Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 5, Problem 5.2CP
To determine

To rewrite:

This function in dimensionless form, using dimensional analysis.

Expert Solution
Check Mark

Answer to Problem 5.2CP

The dimensionless function is π1=m˙RT1p1d22.

Explanation of Solution

Given Information:

For fluid exiting a nozzle which is a gas instead of water, upstream pressure p1is large and d2,the exit diameter is small. The difference p1p2 is no longer controlling and the gas mass flow m˙ reaches a maximum value that depends on p1and d2 and also on the absolute upstream temperature T1 and the gas constant R. Thus, functionally it can be written as:

m˙=fcn(p1,d2,T1,R)

Ref Fig 3.49:

Fluid Mechanics, Chapter 5, Problem 5.2CP , additional homework tip  1

Concept Used:

The number of pi groups are to be calculated:

N=kr

Where, k is the number of variables and r is the number of fundamental references.

On substituting 5 for k and 4 for r ,

N=1

Calculation:

Dimensional analysis is applied to find the pi groups.

π1=Rad2bT1cp1dm˙

Where m˙ is the mass flow rate, R is the gas constant, d2 is the diameter, T1 is the temperature and p1 is the pressure.

On substituting [M0L0T0θ] for π1, [MT1] for m˙, [ML2T2θ1] for R, [L] for d2, [θ1] for T1 and [ML1T2] for p1

[M0L0T0θ]=[ML2T2θ1]a[L]b[θ]c[ML1T2]d[MT1]

[M0L0T0θ0]=[Ma+d+1L2a+bdT12a2dθa+c]

On equating M coefficients:

a+d+1=0

On equating θ coefficients:

a+c=0c=a

On equating T coefficients:

12a2d

On equating L coefficients:

2a+bd=0

Therefore, the following values are obtained:

a=1/2b=2c=1/2d=1

Therefore, the pi group is as follows:

π1=R12d22T112p11m˙

π1=m˙RT1p1d22

Hence, the pi group is: π1=m˙RT1p1d22

Conclusion:

The dimensionless function is π1=m˙RT1p1d22.

To determine

To plot:

Data using the dimensionless form obtained, a curve fit formula and a single value of a range.

Expert Solution
Check Mark

Answer to Problem 5.2CP

The data is plotted as above and the measured value of π1=m˙RT1p1d22 is about 0.54.

Explanation of Solution

Given Information:

For fluid exiting a nozzle which is a gas instead of water, upstream pressure p1 is large and d2, the exit diameter is small. The difference p1p2 is no longer controlling and the gas mass flow m˙ reaches a maximum value that depends on p1 and d2 and also on the absolute upstream temperature T1 and the gas constant R. Thus, functionally it can be written as:

m˙=fcn(p1,d2,T1,R)

Ref Fig 3.49:

Fluid Mechanics, Chapter 5, Problem 5.2CP , additional homework tip  2

Diameter of pipe, d2 = 1 cm

The following values of mass flow through the nozzle are shown by the measurements for flow of air:

Fluid Mechanics, Chapter 5, Problem 5.2CP , additional homework tip  3

Concept Used:

The value of π1 is calculated for the various points provided:

T( K) π1=m˙RT1p1d22
300 0.543
300 0.54
300 0.538
500 0.543
800 0.543

Calculation:

The pi value is calculated by substituting 0.037 kg/s for m˙, 287 J/kg.K for R, 300 K for T1 ,200 × 103 Pa for p1 and 1 cm for d2 in the equation:

π1=0.037287×300p1d22

π1=0.03786100200×103×(1× 10 2)2

=0.543

And

The values obtained are plotted on the graph as follows:

Fluid Mechanics, Chapter 5, Problem 5.2CP , additional homework tip  4

Hence, the measured value of π1=m˙RT1p1d22 is about 0.54 and is always a constant.

Conclusion:

The data is plotted as above and the measured value of π1=m˙RT1p1d22 is about 0.54.

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Chapter 5 Solutions

Fluid Mechanics

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