Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 11, Problem 11.6CP
<
To determine

(a)

Non-dimensionalize the given function into two pi where one is directly proportional to T.

<
Expert Solution
Check Mark

Answer to Problem 11.6CP

The function that is dimensionless is TρD5ωP2=fcn(s)

Explanation of Solution

Given Information:

Diameter=1ft

Coupling rate=2500r/min

Density=56lbm/ft3

Concept Used:

π1=ρaDbωpcT

where,ρ=densityD=diameterωp=speedT=torque

Calculation:

The number of pi in the groups are calculated.

N=kr

Where,

k=total number of variables

r=total number if references

Calculating we have,

N=kr

=532

Dimensional analysis of the first group pi,

π1=ρaDbωpcT

(1)

Equating M coefficients

a+1=0a=1

Equating T coefficients

c2=0c=2

Equating T coefficients

3a+b+23(1)+b+2

b=5

From (1)

π1=ρaDbωpcT

π1=ρ1D5ωp2TTρD5ωp2

(2)

For second pi group

π1=ρaDbωpcs

(3)

M0L0T0=(ML3)aLb(T1)c(M0L0T0)

Substituting we have,

M0L0T0=(ML3)aLb(T1)c(ML2T2)

=(Ma+1L3a+b+2Tc2)

=(MaL3a+bTc)

By equating M and T coefficients we get

a=0,c=0

And so b=0

From (3) equation

π2=ρ0D0ωp0ss

(4)

For the given function:

π1=fcn(π2)

From equation (3), (4):

TρD5ωP2=fcn(s).

<
To determine

(b)

The slip from the table.

<
Expert Solution
Check Mark

Answer to Problem 11.6CP

Slip = 15%

Explanation of Solution

Given Information:

Coupling runs at 3600r/min and torque is given as 900ft.lbf

Concept Used:

TρD5ωP2where,ρ=densityωP=angular velocitys=slip

Calculation:

Angular velocity unit conversion:

ωp=2500×16041.7r/s

Density unit conversion:

ρ=56lbm/ft31.74slugs/ft3

With the equation:

TρD5ωP2=901.74×15×41.72903025.67

=0.0298

From the table the slip is given as 5%:

s% 0 5 10 15 20 25
TρD5ωP2 0 0.0298 0.0911 0.146 0.192 0.225

Calculating the given values:

TρD5ωP2=9001.74×58.32×159005914.069

=0.152

From the table we have, 15%.

Conclusion:

The slip using the table is calculated.

<
To determine

(c)

The diameter for a same coupling that runs at 3000r/min.

<
Expert Solution
Check Mark

Answer to Problem 11.6CP

Diameter = 1.36ft.

Explanation of Solution

Given Information:

Power=3000r/min

s=5%

Torque=600ft-lbf

Concept Used:

TρD5ωP2where,ρ=densityωP=angular velocitys=slip

Calculation:

With the equation:

TρD5ωP2=6001.74*D5*5020.0298=6004350*D5D=1.36ft

Conclusion:

Thus, the diameter is calculated.

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Chapter 11 Solutions

Fluid Mechanics

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