Fox and mcdonalds introduction to fluid mechanics 9th edition pritchard solutions manual

Fox and McDonalds Introduction to Fluid Mechanics 9th Edition Pritchard Solutions Manual Visit to download the full and correct content document: https://testbankdeal.com/dow nload/fox-and-mcdonalds-introduction-to-fluid-mechanics-9th-edition-pritchard-solutio ns-manual/

Problem 7.1 Problem 7.3

[Difficulty: 2]

7.1

Given: Find:

Equation describing the slope of a steady wave in a shallow liquid layer

Solution:

To nondimensionalize the equation all lengths are divided by the reference length and all velocities are divided by the reference velocity. Denoting the nondimensional quantities by an asterisk:

Nondimensionalization for the equation using length scale L and velocity scale V o. Obtain the dimensionless groups that characterize the flow.

h* 

Substituting into the governing equation:

The dimensionless group is

V0

h L

   

x* 

x L

u* 

u V0

   

 h* L u*V0  u *V0    x* L g  x* L

2

g L

which is the square of the Froude number.

h* V02 * u * u   x* gL x*

Problem 7.2 Problem 7.4

[Difficulty: 2]

7.2

Given: Find:

Equation describing one-dimensional unsteady flow in a thin liquid layer

Solution:

To nondimensionalize the equation all lengths are divided by the reference length and all velocities are divided by the reference velocity. Denoting the nondimensional quantities by an asterisk:

Nondimensionalization for the equation using length scale L and velocity scale Vo. Obtain the dimensionless groups that characterize the flow.

x* 

x L



h* 



h L

u* 

u V0

   

t* 

t L V0

   

 u *V0  u *V0  h* L *  u V   g Substituting into the governing equation: 0  t * L V0  x* L  x* L





V02 u * V02 * u * h *  u   g L t * L x * x *

The dimensionless group is

g L V0

2

Simplifying this expression:

Thus:

which is the reciprocal of the square of the Froude number.

* gL h * u * * u u    t * x * V02 x *

Problem 7.3 Problem 7.6

[Difficulty: 2]

7.3

Given:

Equations for modeling atmospheric motion

Find:

Non-dimensionalized equation; Dimensionless groups

Solution: Recall that the total acceleration is

   DV V    V  V t Dt Nondimensionalizing the velocity vector, pressure, angular velocity, spatial measure, and time, (using a typical velocity magnitude V and angular velocity magnitude ):

Hence

  V V*  V

p p*  p

   *  

  V VV *

p  p p *

    *

x* 

x L

t*  t

x  Lx*

V L

t

L t* V

Substituting into the governing equation     V V * V  1 p V  V V *  * V * 2V  * V *   p * L t * L  L

The final dimensionless equation is

The dimensionless groups are

   p V *   L   p *  V *  * V * 2    * V   V t *  V2  

p

V

2

L V

The second term on the left of the governing equation is the Coriolis force due to a rotating coordinate system. This is a very significant term in atmospheric studies, leading to such phenomena as geostrophic flow.

Problem 7.4 (Difficulty 2)

7.4 Fluid fills the space between two parallel plates. The differential equation that describes the instantaneous fluid velocity for unsteady flow with the fluid moving parallel to the walls is 𝜌

𝜕 2𝑢 𝜕𝜕 =𝜇 2 𝜕𝜕 𝜕𝜕

The lower plate is stationary and the upper plate oscillates in the x-direction with a frequency 𝜔 and an amplitude in the plate velocity of 𝑈. Use the characteristic dimensions to normalize the differential equation and obtain the dimensionless groups that characterize the flow.

Find: Use the characteristic dimensions to normalize the equation and obtain the dimensionless groups.

Solution: This is unidirectional flow with oscillating boundary with characteristic dimensions U and H. We will normalize the variables using these characteristic dimensions as: 𝑢∗ =

𝑦∗ =

𝑡∗ =

𝑢 𝑈

𝑦 𝐻

𝑡 𝐻 𝑈

Substituting these dimensions into the differential equation we have: 𝜌

𝜕 2 (𝑈𝑈∗ ) 𝜕(𝑈𝑈∗ ) =𝜇 𝐻 𝜕(𝐻𝐻 ∗ )2 𝜕 � 𝑡 ∗� 𝑈

𝜇 𝜕 2 𝑢∗ 𝜕𝑢∗ = 𝜕𝑡 ∗ 𝜌𝜌𝜌 𝜕𝜕 ∗2 On the boundary we have: At 𝑦 = 0, 𝑢 = 0 so we have: At 𝑦 = 𝐻, 𝑢 = 𝑈 cos 𝜔𝜔 so we have:

𝑢∗ = 0 at 𝑦 ∗ = 0

We define the Reynolds number as:

𝑢∗ = cos 𝜔𝜔 at 𝑦 ∗ = 1

The differential equation becomes:

𝑅𝑅 =

𝜌𝜌𝜌 𝜇

1 𝜕 2𝑢∗ 𝜕𝑢∗ = 𝜕𝑡 ∗ 𝑅𝑅 𝜕𝜕 ∗2

The fluid flow is governed by the following two dimensionless group: 𝑅𝑅 =

It can be written as:

𝜌𝜌𝜌 𝜇

П = cos 𝜔𝜔 𝜌𝜌𝜌 𝑢 = 𝑓� , cos 𝜔𝜔� 𝜇 𝑈

Problem 7.5 Problem 7.7

[Difficulty: 4]

7.5

Given: Find:

The Prandtl boundary-layer equations for steady, incompressible, two-dimensional flow neglecting gravity

Solution:

To nondimensionalize the equation all lengths are divided by the reference length and all velocities are divided by the reference velocity. Denoting the nondimensional quantities by an asterisk:

Nondimensionalization for the equation using length scale L and velocity scale V 0. Obtain the dimensionless groups that characterize the flow.

x* 

Substituting into the continuity equation:

x L

y* 

y L

u* 

u V0

v* 

v V0

       

* *  u *V0  v *V0 Simplifying this expression: V0 u  V0 v  0   0 L x * L y *  x* L  y*L

u * v *  0 x * y * We expand out the second derivative in the momentum equation by writing it as the derivative of the derivative. Upon substitution:

   

   

     

 u *V0  u *V0 1 p   u *V0 * u V0  v V0     x* L  x* L  y*L  y*L  y*L *

u*

 

Simplifying this expression yields:

* 1 p   2 u * Now every term in this equation has been non-dimensionalized except the u * * u v     V02 x * V0 L y * 2 pressure gradient. We define a dimensionless pressure as: y * x *

p* 

p V02

Substituting this into the momentum equation:

u*

Simplifying this expression yields:

The dimensionless group is

ν V0  L





* 1  p * V02   2u * u * * u u v     V0 L y * 2 V02 x * y * x * *

which is the reciprocal of the Reynolds number.

*   2u * u * p * * u v     x * y * x * V0 L y * 2

Problem 7.6 (Difficulty 2)

7.6 Consider a disk of radius 𝑅 rotating in an incompressible fluid at a speed 𝜔. The equations that describe the boundary layer on the disk are:

𝜌 �𝑣𝑟

𝜕𝑣𝑧 1 𝜕(𝑟𝑣𝑟 ) � �+ =0 𝑟 𝜕𝜕 𝜕𝜕

𝜕 2 𝑣𝑟 𝜕𝑣𝑟 𝑣𝜃2 𝜕𝑣𝑟 − + 𝑣𝑧 �=𝜇 𝜕𝜕 𝑟 𝜕𝜕 𝜕𝑧 2

Use the characteristic dimensions to normalize the differential equation and obtain the dimensionless groups that characterize the flow.

Find: Use the characteristic dimensions to normalize the equation and obtain the dimensionless groups.

Solution: This is unidirectional flow with oscillating boundary. We will normalize the variable with the following characteristic dimensions. Because there is no characteristic velocity in the problem, we will use the product 𝜔𝜔 as the characteristic velocity: 𝑣𝑟∗ =

𝑣𝑟 𝜔𝜔

𝑣𝑧∗ =

𝑣𝑧 𝜔𝜔

𝑣𝜃∗ =

𝑣𝜃 𝜔𝜔

𝑟∗ =

𝑧∗ =

𝑟 𝑅

𝑧 𝑅

Substitute these equations in to the differential equation we have: For the continuity equation: 𝜕𝜔𝜔𝑣𝑧∗ 1 1 𝜕(𝑅𝑟 ∗ 𝜔𝜔𝑣𝑟∗ ) � � + =0 𝑅 𝑟∗ 𝜕𝑅𝑧 ∗ 𝜕𝜕𝑟 ∗

Thus

𝜔

Or

1 𝜕(𝑟 ∗ 𝑣𝑟∗ ) 𝜕𝑣𝑧∗ � � + 𝜔 =0 𝑟∗ 𝜕𝑧 ∗ 𝜕𝑟 ∗ 𝜕𝑣𝑧∗ 1 𝜕(𝑟 ∗ 𝑣𝑟∗ ) � � + =0 𝑟∗ 𝜕𝑧 ∗ 𝜕𝑟 ∗

For the r-momentum equation: 𝜌 �𝜔𝜔𝑣𝑟∗

𝜕 2 𝜔𝜔𝑣𝑟∗ 𝜕𝜔𝜔𝑣𝑟∗ (𝜔𝜔𝑣𝜃∗ )2 𝜕𝜔𝜔𝑣𝑟∗ ∗ − + 𝜔𝜔𝑣 � = 𝜇 𝑧 𝜕𝜕𝑟 ∗ 𝑅𝑟 ∗ 𝜕𝑅𝑧 ∗ 𝜕𝑅 2 𝑧 ∗2

𝜌 �𝜔2 𝑅𝑣𝑟∗

(𝑣𝜃∗ )2 𝜇𝜔 𝜕 2 𝑣𝑟∗ 𝜕𝑣𝑟∗ 𝜕𝑣𝑟∗ 2𝑅 2 𝑅𝑣 ∗ − 𝜔 + 𝜔 � = 𝑧 𝑅 𝜕𝑧 ∗2 𝜕𝑟 ∗ 𝑟∗ 𝜕𝑧 ∗

𝜌𝜔2 𝑅 �𝑣𝑟∗ �𝑣𝑟∗

Defining the Reynolds number as:

𝜇𝜔 𝜕 2 𝑣𝑟∗ 𝜕𝑣𝑟∗ (𝑣𝜃∗ )2 𝜕𝑣𝑟∗ ∗ − + 𝑣 � = 𝑧 𝑅 𝜕𝑧 ∗2 𝜕𝑟 ∗ 𝑟∗ 𝜕𝑧 ∗

𝜇 𝜕 2 𝑣𝑟∗ 𝜕𝑣𝑟∗ (𝑣𝜃∗ )2 𝜕𝑣𝑟∗ ∗ − + 𝑣 � = 𝑧 𝜌𝜔𝑅 2 𝜕𝑧 ∗2 𝜕𝑟 ∗ 𝑟∗ 𝜕𝑧 ∗

The differential equation becomes: �𝑣𝑟∗

𝑅𝑅 =

𝜌𝜔𝑅 2 𝜇

1 𝜕 2 𝑣𝑟∗ 𝜕𝑣𝑟∗ (𝑣𝜃∗ )2 𝜕𝑣𝑟∗ ∗ − + 𝑣 � = 𝑧 𝑅𝑅 𝜕𝑧 ∗2 𝜕𝑟 ∗ 𝑟∗ 𝜕𝑧 ∗

The Reynolds number characterizes the fluid flow as:

�⃗ 𝜌𝜔𝑅 2 𝑉 = 𝑓� � 𝜔𝜔 𝜇

Problem 7.7 Problem 7.8

[Difficulty: 2]

7.7

Given:

Equation for unsteady, 2D compressible, inviscid flow

Find:

Dimensionless groups

Solution: Denoting nondimensional quantities by an asterisk

x* 

x L

y* 

y L

u* 

u c0

v* 

v c0

c* 

c c0

t* 

t c0 L

* 

 L c0

Note that the stream function indicates volume flow rate/unit depth!

Hence

x  Lx*

y  Ly*

u  c0 u *

v  c0 v *

c  c0 c *

t

Lt * c0

  L c0  *

Substituting into the governing equation 2 2  c03   2 *  c03   u *2  v *2   c03  2  c03  2  c03   2 * 2   * 2   *               u c v c u v       * * * * 2 * * 0 2 L L t x *2  L  y *2  L  x * y *  L  t *    

The final dimensionless equation is 2 2  2 *  2 *  u *2  v *2  2 2  * 2 2  *     0  2 * *  *  *  *  *  u c v c u v x * y * y *2 x *2 t t *2

No dimensionless group is needed for this equation!

Problem 7.10 7.8 Problem

[Difficulty: 2]

7.8

Given: Find: Solution:

Functional relationship between pressure drop through orifice plate and physical parameters Appropriate dimensionless parameters We will use the Buckingham pi-theorem.





1

p

2

Select primary dimensions M, L, t:

3

p





V

D

d

M Lt 2

M L3

M Lt

L t

L

L



V

D

4 5

V

D

d

n = 6 parameters

r = 3 dimensions

m = r = 3 repeating parameters

We have n - m = 3 dimensionless groups. Setting up dimensional equations: a

b

c

Π1  Δp ρ  V  D

a

Thus:

b

 M    M    L   Lc  M0 L0 t0  2   3   t   L t   L 

Summing exponents: M: 1  a  0

The solution to this system is:

L:

1  3  a  b  c  0

t:

2  b  0

a  1 b  2 c  0

Check using F, L, t primary dimensions:

F L

a

b

c

Π2  μ ρ  V  D

2



L

4

F t a

Thus:

2



t

Π1 

Δp 2

ρ V

2

 1 Checks out.

L 2

b

 M    M    L   Lc  M0 L0 t0    3    L t   L   t 

Summing exponents: M: 1  a  0 L:

1  3  a  b  c  0

t:

1  b  0

The solution to this system is: a  1 b  1 c  1

μ Π2  ρ V D (This is the Reynolds number, so it checks out)

a

b

c

Π3  d  ρ  V  D

Thus:

L 

M

a

 

L

b

c

0

0 0

 3   t   L  M  L  t L 

Summing exponents: M: a  0 L:

1c0

t:

b0

The solution to this system is: a0

b0

c  1

d Π3  D (This checks out)

Problem 7.9 Problem 7.12

[Difficulty: 2]

7.9

At low speeds, drag F on a sphere is only dependent upon speed V, viscosity μ, and diameter D

Given: Find: Solution:

Appropriate dimensionless parameters We will use the Buckingham pi-theorem.



1

F

2

Select primary dimensions M, L, t:

3

F

V



ML t2

L t

M Lt

V



D

n = 4 parameters

D

n = 4 parameters

L

r = 3 dimensions

D

4

V

5

We have n - m = 1 dimensionless group. Setting up a dimensional equation: a b

c

Π1  F V  μ  D

m = r = 3 repeating parameters

a

Thus:

b

 M L    L    M   Lc  M0 L0 t0  2   t   L t   t 

Summing exponents: M: 1  b  0 L:

1abc0

t:

2  a  b  0

The solution to this system is: a  1 b  1 c  1

F Π1  μ V D

2

Check using F, L, t primary dimensions:

t L 1 F    1 Checks out.

L F t L

Since the procedure produces only one dimensionless group, it must be a constant. Therefore:

F μ V D

 constant

Problem 7.10 Problem 7.14

[Difficulty: 2]

7.10

Given: Find: Solution:

Functional relationship between buoyant force of a fluid and physical parameters

1

FB

V

2

Select primary dimensions F, L, t:

3

FB

V

γ

F

L

3

F

Buoyant force is proportional to the specific weight as demonstrated in Chapter 3.

We will use the Buckingham pi-theorem.

L 4 5

V

n = 3 parameters

γ

r = 2 dimensions

3

m = r = 2 repeating parameters

γ

We have n - m = 1 dimensionless group. Setting up dimensional equations: a b

Π1  FB V  γ

 3  F3   F0L0 b

a

Thus:

F L

L 

Summing exponents: F:

1b0

The solution to this system is:

L:

3 a  3 b  0

a  1 b  1

2

FB Π1  V γ

2

Check using M, L, t dimensions:

M L 1 t  L   1 2 3 M t L

The functional relationship is:

Π1  C

FB V γ

C

Solving for the buoyant force:

FB  C V γ

Buoyant force is proportional to γ (Q.E.D.)

Problem 7.11 (Difficulty 1)

7.11 Assume that the velocity acquired by a body falling from rest (without resistance) depends on weight of body, acceleration due to gravity, and distance of fall. Prove by dimensional analysis that 𝑉 = 𝐶�𝑔𝑛 ℎ and is thus independent of the weight of the body.

Find: Prove 𝑉 is independent of the weight of the body.

Solution:

(1) There are four dimensional parameters: 𝑉, 𝑊, 𝑔𝑛 , ℎ so 𝑛 = 4 (2) Select primary dimensions 𝑀, 𝐿 𝑎𝑎𝑎 𝑡.

(3) We have the following relations: 𝑉 𝑀 0 𝐿 1 𝑡 −1

𝐶 0 0 0

𝑊 1 1 −2

𝑔𝑛 0 1 −2

ℎ 0 1 0

All three primary dimensions are represented so r = 3. The number of repeating variables will then be m = r = 3.

The number of groups will be n – r = 4 – 3 = 1. All of the dimensional parameters will be combined into one group.

The group is then assumed to be of the form

So we get:

Π = 𝑉 𝑊 𝑎 𝑔𝑛𝑏 ℎ𝑐

𝐿 𝑀𝑀 𝑎 𝐿 𝑏 Π = � 2 � � 2 � (𝐿)𝑐 = 𝑀0 𝐿0 𝑡 0 𝑡 𝑡 𝑡

We evaluate the coefficients by setting the coefficients of each dimension on the left hand side equal to the coefficient for the dimension on the right hand side (which is 0).

Solving for the values of b and c

𝑎=0 1+𝑎+𝑏+𝑐 =0 −1 − 2𝑎 − 2𝑏 = 0 1 2 1 𝑐=− 2

𝑏=−

Thus Π = 𝑉𝑔𝑛−0.5 ℎ−0.5 =

𝑉

�𝑔𝑛 ℎ

Because there is only one group, it is a constant and we then have 𝑉 = 𝐶�𝑔𝑛 ℎ

The coefficient a for the only variable that contains mass (the weight W) is zero, so it is clear that 𝑉 is independent of the weight of the body.

Problem 7.12 (Difficulty 2) 7.12 Derive by dimensional analysis an expression for the local velocity in established pipe flow through a smooth pipe if this velocity depends only on mean velocity, pipe diameter, distance from pipe wall, and density and viscosity of the fluid.

Find: The appropriate dimensionless groups.

Solution: (1) There are 6 dimensional parameters : 𝑉, 𝑉𝑚 , 𝑑, 𝑦, 𝜌, 𝜇 so 𝑛 = 6 (2) Select primary dimensions 𝑀, 𝐿 𝑎𝑎𝑎 𝑡.

(3) We have the following relation: 𝑉 𝑉𝑚 𝑀 0 0 𝐿 1 1 𝑡 −1 −1

𝑑 0 1 0

𝑦 0 1 0

𝜌 1 −3 0

𝜇 1 −1 −1

We have all three dimensions so r = m = 3. We need 3 repeating parameters that include all of the dimensions and we pick the following three repeating parameters: 𝑉𝑚 𝑑 𝜌

The number of dimensionless groups will be n – m = 6 – 3 = 3 dimensionless groups: For the first group we have combine the repeating variables with the dimensional parameter V 𝐿 𝑎 𝑀 𝑐 𝐿 𝑎 𝑏 𝑐 𝑏 П1 = 𝑉𝑚 𝑑 𝜌 𝑉 = � � (𝐿) � 3 � � � = 𝑀0 𝐿0 𝑡 0 𝑡

Equating the exponents for each of the dimensions

Solving for the exponents

So the first group is

𝐿

𝑐=0 𝑎 + 𝑏 − 3𝑐 + 1 = 0 −𝑎 − 1 = 0 𝑎 = −1 𝑏=0 П1 =

𝑉 𝑉𝑚

𝑡

For the second group we combine with the dimensional parameter y: 𝐿 𝑑 𝑀 𝑓 𝑑 𝑒 𝑓 𝑒 П2 = 𝑉𝑚 𝑑 𝜌 𝑦 = � � (𝐿) � 3 � (𝐿) = 𝑀0 𝐿0 𝑡 0 𝑡

Solving for the exponents

𝐿 𝑓=0 𝑑 + 𝑒 − 3𝑓 + 1 = 0 −𝑑 = 0 𝑒 = −1

The second group is

П2 =

𝑦 𝑑

Similarly, for the third group, combining with the dimensional parameter µ 𝐿 𝑔 𝑀 𝑖 𝑀 П3 = 𝑉𝑚 𝑔 𝑑 ℎ 𝜌 𝑖 𝜇 = � � (𝐿)ℎ � 3 � � � = 𝑀0 𝐿0 𝑡 0 𝑡 𝐿 𝐿𝑡

The exponents are then related as

Solving for the exponents

𝑖+1=0 𝑔 + ℎ − 3𝑖 − 1 = 0 −𝑔 − 1 = 0

The third group is then

𝑖 = −1 𝑔 = −1 ℎ = −1 П3 =

𝜇 𝑉𝑚 𝑑𝑑

The functional relation among the groups is then 𝑉 𝑦 𝜇 = 𝑓� , � 𝑉𝑚 𝑑 𝑉𝑚 𝑑𝑑

Problem 7.13 (Difficulty 2)

7.13 The speed of shallow water waves in the ocean (𝑒. 𝑔. 𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠 𝑤𝑤𝑤𝑤𝑤 𝑜𝑜 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡) depends only on the still water depth and the acceleration due to gravity. Derive an expression for wave speed.

Find: The appropriate dimensionless groups.

Solution: (1) There are three dimensional parameters 𝑉, 𝑑, 𝑔 so 𝑛 = 3 (2) Select primary dimensions 𝑀, 𝐿 𝑎𝑎𝑎 𝑡.

(3) We have the following relation: 𝑉 𝑑 𝑀 0 0 𝐿 1 1 𝑡 −1 0

𝑔 0 1 −2

Mass does not appear in any of the parameters so we need only 2 repeating variables 𝑟=2

We pick up the following two repeating parameters that include all of the dimensions: 𝑑 𝑔

The number of dimensionless groups is n – m = n – r = 3 – 2 = 1 This group is then 𝐿 𝑏 𝐿 П1 = 𝑑 𝑎 𝑔𝑏 𝑉 = 𝐿𝑎 � 2 � � � = 𝑀0 𝐿0 𝑡 0 𝑡 𝑡 Equating the coefficients of the dimensions. There is no exponent for the mass Solving for the values

𝑎+𝑏+1=0 −2𝑏 − 1 = 0 1 2 1 𝑎=− 2 𝑏=−

The group is

П1 =

𝑉

�𝑑𝑑

There is only one group and so it is a constant. The velocity is then 𝑉 = 𝐶 �𝑑 𝑔

Problem 7.14 Problem 7.16

[Difficulty: 2]

7.14

Given:

That speed of shallow waves depends on depth, density, gravity and surface tension

Find:

Dimensionless groups; Simplest form of V

Solution: Apply the Buckingham  procedure  V



D



g

n = 5 parameters

 Select primary dimensions M, L, t

 V   L  t





g



D



g

L

M L3

L t2



 

  M t 2 

D

r = 3 primary dimensions

m = r = 3 repeat parameters

 Then n – m = 2 dimensionless groups will result. Setting up a dimensional equation, a

b

L M  c L 1  g a  b D cV   2   3  L   M 0 L0t 0 t t   L  M: b0 b0 1 Summing exponents, L : a  3b  c  1  0 c   Hence 2 1 t:  2a  1  0 a 2 a b  L M  c M  2  g a  b D c   2   3  L  2  M 0 L0t 0 t t   L  M: b 1  0 b  1 Summing exponents, Hence L : a  3b  c  0 c  2  2 a  2  0 a  1 t: L t  Check using F, L, t as primary dimensions    1 1

The relation between drag force speed V is

1

 L 2  2 L t 

1  f  2 

   V  f   2  gD  gD 

1 

V gD

2 

 gD 2

F L 2   1 L Ft 2 2 L t 2 L4

   V  gD f   2   gD 

Problem 7.15 Problem 7.18

[Difficulty: 2]

7.15

Given: Find: Solution:

Functional relationship between boundary layer thickness and physical parameters Appropriate dimensionless parameters We will use the Buckingham pi-theorem.





1



2

Select primary dimensions M, L, t:

3



x





U

L

L

M L3

M Lt

L t



x

U

4 5

x

U

n = 5 parameters

r = 3 dimensions

m = r = 3 repeating parameters

We have n - m = 2 dimensionless groups. Setting up dimensional equations: a b

c

Π1  δ ρ  x  U

Thus:

L 

M

a

 3 L 

 L   b

L

c

0

0 0

  M L t t

Summing exponents: M: 0  a  0

The solution to this system is:

L:

1  3 a  b  c  0

t:

0c0

a0

Check using F, L, t dimensions: ( L)  

b  1 c  0

δ Π1  x

1

1  L

a b

c

Π2  μ ρ  x  U

a

Thus:

c

 M    M   Lb  L   M0 L0 t0    3    L t   L  t

Summing exponents: M: 1  a  0 L:

1  3  a  b  c  0

t:

1  c  0

Check using F, L, t dimensions:

The solution to this system is: a  1 b  1 c  1 4 2  F t    L    1    t   1  2   2   L   L   L   F t 

μ Π2  ρ x  U

The functional relationship is:

 

Π1  f Π2

Problem 7.16 Problem 7.20

[Difficulty: 2]

7.16

Given: Find: Solution:

Functional relationship between the speed of a free-surface gravity wave in deep water and physical parameters

1

V

λ

2

Select primary dimensions M, L, t:

3

V

The dependence of the speed on the other variables We will use the Buckingham pi-theorem.

L t 4 5

D

D

λ

D

L

L

ρ

ρ

g

M

L

3

2

L ρ

n = 5 parameters

g

t

r = 3 dimensions

m = r = 3 repeating parameters

g

We have n - m = 2 dimensionless groups. Setting up dimensional equations: a b c

Π1  V D  ρ  g

L

Thus:

t

L  a

M

b



L

c

0

0 0

 3   2   M L t L  t 

Summing exponents: M: b  0 L:

1  a  3 b  c  0

t:

1  2  c  0

The solution to this system is: 1 1 a b0 c 2 2

Check using F, L, t dimensions:

a b c

Π2  λ D  ρ  g

Thus:

Π1 

V g D

 L   t   1     t   L L L   a

M

b



L

c

0

0 0

 3   2   M L t L  t 

Summing exponents: M: b  0 L:

1  a  3 b  c  0

t:

2  c  0

The solution to this system is: a  1 b  0

Check using F, L, t dimensions: L

The functional relationship is:

1 L

c0

λ Π2  D

1

 

Π1  f Π2

V g D

 f 

λ

  D

Therefore the velocity is:

V

g  D f 

λ

  D

Problem 7.17 (Difficulty 2)

7.17 Derive an expression for the velocity of very small ripples on the surface of a liquid if this velocity depends only on ripple length and density and surface tension of the liquid.

Find: The appropriate dimensionless groups.

Solution: (1) The dimensional parameters are 𝑉, 𝐿, 𝜌, 𝜎 so 𝑛 = 4 (2) Select primary dimensions 𝑀, 𝐿 𝑎𝑎𝑎 𝑡.

(3) We have the following relation: 𝑉 𝑀 0 𝐿 1 𝑡 −1

𝐿 0 1 0

𝜌 1 −3 0

All three dimensions appear so 𝑟 = 3.

𝜎 1 0 −2

We need 3 repeating parameters that include all of the dimensions. We pick up the following three repeating parameters: 𝐿 𝜌 𝜎

The number of groups is n – m = n – r = 4 – 3 = 1 Thus this group is 𝑀 𝑏 𝑀 𝑐 𝐿 П1 = 𝐿𝑎 𝜌 𝑏 𝜎 𝑐 𝑉 = 𝐿𝑎 � 3 � � 2 � � � = 𝑀0 𝐿0 𝑡 0 𝐿 𝑡 𝑡 Equating the coefficients of the dimensions

Solving for the coefficients

The group is then

𝑏+𝑐 = 0 𝑎 − 3𝑏 + 1 = 0 −2𝑐 − 1 = 0 1 𝑐=− 2 1 𝑏= 2 1 𝑎= 2

П1 =

𝑉 𝜎 �𝜌𝜌

Since there is only one group, it is a constant and then the velocity is given by 𝜎 𝑉 = 𝐶� 𝜌𝜌

Problem 7.18 (Difficulty 3)

7.18 Derive an expression for the axial thrust exerted by a propeller if the thrust depends only on forward speed, angular speed, size, and viscosity and density of the fluid. How would the expression change if gravity were a relevant variable in the case of a ship propeller?

Find: The appropriate dimensionless groups.

Solution: (1) There are 6 dimensional parameters 𝐹, 𝑉, 𝜔, 𝑑, 𝜇, 𝜌, g so 𝑛 = 7 (2) Select primary dimensions 𝑀, 𝐿 𝑎𝑎𝑎 𝑡.

(3) We have the following relation: 𝐹 𝑉 𝑀 1 0 𝐿 1 1 𝑡 −2 −1

𝜔 0 0 −1

𝑑 0 1 0

𝜇 1 −1 −1

𝜌 1 −3 0

𝑔 0 1 −2

All dimensions are present so the number of repeating variables is 𝑟 = 3. We pick the following three repeating parameters that include all of the dimensions:

𝑉 𝑑 𝜌 We will have n – m = n – r = 7 – 3 = 4 nondimensional groups if we include gravity. If we do not include gravity, we will have only three groups. For the first group we will combine the force F with the repeating variables П1

𝐿 𝑎 𝑏 𝑀 𝑐 𝑀𝑀 � 𝐿 � 3 � 2 = 𝑀0 𝐿0 𝑡 0 𝑡 𝐿 𝑡

= 𝑉 𝑎 𝑑𝑏 𝜌𝑐 𝐹 = �

Equating the exponents of the dimensions

𝑐+1=0 𝑎 + 𝑏 − 3𝑐 + 1 = 0 Solving for the values of the exponents 𝑐 = −1 −𝑎 − 2 = 0 𝑎 = −2 𝑏 = −2 The first group is then 𝐹 П1 = 𝜌𝜌 2 𝑑 2

We combine the viscosity with the repeating variables for the second group:

П2

Solving for the exponents

= 𝑉𝑑 𝑑𝑒 𝜌 𝑓 𝜇 = �

𝐿 𝑑 𝑒 𝑀 𝑓 𝑀 � 𝐿 � 3 � � � = 𝑀0 𝐿0 𝑡 0 𝑡 𝐿 𝐿𝐿

𝑓+1=0 𝑓 = −1 𝑑 + 𝑒 − 3𝑓 − 1 = 0 −𝑑 − 1 = 0 𝑑 = −1 𝑒 = −1 П2 =

Similarly for the third group that combines ω

𝜇 𝜌𝜌𝜌

𝐿 𝑔 𝑀 𝑖1 П3 = 𝑉𝑔 𝑑 ℎ 𝜌 𝑖 𝜔 = � � 𝐿ℎ � 3 � = 𝑀0 𝐿0 𝑡 0 𝑡 𝐿 𝑡 Solving for the exponents 𝑖=0 𝑔 + ℎ − 3𝑖 = 0 −𝑔 − 1 = 0 𝑔 = −1 ℎ=1 𝜔𝜔 П3 = 𝑉 For the three groups that do not include gravity, we have the relation 𝜔𝜔 𝜇 𝐹 = 𝑓� , � 2 2 𝑉 𝜌𝜌𝜌 𝜌𝜌 𝑑 For the fourth group with the addition of gravity 𝑔, we have: The exponents are then

𝐿 𝑗 𝑀 𝑙 𝐿 П3 = 𝑉𝑗 𝑑 𝑘 𝜌 𝑙 𝑔 = � � 𝐿𝑘 � 3 � � 2 � = 𝑀0 𝐿0 𝑡 0 𝑡 𝐿 𝑡 𝑙=0 𝑗 + 𝑘 − 3𝑙 + 1 = 0 −𝑗 − 2 = 0 𝑗 = −2 𝑘=1 𝑔𝑔 П4 = 2 𝑉

So if 𝑔 is also the variable, the expression for thrust becomes:

𝜔𝜔 𝜇 𝑔𝑔 𝐹 = 𝑓� , , � 2 2 𝑉 𝜌𝜌𝜌 𝑉 2 𝜌𝜌 𝑑

Problem 7.19 (Difficulty 2)

7.19 Derive an expression for drag force on a smooth submerged object moving through incompressible fluid if this force depends only on speed and size of object and viscosity and density of the fluid.

Find: The appropriate dimensionless groups.

Solution: (1) We have the dimensional parameters 𝐷, 𝑉, 𝐿, 𝜇, 𝜌 so 𝑛 = 5 dimensional parameters (2) Select primary dimensions 𝑀, 𝐿 𝑎𝑎𝑎 𝑡.

(3) We have the following relation: 𝐷 𝑉 𝑀 1 0 𝐿 1 1 𝑡 −2 −1

𝐿 0 1 0

𝜇 1 −1 −1

𝜌 1 −3 0

All three dimensions are present and r = 3 so we need 3 repeating parameters that include all of the dimensions. We pick up the following three repeating parameters: 𝑉 𝐿 𝜌

We will have n – m = n – r = 5 – 3 = 2 dimensionless groups. For the first group we will combine the repeating variables with the parameter D: 𝐿 𝑎 𝑀 𝑐 П1 = 𝑉 𝑎 𝐿𝑏 𝜌 𝑐 𝐷 = � � 𝐿𝑏 � 3 � 𝐿 = 𝑀0 𝐿0 𝑡 0 𝑡 𝐿

Equating the coefficients for each dimension and solving for the values

The first group is

𝑐+1=0 𝑐 = −1 𝑎 + 𝑏 − 3𝑐 + 1 = 0 −𝑎 − 2 = 0 𝑎 = −2 𝑏 = −2 П1 =

𝐷 𝜌𝜌 2 𝐿2

For the second group we combine the repeating variables with the viscosity:

The exponents are then

П2

𝐿 𝑑 𝑒 𝑀 𝑓𝑀 � 𝐿 � 3� = 𝑀0 𝐿0 𝑡 0 𝑡 𝐿 𝐿𝐿

= 𝑉 𝑑 𝐿𝑒 𝜌 𝑓 𝜇 = �

The second group is

The relation between the groups is

𝑓+1=0 𝑓 = −1 𝑑 + 𝑒 − 3𝑓 − 1 = 0 −𝑑 − 1 = 0 𝑑 = −1 𝑒 = −1 П2 =

𝜇 𝜌𝜌𝜌

𝜇 𝐷 = 𝑓� � 2 2 𝜌𝜌𝜌 𝜌𝜌 𝐿

Problem 7.20 Problem 7.22

[Difficulty: 2] (The solution to this problem was first devised by G.I. Taylor in the paper "The formation of a blast wave by a very intense explosion. I. Theoretical discussion," Proceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences, Vol. 201, No. 1065, pages 159 - 174 (22 March 1950).)

7.20

Given: Find:

Functional relationship between the energy released by an explosion and other physical parameters

Solution:

We will use the Buckingham pi-theorem.

Expression for E in terms of the other variables

1

E

2

Select primary dimensions M, L, t:

3

E

t

M L t 4 5

t

R

t

L

ρ

p M

M

2

3

L t t

n = 5 parameters

ρ

p

2

2

ρ

R

L

r = 3 dimensions

m = r = 3 repeating parameters

R

We have n - m = 2 dimensionless groups. Setting up dimensional equations: a b

Π1  E ρ  t  R

c

Thus:

M L

2  3 a  c  0

t:

2  b  0

a  1

Check using F, L, t dimensions:

a b

Π2  p  ρ  t  R

c

Thus:

F L

t:

2  b  0

b

c

0

0 0

4

2 1

t 

M



M

L t

L 

L a

2  3

5

b

c  5

a  1

F L

2



L

4

F t

 

Π1  f Π2

2

Π1 

E t

2

ρ R

5

1

c

0

0 0

t L  M L t

The solution to this system is:

Check using F, L, t dimensions:

The functional relationship is:

L

F t

M: 1  a  0 1  3  a  c  0

a

b2

2

Summing exponents:

L:

M

The solution to this system is:

M: 1  a  0 L:



 3  t L  M L t L 

2

t Summing exponents:

2

b2

2 1

t 

L

2

c  2

Π2 

p t

2

ρ R

2

1

 p t2    f 5  2 ρ R  ρ R  E t

2

 p t2   E f  2  2 t  ρ R  ρ R

5

Problem 7.24 7.21 Problem

[Difficulty: 2]

7.21

Given: Find: Solution:

Functional relationship between the flow rate over a weir and physical parameters

1

Q

h

2

Select primary dimensions L, t:

3

Q L

An expression for Q based on the other variables We will use the Buckingham pi-theorem.

h

5

g

3

h

L

L

t 4

g

t

2

n = 5 parameters

b

b r = 2 dimensions

L

m = r = 2 repeating parameters

g

We have n - m = 2 dimensionless groups. Setting up dimensional equations: a b

Π1  Q h  g

Thus:

L

3

t

L  a

L

3ab0

t:

1  2  b  0

Check:

The solution to this system is: 5 1 a b 2 2

Π1 

Q 2

h  g h

 L3   1  2  t         1  t   L  L a b

Π2  b  h  g

Thus:

L L   a

L

L:

1ab0

t:

2  b  0 L

1 L

b

0 0

 2   L t t 

Summing exponents:

Check:

0 0

 2   L t t 

Summing exponents: L:

b

The solution to this system is: a  1

b0

b Π2  h

1

The functional relationship is:

 

Π1  f Π2

Q 2

h  g D

 f 

b

 h

Therefore the flow rate is:

Q  h  g  h  f  2

b

 h

Problem 7.22 Problem 7.27

[Difficulty: 2]

7.22

Given: Find: Solution:

Functional relationship between the load bearing capacity of a journal bearing and other physical parameters

1

W

D

2

Select primary dimensions F, L, t:

3

W

D

l

c

F

L

L

L

4 5

D

Dimensionless parameters that characterize the problem.

We will use the Buckingham pi-theorem.

ω

l

c

n = 6 parameters

ω

μ

ω

μ

1

F t

t

2

L

r = 3 dimensions

m = r = 3 repeating parameters

μ

We have n - m = 3 dimensionless groups. Setting up dimensional equations: a

b

c

Π1  W D  ω  μ

Thus:

a

Summing exponents: F:

1c0

L:

a  2 c  0

t:

b  c  0

1

c

The solution to this system is: a  2

Check using M, L, t dimensions:

The functional relationship is:

b

0 0 0  F t    2   F L t  t  L 

F L  

b  1

L t M L 1   t 1 2 2 M t L



Π1  f Π2 Π3



c  1

Π1 

W 2

D  ω μ

By inspection, we can see that:

l Π2  D

W 2

D  ω μ

c Π3  D

 f 

l

c

  D D 

Problem 7.23 (Difficulty 2)

7.23 Derive an expression for the drag force on a smooth object moving through compressible fluid if this force depends only on speed and size of object, and viscosity, density and modulus of elasticity of the fluid.

Find: The appropriate dimensionless groups.

Solution: (1) We have six dimensional parameters 𝐷, 𝑉, 𝐿, 𝜇, 𝜌, 𝐸 so 𝑛 = 6 dimensional parameters (2) Select primary dimensions 𝑀, 𝐿 𝑎𝑎𝑎 𝑡.

(3) We have the following relation: 𝐷 𝑉 𝑀 1 0 𝐿 1 1 𝑡 −2 −1

𝐿 0 1 0

𝜇 1 −1 −1

𝜌 1 −3 0

𝐸 1 −1 −2

All three dimensions are present so 𝑟 = 3. We need 3 repeating parameters that include all of the dimensions and we will pick the following three repeating parameters: 𝑉 𝐿 𝜌

We will have n – m = n – r = 6 – 3 = 3 dimensionless variables. The first group will combine the dimension D 𝐿 𝑎 𝑏 𝑀 𝑐 𝑎 𝑏 𝑐 П1 = 𝑉 𝐿 𝜌 𝐷 = � � 𝐿 � 3 � 𝐿 = 𝑀0 𝐿0 𝑡 0 𝑡 𝐿 Equating the exponents of the dimensions and solving for their values

The first group is then

𝑐+1=0 𝑐 = −1 𝑎 + 𝑏 − 3𝑐 + 1 = 0 −𝑎 − 2 = 0 𝑎 = −2 𝑏 = −2 П1 =

𝐷 𝜌𝜌 2 𝐿2

For the second group we will combine the viscosity: П2

𝐿 𝑑 𝑒 𝑀 𝑓𝑀 � 𝐿 � 3� = 𝑀0 𝐿0 𝑡 0 𝑡 𝐿 𝐿𝐿

= 𝑉 𝑑 𝐿𝑒 𝜌 𝑓 𝜇 = �

The exponents are related as

𝑓+1=0 𝑓 = −1 𝑑 + 𝑒 − 3𝑓 − 1 = 0 −𝑑 − 1 = 0 𝑑 = −1 𝑒 = −1

The second group is then

П2 =

𝜇 𝜌𝜌𝜌

The third group combines the modulus of elasticity

𝐿 𝑔 𝑀 𝑖 𝑀 П3 = 𝑉𝑔 𝐿ℎ 𝜌 𝑖 𝐸 = � � 𝐿ℎ � 3 � 2 = 𝑀0 𝐿0 𝑡 0 𝑡 𝐿 𝐿𝑡

Solving for the exponents

𝑖+1=0 𝑖 = −1 𝑔 + ℎ − 3𝑖 − 1 = 0 −𝑔 − 2 = 0 𝑔 = −2 ℎ=0

The third group is

The relation among the groups is then

П3 =

𝐸 𝜌𝑉 2

𝜇 𝐸 𝐷 = 𝑓� , 2� 2 2 𝜌𝜌𝜌 𝜌𝑉 𝜌𝜌 𝐿

Problem 7.24 (Difficulty 2)

7.24 A circular disk of diameter 𝑑 and of negligible thickness is rotated at a constant angular speed, 𝜔, in a cylindrical casing filled with a liquid of viscosity 𝜇 and density 𝜌. The casing has an internal diameter 𝐷, and there is a clearance 𝑦 between the surfaces of disk and casing. Derive an expression for the torque required to maintain this speed if it depends only on the foregoing variables.

Find: The appropriate dimensionless groups.

Solution: (1) There are seven dimensional parameters 𝑇,𝑑, 𝜔, 𝜇, 𝜌, 𝐷, 𝑦 so 𝑛 = 7 dimensional parameters (2) Select primary dimensions 𝑀, 𝐿 𝑎𝑎𝑎 𝑡.

(3) We have the following relation: 𝑇 𝑑 𝑀 1 0 𝐿 2 1 𝑡 −2 0

𝜔 0 0 −1

𝜇 1 −1 −1

𝜌 1 −3 0

𝐷 0 1 0

All three dimensions are present so r = 3. We need 3 repeating parameters that include all of the dimensions and we will pick the following three repeating parameters: 𝜔 𝑑 𝜌

We will have n – m = n – r = 7 – 3 = 4 dimensionless groups. For the first group we will combine the repeating variables with the torque T П1

1 𝑎 𝑏 𝑀 𝑐 𝑀𝐿2 � 𝐿 � 3� = 𝑀0 𝐿0 𝑡 0 𝑡 𝐿 𝑡2

= 𝜔𝑎 𝑑 𝑏 𝜌 𝑐 𝑇 = �

Equating the exponents of the dimensions and solving for the exponents 𝑐+1=0 𝑐 = −1 𝑏 − 3𝑐 + 2 = 0 𝑏 = −5 −𝑎 − 2 = 0 𝑎 = −2 𝑇 П1 = 𝜌𝜌 2 𝑑 5

𝑦 0 1 0

For the second group, we will combine the repeating variable with the viscosity 1 𝑎 𝑀 𝑐𝑀 П2 = 𝜔𝑑 𝑑 𝑒 𝜌 𝑓 𝜇 = � � 𝐿𝑏 � 3 � = 𝑀0 𝐿0 𝑡 0 𝑡 𝐿 𝐿𝐿

Equating the exponents of the dimensions

𝑓+1=0 𝑓 = −1 𝑒 − 3𝑓 − 1 = 0 𝑒 = −2 −𝑑 − 1 = 0 𝑑 = −1

The second group is

П2 =

𝜇 𝜌𝜌𝑑 2

For the third group we combine the internal diameter

1 𝑎 𝑀 𝑐 П3 = 𝜔 𝑔 𝑑 ℎ 𝜌 𝑖 𝐷 = � � 𝐿𝑏 � 3 � 𝐿 = 𝑀0 𝐿0 𝑡 0 𝑡 𝐿

Solving for the exponents

𝑔=0 𝑖=0 ℎ=1

The third group is

The last group combines the clearance y:

The exponents are

П3 =

𝐷 𝑑

1 𝑎 𝑀 𝑐 П4 = 𝜔 𝑗 𝑑 𝑘 𝜌 𝑙 𝑦 = � � 𝐿𝑏 � 3 � 𝐿 = 𝑀0 𝐿0 𝑡 0 𝑡 𝐿 𝑗=0 𝑙=0 𝑘=1 𝑦 П4 = 𝑑

The relation among the groups is 𝜇 𝐷 𝑦 𝑇 = 𝑓� , , � 2 5 𝜌𝜌𝑑 2 𝑑 𝑑 𝜌𝜌 𝑑

Problem 7.25 (Difficulty 2)

7.25 Two cylinders are concentric, the outer one fixed and the inner one movable. A viscous incompressible fluid fills the gap between them. Derive an expression for the torque required to maintain constant-speed rotation of the inner cylinder if this torque depends only on the diameters and lengths of the cylinders, the viscosity and density of the fluid, and the angular speed of the inner cylinder.

Find: The appropriate dimensionless groups.

Solution: (1) There are eight dimensionless parameters 𝑇,𝑑1 , 𝑑2 , 𝐿1 , 𝐿2 , 𝜇, 𝜌 𝜔 so 𝑛 = 8 (2) Select primary dimensions 𝑀, 𝐿 𝑎𝑎𝑎 𝑡.

(3) We have the following relation: 𝑇 𝑑1 𝑑2 𝑀 1 0 𝐿 2 1 𝑡 −2 0

𝐿1 𝐿2 0 1 0

𝜇 1 −1 −1

𝜌 1 −3 0

𝜔 0 0 −1

All of the primary dimensions appear and we have r= 3. We need 3 repeating parameters that include all of the dimensions and we pick the following three repeating parameters: 𝜔 𝑑1 𝜌

There will be n – m = n – r = 8 – 3 = 5 dimensionless groups. For the first group we will combine the torque T: 1 𝑎 𝑏 𝑀 𝑐 𝑀𝐿2 � 𝐿 � 3� = 𝑀0 𝐿0 𝑡 0 𝑡 𝐿 𝑡2

П1 = 𝜔𝑎 𝑑1𝑏 𝜌 𝑐 𝑇 = �

Equating the exponents of the dimensions and solving for the value of the exponents 𝑐+1=0 𝑐 = −1 𝑏 − 3𝑐 + 2 = 0 𝑏 = −5 −𝑎 − 2 = 0 𝑎 = −2

The first group is then П1 =

𝑇 𝜌𝜔 2 𝑑15

For the second group, we will combine the repeating variables with the viscosity: 1 𝑑 𝑀 𝑓𝑀 = 𝑀0 𝐿0 𝑡 0 П2 = 𝜔𝑑 𝑑1𝑒 𝜌 𝑓 𝜇 = � � 𝐿𝑒 � 3 � 𝑡 𝐿 𝐿𝐿

Equating the exponents of the dimensions

The second group is

𝑓+1=0 𝑓 = −1 𝑒 − 3𝑓 − 1 = 0 𝑒 = −2 −𝑑 − 1 = 0 𝑑 = −1

The third group combines the diameter d2:

П2 =

𝜇 𝜌𝜌𝑑12

1 𝑔 𝑀 𝑖 П3 = 𝜔 𝑔 𝑑1ℎ 𝜌 𝑖 𝑑2 = � � 𝐿ℎ � 3 � 𝐿 = 𝑀0 𝐿0 𝑡 0 𝑡 𝐿

Equating the exponents and solving for their value

So the third group is

𝑔=0 𝑖=0 ℎ = −1 П3 =

𝑑2 𝑑1

The fourth group is similar to the third with the parameter L1: 1 𝑗 𝑀 𝑙 П4 = 𝜔 𝑗 𝑑1𝑘 𝜌 𝑙 𝐿1 = � � 𝐿𝑘 � 3 � 𝐿 = 𝑀0 𝐿0 𝑡 0 𝑡 𝐿

The solution is the same as for the third group

𝑗=0 𝑙=0 𝑘 = −1

And the fourth group is П4 =

𝐿1 𝑑1

The fifth group is similar to the previous two groups, and could be obtained by inspection 1 𝑚 𝑀 𝑜 П5 = 𝜔𝑚 𝑑1𝑛 𝜌 𝑜 𝐿2 = � � 𝐿𝑛 � 3 � 𝐿 = 𝑀0 𝐿0 𝑡 0 𝑡 𝐿

Solving for the exponents

The fifth group is

𝑚=0 𝑜=0 𝑛 = −1 П5 =

The functional relation between the groups is

𝐿2 𝑑1

𝜇 𝑑2 𝐿1 𝐿2 𝑇 5 = 𝑓 �𝜌𝜌𝑑 2 , 𝑑 , 𝑑 , 𝑑 � 2 𝜌𝜌 𝑑1 1 1 1 1

Problem Problem7.26 7.28 (In Excel)

[Difficulty: 2]

7.26

Given: That drain time depends on fluid viscosity and density, orifice diameter, and gravity Find: Functional dependence of t on other variables Solution: We will use the workbook of Example 7.1, modified for the current problem n =5 r =3 m =r =3 n -m =2

The number of parameters is: The number of primary dimensions is: The number of repeat parameters is: The number of  groups is:

Enter the dimensions (M, L, t) of the repeating parameters, and of up to four other parameters (for up to four  groups).

The spreadsheet will compute the exponents a , b , and c for each.

REPEATING PARAMETERS: Choose , g , d M 1

 g d

L -3 1 1

t -2

 GROUPS: t  1:

M 0

L 0

t 1

a = b = c =

0 0.5 -0.5

  2:

M 1

L -1

a = b = c =

-1 -0.5 -1.5

M 0

L 0

a = b = c =

0 0 0

t -1

The following  groups from Example 7.1 are not used:

Hence

1  t

The final result is

g d

t

M 0

L 0

 3:

a = b = c =

0 0 0

and

2 

d   2  f g   2 gd 3 

 1 3 g 2 d 2



t 0  4:

2  gd 3 2

with  1  f  2 

t 0

Problem 7.27 Problem 7.30

[Difficulty: 2]

7.27

Given:

Functional relationship between the time needed to drain a tank through an orifice plate and other physical parameters

Find:

(a) the number of dimensionless parameters (b) the number of repeating variables (c) the Π term which contains the viscosity

Solution:

We will use the Buckingham pi-theorem.

1

τ

h0

2

Select primary dimensions M, L, t:

3

τ

h0

D

d

T

L

L

L

D

d

g

ρ

μ

g

ρ

μ

L

M

M

2

3

L t

t 4 5

ρ

d

L

n = 7 parameters

r = 3 dimensions

m = r = 3 repeating parameters

g

We have n - m = 4 dimensionless groups.

Setting up dimensional equation including the viscosity: a b c

Π1  μ ρ  d  g

a

Thus:

Summing exponents: M: 1  a  0 L:

1  3  a  b  c  0

t:

1  2  c  0

c

M  M b  L 0 0 0  L   M L t     3 2 L t L  t  The solution to this system is: 3 1 a  1 b c 2 2

4

Check using F, L, t dimensions:

1 t F t L    1 2 2 3 1 L F t L

2

L

2

Π1 

μ 3

1

2

2

ρ d  g

Problem 7.28 Problem 7.31

[Difficulty: 3]

7.28

Given: Find:

Functional relationship between the flow rate of viscous liquid dragged out of a bath and other physical parameters Expression for Q in terms of the other variables

Solution:

We will use the Buckingham pi-theorem.

1

Q

μ

2

Select primary dimensions M, L, t:

3

Q L

ρ

M

M

L

L t

3

2

3

5

ρ

V

g

μ

t 4

ρ

L

g

t

h

n = 6 parameters

V

h

V L

L

r = 3 dimensions

t m = r = 3 repeating parameters

h

We have n - m = 3 dimensionless groups. Setting up dimensional equations: a

b c

Π1  Q ρ  V  h

Thus:

L

3



a

 

a0

L:

3  3 a  b  c  0

t:

1  b  0

Check using F, L, t dimensions:

b c

b

c

0

0 0

The solution to this system is:

M: a  0

Π2  μ ρ  V  h

L

 L  M L t t  3  t  L  

Summing exponents:

a

M

b  1

Summing exponents: M: 1  a  0 L:

1  3  a  b  c  0

t:

1  b  0

V h

2

3

b

M  M   L c 0 0 0    L  M L t   3 t L t L    The solution to this system is: a  1

4

Check using F, L, t dimensions:

Q

t 1   1 t L 2 L

L

a

Thus:

c  2

Π1 

b  1

t 1 F t L    1 2 2 L L L F t

c  1

μ Π2  ρ V h

a

b c

Π3  g  ρ  V  h

a

Thus:

Summing exponents:

The solution to this system is: a0

M: a  0 L:

1  3 a  b  c  0

t:

2  b  0

Check using F, L, t dimensions:

L t

The functional relationship is:

b

L  M   L c 0 0 0    L  M L t 2  3  t  t L 

2

 L

t

c1

g h 2

V

2

L



b  2

Π3 

2

1

Π1  f Π2 Π3



Q V h

2

 ρ V h V2     μ g h 

 f

 ρ V h V2     μ g h 

Q  V h  f  2

Problem 7.29 (Difficulty 2)

7.29 Derive an expression for the frictional torque exerted on the journal of a bearing if this torque depends only on the diameters of journal and bearing, their axial lengths (these are the same), viscosity of the lubricant, angular speed of the journal, and the transverse load (force) on the bearing.

Find: The appropriate dimensionless groups.

Solution: (1) There are seven dimensional parameters 𝑇,𝑑1 , 𝑑2 , 𝐻, 𝜇, 𝜔, 𝐹 so 𝑛 = 7s (2) Select primary dimensions 𝑀, 𝐿 𝑎𝑎𝑎 𝑡.

(3) We have the following relation: 𝑇 𝑑1 𝑑2 𝑀 1 0 0 𝐿 2 1 1 𝑡 −2 0 0

𝐻 0 1 0

𝜇 1 −1 −1

𝜔 0 0 −1

𝐹 1 1 −2

All of the dimensions are present so 𝑟 = 3. We will have 3 repeating parameters and we pick the following three repeating parameters that include all of the dimensions: 𝑑1 𝜔 𝐹

There will be n – m = n – r = 7 – 3 = 4 dimensionless groups. For the first group we will combine the repeating variables with the torque: 1 𝑏 𝑀𝑀 𝑐 𝑀𝐿2 = 𝑀0 𝐿0 𝑡 0 П1 = 𝑑1𝑎 𝜔𝑏 𝐹 𝑐 𝑇 = 𝐿𝑎 � � � 2 � 𝑡 𝑡 𝑡2

We equate the exponents of the dimensions and solve for their values

The first group is then

𝑐+1=0 𝑐 = −1 𝑎+𝑐+2=0 𝑎 = −1 −𝑏 − 2𝑐 − 2 = 0 𝑏=0 П1 =

𝑇 𝐹𝑑1

For the second group we will combine the repeating variables with the diameter d2: 1 𝑒 𝑀𝑀 𝑓 � � 2 � 𝐿 = 𝑀0 𝐿0 𝑡 0 𝑓 = 0 𝑡 𝑡

П2 = 𝑑1𝑑 𝜔𝑒 𝐹 𝑓 𝑑2 = 𝐿𝑑 �

Equating the exponents of the dimensions and solving for their values

The second group is

𝑑+𝑓+1=0 𝑑 = −1 −𝑒 − 2𝑓 = 0 𝑒=0 П2 =

Similarly, for the third group with the length H

𝑑2 𝑑1

1 ℎ 𝑀𝑀 𝑖 𝑔 П3 = 𝑑1 𝜔ℎ 𝐹 𝑖 𝐻 = 𝐿𝑔 � � � 2 � 𝐿 = 𝑀0 𝐿0 𝑡 0 𝑡 𝑡 𝑔 = −1 ℎ=0 𝑖=0

The third group is similar to the second group and could have been determined by inspection П3 =

𝐿 𝑑1

For the fourth group we combine the viscosity with the repeating variables: 1 𝑏 𝑀𝑀 𝑐 𝑀 𝑗 = 𝑀0 𝐿0 𝑡 0 П4 = 𝑑1 𝜔𝑘 𝐹 𝑙 𝜇 = 𝐿𝑎 � � � 2 � 𝑡 𝑡 𝑡𝑡

Equating the exponents of the dimensions and solving for them

The fourth group is

𝑙+1=0 𝑙 = −1 𝑗+𝑙−1= 0 𝑗=2 −𝑘 − 2𝑙 − 1 = 0 𝑘=1 П4 =

𝜇𝜔𝜔12 𝐹