Chapter 14, Problem 103RQ

To determine

The expression for the loss coefficient, its value for the given case and the pressure at the larger pipe.

Explanation of Solution

Given:

Diameter of the smaller pipe is $8\text{ cm}$.

Diameter of the larger pipe is $24\text{ cm}$.

Pressure in the smaller pipe is $135\text{ kPa}$.

Velocity in the smaller pipe is $10\text{m}/\text{s}$.

Calculation:

From the continuity equation,

  $\begin{array}{l}{V}_1{A}_1=V_2{A}_2\\ V_2=\frac{A_1}{A_2}{V}_1\end{array}$        (I)

Applying the law of conservation of momentum,

  $\begin{array}{l}{\displaystyle \sum \overrightarrow{F}}={\displaystyle \sum _{out}\beta \dot{m}V}-{\displaystyle \sum _{in}\beta \dot{m}V}\\ \dot{m}\left(V_2-V_1\right)=P_1{A}_1+P_1\left(A_x-A_1\right)-P_2{A}_2\\ \dot{m}\left(V_2-V_1\right)=P_1{A}_x-P_2{A}_2\\ \rho A_2{V}_2\left(V_2-V_1\right)=P_1{A}_2-P_2{A}_2\\ \rho A_2\left(\frac{A_1}{A_2}{V}_1\right)\left(\frac{A_1}{A_2}{V}_1-V_1\right)=P_1{A}_2-P_2{A}_2\\ \rho A_2\left(\frac{A_1}{A_2}\right)\left(\frac{A_1}{A_2}-1\right)V_1{}^2=\left(P_1-P_2\right)A_2\\ \frac{P_1-P_2}{\rho }=\left(\frac{A_1}{A_2}\right)\left(\frac{A_1}{A_2}-1\right)V_1{}^2\end{array}$        (II)

Apply energy equation,

  $\begin{array}{l}\frac{P_1}{\rho g}+{\alpha }_1\frac{V_1{}^2}{2g}+z_1+h_{pump}=\frac{P_2}{\rho g}+{\alpha }_2\frac{V_2{}^2}{2g}+z_2+h_{turbine}+h_L\\ h_L=\frac{P_1-P_2}{\rho g}+\frac{V_1{}^2-V_2{}^2}{2g}\end{array}$        (III)

Substitute Equation (I) and Equation (II) in Equation (III).

  $\begin{array}{l}{h}_L=\left(\frac{A_1}{A_2}\right)\left(\frac{A_1}{A_2}-1\right)\frac{V_1{}^2}{g}+\frac{V_1{}^2-{\left(\frac{A_1}{A_2}{V}_1\right)}^2}{2g}\\ K_L\frac{V_1{}^2}{2g}=\left(\frac{A_1}{A_2}\right)\left(\frac{A_1}{A_2}-1\right)\frac{V_1{}^2}{g}+\frac{V_1{}^2-{\left(\frac{A_1}{A_2}{V}_1\right)}^2}{2g}\\ K_L=\frac{2A_1}{A_2}\left(\frac{A_1}{A_2}-1\right)+\left[1-{\left(\frac{A_1}{A_2}\right)}^2\right]\\ K_L=2{\left(\frac{A_1}{A_2}\right)}^2-2\frac{A_1}{A_2}+1-{\left(\frac{A_1}{A_2}\right)}^2\\ K_L=1+{\left(\frac{A_1}{A_2}\right)}^2-2\frac{A_1}{A_2}\\ K_L={\left[1-\frac{A_1}{A_2}\right]}^2\\ K_L={\left[1-\frac{\frac{\pi }{4}{D}_1{}^2}{\frac{\pi }{4}{D}_2{}^2}\right]}^2\\ K_L={\left[1-\frac{D_1{}^2}{D_2{}^2}\right]}^2\end{array}$

Thus, the expression for the loss coefficient is $K_L={\left[1-\frac{D_1{}^2}{D_2{}^2}\right]}^2$.

Substituting the respective values in the above equation,

  $\begin{array}{l}{K}_L={\left[1-\frac{{0.08}^2}{{0.24}^2}\right]}^2\\ K_L=0.7901\end{array}$

Thus, the value of the loss coefficient is $0.7901$.

The head loss is,

  $\begin{array}{c}{h}_L=K_L\frac{V_1{}^2}{2g}\\ =0.7901\frac{{\left(10\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^2\right)}\\ =4.027\text{m}\end{array}$

The velocity in the larger pipe is,

  $\begin{array}{c}{V}_2=\frac{A_1}{A_2}{V}_1\\ =\frac{D_1{}^2}{D_2{}^2}{V}_1\\ =\frac{{0.08}^2}{{0.24}^2}\left(10\text{m}/\text{s}\right)\\ =1.111\text{m}/\text{s}\end{array}$

From Equation (III),

  $\begin{array}{c}{h}_L=\frac{P_1-P_2}{\rho g}+\frac{V_1{}^2-V_2{}^2}{2g}\\ P_2=P_1+\left[\frac{V_1{}^2-V_2{}^2}{2g}-h_L\right]\rho g\\ P_2=P_1+\left[\frac{V_1{}^2-V_2{}^2}{2}-g{h}_L\right]\rho \\ =\left(135\text{ kPa}\right)+\left[\frac{{\left(10\text{m}/\text{s}\right)}^2-{\left(1.111\text{m}/\text{s}\right)}^2}{2}-\left(9.81\text{m}/{\text{s}}^2\right)\left(4.027\text{ m}\right)\right]\left(1000\text{kg}/{\text{m}}^3\right)\\ =144.9\text{ kPa}\end{array}$

Thus, the pressure in the larger pipe is $144.9\text{ kPa}$.