Chapter 8, Problem 66P

A horizontal pipe has an abrupt expansion from $D_1=8$ cm to $D_2=16$ cm. The water velocity in the smaller section is 10 m/s and the flow is turbulent. The pressure in the smaller section is $P_1=410$ kPa. Taking the kinetic energy correction factor to be 1.06 at both the inlet and the outlet, determine the downstream pressure $P_2$ , and estimate the error that would have occurred if Bernoulli’s equation had been used.

To determine

The Downstream pressure and the error that would have occurred if Bernoulli’s equation had been used.

Answer to Problem 66P

The Downstream pressure is $431.5\text{kPa}$, the error occurred is $25.375\text{kPa}$, and the error percentage is $5.88\%$ .

Explanation of Solution

Given information:

The diameter of the smaller pipe that is pipe 1 is $8\text{cm}$, the diameter of the bigger pipe that is pipe 2 is $16\text{cm}$, the kinetic energy correction factor is $1.06$, the pressure in the smaller section is $410\text{kPa}$, and the water velocity in the smaller section is $10\text{m}/\text{s}$.

The flow is assumed to be steady, fully developed and incompressible, and the kinetic energy correction factor for both inlet and outlet is equal.

Write the expression for the cross section flow area of pipe 1.

   $A_{C1}=\frac{\pi }{4}{D}_1^2$ ...... (I)

Here, the diameter of the smaller pipe is $D_1$.

Write the expression for the cross section flow area of pipe 2.

   $A_{C2}=\frac{\pi }{4}{D}_2^2$ ...... (II)

Here, the diameter of the bigger pipe is $D_2$.

Write the expression for the velocity of water using the conservation of mass flow.

   $\begin{array}{c}\rho V_1{A}_{C1}=\rho V_2{A}_{C2}\\ V_2=\frac{A_{C1}}{A_{C2}}{V}_1\end{array}$ ...... (III)

Here, the cross section area of the smaller pipe is $A_{C1}$, the cross section area of the bigger pipe is $A_{C2}$, the water velocity in the smaller pipe is $V_1$, and the water velocity in the bigger pipe is.

Write the expression of the loss coefficient for sudden expansion.

   $K_L={\left(1-\frac{A_{C1}}{A_{C2}}\right)}^2$ ...... (IV)

Here, the cross section area of the smaller pipe is $A_{C1}$, and the cross-section area of the bigger pipe is $A_{C2}$.

Write the expression for the head loss.

   $h_L=K_L\frac{V_1^2}{2g}$ ...... (V)

Here, the loss coefficient for sudden expansion is $K_L$, the acceleration due to gravity is $g$, and the water velocity in the smaller pipe is $V_1$.

Write the expression for the energy equation for the horizontal expansion section in terms of heads excluding the wok devices.

   $\begin{array}{l}\frac{P_1}{\rho g}+{\alpha }_1\frac{V_1^2}{2g}=\frac{P_2}{\rho g}+{\alpha }_2\frac{V_2^2}{2g}+h_L\\ \frac{P_2}{\rho g}=\frac{P_1}{\rho g}+{\alpha }_1\frac{V_1^2}{2g}-{\alpha }_2\frac{V_2^2}{2g}-h_L\\ P_2=P_1+\rho \left(\frac{{\alpha }_1{V}_1^2}{2}-\frac{{\alpha }_2{V}_2^2}{2}-g{h}_L\right)\end{array}$ ...... (VI)

Here, the pressure in the bigger section is $P_2$, the pressure in the smaller section is $P_1$, the density of water is $\rho$, the acceleration due to gravity is $g$, the water velocity in the smaller pipe is $V_1$, the water velocity in the bigger pipe is $V_2$, the kinetic energy correction factor for the inlet is ${\alpha }_1$, the kinetic energy correction factor for the outlet is ${\alpha }_2$ and the head loss is $h_L$.

Write the expression for the Bernoulli’s equation for the horizontal pipe.

   $\begin{array}{l}\frac{P_1}{\rho g}+\frac{V_1^2}{2g}=\frac{P_{2,Bernouli}}{\rho g}+\frac{V_2^2}{2g}\\ \frac{P_{2,Bernouli}}{\rho g}=\frac{P_1}{\rho g}+\frac{V_1^2}{2g}-\frac{V_2^2}{2g}\\ P_{2,Bernouli}=P_1+\rho \left(\frac{V_1^2}{2}-\frac{V_2^2}{2}\right)\end{array}$ ...... (VII)

Here, the pressure in the bigger pipe is $P_{2,Bernouli}$.

Write the expression for finding the error in the Bernoulli’s equation.

   $Error=P_{2,Bernouli}-P_2$ ...... (VIII)

Write the expression for finding the error percentage.

   $Error\%=\frac{P_{2,Bernouli}-P_2}{P_2}$ ...... (IX)

Calculation:

Substitute $8\text{cm}$ for $D_1$ in the Equation (I).

   $\begin{array}{c}{A}_{C1}=\frac{\pi }{4}{\left(8\text{cm}\right)}^2\\ =\frac{\pi }{4}{\left(8\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2\\ =\frac{\pi }{4}{\left(0.08\text{m}\right)}^2\\ =5.026\times {10}^{-3}{\text{m}}^{\text{2}}\end{array}$

Substitute $16\text{cm}$ for $D_2$ in the Equation (II).

   $\begin{array}{c}{A}_{C2}=\frac{\pi }{4}{\left(16\text{cm}\right)}^2\\ =\frac{\pi }{4}{\left(16\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2\\ =\frac{\pi }{4}\left(0.16\text{m}\right)\\ =20.106\times {10}^{-3}{\text{m}}^{\text{2}}\end{array}$

Substitute $5.026\times {10}^{-3}{\text{m}}^{\text{2}}$ for $A_{C1}$, $20.106\times {10}^{-3}{\text{m}}^{\text{2}}$ for $A_{C2}$, and $10\text{m}/\text{s}$ for $V_1$ in the Equation (III).

   $\begin{array}{c}{V}_2=\frac{5.026\times {10}^{-3}{\text{m}}^{\text{2}}}{20.106\times {10}^{-3}{\text{m}}^{\text{2}}}\left(10\text{m}/\text{s}\right)\\ =0.24997\left(10\text{m}/\text{s}\right)\\ =2.5\text{m}/\text{s}\end{array}$

Substitute $5.026\times {10}^{-3}{\text{m}}^{\text{2}}$ for $A_{C1}$, and $20.106\times {10}^{-3}{\text{m}}^{\text{2}}$ for $A_{C2}$ in the Equation (IV).

   $\begin{array}{c}{K}_L={\left(1-\frac{5.026\times {10}^{-3}{\text{m}}^{\text{2}}}{20.106\times {10}^{-3}{\text{m}}^{\text{2}}}\right)}^2\\ ={\left(1-0.24997\right)}^2\\ =0.5625\end{array}$

Substitute $0.5625$ for $K_L$, $10\text{m}/\text{s}$ for $V_1$, and $9.81\text{m}/{\text{sec}}^{\text{2}}$ for the $g$ in the Equation (V).

   $\begin{array}{c}{h}_L=0.5625\frac{{\left(10\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{sec}}^{\text{2}}\right)}\\ =0.5625\times 0.5096\text{m}\\ =0.2867\text{m}\end{array}$

Substitute $410\text{kPa}$ for $P_1$, $9.81\text{m}/{\text{sec}}^{\text{2}}$ for the $g$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$

   $1.06$ for ${\alpha }_1$ and ${\alpha }_2$

   $0.2867\text{m}$ for $h_L$

   $10\text{m}/\text{s}$ for $V_1$, and $2.5\text{m}/\text{s}$ for $V_2$ in the Equation (VI).

   $\begin{array}{c}{P}_2=410\text{kPa}+1000\text{kg}/{\text{m}}^{\text{3}}\left(\begin{array}{l}\frac{1.06\times {\left(10\text{m}/\text{s}\right)}^2}{2}-\frac{1.06\times {\left(2.5\text{m}/\text{s}\right)}^2}{2}\\ -\left(9.81\text{m}/{\text{sec}}^{\text{2}}\right)\left(0.2867\text{m}\right)\end{array}\right)\\ =410\text{kPa}\left(\frac{1000\text{Pa}}{1\text{kPa}}\right)\text{+}1000\text{kg}/{\text{m}}^{\text{3}}\left[\begin{array}{l}\left(53{{\text{m}}^2/\text{s}}^2\right)-\left(3.3125{{\text{m}}^2/\text{s}}^2\right)\\ -2.812527{{\text{m}}^2/\text{s}}^2\end{array}\right]\\ =410\times {10}^3\text{Pa}+1000\text{kg}/{\text{m}}^{\text{3}}\left(46.874973{{\text{m}}^2/\text{s}}^2\right)\\ =410\times {10}^3\text{Pa}\left(\frac{\text{1}\text{kg}/{\text{ms}}^{\text{2}}}{1\text{Pa}}\right)\text{+}1000\text{kg}/{\text{m}}^{\text{3}}\left(46.874973{{\text{m}}^2/\text{s}}^2\right)\end{array}$

   $\begin{array}{c}{P}_2=410\times {10}^3\text{kg}/{\text{ms}}^{\text{2}}+1000\text{kg}/{\text{m}}^{\text{3}}\left(46.874973{{\text{m}}^2/\text{s}}^2\right)\\ =410\times {10}^3\text{kg}/{\text{ms}}^{\text{2}}+46874.973\text{kg}/{\text{ms}}^{\text{2}}\\ =431.562\times {10}^3\text{kg}/{\text{ms}}^{\text{2}}\left(\frac{1\text{Pa}}{1\text{kg}/{\text{ms}}^{\text{2}}}\right)\\ =431.562\times {10}^3\text{Pa}\end{array}$

   $\begin{array}{c}=431.562\times {10}^3\text{Pa}\left(\frac{\text{1}\text{kPa}}{\text{1000}\text{Pa}}\right)\\ =\text{431}\text{.5}\text{kPa}\end{array}$

Substitute $410\text{kPa}$ for $P_1$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $10\text{m}/\text{s}$ for $V_1$, and $2.5\text{m}/\text{s}$ for $V_2$ in the Equation (VII).

   $\begin{array}{c}{P}_{2,Bernouli}=410\text{kPa}+1000\text{kg}/{\text{m}}^{\text{3}}\left(\frac{{\left(10\text{m}/\text{s}\right)}^2}{2}-\frac{{\left(2.5\text{m}/\text{s}\right)}^2}{2}\right)\\ =410\text{kPa}\left(\frac{1000\text{Pa}}{1\text{kPa}}\right)+1000\text{kg}/{\text{m}}^{\text{3}}\left(46.875{{\text{m}}^2/\text{s}}^2\right)\\ =410\times {10}^3\text{Pa}+1000\text{kg}/{\text{m}}^{\text{3}}\left(46.875{{\text{m}}^2/\text{s}}^2\right)\\ =410\times {10}^3\text{Pa}\left(\frac{1\text{kg}/{\text{ms}}^{\text{2}}}{\text{1}\text{Pa}}\right)+1000\text{kg}/{\text{m}}^{\text{3}}\left(46.875{{\text{m}}^2/\text{s}}^2\right)\end{array}$

   $\begin{array}{c}{P}_{2,Bernouli}=410\times {10}^3\text{kg}/{\text{ms}}^{\text{2}}+46875\text{kg}/{\text{ms}}^{\text{2}}\\ =456.875\text{kg}/{\text{ms}}^{\text{2}}\left(\frac{1\text{Pa}}{1\text{kg}/{\text{ms}}^{\text{2}}}\right)\\ =456.875\text{Pa}\\ =456.875\text{Pa}\left(\frac{1\text{kPa}}{1000\text{Pa}}\right)\end{array}$

   $P_{2,Bernouli}=456.875\times {10}^{-3}\text{kPa}$

Substitute $456.875\text{kPa}$ for $P_{2,Bernouli}$, and $\text{431}\text{.5}\text{kPa}$ for $P_2$ in the Equation (VIII).

   $\begin{array}{c}\text{Error}=456.875\text{kPa}-\text{431}\text{.5}\text{kPa}\\ =25.375\text{kPa}\end{array}$

Substitute $456.875\text{kPa}$ for $P_{2,Bernouli}$, and $\text{431}\text{.5}\text{kPa}$ for $P_2$ in the Equation (IX).

   $\begin{array}{c}\text{Error\%}=\frac{456.875\text{kPa}-\text{431}\text{.5}\text{kPa}}{\text{431}\text{.5}\text{kPa}}\\ =\frac{25.375\text{kPa}}{\text{431}\text{.5}\text{kPa}}\\ =0.0588\times 100\\ =5.88\%\end{array}$

Conclusion:

The Downstream pressure is $431.5\text{kPa}$, the error occurred is $25.375\text{kPa}$, and the error percentage is $5.88\%$ .