient because it has a built-in Newton solver already called fzero.

Public Function PMF(gamma, M) 'Computes the value of the Prandtl-Meyer function in RADIANS. 'Call this iteratively to find M for a given deflection angle A = (gamma + 1) / (gamma − 1) B = (gamma − 1) / (gamma + 1) C = (M * M) - 1 PMF = Sqr(A) * Atn(Sqr(B * C)) − Atn(Sqr(C)) End Function Public Function FindM_PMF(theta, gamma) 'Uses Newton iteration to finds the value of M associated with a 'particular value of theta in RADIANS. 'Works with the function PMF. Etol = 0.0000001 'error tolerance dM = 0.2 'Initial perturbation in M used to compute slope M1 = 1.1 'Initial guess for M1 V1 = PMF(gamma, M1) Err1 = Abs(theta − V1) M2 = M1 + dM V2 = PMF(gamma, M2) Err2 = Abs(theta − V2) While Err2 > Etol V2 = PMF(gamma, M2) Err2 = Abs(theta − V2) Slope = (Err2 − Err1) / (M2 − M1) Intercept = Err2 − (Slope * M2) M1 = M2 2

If the Developer tab doesn‟t show on your version of Excel, go to File->Options->Customize Ribbon and add „Developer‟.

55 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Err1 = Err2

56 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

M2 = -Intercept

/ Slope V2 = PMF(gamma, M2) Err2 = Abs(theta − V2) Wend FindM_PMF = M2 End Function Figure 1 is a „generic‟ sketch of the flow field valid for   LE . For   LE the shock on the upper surface becomes an expansion wave. The angles associated with the leading and trailing edges are:

 y /2  1  0.06/2  LE  tan 1    tan    5.143  x   1/3   y /2  1  0.06/2  TE  tan 1    tan    2.577 1  x   1  1/3 

Figure 1. Sketch of flow structures associated with angles of attack less than and greater than the half angle of the airfoil leading edge. For clarity, the trailing shocks present at the trailing edge are not shown. Also shown are triangles used to determine  LE and TE from x and y.

The overall approach is to compute the normal and axial components of the force acting on the airfoil at each angle of attack and then use these to compute the lift and drag forces using:

L  Ncos  Asin 

(1)

D  N sin  A cos 

(2)

The normal and axial components of the force per unit span are:

N   p2'  p2  xc   p3'  p3  1  x  c

(3)

A   p2'  p2  yc   p3'  p3  yc

(4)

57 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Our task now is to calculate the pressures on each surface of the airfoil. Let‟s start with the case illustrated on the LHs of Figure 1 where  LE . The angle of the flow behind the upper oblique shock is given by: 2  LE  

(5)

Find 1 from  2 and M1 using the function you created in Problem 9.33. Next, find the normal component of the Mach number entering the shock ( M n1 ) using: M n1  M1sin1

(6)

Compute the Mach number normal to the wave downstream of the shock

M n2 

1

 1 2

 M 2n1 

M 2n1

(7)

  1 2

Now, use M n1 in the normal shock relations to find the pressure downstream of the shock:

  2 2 p 2  p1 1  M n1 1    1 





(8)

Next, compute M 2 from M n1 using:

M2 

Mn2 sin     2 

(9)

The angle that the flow turns through as it passes from surface 2 to 3 is given by:  23   LE  TE  7.719

(10)

Conveniently, this angle is independent of . Use the function you created in Problem 9.34 to find the value of M 3 that satisfies:

  M3     M 2   Δ 23

(11)

The pressure in region 3 is given by:  p   p  p  p3  p 2  O2   O3  3   p 2  pO2  pO3 

(12)

Or, 



  1 2   1    1 2   1  p3  p 2  1  M 2  1 1  M3  2 2     58 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

(13)

59 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Turning our attention to the lower surface:  2 '   LE  

(14)

Find 1' from  2 ' and M1 using the function you created in Problem 9.33. Next, find the normal component of the Mach number entering the shock ( M n1' ) using: M n1'  M1sin1'

(15)

Compute the Mach number normal to the wave downstream of the shock

Mn2' 

1

 1 2

 M 2n1' 

M 2n1'

(16)

  1 2

Now, use M n1' in the normal shock relations to find the pressure in region 2‟ downstream of the shock:

  2 2 p 2 '  p1 1  M n1' 1    1 





(17)

Next, compute M 2 ' from M n1' using:

M2' 

Mn2 ' sin     2 ' 

(18)

The angle that the flow turns through as it passes from surface 2‟ to 3‟ is the same as on the upper surface of the airfoil so:  23'   LE  TE  7.719

(19)

As before, use the function you created in Problem 9.34 to find the value of M 3 ' that satisfies:

  M3'     M 2'   Δ23' The pressure in region 3‟ is given by:  p  p  p  p3'  p 2 '  O2 '   O3'   3'   p 2 '   pO2 '   pO3' 

(20)

Or, 



  1 2   1    1 2   1  p3'  p 2 ' 1  M 2 '  1 1  M 3'  2 2     60 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

(21)

Substituting the pressures from Equations 21 and 13 into Equations 3 and 4 gives the normal and axial forces. These, in turn, are used in Equations 1 and 2 to compute the lift and drag.

The solution process for   LE is basically the same as above except that the oblique shock on the upper surface is replaced by an expansion wave (see the RHS of Figure 1). The results of these calculations are summarized below in Tables 1 through 3 and the lift/drag ratio is plotted in Figure 2.

The maximum value of  that can be considered in the context of the planar shock waves assumed in shock-expansion theory is set by the maximum turning angle associated with a planar wave. For M1  2, max  23. Maximum flow turning occurs on the lower leading edge so from Equation 14:

 max  max  LE  23  5.143  17.857 Table 1. Flow properties along the upper surface of the airfoil at various angles of attack. Note that the upper oblique shock is replaced by an expansion wave between   5, and   6.

As a result, different methods – and different table entries – are used to compute the flow properties in region 2. An expansion wave always separates regions 2 and 3 so the angle of attack does not affect the methods used to calculate flow properties in region 3.

2, upper alpha (deg)

Theta_2 (deg)

3, upper Beta

p2 (atm)

PMF2 (deg)

p3 (atm)

Mn1

Mn2

M2

1.13

0.89

1.82

1.33

21.19

2.09

0.86

M3

0

5.14

34.43

1

4.14

33.52

1.10

0.91

1.85

1.26

22.21

2.13

0.81

3.14

32.63

1.08

0.93

1.89

1.19

23.23

2.17

0.77

3

2.14

31.77

1.05

0.95

1.92

1.13

24.23

2.21

0.72

4 5

1.14 0.14

30.93

1.03

0.97

30.11

1.00

1.00

1.96 1.99

1.07 1.01

25.24 26.24

2.25 2.29

0.68 0.64

2

Theta_2 (deg)

PMF1 (deg)

6

–0.86

26.38

2.03

0.95

27.24

2.33

0.60

7

–1.86

26.38

2.07

0.90

28.24

2.37

0.56

8

–2.86

26.38

2.11

0.85

29.24

2.41

0.53

9

–3.86

26.38

2.14

0.80

30.24

2.45

0.49

–4.86

26.38

2.18

0.75

31.24

2.49

0.46

10

61 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Table 2. Flow properties along the lower surface of the airfoil. The structure of the flow remains the same for all angles of attack so there are no discontinuities as in Table 1. The maximum possible angle of attack that still leads to planar shock waves is 17.857. Therefore, the use of shock-expansion theory in the range 0    10 is justified.

2, lower

3, lower

alpha (deg)

Theta_2 (deg)

Beta

Mn1

Mn2

M2

p2 (atm)

PMF2 (deg)

M3

p3 (atm)

0

5.14

34.43

1.13

0.89

1.82

1.33

21.19

2.09

0.86

1

6.14

35.38

1.16

0.87

1.78

1.40

20.16

2.05

0.91

2

7.14

36.35

1.19

0.85

1.74

1.47

19.12

2.02

0.97

3

8.14

37.36

1.21

0.83

1.71

1.55

18.06

1.98

1.03

4

9.14

38.39

1.24

0.82

1.67

1.63

16.99

1.94

1.08

10.14

39.47

1.27

0.80

1.64

1.72

15.90

1.90

1.15

6

11.14

40.58

1.30

0.79

1.60

1.81

14.79

1.86

1.21

7

12.14

41.74

1.33

0.77

1.56

1.90

13.67

1.82

1.28

8

13.14

42.95

1.36

0.76

1.52

2.00

12.52

1.78

1.35

9

14.14

44.21

1.39

0.74

1.48

2.10

11.34

1.74

1.43

15.14

45.54

1.43

0.73

1.44

2.21

10.14

1.70

1.50

5

10

Table 3. Forces calculated from the results of Tables 1 and 2 using Equations 1-4.

Forces alpha (deg)

N (N)

A (N)

L (N)

D (N)

L/D

0

0.00

2,811.57

0.00

2,811.57

0.00

1

11,584.98

2,813.56

11,534.11

3,015.32

3.83

2

23,190.73

2,819.56

23,078.20

3,627.18

6.36

3 4

34,838.28 46,549.17

2,829.61

34,642.44

4,649.03

5

58,345.77

2,843.83 2,862.36

46,237.40 57,874.28

6,084.00 7,936.63

7.45 7.60

2,885.42

69,565.11

6

70,251.57

7

82,291.09

8

94,490.99 106,881.2 8

10,212.9 0 12,920.3 7 16,068.4 6 19,668.8 3

2,913.33

9

81,322.66

2,946.54

93,161.33

2,985.69

105,098.33

7.29 6.81 6.29 5.80 5.34

62 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

10

119,496.5 9

3,031.72

117,154.72

23,736.0 1

4.94

63 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Figure 2. Plot of lift to drag ratio as a function of angle of attack.

The flow through an expansion wave is normally isentropic. If condensation occurs, however, the flow is no longer isentropic. We can assume, never-the-less, that Eq. (7.22) holds just as the fluid element enters the front of the expansion, and use this equation to obtain an initial rate of change of pressure with temperature. From Eq. (7.22)

ds  c p

dT dp R 0 T p

Also, from Eq. (7.9),

Cp   R  1 Thus,

 R dT  R dp  0  1 T p  dT dp  0  1 T p dp p       dT T    1  For an ambient pressure of 1 atm  2116 lb/ft 2, and a temperature of 520R, we have

dp    p  1.4  2116 lb   14.24 2    dT    1  T  0.4  520 ft R 64 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

From the given saturation curve information, at point A, pH 2O  38.16 lb/ft .

2

Thus, the mole fraction of water vapor in the air is

X H2O 

p H 2O p

Since dp/dT  14.24 dp H2O dT



38.16  0.018 2116

lb then ft R

 X H2O

2

dp lb  (0.018)(14.24)  0.256 2 dT ft R

This slope is smaller than the slope of the saturation curve at point A, given as 1.08

lb . ft R 2

Hence, the fluid element will penetrate into the liquid-vapor region of the phase diagram as sketched below, and water condensation will occur in the expansion wave. (b) Considering a normal shock with conditions upstream of the wave being M1  1.2, p1  2116

lb lb , pH 2O  38.16 2 , and T1  520R. Assume the upstream flow is in the 2 ft ft

liquid-vapor portion of the phase diagram, just to the left of point A. From Appendix B for normal shock properties,

p2 lb  1.513; hence p2  (1.513)(2116)  3202 2 p1 ft T2  1.128; hence T2  (1.128)(520) = 586.6R T1 Thus, behind the shock, p H2O  X H2O p 2  (0.018)(3202)  57.63

lb ft 2

Across the shock, we have

p H2 O T



57.63  38.16 lb  0.2923 2 586.6  520 ft R

This slope is smaller than the slope of the saturation curve at point A, given as 1.08

lb . ft R 2

Hence, the fluid element initially with properties just to the left of point A in the phase diagram, hence with some condensed water, will exit into the purely vapor region, and evaporation will occur across the shock wave.

65 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

CHAPTER 10 From Table A.1, for A e /A  2.193,

p oc pe

 12.5,

Toe Te

M e  2.3

 2.058.

For isentropic flow, To  constant and po  constant. Hence, poc  po  5 atm , and Toc  To  520R

pe 

pe  1  po    (5 atm)  0.4 atm p oc  12.5 

Te 

Te  1  To    (520)  252.7R Toc  2.058 

e 

pe (0.4)(2116)   0.00195 slug/ft 3 RTe (1716)(252.7)

a e   RTe  (1.4)(1716)(252.7)  779.2 ft/sec u e  Mea e  (2.3)(779.2)  1792 ft/sec

po 1   3.182. From Table A.1, we see that M e  1.4 , and Ae /A  1.115 . pe 0.3143

Ahead of the normal shock in front of the Pitot tube, po1  po  2.02  105 N/m 2

66 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

From Table A.2: M e  2.65 From Table A.1: A e /A  3.036

p (5)(2116) slug m   u A; o  o   0.01186 3 RTo (1716)(520) ft

 o  (0.634)(0.01186)  0.007519 slug/ft3 o T T  To  (0.833)(520)  433.2R To

 

u  a  (1.4)(1716)(433.2)  1020 ft/sec slug  4  m   u A  (0.007519)(1020)    0.213 sec  144 

m   u A p u  RT and   RT Hence,

m

 p    p A  A  RT RT RT

Since, M  1, then

To   1 2   1  1  M  T 2 2  / ( 1)

po    1    P*  2 

15 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Thus,

m



   1 A   R  2 

 / ( 1) / 2( 1)

po To

or,

m

po A To

  2    R    1

( 1) / ( 1)

po  5 atm  5(2116)  10580 lb/ft 2

A  4/144  0.02778 ft 2 6

m

(105809)(0.02778) (1.4)  2  slug    0.213 (1716)  2.4  sec 520 which is the same as obtained in Problem 10.4

16 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

First, check to see if the flow is sonic at the throat.

po 1   1.056 pe 0.947 From Table A., for

po  1.056: Me  0.28 and Ae /A  2.166 pe

Ae A  1.616  e  2.166, then A t  A. The throat size is larger than that for sonic At A flow, hence the throat Mach number, Mt, is subsonic.

Since

At A A 1  t e  (2.166)  1.34 A A A 1.616 e

From Table A.1, for pt 

At p  1.34; M t  0.5 , o  1.186  A pt p t po  1  pe    (1.056)(0.947)  0.843 atm po pe  1.186 

Note: The equation for m given in Problem 10.5 can not be used here because the flow is not choked, i.e., the throat Mach number is not sonic.

m  eAe u e From Table A.1, for

po T   1.056: M e  0.28, o  1.016 pe Te

Te  To /1.016  288/1.016  283.5K e 

pe (0.947)(1.01  105 )   1.176 kg/m3 RTe (287)(283.5)

a e   RTe  (1.4)(287)(283.5)  337.5 m/sec

u e  Mea e  (0.28)(337.5)  94.5 m/sec

17 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

A  A e  A t  e   (0.3)(1.616)  0.4848 m 2  At  m  e Ae u e  (1.176)(0.4848)(94.5)  53.88 kg/sec

po 1   1.064. pe 0.94 From Table A.1: M e  0.3 and A e /A  2.035.

A t A t Ae  1    (2.035)  1.33.

A A e A  1.53 

Since A t  A, then the flow is completely subsonic. No shock wave exists. Hence, from p Table A.1, for o  1.064, M e  0.3 . pe

po 1   1.129. pe 0.886 From Table A.1, for

po A  1.129: Me  0.42 and e  1.539. pe A

At A A  1   t e   (1.529)  0.999  1.0.   A Ae A  1.53 

Hence, A t  A, and the flow is precisely sonic at the throat. It is subsonic everywhere else.

Hence, from the above M e  0.42 .

From the above results, clearly when pe is reduced below 0.866 atm, sonic flow will occur at the throat, and the nozzle will be choked. Since pe  0.75 atm is far above the supersonic exit pressure, we suspect that a normal shock wave exists within the nozzle. Note that, if we run the same calculation as in parts (a) and (b) above, we find: po  1     1.333. pe  0.75  From Table A.1, for

po  1.333, we have pe

Ae  1.127 A

18 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

At A A  1    t e    (1.127)  0.7366. Since it is impossible for A t  A , then clearly  A Ae A  1.53  the flow can not be completely isentropic. There must be a shock wave inside the nozzle, with a consequent change in both po and A* across the shock. Hence, the above calculation is meaningless. Instead, set up the following trial-and-error process as follows:

Assume a normal shock exists inside the nozzle, say at a location where A2/At  1.024.

Let: A1*  sonic throat area for the flow ahead of the shock.

A2*  sonic throat area for the flow behind the shock.

po1  total pressure for the flow ahead of shock. p o2  total pressure for the flow behind shock.

Note that po2  po1

A 2 *  A1*

which comes from the shock wave theory discussed in the text.

Key equation:

19 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

pe po2 po po2 po1 1

pe 

To find the values of the ratios in Eq. (1): From Table A.1 for A2 /A1*  1.204: M1  1.54 po From Table A.2 for M1  1.54: M 2  0.6874, 2  0.9166 po1 From Table A.1, for M 2  0.6874:

A2  1.1018 A

Ae A A A  1   e t 2  (1.53)   (1.1018)  1.4 A2* At A2 A2*  1.204 

po Ae  1.4: M e  0.47, 2  1.163 A2 * pe

From Table A.1, for Returning to Eq. (1): pe 

pe po2  1  po1    (0.9166)(1 atm)  0.788 atm. po2 po1  1.163 

This is slightly higher than the given pe  0.75. Hence, move the shock wave slightly downstream.

A /A  1.301

Assume 2 t From Table A.1: M1  1.66 From Table A.1, for M1  1.66:

po2 po1

From Table A.1, for M 2  0.6512:

 0.872, M 2  0.6512

A2  1.1356 A2 *

Ae A A A  1   e t 2  (1.53)   (1.1356)  1.335 A2* At A2 A2*  1.301 

20 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

From Table A.1, for

po Ae  1.335: M e  0.50, 2  1.1862 A2 * pe

From Eq. (1): pe 

Interpolate:

pe po2  1  po1    (0.872)(1 atm)  0.735 atm. po2 po1  1.1862 

A2 0.75  0.735  1.301  (1.301  1.204)  1.274 At 0.788  0.735

Thus, Assume A 2 /A t  1.274 From Table A.1: M1  1.63 From Table A.2: M 2  0.6596, From Table A.1:

po2 po1

 0.8838

A2  1.1265 A2 *

Ae A A A2  1   e t  (1.53)   (1.1265)  1.353 A2* At A2 A2*  1.274  po From Table A.1: M e  0.49, 2  1.178 pe p po  1  pe  e 2 po1    (0.8838)(1 atm)  0.75 atm po2 po1  1.178 

Hence, pe calculated agrees with pe given. Thus, M e  0.49

po1 pe



1 atm  6.49 0.154 atm

Ae  1.53, which is precisely the given area ratio of the nozzle. Hence, for A* this case, we have a completely isentropic expansion, where,

From Table A.1:

M e  1.88

From the --M diagram, for   20 and   41.8, we have M1  2.6. From Table A.1, Ae  2.896 A*

21 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

From Table A.1, for

Ae  6.79, M e  3.5 A*

From Table A.2, for M e  3.5: po1 

po2 po1

 0.2129

 1  po2    (1.448)  6.8 atm po2  0.2129  po1

From Table A.1, for M e  2.8:

p oe pe

 27.14,

Toe Te

 2.568

At standard sea level: p  2116 lb/ft 2 , T  519R

po  To 

p oc pe Toc

pe  (27.14)(2116)  57,430 lb/ft 2  27.14 atm

Te  (2.568)(519)  1333R Te p 57430 o  o   0.251 slug/ft 3 RTo (1716)(1333)

  (0.6339)(0.0251)  0.0159 slug/ft3 T*  0.833 (1333)  1110R a*   RT*  (1.4)(1716)(1110)  1633 ft/sec  u*

22 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

m  *u*A1* m 1   0.0385 ft 3  * u* (0.0159)(1633) From Table A.1: Ae /A*  3.5 A1* 

Ae A*  (3.5)(0.0385)  0.1348 ft 2 A* At2 A * po From Eq. (10.38) in text:  2  1 A t1 A1* po2 Ae 

From Table A.2: for M e  2.8:

po2 po1

 0.3895

 po   po   1  2 A t 2  A t1  1   A1*  1   (0.0385)    0.0988 ft  po   po   0.3895   2  2 m  *u*A*

(1)

Also, R  R /M 

8314 joule  519.6 16 kg K 1

1

 2   1 po po *  2  0.2 *  o     3.319  107 po   o    1  RTo  2.2  (519.6)(3600)  2  T* T*  To    (3600)  3273 K To    1 u*  a*   RT* 

Hence, from Eq. (1), with m  287.2

(1.2)(519.6)(3273)  1428.6 m/sec

kg , sec

287.2  (3.319  107 po )(1428.6)(0.2) or,

23 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

po 

287.2 6 N  3.029  10 (3.319  107 )(1428.6)(0.2) m2

po 

3.029  106  30 atm 1.01  105

or,

We assume the flow velocity is low at the diffuser exit; hence the total pressure at the exit is 1 atm. From Appendix B, for M  3,

D 

po2 po1

 0.3283.

p B /po  1.2 po2 /po1

po pB  1.2 2  1.2 (0.3283)  0.394 po po1

po 

pB 1   2.54 atm 0.394 0.394

Let h1 be the height of the duct entrance, and h2 be the local height at location AB. Let h* be the reference sonic throat area. For M1  2, from Appendix A, h1  1.687 at M1  2 h*

During the solution of Problem 18 of Chapter 9, we found that h2  1.0908 h1.

Hence, h 2  h 2   h1        (1.0908)(1.687)  1.84 h*  h1   h* 

From Appendix A, for

h 2  A1     1.84, h*  A* 

M 2  2.10

24 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Note: This value of M2 obtained from our quasi-one-dimensional analysis agrees almost exactly (within roundoff error) with the average Mach number of Mave  2.10 calculated in Problem 18 of Chapter 9. This agreement speaks well of the viability of the assumption of quasi-one-dimensional flow.

From Appendix A, h1  1.687 at M1  2 h*

For this case, h2  2h1. h 2  h 2   h1        2(1.687)  3.374 h*  h1   h* 

From Appendix A, for

h 2  A1     3.374, h*  A* 

M 2  2.76

Comparing with the more exact two-dimensional result of Mave  2.89 from Problem 19 of Chapter 9, we see the more approximate value of 2.76 obtained from the quasi-onedimensional calculation agrees to within 4.5%.

Observation: For this problem, where   30, the deviation of the quasi-one-dimensional result from the two-dimensional result is larger than obtained for Problem 15, where   3.

This is to be expected, because the two-dimensional aspects for   30 are much stronger than for   3.

Consider the point at (r, )  (r*, /2) at a distance r* directly above the corner. The streamline through this point, and the streamline along the surface, form the upper and lower surfaces of a streamtube. Assume a unit depth of the streamtube perpendicular to the flow. Thus, the cross-sectional area of the streamtube as it enters the expansion fan at Mach one is simply A*  h*. Take a point on the upper streamline when it is in the expansion wave, such as point P shown below. The radial coordinate of this point is r. The radius vector, originating at the corner of the centered expansion wave, is at an angle relative to the local streamline direction equal to the local mach angle, . At that point, the local deflection angle of the streamline relative to the horizontal is the local Prandtl-Meyer expansion angle, .

25 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Both μ and ν are functions of the local Mach number M at point P. Thus, the polar angle coordinate of point P is ( –  ), which is a function of the local Mach number M at point P. The cross-sectional area of the streamtube as it enters the expansion wave is A*. Its crosssectional area at the location of point P in the streamline is A. From the geometry of the above sketch,

A r sin   A* r* From the isentropic relation for area ratio given by Equation (10.32), we have  1

r sin  A 1  2    1 2   2( 1)   1  M   r* A* M    1  2  Since sin  

1 , this becomes M  1

 2  r   1 2   2( 1)  M  1  r    1  2 

(1)

Repeating, the polar angle of point P is v

(2)

By using M as a parameter in equations (1) and (2), various points on the streamline located by r and  can be calculated, and thus the shape of the streamline can be plotted.

Using local Mach number as a parameter, and calculating r/r* and  from Eq. (1) and (2) respectively in the solution of Problem 10.17, we have the following tabulation (note that Appendix A is used to evaluate Eq. (1)) A r A M (App A) M  (deg)  (deg)      (deg) A* r* A*

26 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

1.0

1.0

1.0

90

0

90

1.1

1.008

1.109

65.38

1.336

64.04

1.2

1.03

1.236

56.44

3.558

52.88

1.3

1.066

1.386

50.28

6.17

44.11

1.4

1.115

1.56

45.58

8.987

36.6

1.5

1.176

1.76

41.81

11.91

29.9

1.6

1.25

2

38.68

14.86

23.82

The plot of the streamline is on the next page.

27 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Since the flow is isentropic, we know that: 

pO    1 2   1  1  Me  pe  2 

where p O is the reservoir pressure and p e is the pressure of the environment. Solving for M e gives:

Me 

 1     2  pO   1  1.709      1  pe   

Similar reasoning can be used to determine the temperature at the nozzle exit:

TO  1 2 1 Me Te 2 So Te 

TO  1   TO    361.8K  1 2  1.5838  1 Me 2

The density at the exit is found using the ideal gas law: pe 100, 000 N/m 2 kg e    0.960 3 RTe  288 J/kgK  361.8 K  m

The mass flow rate at the nozzle exit is given by: m  e u e Ae  e Me  RTe Ae

Solving for the exit area gives:

Ae 

m

e M e  RTe

 kg 2  100 cm  2 s    1.597 cm m    J  1.4   288  361.8K   kgK  

0.1 kg    0.960 3  1.709  m  

28 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

From the Area-Mach number relation: M e  1.71  A*  Ae



Ae  1.346 A*



A*  1  2  1.597 cm2    1.187 cm Ae 1.346  

The thrust is found by applying the integral form of the momentum equation to the control volume illustrated in Figure 1:

Figure 1. Control volume used to apply momentum conservation to the thruster.

The integral form of momentum conservation is given by:   udV    u  u  dA     pdA  FT A A t V

(1)

Let‟s consider each term. The first term is the rate of change of momentum of the control volume as a whole. We are told that temperature and pressure inside the tank remain approximately constant so the density must remain approximately constant. Since the tank is relatively large compared to the thruster, the velocity of the gas within it is expected to be relatively small and not to change much over the course of firing. Thus this term.is expected to be small compared to the others so:   udV  0 t V

(2)

The second term is the net momentum flux leaving the control volume. The nozzle exit is the only place where gas flows across the control volume so:

A u  u  dA   e u e  u e Ae  ˆi  mu eˆi

(3)

where î is a unit vector pointing along the nozzle axis in the x-direction. The third term is the net pressure force acting on the control volume. The only place on the control volume surface where the pressure does not equal p  is along the nozzle exit plane where the pressure is pe .

Therefore:   pdA    p e  p   A e ˆi A

29 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

(4)

Substituting Equations 8-10 back into Equation 7 and rearranging gives: FT   mu e   pe  p  A e  î

(5)

The mass flow rate is calculated using Equation 6.  1

m  po A t

  2   1   RTO    1 



N     2, 026, 480 2  0.0005 m 2 m  



1.4 1

1.4 kg  2 1.4 1  2.324   s  J   1.4  1   297 kgK   300 K   

(6) The exit pressure is determined from the exit Mach number. Since we are assuming that the flow through the nozzle is isentropic, Ae /A t  5.9  Me  3.35 . So,

po

pe 





2, 026, 480

N m2 1.4 1.4 2  1

 32,940

  1 2   1  1.4  Me   3.35  1  1  2    1.4  1  To 300 K Te    92.46 K  1 2 1.4  1 2 1 Me 1  3.35   2 2

N m2

(7)

(8)

and

u e  M e  RTe   3.35 



1.4   297 

J  m   92.46 K   656.9 kgK  s

(9)

Substituting the results of Equations 6-9 into Equation 5 gives:

  kg  m  N N  FT   2.324  656.9    32,940 2  101,324 2  0.0005m2  ˆi s  s  m m    F  1527 N  201.7 N ˆi  1325 Niˆ





T

This is the thrust force on the ground. In space, where p  0 atm,

FT  1527 N  97.19 N ˆi  1624 Niˆ

30 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

This difference between performance on the ground and in space arises because the pressure at the nozzle‟s exit is lower than atmospheric pressure. On the ground, this pressure difference leads to a force acting on the control volume in a direction opposite to the force associated with the flow of momentum across the nozzle exit plane. In space, the situation is reversed.

The pressure of the environment is lower than the pressure at the nozzle exit and this exerts a force in the same direction as that associated with the flow of momentum across the nozzle exit plane.

The thrust is found by applying the integral form of the momentum equation to the control volume illustrated in Figure 1:

Figure 1. Control volume used to apply momentum conservation to the thruster.

The integral form of momentum conservation is given by:   udV    u  u  dA     pdA  FT A A t V

(1)

Let‟s consider each term. The first term is the rate of change of momentum of the control volume as a whole. We are told that temperature and pressure inside the tank remain approximately constant so the density must remain approximately constant. Since the tank is relatively large compared to the thruster, the velocity of the gas within it is expected to be relatively small and not to change much over the course of firing. Thus this term.is expected to be small compared to the others so:   udV  0 t V

(2)

The second term is the net momentum flux leaving the control volume. The nozzle exit is the only place where gas flows across the control volume so:

A  u  u  dA   e u e  u e Ae  ˆi  mu eˆi

(3)

where iˆ is a unit vector pointing along the nozzle axis in the x-direction. The third term is the net pressure force acting on the control volume. The only place on the control volume surface where the pressure does not equal p  is along the nozzle exit plane where the pressure is pe .

31 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Therefore:   pdA    p e  p   A e ˆi A

(4)

Substituting Equations 8-10 back into Equation 7 and rearranging gives: FT   mu e   pe  p  A e  ˆi

(5)

The mass flow rate is calculated using Equation 6.  1

m  po A t

  2   1   RTO    1 

(6)

The exit pressure is determined from the exit Mach number.

pe 

Te 

po



  1 2   1  Me  1  2  

(7)

To  1 2 1 Me 2

(8)

and u e  Me  RTe

(9)

Use Equations 6-9 to compute the values of m, pe , and u e for a range of M e and then substitute their values into Equation 5 to calculate the thrust. The results of these calculations are presented in Figure 2.

Figure 2. Thrust force (blue) and net pressure force (orange) as a function of nozzle area ratio when the ambient pressure is 1 atm (solid lines) and 0 atm (dashed lines).

32 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

On the earth‟s surface (ie. at 1 atm) the net pressure force acting on the control volume (ie. the second term on the RHS of equation 5) decreases with increasing nozzle exit area due to the associated decrease in nozzle exit pressure. When the pressure at the nozzle exit falls below atmospheric pressure, the net pressure force becomes negative and begins to reduce the overall thrust produced. In space, the net pressure force decreases as the nozzle exit area increases but the net pressure force never becomes zero so the net force continues to increase with increasing area ratio.

Figure 3 shows specific impulse as a function of nozzle area ratio. While the thruster reaches a maximum specific impulse of about 110 s on the ground, in theory it can achieve more than twice that in space. However, boundary layer growth limits the degree of expansion that can be achieved in practice. The limit depends on many factors but if this limit is Ae /A t ~ 10 , then this thruster would provide Isp ~ 110 s on the ground and Isp ~ 134 s in space. While chemical rockets achieve more than twice these values, it isn‟t bad for a simple device that doesn‟t get hot and can be turned on and off easily.

Figure 3. Specific impulse for the thruster when the ambient pressure is 1 atm (solid line) and at 0 atm (dashed line).

Sketch the tank and think about what is happening. N2 leaves the tank through a choked nozzle as illustrated in Figure 1. The pressure of the N2 in the tank decreases as more N2 leaves the tank through the nozzle.

Figure 1. Control volume used to apply mass conservation to the cold gas thruster.

The nozzle is choked so from the results of this chapter (and problem 10.5), we know that the mass flow rate through the nozzle is given by: 33 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

m  pA t

 1   1

  2   RTO    1 

(1)

Thinking back to Chapter 2, the „system‟ to which we apply nature‟s laws is the N2 in the tank.

Applying conservation of mass to the N2 trapped in the tank gives:   dV    u  dA  0 A t V

(2)

Next, recognize that the first term is the time rate of change of mass in the control volume and the second term is the rate at which mass is leaving the control volume. Remember that u  dA is positive when the unit normal and the velocity are pointing in the same direction.

So, substituting Equation 1 into 2 gives:  1

m   2   1  pA t 0   t RTO    1 

(3)

where m is the mass of N2 in the tank. Re-writing the mass of N2 in the tank using the ideal gas law gives:

m   pV  V dp p dV pV dT      t t  RT  RT dt RT dt RT 2 dt Recognizing that the N2 volume and temperature don‟t change with time allows one to eliminate the second and third terms on the RHS leading to:

  pV  V dp   t  RT  RT dt

(4)

Substituting Eq. 4 into 3 gives:  1

V dp   2   1  pA t 0   RT dt RTO    1  Rearranging gives an ordinary differential equation for the time evolution of the pressure:  1

dp RT   2   1  At   dt p V RTO    1 

(5)

34 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

This can be solved by integrating both sides:  1

RT   2   1   At   dt 0 p V RTO    1 

po dp

p

t

This gives:  1

 p  RT   2   1 ln     At   t V RTO    1   po  Or,  1

p e po

RT   2   1  At   t V RTO   1 

(6)

Equation 6 is plotted in Figure 2. Finally, the minimum pressure ratio across the nozzle required to achieve sonic conditions in the throat is given by: 

1.4

p   1 2   1  1.4  1 2 1.4 1   1  M   1   1.893 1  p  2 2   

The pressure ratio across the nozzle is given by: p  p  p  o  p p  po 

So the range of p/po over which the solution is valid is given by:

1

p p  1.893  po po

where p is the tank pressure at time t, p o is the initial tank pressure, and p  is the pressure downstream of the nozzle.

35 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Figure 2. Pressure ratio ( p/po ) as a function of time for a tank full of N2 blowing down via a choked nozzle where the temperature of the N2 in the tank is assumed to be constant. The horizontal line indicates the tank pressure ratio when the nozzle becomes unchoked indicating that the solution is valid until about 23.3 s.

Start by sketching the problem and thinking about what is happening. Air leaves the tank through a choked nozzle. The pressure and temperature of the N2 in the tank decrease as more N2 leaves the tank through the nozzle.

Figure 1. Control volume used to apply mass conservation to the cold gas thruster.

From the results of this chapter, we know that the mass flow rate through a choked nozzle is given by:  1

m  pA t

  2   1   RT    1 

(1)

Thinking back to Chapter 2, the „system‟ to which we apply nature‟s laws is the air in the tank.

Applying conservation of mass to the air trapped in the tank gives:   dV    u  dA  0 A t V

(2)

36 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Next, recognize that the first term is the time rate of change of mass in the control volume and the second term is the rate at which mass is leaving the control volume. Remember that u  dA is positive when the unit normal and the velocity are pointing in the same direction.

So, substituting Equation 1 into 2 gives:  1

m   2   1  pA t 0   t RT    1 

(3)

where m is the mass of air in the tank. Re-writing the mass of air in the tank using the ideal gas law gives:

m   pV  V dp p dV pV dT      t t  RT  RT dt RT dt RT 2 dt Recognizing that the air volume doesn‟t change with time allows one to eliminate the second term on the RHS leading to:

  pV  V dp pV dT    t  RT  RT dt RT 2 dt

(4)

Since the process is isentropic, we can write the temperature in terms of the pressure using: 

p o  TO   1  p  T 

where p o and To are the initial temperature and pressure in the tank. Rearranging gives:  1  p 

T  To    po 

(5)

So, 1 

1

dT   1   dp  To po  p dt  dt

(6)

Substituting Eq. 6 into Eq. 4 gives: 1 

1

 dp   pV  V dp pV   1    T p p   2 o o t  RT  RT dt RT  dt

37 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Simplifying gives: 1   1    pV  V  1 1 dp   1  p       2 To po t  RT  R  T T   dt 

Now, substituting Eq. 5 gives:   1   1     1 1  2  2  1   1           dp po p   pV  V po p   1    T p p   o o   t  RT  R  To  To2  dt  

Simplifying gives:  1 

1 

  pV  1 V po dp p   t  RT   R To dt

(7)

Substituting Eq. 7 into 3gives:  1 

1 

 1 

1 

 1

1 V po dp   2   1 p   pA t 0    R To dt RT    1 

Rearranging gives:  1   1

1 V po dp 1   2 p    pA t    R To dt T R    1

Again, substituting (5) gives:  1 

1   p  2

1 

1 V po dp p    pA t    R To dt  po 

 1

  2   1   R    1

1 To

Rearranging gives:

p

1 3 2

p

p

p o

1  dp  R 2   At po

dt

V

1 3 2

1   R 2 dp   A t po

V

To

To

 1   1

  2   R    1

 1   1

  2   R    1

t

 0dt

38 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

2 p 1

1  p 2

1   R 2   At po

V

 1

To

po

  2   1   t R    1

1   1    1   2  p  2  R  2  1   A t To     t  1    po  V R    1  

1 

 1

 p  2 1 R   2   1 1 At To     t 2 V R    1  po    1      R  p   2  1    At     To po   2   V  R    1 

 1   1

2 1  

 t  

2 p  1  Ct 1 po

(8)

where  1

1      R    2   1 C A T  t    o R    1  2   V 

(9)

Since the flow is isentropic, the time evolution of the temperature ratio is given by:  1  p 

T   To  po 

(10)

Finally, this solution is valid as long as the nozzle remains choked. The minimum pressure ratio across the nozzle required to achieve sonic conditions at the throat is given by: 

1.4

p   1 2   1  1.4  1 2 1.4 1   1  M   1   1.893 1  p  2 2   

39 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

The pressure ratio across the nozzle is given by: p  p  p  o  p p  po 

So the range of p/po over which the solution is valid is given by:

1

p p  1.893  po po

where p is the tank pressure at time t, p o is the initial tank pressure, and p  is the pressure downstream of the nozzle.

The results (temperature and pressure ratio) are plotted in Figure 1. The pressure falls faster in the isentropic case which means that the nozzle becomes unchoked after about 19.9 s instead of 23.3 in the isothermal case. This is because the gas is denser due to the substantial decrease in its temperature as the tank blows down (clearly the isothermal assumption is not very good). The lower gas temperature and higher gas density indicate that the temperature change lowers the exit velocity.

Figure 2. Pressure ratio ( p/po ) as a function of time for an air tank blowing down via a choked nozzle. The horizontal line indicates the tank pressure ratio when the nozzle becomes unchoked. The solution for the isothermal case is valid until about 23.3 s whereas the solution for the isentropic case is only valid until about 19.9 s. The temperature of the tank drops substantially during the blowdown indicating that the isothermal assumption is not very good.

40 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Start by finding  for the situation when the nozzle is choked. The expression for the mass flow rate through a choked nozzle using the notation of Figure 10.27:  1

mchoked  p0 A t

  2   1   RT0    1 

(1)

From the ideal gas law:

RT0 

p0

0

(2)

Substituting Equation 2 into Equation 1 gives:  1

m  p0 A t

0  2   1   p0    1 

Or m  At choked 20 p0

(3)

where  1

  2   1  choked    2    1

(4)

Now, consider the case when the flow is not choked. The mass flow rate is given by: m   uA t

(5)

From conservation of energy along a streamline:

u2 cp T0  cp T  2 Solving for u gives: u

Since

 T 2c p T0 1   T0  

cp   R/  1 u

2 RT0  T 1    1  To 

41 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Inserting Equation 2 gives: u

2 p0  T 1     1 0  T0 

(6)

Since the flow is isentropic:  1

T  p    T0  p0 

(7)

And 1

 p 

   0  p0 

(8)

Substituting Equations 6, 7, and 8 into 5 gives: 1

 p  m  A t 0    p0 

 1   2 p0   p    1     1 0   p0    

Rearranging gives:

m  At

2   1   p    p    2p    1      1 0 0  p0    p0    



or, m  At unchoked 2p0 0

where

 unchoked 

2  1         p p            1  p0   p0    



The minimum pressure ratio required to achieve sonic flow at the throat (ie. choked flow) is given by: 



p0    1 2   1    1   1  1  1     p  2   2 

42 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Therefore: m  At 2p0 0

where

 1     2   1    2    1    2  1    p    p             1  p0   p0     

a.



p0    1   1   p    

p0    1   1   p   

Since the tunnel draws air in from the room, the total pressure and temperature are

simply the ambient conditions in the room ie: p0  101324 N/m2 and T0  300 K b. The Mach number in the test section is given by the ratio of the test section area to throat area so:  0.1651 m  0.1524 m   2  M  2.2 A  ts A*  0.0826 m  0.1524 m  c.

From the isentropic flow relations: p0 101324 N/m 2 N p ts    9478.4 2  1.4 m   1 2   1  1.4  1 2 1.4 1  1  M 1  2.2      ts  2 2    

d. The valve opens and supersonic flow ( M ts  2.2 ) starts in the test section. Air flows into the test section at a constant rate because the nozzle is choked and the temperature and pressure in the room remain constant. Initially the pressure in the vacuum chamber is lower than in the test section and expansion waves form at the diffuser entrance to lower the pressure of the air exiting the test section to the chamber pressure. As air flows into the vacuum chamber, the pressure in the chamber rises causing the expansion waves get weaker and weaker. Eventually, the pressure in the vacuum chamber equals the pressure in the test section and the flow exiting the test section can be thought of as perfectly expanded. As flow continues into the vacuum chamber, the chamber pressure continues to rise and progressively stronger shock waves form to raise the pressure of the flow exiting the test section to the pressure in the vacuum chamber. Eventually a normal shock wave is stabilized at the diffuser entrance. Further chamber pressure rise destabilizes the wave which moves upstream through the test section causing the pressure in the test section to decrease. The tunnel has become unstarted and the experiment is over. Flow continues until the pressure in the vacuum chamber equals the pressure in the room. This process is illustrated schematically in Figure 1.

43 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Figure 1. Vacuum chamber pressure as a function of time (notional) with illustrations of flow structure at the diffuser entrance at various points during the blowdown. The tunnel unstarts when the pressure difference between the vacuum chamber and test section is greater than what can be produced by a normal shock. e. Based on the discussion above, the tunnel unstarts when the pressure in the vacuum chamber reaches the pressure downstream of a normal shock stabilized at the diffuser entrance. So, p02 p (1) pu  p ts 2 p ts NS p2 From part c:

p0

p ts   1  

 1 2

  2  1

(2)

M ts  

The pressure ratio across the normal shock is given by:

p2 2 1 M2  1 p ts NS   1 ts





(3)

The Mach number downstream of the normal shock is:

M 22 

1

 1 2

 M 2ts 

M 2ts

 1

(4)

2

The pressure ratio downstream of the normal shock is: 

p02    1 2   1  1  M2  p2 2  

44 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Re-writing M 2 using Equation 4 gives: 

   1 2    1  1  M ts   p02    1  2  1    p2 2   M2    1   ts 2    

(5)

Inserting Equations 2, 3, and 5 into 1 gives:       1 2    1          1  1  2 M ts   1 2 2 p u  p0  1  M  1 ts  1  2      1     1     M 2ts     1 2   1  M ts   2      1  2   





(6) Plugging in numbers gives: N  N  pu  101324 2   0.09352  5.48 1.2256   63, 645 2 m  m 

Note that M 2  0.5471 f. Start by making a sketch of the vacuum tank. Air exiting the test section enters the tank as illustrated in Figure 2. The total enthalpy of the air entering the vacuum tank is the same as the enthalpy of the room air. You can prove this to yourself by following a streamline from the room, through the test section, and into the vacuum chamber. The flow is assumed to be inviscid and adiabatic so the total temperature (and thus total enthalpy) is constant.

Figure 2. Schematic illustration of wind tunnel showing control volume used to analyze the flow into the vacuum tank.

The time over which supersonic flow is sustained equals the time at which the tunnel unstarts.

Since the mass flow rate into the vacuum tank is constant, the time to unstart is given by: 45 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

tu 

mu  mo m

(7)

where m u is the mass of air in the vacuum chamber at unstart and mo is the initial mass of air in the tank. The mass flow rate into the tank is given by:  1

m  p0 A t

  2   1   RT0    1 

and is constant because the temperature ( T0 ) and pressure ( p 0 ) of the room are constant.

Therefore: 1.4 1

N  1.4 kg   2 1.4 1 m  101324 2   0.1651 m  0.0826 m    2.971   s  J  m    1.4  1   288 kgK   300 K    (8) The mass of air in the tank initially is found using the ideal gas law:





N   5066 2  62.77 m3  P V m  m o  V,o    3.680 kg RTV,o  J   288 kgK   300 K   

(9)

Finding the mass of air in the tank when the tunnel is more complex because the temperature of the air in the tank is not constant. We start by applying conservation of energy to the vacuum tank illustrated in Figure 2. The integral form (Equation 2.95) is most useful here.

Recognizing that the flow is adiabatic and inviscid and that body forces can be neglected eliminates the first, second, fourth and fifth terms on the LHS of Equation 2.95 leaving us with:   pV  dS  Svac

   u2  u2   e  dV   e       2  V  dS t V  2  S vac

vac

Rearranging gives:    u2  p u2   e  dV    e         2  V  dS t V  2  S vac vac Incorporating the definition of enthalpy h  e 

p



leads to:

46 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

   u2  u2   e  dV    h      2  V  dS t V  2  S vac

(10)

vac

Now let‟s return to Figure 2 and think a little more about the terms in the energy equation and how they are related to our problem. The LHS of Equation 10 is the time rate of change of energy contained in the vacuum tank. Since material flows into the tank but does not leave, the velocity inside the tank will be relatively small so,

u2 (11) e 2 The RHS of Equation 10 is the net rate of energy increase in the control volume ( Vvac ) due to the flow into the tank from the diffuser. The flow originates in the room and is adiabatic so applying conservation of energy along a streamline originating in the room gives:  u2   ho h   (12) 2 S  e

vac

Note that h o is a constant and so it can be pulled outside the integral. So, incorporating Equations 11 and 12 into Equation 10 gives:   edV  h o   V  dS t Vvac S

(13)

vac

Recognizing that    edV   me   t V t vac

  V  dS  min Svac

Leads to:

d  me   h o min dt Equation 14 can be integrated leading to:

me  mo eo  h o  m  mo 

47 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

(14)

Incorporating the definitions of internal energy and enthalpy: mcV T  mo,v cV To,v  cp To  m  mo,v 

where m is the mass of air in the vacuum tank at some time, mo,v is the initial mass of air in the vacuum tank, To,v is the initial temperature of the air in the vacuum tank, and To is the temperature of the air in the room. In this experiment, To,v  To . Rearranging to find T (the temperature of the gas in the vacuum tank at some time t) gives:

m T    1    o,v To m

(15)

Now, we can use the ideal gas law to show that:

mo,v m



po,v T

(16)

p To

Inserting Equation 16 into 15 and solving for the temperature at unstart gives:

Tu 

 To 1     1

po,v pu

Now we can find the temperature at unstart: Tu 

1.4  300 K  5, 066 N/m 2 1  1.4  1 63, 645 N/m 2

 407.0 K

(17)

The mass of air in the tank at unstart is given by:







63, 645 N/m 2 62.77 m3 pu V mu    34.08 kg RTu  J   288 kgK   407 K   

Finally, we calculate the time that the tunnel unstarts by substituting the results of Equations 8, 9, and 17 into Equation 7:

tu 

mu  mo 34.08 kg  3.680 kg   10.23 s m 2.971 kg/s

CHAPTER 11

48 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

1.30

u

2  2 1 M  y  V  2 (70) e cos (2x) 2 x 1 M



v





2  70 2 1 M  y  2 1  M 2 e sin (2x) 2 y 1  M

 140  e

2 2 1 M y

sin (2x)

a    RT  (1.4)(1716)(519)  1116.6 ft/sec V 700 M     0.6269 a  1116.6

Thus, at (x, y)  (0.2, 0.2) u  700 

2 (70) 2 (0.779)(0.2) e cos [2(.2)]  765.6 ft/sec 0.779

v  140  e2 (.779)(.2) sin [2(.2)]  157.2 ft/sec

V  u 2  v2  (765.6)2  (157.2)2  781.6 ft/sec From Table A.1, for M  0.6269,

To  1.079 T

To  1.079 T  0.079 (519)  560R a o   RTo  (1.4)(1716)(560)  1160 ft/sec

a  ao  2

2

 1

2 a  1106 ft/sec M

V  2

ft   1.345  10  (.2)(781.6)  1.223  10    sec  6

2

6 

2

V 781.6   0.7067 a 1106

From Table A.1, for M  0.6269:

For M  0.7067 

po T  1.3065, o  1.079 p T

po T  1.400, o  1.101 p T

p

p po  1  p    (1.3065)(1 atm )  0.933 atm po p  1.4 

T

T To  1  T    (1.079)(519)  508.6R To T  1.101 

49 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

1.31

The results of Fig. 4.5 are for low-speed, incompressible flow. Hence, from Fig. 4.5, at

  5, at   5, c o  0.75 co

c 

1.32

Cp 

1  M 2

C po 1  M 2



0.75 1  (0.6) 2

0.54 1  (.58)

(a) Cp 

d.



2



 0.938

0.54  0.663 0.8146

C po

 C M 2  po 1  M 2    1  1  M2  2    Cp  0.7063



0.54  0.3364  (0.54) 0.8146   1  0.8146  2

C po

(b) Cp 

   1 2   1  M 2   M 2 1  M   / 2 1  M 2  C po 2     0.54 e. Cp  0.8146  [0.3364(1.067)/1.6292](0.54)

f.

Cp  0.7763

Note the differences: There is a 17% discrepancy between the three compressibility corrections. Of the three, experience has shown the Karman-Tsien rule to be more accurate.

1.33

For the pressure coefficient on the airfoil: C po 0.41 Cp   1  M 2 1  M 2

M

Cp

0.3

0.4

0.5

0.6

0.7

0.8

− 0.4 3

−0.

447

−0.

473

−0.

513

−0.

574

−0.

683

50 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

51 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

1.34

When M  Mcr , then p at the minimum pressure point is clearly pcr .  p  p  p p  cr   cr   o   (0.528)(1.524)  0.805 p p  po   p 

Evaluated Evaluated at M = 1 at M = 0.8 1.35

From Appendix A:

For M  0.5,

po  1.186 p

For M  0.86,

po  1.621 p

Cp 

 p  p p  p 2  p    1  2  q  M   p  p M 2 2

p /p p 1.186  o    0.7316 p po /p 1.621

Cp 

2  (0.7316  1)  1.53 (1.4)(0.5)2

Check: Using Eq. (11.58)     1    1 M 2   1    2   2 Cp   1  2   M  1    1 M 2    2   

 1  0.2(0.5)2  2    (1.4)(0.5)2  1  0.2(0.86) 2  

3.5

  1  1.53  

It checks!

1.36 First, calculate pressure coefficient is Cp,A 

Cp,o

at point A from the information in Figure 11.5(a). The actual

 2  pA  1    M 2  p 

52 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

where

p A p A po  p po p From Appendix A (interpolating between entries for more accuracy for this problem), For M  0.3:

po  1.064 p

For M A  0.435:

po  1.139 pA

Thus, Cp,A 

2  1.064   1  1.045 2  (1.4)(0.3)  1.139 

From the Prandtl-Glauert rule, Cp, o  Cp,A 1  M 2  (1.045) 1  (0.3) 2  0.9969

For the case of part (c) where M   0.61, again using the Prandtl-Glauert rule, CP, A 

Cp,o 1  M 2



0.9969 1  (0.61) 2

 1.258

To find the local Mach number, Ma , which corresponds to this value of CP, A , note that Cp,A 

 2  pA  1    M 2  p 

15 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

or, p A  M  Cp,A (1.4)(0.61) 2 ( 1.258)  1   1  0.6723 p 2 2 2

However,

p p A p A po  where o for M  0.61 is 1.286 p po p p Thus,

pA pA /p 0.6723    0.523 po po /p 1.286 Hence,

po  1.912. pA From Appendix A, for

po  1.912, M A  1.01 pA

This is close enough. Hence, given the numbers in Figure 11.5(a), the numbers in Figure 11.15(c) are consistent with the laws of physics.

1.37 There is a three-dimensional relieving effect for the flow over a sphere. The flow over a cylinder is two-dimensional – in order to get out of the way of the cylinder, the flow can move only upwards or downwards. This means it must greatly accelerate to get out of the way of the cylinder. In contrast, the flow over a sphere is three-dimensional – it can move not only upward or downward but also sideways. This extra degree of freedom means that the flow does not have to speed up so much in flowing over the sphere. Hence, the freestream Mach number of the sphere is higher in order to achieve sonic flow on the sphere – i.e., the critical Mach number is higher.

16 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Cp  1.38 For the incompressible flow over a circular cylinder, from Eq. (3.101), 1  4 sin 2 . Therefore, at the point of maximum velocity, Cp  3. From the PrandtlGlauert rule, Cp 

3 2 1  M cr

(1)

From Eq. (11.60) 3.5      1 2  M  1     cr  2  2     1 Cp,cr  2   1  (   1)/2  M cr        

(2)

Equating Eqs. (1) and (2): 3.5      1 2  M  1    cr   3 2  2     1  2   2  M cr  1  (  1)/2  1  M cr      

Solve for Mcr by trial-and-error.

M cr

Eq (1)

Eq (2)

0.4

−3.2733

−3.66

0.41

−3.289

−3.457

0.42

−3.306

−3.268

0.418

−3.3025

−3.305 M cr  0.418

17 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

1.39

From the Prandtl-Glauert rule we know that: CP 

C p,o

(1)

1  M 2

The textbook also developed an expression for the pressure coefficient at the point on the airfoil where the flow first goes sonic (Equation 11.59):     1     1   2  1   2  M    2    Cp,cr   1 2   M   1    1     2     

(2)

The critical Mach number (Mcr ) is the flight Mach number (M  ) where the pressure coefficients predicted by equations 1 and 2 are the same. We can find the value of Mcr in this problem in several ways.

One approach is to set equations 1 and 2 equal leaving us with an analytical equation that can be solved iteratively for M :     1   1   2  1  M   Cp,o 2  2   1  2   2  M  1    1  1  M   2   

(3)

Defining the error as:

 LHS  RHS  Error  %   100   LHS   We can look for the value of M  that makes the error zero. We can do this via a series of manual trials (as illustrated in Table 1) or using the „goal seek‟ function in Excel.

Table 1. Trials in the iterative solution to equation 3.

M_inf

LHF

RHS

Error(%)

0.5

−0.9238

−2.1334

−130.9476

0.6

−1.0000

−1.2943

−29.4344

0.7

−1.1202

−0.7791

30.4545

18 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

0.65

−1.0527

−1.0085

4.1983

0.64

−1.0412

−1.0606

−1.8679

0.645

−1.0469

−1.0343

1.2028

0.643

−1.0446

−1.0447

−0.0163

Another approach is to plot Equations 1 and 2 and find the point at which they intersect:

Figure 1. Graphical solution to the problem of finding the critical Mach number.

Either way, Cp  Cp,cr when M  0.643 so: M cr  0.643

1.40

Lift equals weight in steady level flight so: W  C L q S 

C L 2 M PS 2

From the Prandtl-Glauert rule: Cl 

Cl,o 1  M 2

So,

19 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.



cl,o

W

1  M2 2

M 2 PS 

0.2 *1.4 * 0.552 *10, 000 *15 2 1  0.552

W  7606 kg

CHAPTER 12 12.1

Consider   5  0.0873 rad. c 

4 M 2  1



4(0.0873) (2.6) 2  1

 0.1455

From exact theory (Prob. 9.13): c  0.148

% error  cd 

0.148  0.1455  100  1.69% 0.148

4 2 M2  1

 c   (0.1455)(0.0873)  0.0127

From exact theory (Prob. 9.13): cd  0.0129

0.0129  0.0127  100  1.53% 0.0129 (b)   15  0.2618 rad 4 c   0.436 M 2  1 % error 

From exact theory (Prob. 9.13): c  0.452

% error 

0.452  0.426  100  3.47% 0.452

cd  c   (0.436)(0.2618)  0.114

From exact theory (Prob. 9.13): cd  0.121

% error 

0.121  0.114  100  5.7% 0.121

20 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

(c)   30  0.5236 rad 4 4(0.5236) c    0.873 M 2  1 (2.6) 2  1 From exact theory (Prob. 9.13): c  1.19

1.19  0.873  100  26.7% 1.19 cd  c   (0.873)(0.5236)  0.457

% error 

From exact theory (Prob. 9.13): cd  0.687

% error 

0.687  0.457  33.5% 0.687

Conclusion: At low  , linear theory is reasonably accurate. However, its accuracy deteriorates rapidly at high . This is no surprise; we do not expect linear theory to hold for large perturbations. It appears that linear theory is reasonable to at least 5, and that it is acceptable as high as 15. At 30 it is unacceptable. Keep in mind that the above comments pertain to the lift and wave drag coefficients only. They say nothing about the accuracy of the pressure distributions themselves.

12.2

21 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

(a)

Cp 

 p  p p  p 2  p    1  2  q  M   p  p M 2 2

 M  Cp p  1 p 2 2 2 2 Cp     2.4 M 2  1 (2.6)2  1 2

or, Cp   0.8333

 M  Cp p (1.4)(2.6)2 (0.8333)  1   1 p 2 2 2

p  3.943  1 p

Hence: Examining the physical picture: recalling   5  0.873 rad. p2  3.943(.0873)  1  0.6558 p

From exact theory (Prob. 9.13):

% error 

p2  0.7022 p

0.7022  0.6558  100  6.6% 0.7022

p3  3.943  1  3.943 (.0873)  1  1.344 p

From exact theory (Prob. 9.13):

% error 

p3  1.403 p

1.403  1.344  100  4.2% 1.403

22 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

(b)

For   15  0.2618 rad:

p2  3.943  1  3.943(.2618)  1  0.0322 (physically impossible) p

p2  0.315 p

The result from exact theory (Prob. 9.13) is

p3  3.943  1  3.943 (.2618)  1  2.032 p

From exact theory (Prob. 9.13):

p3  2.529 p

2.529  2.032  100  19.7% 2.529 For   30  0.5236 rad

% error  (c)

p2  3.943  1  3.943 (0.5236)  1  1.064 (physically impossible) p The result from exact theory (Prob. 9.13) is

p2  0.0725 p

p3  3.943  1  3.943 (0.5236)  1  3.065 p

From exact theory (Prob. 9.13):

% error 

p3 5.687 p

5.687  3.065  100  46% 5.687

Conclusions: (1) Pressures predicted by linear theory rapidly become inaccurate as  increases. (2) Pressures predicted by linear theory are reasonable only at low values of  , say below 5. (3) At each value of  , the % error is much greater for pressure than for lift and wave drag coefficients. (See Prob. 12.1). Hence, linear theory works better for c and c d than it does for p. What happens is that the inaccuracies in p on the top and bottom surfaces tend to compensate, yielding a more accurate aerodynamic force coefficient.

23 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

 M  Cp p 2   1 where Cp   p 2 M 2  1 2

12.3

Cp  

2 (3)  1 2

 0.7071

p (1.4)(3)2 (0.7071)  1 p 2 p  4.455  1 p

Surface 2:   5  0.08727 rad. p2  4.455 (.08727)  1  0.6112 p

Note: Although a negative pressure is not physically possible, in order to calculate the net force, we must carry it as such.

Surface 3:   25  0.4663 rad p3  4.455 (.4363)  1  0.9439 p

Surface 4:   25  0.4363 rad p4  4.455 (.4363)  1  2.944 p

24 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Surface 5:   5  0.08727 rad p5  4.455 (.08727)  1  1.3888 p

c 

 p 4  p p  p  2  3  cos 25   5  2  cos 5  (From Prob. 9.14)  2  M1 c  p p   p p  

c 

2 (2.944  0.9439) cos 25  (1.3888  0.6112) cos 5 (1.4)(3)2 c

c  0.682 . However,  0.5077 (From Prob. 9.14) c c c  (0.682)(.5077)  0.346 cd 

 p 4  p p  p  2  3  sin 25   5  2  sin 5   2  M1 c  p p   p p  

cd 

2 (.5077)  (2.944  0.9439) sin 25  (1.3888  0.6112) sin 5 (1.4)(3)2

cd  0.1089 Comparison Exact (Prob. 9.14)

Linear Theory

% Error

c

0.418

0.346

17.2%

cd

0.169

0.1089

35.6%

211 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

12.4

cd 

4 2 M 2  1

q 

 p M 2 2

D  q  S cd 

Y 4 2 p M 2 S 1/ 2 2 M2  1







or, DC

M 2

 M  1

1/ 2

2 

where

 

Y pS 4 2 2

C

Look for a local minimum in this variation, i.e., is D a local minimum at some value of M  ?

1/ 2 1/ 2   2 21 2 M  1 2M  M M  1 2M          dD 2 0  C 2 dM  M  1    







 M  1  2 M   M  M  1 2 M  M  1  M  0 2 



1/ 2

3 



2 

2 

 12



0

3 

2 M3  2 M  M3  0 M3  2 M   0 M2  2  0 M  2 The flat plate drag, based on linear theory, first decreases with increasing M , reaches a local minimum at M  

2, and then increases for increasing M  above

decrease in drag for 1  M  

2. The initial

2 is somewhat counterintuitive; we usually expect the

wave drag to increase with increasing Mach number. However, keep in mind that: 212 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

1. The linear theory result is unreliable at low supersonic Mach number near 1.

2. In real life, there is skin friction drag added to the wave drag, and the skin friction drag will increase as the flow velocity (hence M  ) increases.

So the results of this problem, at low supersonic Mach numbers, are a little misleading until you put them in perspective with the limiting assumptions behind linearized theory.

12.5

From Eq. (12.23)

c 

4 M 2  1

From Eq. (12.24) Cd 

4 2 Μ 2  1

Thus, c  cd

4 M 2  1

/

4 2 M 2  1



1 

The maximum value of cd /c is infinity at a zero angle of attack.

213 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

12.6

(a)

4

c 

M 2  1 4 2

cd 

M 2  1

 Cf

4 M 2  1

c  cd

4

2

M 2  1

1



 Cf



Cf M 2  1 4

For maximum c /cd , 2 2      c    Cf M   1  (0)  (1) 1  Cf M   1  d  4 4 2      cd       0 2 d  Cf M 2  1       4  

Thus,

1  2 



Cf M 2  1 4 2

0

Cf M2  1 4





Cf M2  1

1/ 4

2

214 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

This is the angle of attack at which maximum lift-to-drag ratio is achieved. (b) As shown in part (a), c  cd

For maximum

1

(1)

C M 2  1  f 4

c , we proved in part (a) that cd

 





Cf M2  1

1/ 4

(2)

2

Substitute (2) into (1), c      cd max

1





1/ 4

Cf M 2  1



2



c      cd max

12.7





 C M 2  1 1/ 4  f   4   2  

1





1/ 4

Cf M 2  1



2 1





1/ 4

2

1/ 4

3

1





Cf M 2  1



Cf M 2  1

From Example 12.3, Cf  4.3  10 c      cd max

Cf M 2  1



Cf M 2  1

1/ 4

and M   2. From the results of Problem 12.6, 

1

 4.3  10  (4  1) 3

1/ 2

1/ 4

215 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.



1  11.58 (0.0656)(1.316)

This occurs at 







Cf M 2  1 2

1/ 4



1 c  2   cd max

1  0.044 rad.

2(11.58)

 0.044 (57.3)  2.52

12.8 Shock waves and therefore wave drag are unavoidable in supersonic flow over any object, let alone airplanes. Everything else being equal, the strength of a shock wave over a given aerodynamic configuration increases with M  , and therefore the wave drag increases. This serves to increase the denominator of the ratio L/D. Hence, L/D will decrease as M  increases. The hope of the supersonic airplane designer, however, is to create a configuration that, at a given Mach number, has an already lower wave drag by decreasing the strength of the shock waves. This can be done by using more slender shapes with sharp noses and leading edges, or by using highly swept wings. For such configurations, (L/D)max will still decrease as M  increases, but at least they start out with a higher base value for (L/D)max.

12.9

Part a)

Lift equals weight in steady level flight so: L  gWSS

(1)

where g is the acceleration due to gravity ( 9.8 m/s2 ) and S is the total wing area. Linearized airfoil theory gives expressions for lift and drag coefficients so re-writing Equation 1 as a lift coefficient gives:

CL 

 2

gWS p M 2

(2)

216 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

From linearized supersonic airfoil theory:

4

CL 

(3)

M 2  1

Setting Equations 2 and 3 equal gives:

gWS



 p M2 2

4 M2  1

After some algebra we are left with: 2

 M4  

2

 gWS  gWS  2  M    0  2 p   2 p 

The solutions to this quadratic equation are given by: 

2 4 2  gWS   gWS   gWS       4  2  2 p   2 p    2 p     

1  M 2  

(4)

Evaluating the constant term gives:







2

2   9.8 m/s 2 400 kg/m 2  gWS     7.389     2  0.227 rad 1.4  22700 N/m 2   2 p   





(5)

Note that the angle of attack must be in radians.

Substituting Equation 5 into Equation 4 gives the numerical values of the two solutions:

M,1 

1 7.389  2 

 7.389   4  7.389    2.49

M,2 

1 7.389  2 

 7.389   4  7.389    1.09

2



2



Part b) Equation 4 shows that the flight Mach number will remain constant at 2.49 or 1.09 as long as gWS  2 p

7.389

217 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

So, the pilot should adjust the angle of attack such that:



g

  

 2 p

  WS 7.389 

where WS is the current wing loading.

12.10 Figure 1 shows the airfoil in question where p1  1 atm , M1  2 , c  1 m , x  1/3 and y  0.06 .

Figure 1. Airfoil of Problem 9.35.

Our approach will be to use the linearized pressure coefficient to calculate the pressures on each surface of the airfoil and from these determine the contributions to lift and drag from each surface. Adding these together give the total lift and drag forces acting on the airfoil.

The linearized pressure coefficient is given by:

Cp 

2 M 2  1

(1)

where M  is the free stream Mach number and  is the angle of the surface relative to the incoming free stream in radians. Note that  is positive if the slope of the surface is positive and negative if the slope of the surface is negative.

The pressure coefficient is defined as:

Cp 

p  p  p M 2 2

(2)

where p  is the free stream pressure. Setting Equation 1 equal to Equation 2 and solving for the pressure gives:

   M2  p  1   p 2  M  1  

(3)

218 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

where the  sign applies to the upper surface and the  sign applies to the lower surface.

Applying Equation 3 to find the forces acting on the airfoil is simple conceptually but requires careful bookkeeping in order to do it correctly.

The first step is to find  for each surface of the airfoil as well as how it varies with  . We do this using the set of sketches presented in Figure 2.

Figure 2. Sketches of airfoil showing angles of attack of each surface relative to the freestream, the principal coordinate directions, the lengths of each surface ( ), the unit normal vectors to each surface, the projections of each surface on the x and y axes (or î and ĵ directions), and the triangles used to determine  LE and TE from x and y.

The half angles of the leading and trailing edges of the airfoil are:  0.06   y  1  tan    5.143   2x   2 1/3 

 LE  tan 1 



  0.06  y 1   tan    2.577  2 1  x    2 1  1/3 

TE  tan 1 

The angles of each airfoil surface with respect to the horizontal are also shown in Figure 2.

These angles are given by:

 2   LE   3   TE     2    LE    3   TE  

(4)

219 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Substituting the appropriate angles in Equation (4) into Equation 3 gives expressions for the pressures on each surface:  p 2  p 1     p3  p 1   

 M 2 M 2  1

 M 2 M 2  1



 LE     



TE     

 p 2  p 1   

M 2  1

 p3  p 1   

 TE    2  M  1 

 M 2



 LE     

(5)

 M 2

The net pressure force acting on the airfoil is given by:

Fnet   pnˆ dA A

(6)

where p is the surface pressure and n̂ is the unit normal to the surface. If we assume that the coordinate axes are pointing parallel to the flow direction and perpendicular to the flow direction then: Fnet  Liˆ  Djˆ

(7)

where L is the net lift force, D is the net drag force, and î and ĵ are unit vectors pointing in the coordinate directions.

There are two methods for evaluating the integral in Equation 6. The first involves breaking the integral into parts associated with each surface and adding everything together vectorially. The second involves using some physical insight into how the integral actually works. We will show both methods.

Method 1: Re-write Equation 6 as:

  Fnet     p2 nˆ 2 dA1   p3nˆ 3 dA3   p2 nˆ 2 dA2   p3 nˆ 3 dA3   A2  A3 A 2 A3

(8)

220 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

where the subscripts indicate the various surfaces Figure 2. The components of the unit normal vectors and the magnitudes of the area vectors are summarized in Table 1 below: Table 1. Unit vectors and areas associated with each surface of the airfoil. n̂i dA i i

î

ĵ

2

- sin(qLE - a )

cos(qLE - a )

l2

3

sin(qTE + a )

cos(qTE + a )

l3

2

- sin(qLE + a )

cos(qLE + a )

l 2¢

3

sin(qTE - a )

- cos(qTE - a )

l 3¢

The lengths of the various airfoil surfaces are given by: 2 

3 

y 2 2  c x    2

3  c

1  x 

2

2

y   2

2

(9)

where c is the airfoil‟s chord. Plugging in numbers from the problem statement gives: 2  2  0.335 m and 3  3  0.667 m .

Method 2: A tricky/tedious aspect of Method 1 is finding the unit normal vectors to each surface because each is influenced by the geometry of the airfoil (  LE and TE ) and the angle of

attack ( ) . from the angles of each surface as well as the angle of attack. This can be avoided by recognizing by imagining that the pressure forces act instead on the chord (leading to the normal force) and the thickness (leading to the axial force). Therefore, we can calculate the normal and axial forces using just the airfoil‟s geometry and then relate these to angle of attack later. This is the method used to solve problem 9.35. Of course, the expressions for the pressures (Equations 3) still depend on geometry and angle of attack.

Figure 3 illustrates how this works:

221 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Figure 3. Normal and axial forces acting on an airfoil at angle of attack  .

From the figure, it is clear that: N   p2  p2  xc   p3  p3  1  x  c

y y A   p 2  p3  c   p 2  p3  c 2 2

(10) (11)

The lift and drag forces are calculated using: L  Ncos  Asin  D  Nsin  Acos 

(12) (13)

Regardless of what method you use, the lift and drag coefficients are given by:

CL 

 2

CD 

 2

L

(14)

p M 2 c

D

(15)

p M2 c

Note that lift and drag are per unit span. The results of the calculations are reported below and summarized in the tables on the following pages. They show that shock-expansion and linearized theory are consistent within 8% for 0    10 .

222 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

(a)

(b)

Figure 4. Lift coefficient (a) and drag coefficient (b) as functions of angle of attack calculated using two methods: Linearized flow theory (solid lines) and shock-expansion theory (dashed lines). The dotted line shows the fractional difference between the two methods in percent.

Figure 5. Lift to drag ratio as a function of angle of attack calculated using two methods: Linearized flow theory (solid lines) and shock-expansion theory (dashed lines). The dotted line shows the fractional difference between the predictions of the two methods in percent.

Table 2. Summary of results of calculations using linearized theory. A summary of the results from shock expansion theory are available in the solution to problem 9.35. α

Surface angles wrt. Horizontal

(d θ2 θ3 eg (de (de ) g) g) – 5.1 0 2.5 43 77 – 4.1 1 3.5 43 77 – 3.1 2 4.5 43 77 – 2.1 3 5.5 43 77 – 1.1 4 6.5 43 77 – 0.1 5 7.5 43 77

Pressures

Surface area vectors

Forces

Force Coefficients

L/D

θ2′ θ3′ θ2′ θ3′ θ2 N2 N3 N2′ N3′ Drag θ2 θ3 θ3 θ2′ θ3′ Lift (de (de (rad) (rad) (atm) (N/m CD 12.10 9.35 (rad) (rad) (atm) (atm) (atm) (N/m) CL x y x y x y x y g) g) ) – – – – – – 5.1 0.08 0.08 1.290 0.85 1.29 0.85 – 0.33 0.666 – 2648 –1.1E- –4E- 0.00 2.5 0.04 0.044 0.03 0.33 0.03 0.66 4.2479E0 43 9758 9758 203 4606 0203 4606 0.03 3333 67 0.03 .185 11 17 9334 77 497 97 333 667 15 – – – – – – – 6.1 0.07 0.10 1.233 0.79 1.34 0.91 0.33 0.04 0.666 0.01 2847 11387 0.040 0.01 3.99926 3.82 1.5 0.06 0.027 0.02 0.03 0.33 0.66 43 2305 7211 773 8177 6632 1035 3806 163 042 8361 .355 .32 138 0036 2774 5175 77 242 52 418 581 276 709 – – – – – – – 7.1 0.05 0.12 1.177 0.74 1.40 0.96 0.33 0.05 0.665 0.00 3444 22764 0.080 0.01 6.60839 6.36 0.5 0.07 0.010 0.01 0.04 0.33 0.66 43 4852 4665 344 1747 3061 7464 4177 3248 214 6715 .742 .2 238 2142 1503 2571 77 988 06 835 161 208 731 – – – – – – 8.1 0.4 0.03 0.14 0.007 1.120 0.68 1.45 1.02 0.33 0.06 0.664 4439 34120 0.120 0.01 7.68476 7.45 0.09 0.01 0.04 0.33 0.00 0.66 43 23 7398 2118 39 915 5318 949 3894 4447 4849 183 .983 .22 266 565 483 1548 733 251 74 131 493 732 – – – – – – 9.1 1.4 0.01 0.15 0.024 1.064 0.62 1.51 1.08 0.33 0.07 0.662 5832 45444 0.160 0.02 7.79171 7.59 0.11 0.00 0.05 0.33 0.01 0.66 43 23 9945 9571 843 486 8889 592 0323 4614 6431 95 .47 .96 183 0558 71804 9833 478 667 318 043 658 714 – – – – – – 10. 2.4 0.00 0.17 0.042 1.008 0.57 1.57 1.13 0.33 0.08 0.661 7621 56728 0.199 0.02 7.44329 7.29 0.13 0.00 0.05 0.32 0.02 0.66 143 23 2492 7025 297 056 2459 2349 6752 468 799 515 .354 .01 953 6863 8107 2048 224 083 894 945 822 674

223 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

6

7

8

9

10

11

12

13

14

15

– – 0.8 8.5 57 77 – – 1.8 9.5 57 77 – – 2.8 10.

57 577 – – 3.8 11.

57 577 – – 4.8 12.

57 577 – – 5.8 13.

57 577 – – 6.8 14.

57 577 – – 7.8 15.

57 577 – – 8.8 16.

57 577 – – 9.8 17.

57 577

– 11. 3.4 0.01 143 23 496 – 12. 4.4 0.03 143 23 241 – 13. 5.4 0.04 143 23 987 – 14. 6.4 0.06 143 23 732 – 15. 7.4 0.08 143 23 477 – 16. 8.4 0.10 143 23 223 – 17. 9.4 0.11 143 23 968 – 18. 10.

0.13 143 423 713 – 19. 11.

0.15 143 423 459 – 20. 12.

0.17 143 423 204

– 0.19 0.059 0.951 0.51 0.14 4478 75 627 603 969 – 0.21 0.077 0.895 0.45 0.16 1931 203 198 9601 714 – 0.22 0.094 0.838 0.40 0.18 9384 657 769 3172 46 – 0.24 0.112 0.782 0.34 0.20 6838 11 339 6742 205 – 0.26 0.129 0.725 0.29 0.21 4291 563 91 0313 95 – 0.28 0.147 0.669 0.23 0.23 1744 016 481 3884 696 – 0.29 0.164 0.613 0.17 0.25 9198 47 051 7455 441 – 0.31 0.181 0.556 0.12 0.27 6651 923 622 1025 186 – 0.33 0.199 0.500 0.06 0.28 4104 376 193 4596 932 – 0.35 0.216 0.443 0.00 0.30 1557 83 764 8167 677

– – 1.62 1.19 0.00 0.33 0.09 0.659 0.06 0.32 8778 3181 5007 4643 9521 879 468 837 – – 1.68 1.24 0.01 0.33 0.11 0.658 0.07 0.32 5208 9611 0847 4505 1023 041 04 719 – – 1.74 1.30 0.01 0.33 0.12 0.656 0.07 0.32 1637 604 6683 4265 249 004 61 591 – – 1.79 1.36 0.02 0.33 0.13 0.653 0.08 0.32 8066 2469 2514 3922 392 766 178 454 – – 1.85 1.41 0.02 0.33 0.14 0.651 0.08 0.32 4495 8898 8338 3479 531 329 743 306 – – 1.91 1.47 0.03 0.33 0.15 0.648 0.09 0.32 0925 5328 4154 2933 6655 694 305 148 – – 1.96 1.53 0.03 0.33 0.16 0.645 0.09 0.31 7354 1757 9959 2287 7952 861 865 981 – – 2.02 1.58 0.04 0.33 0.17 0.642 0.10 0.31 3783 8186 5753 1539 9198 832 421 804 – – 2.08 1.64 0.05 0.33 0.19 0.639 0.10 0.31 0212 4616 1532 069 039 606 975 617 – – 2.13 1.70 0.05 0.32 0.20 0.636 0.11 0.31 6642 1045 7295 974 1524 168 525 421

– – 0.03 0.66 985 615 – – 0.05 0.66 147 535 – – 0.06 0.66 307 435 – – 0.07 0.66 466 315 – – 0.08 0.66 622 175 – – 0.09 0.66 776 014 – – 0.10 0.65 926 834 – – 0.12 0.65 074 633 – – 0.13 0.65 217 412 – – 0.14 0.65 357 172

9805 0.239 0.03 67959 .544 539 4562

6.93067 6.81 058 1497

1238 79127 0.278 0.04 3.71 .56 906 365

6.38965 6.29 0601 4143

1535 90223 0.318 0.05 4.27 .37 016 412

5.87611 5.79 053 7777

1871 10123 0.356 0.06 5.41 6.2 833 5967

5.40923 5.34 972 3395

2246 11215 0.395 0.07 5.09 5.7 322 9184

4.99244 4.93 2189 5738

2660 12297 0.433 0.09 1 1.7 446 3762

4.62282 1257

3112 13367 0.471 0.10 0.63 4.1 169 9693

4.29535 3361

3602 14425 0.508 0.12 1.2 2.8 457 6966

4.00466 4265

4129 15469 0.545 0.14 9.71 7.8 273 5572

3.74573 5251

4695 16499 0.581 0.16 2.93 9.1 582 5498

3.51413 7793

CHAPTER 13 1.41

At point 1:

a1   RT1  (1.4)(287)(288)  340 m/sec

V1  u12  v12  (639)2  (232.6)2  680 m/sec M1 

V1 680  2 a1 340

1  tan 1

v1  232.6   Tan 1    20 u1  639 

1   M1   26.38 15 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

K       20  26.38  46.38

At point 2:

a 2   RT2  (1.4)(287)(288)  340 m/sec V2  680 m/sec

16 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

M2 

V1 680  2 a1 340

2  0  2   K       26.38

At point 3:

3  1 2  K  1   K  2 ]  1 2 (46.38  26.38)  10 3  1 2  K  1   K  2 ]  1 2 (46.38  26.38)  36.38 M3  2.4

To obtain the other flow variables at 3, note that: po1 p1

 7.824 and

p3  To1 T1

p3

 14.62

p3 po3 po1  1  p1    (1)(7.824)(1 atm)  0.535 atm po3 po1 p1  14.62 

 1.8 and

T3 

p o3

To3 T3

 2.152

T3 To3 To1  1  T1    (1)(1.8)(288)  240.9K To3 To1 T1  2.152 

a 3   RT3  (1.4)(287)(240.9)  211.1 m/sec V3  M3a 3  2.4 (311.1)  746.6 m/sec

u 3  V3 cos 3  746.6 cos 10  735.3 m/sec v3  V3 sin 3  746.6 sin 10  129.6 m/sec To locate point 3: Along the C+ Characteristic:

17 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

ave  1 2  2  3   1 2 (0  10)  5 ave  1 2   2  3   1 2  30  24.62   27.31

dy  Tan  ave  ave   Tan  5  27.31   0.6324 dx Thus: y  0.6324 x  0.00765

(1)

Along the C characteristic:

ave  1 2  θ1  θ3   1 2  20  10   15 ave  1 2  1  3   1 2 (30  24.62)  27.31

dy  Tan  ave  ave   Tan 15  27.31   0.2182 dx y  0.2182 x  0.0684

(2)

Point 3 lies at the intersection of Eqs. (1) and (2) y  0.6324 x  0.00765 y  0.2182 x  0.0684

Solving simultaneously:

x  0.0894 y  0.0489

Thus:

 x 3 , y3   (0.0894, 0.0489) CHAPTER 14





c  C p3  C p2 cos 

18 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.





cd  Cp3  Cp2 sin 

(a)

Using straight Newtonian theory:

Cp  2 sin 2  For   5:

Cp3  2 sin 2 5  0.0152

Cp2  0 c  0.0152 cos 5  0.0151 cd  0.0152 sin 5  0.00132 For   15:

Cp3  2 sin 2 15  0.1340, Cp2  0

c  0.1340 cos 15  0.129 cd  0.1340 sin 15  0.0347

19 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

For   30:

Cp3  2 sin 2 30  0.5 c  0.5 cos 30  0.433

cd  0.5 sin 35  0.25 (b)

Using modify Newtonian:

Cp  Cpmax sin 2  Cpmax 

C pmax 

 po  p p  p 2  po  o   1 2   2 q  M  p  M   p 2 2 (9.181  1)  1.729 (1.4)(2.6) 2

For   15

Cp3  1.729 sin 2 5  0.0131 c  0.0131 cos 5  0.0131 cd  0.0131 sin 5  0.00114

For   15 Cp3  1.729 sin 2 15  0.1158 c  0.1158 cos 15  0.1119 cd  0.1158 sin 15  0.030

For   30 Cp3  1.729 sin 2 30  0.4323 c  0.4323 cos 30  0.374

cd  0.4323 sin 30  0.216

20 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Comparison:

 5 15 30

 5 15 30

Exact c (Prob. 9.13)

Newtonian c

% error

Mod. Newtonian c

0.148 0.452 1.19

0.0151 0.129 0.433

90 71 63.6

0.0131 0.1119 0.374

91 75.2 68.6

Exact cd (Prob. 9.13)

Newtonian c

% error

Mod. Newtonian cd

% error

0.0129 0.121 0.687

0.00132 0.0347 0.25

90 71 63.6

0.00114 0.03 0.216

91 75.2 68.6

% error

Conclusion: Newtonian theory gives terrible results for a flat plate a moderate  at low Supersonic Mach numbers.

From Newtonian theory:

Cp  2 sin 2   2 sin 2 20  0.234

21 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

c  0.234 cos   0.220 cd  0.234 sin   0.08

From shock-expansion theory: On the top surface:  2  1    116.2  20  136.20 This is beyond the maximum expansion angle. Hence, a “void” exists on the top surface, i.e., p 2  0.

On the bottom surface: From the --M diagram,   24.9

Mn1  M1 sin   20 sin 24.9   p3  82.15 p1 From Prob. 9.13: c 

2  p3 p 2     cos   M 2  p1 p1 

and

cd  c c 

sin  cos 

2 (82.15  0) cos 20  0.2757 (1.4)(20) 2

cd  0.2757 Tan 20  0.100

For c : % error 

0.2757  0.220  20% 0.2757

For cd : % error 

0.100  0.08  20% 0.10

Note: Newtonian theory works much better for blunt bodies, i.e., for large value of .

22 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

(a) Use Eq. (14.7) to estimate the pressure at point A. We first need to obtain Cp,max , which is a function of po,2 /p. From appendix B for M  20, po,2 /p  0.5155 103. Hence, Cp,max 

2  po,2  2  1  (515.3  1)  1.837 2  2  M   p  (1.4)(20)

From Eq. (14.7) at point A on the surface

CpA  Cp,max sin 2   (1.837) sin 2 20  0.2149 Since C pA 

 2  pA  1 2   M   p 

then, 2 p A  M  CpA   1  (1.4)(20) (0.2149)  1  61.17 p 2 2 2

Hence,

PA  61.17 (3.06)  187.2 lb/ft 2 (b) The stagnation temperature is found from Eq. (8.40) To  1 2  1 M   1  0.2 (20)2  81 T 2

Assuming an isentropic flow from the stagnation point to point A,

23 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

T  pA p /p  A   A  po,2 po,2 /p   To 

  1

or,  1

TA  61.17   0.2857   0.5439   (0.1187) To  515.5 

TA 

TA  To    T  (0.5439)(81)(500)  22, 028R To  T 

(Please note. Relative to our discussion in Problems 8.17 and 8.18, we know this estimate of TA to be too large because we are not taking into account the effect of chemically reacting flow.) (c) At point A, for an isentropic flow, po,A  po,2 

   1 2   1 po,2 /p  515.5  1  MA     8.427 pA  2 p A /p  61.17 

po,A

1

 1 2

M  (8.427) 2 A

M 2A  (1.8385  1)

 1 

 (8.427) 0.2857  1.8385

2  (0.8385)(5)  4.1925  1

M A  2.05

(d) a A   RTA  (1.4)(1716)(22,028)  7275 ft/sec

VA  a A MA  (7275)(2.05)  1.49 104 ft/sec Note: Once again, this estimate of VA is too high because TA , hence a A, is too high.

Also note: The purpose of this problem is to illustrate that, from the Newtonian sine-squared law for pressure variations, the other flow field quantities can also be obtained.

CHAPTER 15

24 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

1.42

(a) Since the plates are infinite in length, u  u(y) only. Also v  0, i.e., the flow is in the x-direction only. The governing equation is Eq. (15.18a), which reduces to the following u  u(y), v  0 and p  const.

0

d  du    dy  dy 

Integrating: 

du  const  c1 dy

u  c1y  c2 At y  0, u  0: c2  0 At y  h, u  u e : u e  c1h

c1 

ue h

Thus:

u

ue

y y, or u  u e   h h

The velocity variation is linear between the plates. (b)

du u e  dy h

25 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.



u du  e dy h

 T   0  To 

3/2

To  110  320    T  110  288.16 



3/2

288.16  110  1.084 320  110



  1.084 o  1.084 1.7894 105  1.94 105

kg m sec

 30  2 2   1.94 105    5.82 10 N/m 0.01  





The shear stress is constant, and hence is the same on the top and bottom walls.

1.43

u  u(y), v  0, p  p(x)

0



dp d  du     dx dy  dy 

du  dp      y  c1 dy  dx 

 dp  y u      c1 y  c 2  dx  2 2

2 c c  dp  y u     1 y 2   dx  2  

At y  0, u  0. Thus c2  0

26 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

At y  h, u  0. Thus, 2 c  dp  h 0     1h  dx  2  

 dp  h c1     dx  2

2  dp  y  dp  h u      y  dx  2   dx  2 

u

1  dp  2   h yy 2  dx 





The velocity profile is parabolic.

du  dp  y  dp  h       dy  dx    dx  2

On the bottom plate, y  0:   

du dy

  dp  0  dp  h  h  dp              2  dx    dx    dx  2 

 du  On the top plate, y  h:       since dy is negative, i.e., the distance away from the top  dy  plate is in the downward (negative direction)   dp  h  dp  h              dx    dx  2  

h  dp    2  dx 

For both the top and bottom walls, 

h  dp    2  dx 

Shear stress varies linearly with the magnitude of the pressure gradient.

Note: Due to the content of chapters 16, 17, and 18, no homework problems are required.

27 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.