Chapter 5, Problem 5.2CP

To determine

To rewrite:

This function in dimensionless form, using dimensional analysis.

Answer to Problem 5.2CP

The dimensionless function is ${\pi }_1=\frac{\dot{m}\sqrt{R{T}_1}}{p_1{d}_2^2}$.

Explanation of Solution

Given Information:

For fluid exiting a nozzle which is a gas instead of water, upstream pressure p₁is large and d_2,the exit diameter is small. The difference $p_1-p_2$ is no longer controlling and the gas mass flow $\dot{m}$ 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:

$\dot{m}=fcn(p_1,d_2,T_1,R)$

Ref Fig 3.49:

Concept Used:

The number of pi groups are to be calculated:

$N=k-r$

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.

${\pi }_1=R^a{d}_2^b{T}_1^c{p}_1^d\dot{m}$

Where $\dot{m}$ 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 $[M^0{L}^0{T}^0\theta ]$ for ${\pi }_1$, $[M{T}^{-1}]$ for $\dot{m}$, $[M{L}^2{T}^{-2}{\theta }^{-1}]$ for R, $[L]$ for d₂, $[{\theta }^1]$ for T₁ and $[M{L}^{-1}{T}^{-2}]$ for p₁

$[M^0{L}^0{T}^0\theta ]={[M{L}^2{T}^{-2}{\theta }^{-1}]}^a{[L]}^b{[\theta ]}^c{[M{L}^{-1}{T}^{-2}]}^d[M{T}^{-1}]$

$[M^0{L}^0{T}^0{\theta }^0]=[M^{a+d+1}{L}^{2a+b-d}{T}^{-1-2a-2d}{\theta }^{-a+c}]$

On equating M coefficients:

$a+d+1=0$

On equating $\theta$ coefficients:

$\begin{array}{l}-a+c=0\\ c=a\end{array}$

On equating T coefficients:

$-1-2a-2d$

On equating L coefficients:

$2a+b-d=0$

Therefore, the following values are obtained:

$\begin{array}{l}a=1/2\\ b=-2\\ c=1/2\\ d=-1\end{array}$

Therefore, the pi group is as follows:

${\pi }_1=R^{\frac{1}{2}}{d}_2^{-2}{T}_1^{-\frac{1}{2}}{p}_1^{-1}\dot{m}$

${\pi }_1=\frac{\dot{m}\sqrt{R{T}_1}}{p_1{d}_2^2}$

Hence, the pi group is: ${\pi }_1=\frac{\dot{m}\sqrt{R{T}_1}}{p_1{d}_2^2}$

Conclusion:

The dimensionless function is ${\pi }_1=\frac{\dot{m}\sqrt{R{T}_1}}{p_1{d}_2^2}$.

To determine

To plot:

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

Answer to Problem 5.2CP

The data is plotted as above and the measured value of ${\pi }_1=\frac{\dot{m}\sqrt{R{T}_1}}{p_1{d}_2^2}$ 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 $p_1-p_2$ is no longer controlling and the gas mass flow $\dot{m}$ 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:

$\dot{m}=fcn(p_1,d_2,T_1,R)$

Ref Fig 3.49:

Diameter of pipe, d₂ = 1 cm

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

Concept Used:

The value of ${\pi }_1$ is calculated for the various points provided:

T( K)	${\pi }_1=\frac{\dot{m}\sqrt{R{T}_1}}{p_1{d}_2^2}$
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 $\dot{m}$, 287 J/kg.K for R, 300 K for T₁ ,200 × 10³ Pa for p₁ and 1 cm for d₂ in the equation:

${\pi }_1=\frac{0.037\sqrt{287\times 300}}{p_1{d}_2^2}$

${\pi }_1=\frac{0.037\sqrt{86100}}{200\times {10}^3\times {(1\times {10}^{-2})}^2}$

$=0.543$

And

The values obtained are plotted on the graph as follows:

Hence, the measured value of ${\pi }_1=\frac{\dot{m}\sqrt{R{T}_1}}{p_1{d}_2^2}$ is about 0.54 and is always a constant.

Conclusion:

The data is plotted as above and the measured value of ${\pi }_1=\frac{\dot{m}\sqrt{R{T}_1}}{p_1{d}_2^2}$ is about 0.54.