AE 2010 Thermodynamics (PROBLEM SET #2 WORK, PATHS, AND IDEAL GAS LAW) WITH SOLUTION Georgia Institu

AE 2010 Thermodynamics (PROBLEM SET #2: WORK, PATHS, AND IDEAL GAS LAW) WITH SOLUTION Georgia Institute Of Technology

Problem Set #2: Work, Paths, and ideal gas law • Always indicate any assumptions you make. If you use any results or equations from the class notes or text in you solutions, please note and reference them (but you better be sure they are applicable to the problem at hand).

• Show all your work, no credit for just answers. When applicable, try to solve the problem algebraically first. Only use numbers/values in the final steps of your solution – and be sure to include units when you insert numbers.

• If the problem statement is given in ENGLISH units, the answer must also be in English units; if the problem statement is in SI units, the answer must be in SI units.

1. Work and Paths A simple compressible substance of fixed mass undergoes an increase in both pressure (p) and specific volume (v) from state 1 (p1, v1) to state 2 (p2, v2). Two ways to accomplish this are being considered. In the first case (path A), the overall path is a constant pressure (isobaric) process followed by a constant-volume (isochoric) process. In the second case (path B), the pressure rises as the square-root of the 2 volume increase, i.e., p  v1 .

Determine: (a) is there net work being done in path (A) to the substance or by the substance on its surroundings; (b) same question as (a) for path (B); (c) which path involves the largest (absolute value) amount of work; and (d) will the change in internal energy be greater for path A than for path B. As part of your answer, you MUST also draw the two paths on a p-v diagram.

2. Expansion Work A pneumatic “blaster” is a device that launches a low mass projectile from a cylindrical tube using pressurized air stored upstream of the projectile. The projectile is held in place while the pressurized air is introduced to the cylinder. After the trigger is released, the projectile accelerates down the tube, while the pressure upstream of the projectile drops according to the relation,  (x d )2   p = pi 1 − 30   where pi is the initial (high) pressure in the tube before the trigger is release, x is the distance traveled by the projectile and d is the diameter of the tube. If we have a 5.0 cm diameter tube, with a projectile initially located 5.0 tube diameters upstream of the tube exit, and the initial pressure upstream of the projectile is 1.2 MPa, determine the work done by the air on the projectile during the time that projectile is in the tube, assuming there is no friction between the tube and the projectile.

3. Enthalpy First, determine the specific enthalpy of the following two substances at the given conditions. Then identify which of the substances is likely to be a gas. Finally, for the substance that is likely to be a gas, determine if it can be considered a calorically perfect gas over the range 300-1000 K Substance A A A A

p (bar) 1.00 20.0 20.0 20.0

T (K) 300.

300.

500.

1000.

 (kg/m3) 1.20 24.0 14.0 7.20

u (kJ/kg) 0.00 0.50 110 440

B B B B

1.00 20.0 20.0 20.0

300.

300.

500.

1000.

500 505 501 490

0.00 1.00 220 810

Solution for Problem Set #2: Work, Paths and Gas State Equations Problem 1. Work and Paths

p

2

Given: Fixed mass system going from state 1 to 2, with p2>p1, v2>v1 in 2 paths A: isobaric then isochoric B: pv1/2 Find: (a) WA in or out (b) WB in or out (c) | WA | vs. | WB | (d) UA vs. UB

B

1

A a v

Assume: only expansion/compression work, paths internally reversible (quasi-equil) Analysis: a) The work in path A can be broken into 2 subpaths

2  v2   a WA = m pdv = m pdv +  pdv = m p1  dv + 0 = mp1 (v2 − v1 ) > 0  v1  1 1 a  2

0

so the work is being DONE BYTHE GAS b) As seen in the figure, the specific volume in path B drops quadratically with the pressure 2

(or pv−1 = C )

2 −1 2 32 32 −1 2  2 3 2 v2 ( ) 2 1 2   WB = m pdv = mC   v dv = mp1v1 v  = mp1v1 v2 − v1 3  v1 3 1 1  2  2 p   3  WB = 2 mp1  v2 − v1  = 2 mp1  v12 2 v2 − v1  = mp1  2 v2 − v1   v1 2  3  v1 2  3 p  3  1   1   1  W = 2 m p v − p v  0 since p  p and v  v B 2 1 2 1 3 2

(

)

so again the work is being DONE BY THE GAS

c) While it can be shown from the equations (see below), it is easier to see from the p-v diagram (above) that the shaded area under path B is larger, therefore |WA | < | WB |, i.e., Path B WILL PRODUCE MORE WORK

========================= A short analytic “proof” of the above.

We need to show 2 v2 v − v  v − v or for z  v1 v2 So

3 2

2

v1

1

2

1

(z −1 − z) 1− z 3 2 −1 2 2 ? z − z  1− z 3 3 2

or

1

or

z1 (z − 3)− 2

or

z1 ( 3 − z)  2

with 0  z  1

2 ?

2 z + z −1 1 3 3 ?

2 2

?

What is the maximum value that F (z) = z1 (3 − z)can obtain for 0  z  1?

1) With z = 0 , F (z) = 0 2) With z = 1, F (z) = 2 3) dF dz = 3 z −1 2 − 3 z1 2  0 for 0  z  1, therefore F(z) monotonically increases for z = 0 →1 .

2 2 It therefore follows that F (z) 2 for 0  z  1…. QED!!

2

Note, for the case of z = 1 (or v1 = v2 ), there is no change in state and both paths are equivalent (i.e., W=0).

d) Since internal energy is a state variable, it does not depend on path. The beginning and ending states are the same in both cases, so UA = UB Implications: 1) While work is path dependent, changes in state properties are path independent.

2) If p is increasing in a process, having it drop first at constant volume will reduce the work produced.

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Problem 2. Expansion Work

x

Given: pneumatic (cylindrical) cannon launching projectile with pressure upstream of the projectile given by  (x d )2   and pi = 1.2 Mpa, d=5.0 cm p = pi 1 − 30  

projectile

air

Find: Work done by air during process Assume: 1) No friction 2) Quasi-equilibrium process 3) No work but compression/expansion (e.g., no gravity field change) pA=force

Analysis:

 (x d )2  dx W12 = pdV = pAdx = pi A  1 − 1 1 30  0 L d 2  1 1 x3   − 1 x 2  L d 3  x 1 90 d 2 = pi 4  x − 30 3 d 2  = pi 4 d  0 0   3 d  25  3  65  6 N  5  = pi 51 −  = 1.2  10 2  (0.050m )   4  90  m  4   90  N 5  65   = 1.2 106 (0.050m)3   = 425Nm     m2 4 90     2

2

L

W12 = Implications: 1) Fdx and PdV are equivalent expressions 2) The pressure variation behind a projectile in a “cannon” can be used to determine the work imparted - this could also be related to the projectile’s velocity if the forces that must be overcome, e.g., friction and ambient pressure, are known.

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Problem 3. Enthalpy Given: Table of properties for two substances Substance A A A

p (bar) 1.00 20.0 20.0

T (K) 300 300 1000

 (kg/m3) 1.20 24.0 7.20

u (kJ/kg) 0.00 0.50 440

B B B

1.00 20.0 20.0

300 300 1000

500 505 490

0.00 1.00 800

Find: (a) enthalpy, h at each condition (b) which substance likely to be gas (c) can gas be considered calorically perfect Assume: none Analysis: a) Enthalpy By definition, h = u + p  . So for the first condition, we have

h=u+

p



= 0 kJ kg +

1.00bar 106 N m2 J = 83.3kJ kg 1.20kg m3 1 bar Nm

Using a similar approach for the other conditions, we get Substance A A A A

p (bar) 1.00 20.0 20.0 20.0

T (K) 300 300 500 1000

 (kg/m3) 1.20 24.0 14.0 7.20

u (kJ/kg) 0.00 0.50 110 440

h (kJ/kg)

B B B B

1.00 20.0 20.0 20.0

300 300 500 1000

500 505 501 490

0.00 1.00 220 800

0.20 5.00 224 804

83.3 83.8 253 718

b) Gas For an ideal gas,  = p RT ; so density should be roughly proportional to pressure for a gas. Since density is nearly independent of pressure for substance B, it is unlikely to be a gas. For substance A, the first two conditions show that the density is proportional to pressure when the temperature is held constant. So substance A is likely to be a gas.

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c) Calorically perfect To be calorically perfect, we would expect a gas to have cv~constant or cp~constant. Since we have a number of values for constant pressure, let’s look at c p=h/T|p. Approximating this from our table of data, e.g., c p 

h (253 − 83.8)kJ kg = = 0.85kJ kg K T (500 − 300)K

Substance A A A

p (bar) 20.0 20.0 20.0

T (K) 300 500 1000

h (kJ/kg) 83.8 253 718

h/T (kJ/kgK) --0.85 0.93

Therefore, cp increases by at least 10% from 300 to 1000 K. To be calorically perfect, cp should be nearly constant.

Strictly speaking, this is NOT a calorically perfect gas…

BUT, is a 10% change significant? Answer – it depends. For example, for a given h, using a calorically perfect assumption could produce a 10% error in calculating T - in our case, 0.10(1000K-300K)=70K This error in temperature may or may not be significant depending on the situation (like how accurately you could measure temperature?).

Implications: 1) For a gas (substance A), enthalpy is mostly a function of temperature, not pressure.

2) For other substances (like B) where density is nearly constant (i.e., an incompressible substance) with changes in pressure, enthalpy is a stronger function of pressure.

3) For high density substances (like B), enthalpies are much less than for for a low density substance at the same internal energy.

4) Whether a substance can be considered calorically perfect depends on the size of the error introduced relative to the “acceptable” error.

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