Chapter 8, Problem 153P

The water at $20°$ C in a l0-m-diameter, 2-m-high aboveground swimming pool is to be emptied by unplugging a 5-cm-diameter, 25-m-long horizontal plastic pipe attached to the bottom of the pool. Determine the initial rate of discharge of water through the pipe and the time (hours) it would take to empty the swimming pool completely assuming the entrance to the pipe is well-rounded with negligible loss. Take the friction factor of the pipe to be 0.022. Using the initial discharge velocity, check if this is a reasonable value for the friction factor.

To determine

The initial discharge rate of water via the pipe.

The time taken to empty the swimming pool.

The value of the friction factor.

Answer to Problem 153P

The discharge rate of water via the pipe is $3.55\text{L}/\text{s}$.

The time taken to empty the swimming pool is $24.7\text{h}$.

The given value of the friction factor is not accurate and the reasonable value of friction factor is $0.0184$.

Explanation of Solution

Given information:

The swimming pool diameter is $10\text{m}$, the swimming pool height is $2\text{m}$, the temperature of water is $20°\text{C}$, the pipe length is $25\text{m}$, the pipe diameter is $5\text{cm,}$ and the friction factor is $0.022$.

The flow is incompressible and turbulent. The difference in elevation between the fountain and the pipe is null. The pressure at the inlet ad the exit is atmospheric in nature.

Write the expression to calculate the energy equation. Assume the effect of kinetic energy correction factor to be negligible.

  $\frac{P_1}{\rho g}+{\alpha }_1\frac{V_1^2}{2g}+z_1+h_{pump}=\frac{P_2}{\rho g}+{\alpha }_2\frac{V_2^2}{2g}+z_2+h_{turbine}+h_L$ (I)

Here, the inlet pressure is $P_1$, the exit pressure is $P_2$, the inlet velocity is $V_1$, the exit velocity is $V_2$, the exit datum is $z_2$, the inlet datum is $z_1$, the gravitational acceleration is $g$, the density of fluid is $\rho$, the correction factor for kinetic energy at the inlet is ${\alpha }_1$, the correction factor for the kinetic energy at the exit is ${\alpha }_2$, the head loss is $h_L$, the loss from the pump is $h_{pump},$ and the loss of head of the turbine is $h_{turbine}$.

Write the expression to calculate the exit velocity when time is $0$.

  $V_{2,t\left(0\right)}=\frac{\dot{\vee }}{A_c}$ (II)

Here, he volume flow rate is $\dot{\vee }$ and the area of cross section is $A_c$.

Write the expression for the area of cross section.

  $A_c=\frac{\pi }{4}{D}^2$ (III)

Here, the diameter of the pipe is $D$.

Substitute $\frac{\pi }{4}{D}^2$ for $A_c$ in Equation (II).

  $V_{2,t\left(0\right)}=\frac{\dot{\vee }}{\frac{\pi }{4}{D}^2}$ (IV)

Write the expression to calculate the head loss.

  $h_L=f\frac{L}{D}\left(\frac{V_{2,t\left(0\right)}^2}{2g}\right)$ (V)

Here, the diameter of the pipe is $D$, the length of the pipe is $L$ and the friction factor is $f$.

Write the expression for the elevation at any given time.

  $\begin{array}{l}z=\frac{V_2^2}{2g}+f\frac{L}{D}\left(\frac{V_2^2}{2g}\right)\\ z=\frac{V_2^2}{2g}\left(1+f\frac{L}{D}\right)\\ V_2=\sqrt{\frac{2gz}{1+f\frac{L}{D}}}\end{array}$ (VI)

Write the expression for the flow rate at any time given.

  $\dot{\vee }=V_2{A}_c$ (VII)

Substitute $\sqrt{\frac{2gz}{1+f\frac{L}{D}}}$ for $V_2$ and $\frac{\pi }{4}{D}^2$ for $A_c$ in Equation (VII).

  $\dot{\vee }=\left(\sqrt{\frac{2gz}{1+f\frac{L}{D}}}\right)\left(\frac{\pi }{4}{D}^2\right)$ (IX)

Write the expression to calculate the water amount flowing b=via the pipe for a differential time.

  $dV=\dot{\vee }dt$ (X)

Here, the differential time is $dt$.

Substitute $\left(\sqrt{\frac{2gz}{1+f\frac{L}{D}}}\right)\left(\frac{\pi }{4}{D}^2\right)$ for $\dot{\vee }$ in Equation (X).

  $dV=\left(\left(\sqrt{\frac{2gz}{1+f\frac{L}{D}}}\right)\left(\frac{\pi }{4}{D}^2\right)\right)dt$ (XI)

Write the expression for the area of the tank.

  $A_{c,\text{tank}}=\frac{\pi }{4}{D}_L^2$

Here, the diameter of the tank is $D_L$.

Write the expression for the decrease of volume of water in the swimming pool.

  $dV=A_{c,\text{tank}}\left(-dz\right)$ (XII)

Here, the negative change in datum of tank is $\left(-dz\right)$.

Substitute $\left(\left(\sqrt{\frac{2gz}{1+f\frac{L}{D}}}\right)\left(\frac{\pi }{4}{D}^2\right)\right)dt$ for $dV$ in Equation (XII).

  $\begin{array}{l}\left(\left(\sqrt{\frac{2gz}{1+f\frac{L}{D}}}\right)\left(\frac{\pi }{4}{D}^2\right)\right)dt=A_{c,\text{tank}}\left(-dz\right)\\ dt=-\frac{D_L^2}{D^2}\sqrt{\frac{1+f\frac{L}{D}}{2g}}\left(z^{\frac{1}{2}}\right)\cdot dz\end{array}$ (XIII)

Integrate Equation (XIII) from $0$ to $t$ in LHS and $z_1$ to $0$ in RHS.

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{t}{\int }}dt}={\displaystyle \underset{z_1}{\overset{0}{\int }}-\frac{D_L^2}{D^2}\sqrt{\frac{1+f\frac{L}{D}}{2g}}\left(z^{\frac{1}{2}}\right)\cdot dz}\\ {\left(t\right)}_0^t=-\frac{D_L^2}{D^2}\sqrt{\frac{1+f\frac{L}{D}}{2g}}\times {\left(\frac{z^{\frac{1}{2}}}{\frac{1}{2}}\right)}_{z_1}^0\\ t-0=-\frac{D_L^2}{D^2}\sqrt{\frac{1+f\frac{L}{D}}{2g}}\times \left(\frac{0^{\frac{1}{2}}}{\frac{1}{2}}-\frac{z_1{}^{\frac{1}{2}}}{\frac{1}{2}}\right)\\ t=\frac{D_L^2}{D^2}\sqrt{\frac{1+f\frac{L}{D}}{2g}}\left(z_1^{\frac{1}{2}}\right)\end{array}$ (XIV)

Write the expression to calculate the Reynolds number.

  $\mathrm{Re}=\frac{\rho V_{2,\left(t\right)}D}{\mu }$ (XV)

Here, the viscosity of the fluid is $\mu$.

Write the expression for the Colebrook equation.

  $\frac{1}{\sqrt{f}}=-2\log \left(\frac{\frac{\epsilon }{D}}{3.7}+\frac{2.5}{\mathrm{Re}\sqrt{f}}\right)$ (XVI)

Here, the friction factor is $f$ and roughness is $\epsilon$.

Calculation:

Refer to Table A-3, "Properties of saturated water" to obtain the value of $\mu$ as $1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$ and $\rho$ as $998\text{kg}/{\text{m}}^{\text{3}}$ at $20°\text{C}$.

The flow is fully developed making the roughness of the plastic null.

Substitute $5\text{cm}$ for $D$ in Equation (IV).

  $\begin{array}{c}{V}_{2,t\left(0\right)}=\frac{\dot{\vee }}{\frac{\pi }{4}{\left(5\text{cm}\right)}^2}\\ =\frac{\dot{\vee }}{\frac{\pi }{4}{\left(5\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}\\ =509.3\dot{\vee }\end{array}$

Substitute $509.3\dot{\vee }$ for $V_{2,t\left(0\right)}$ in Equation (V).

  $\begin{array}{c}{h}_L=0.022\left(\frac{25\text{m}}{5\text{cm}}\right)\left(\frac{{\left(509.3\dot{\vee }\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\right)\\ =0.022\left(\frac{25\text{m}}{5\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)}\right)\left(\frac{{\left(509.3\dot{\vee }\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\right)\\ =145.426\times {10}^3{\dot{\vee }}^2\end{array}$

Consider the time is zero.

Substitute $V_{2,t\left(0\right)}$ for $V_2$, $0$ for $V_1$, $0$ for $h_{pump}$, $0$ for $h_{turbine}$, $0$ for $z_2$, $1$ for ${\alpha }_2$, $1$ for ${\alpha }_1$, $P_{atm}$ for $P_1$ and $P_{atm}$ for $P_2$ in Equation (I).

  $\begin{array}{l}\frac{P_{atm}}{\rho g}+1\frac{0}{2g}+z_1+0=\frac{P_{atm}}{\rho g}+1\frac{V_{2,t\left(0\right)}^2}{2g}+0+0+h_L\\ z_1=\frac{V_{2,t\left(0\right)}^2}{2g}+h_L\end{array}$ (XVII)

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $2\text{m}$ for $z_1$, $509.3\dot{\vee }$ for $V_{2,t\left(0\right)}$ and $145.426\times {10}^3{\dot{\vee }}^2$ for $h_L$ in Equation (XVII).

  $\begin{array}{l}2\text{m}=\frac{{\left(509.3\dot{\vee }\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}+145.426\times {10}^3{\dot{\vee }}^2\\ \dot{\vee }=\sqrt{\frac{2}{158.646\times {10}^3}}{\text{m}}^{\text{3}}/\text{s}\\ \dot{\vee }=3.55\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\end{array}$

Convert ${\text{m}}^{\text{3}}/\text{s}$ into $\text{L}/\text{s}$.

  $\begin{array}{c}\dot{\vee }=3.55\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\left(\frac{1000\text{L}/\text{s}}{1{\text{m}}^{\text{3}}/\text{s}}\right)\\ =3.55\text{L}/\text{s}\end{array}$

Substitute $10\text{m}$ for $D_L$, $5\text{}\text{cm}$ for $D$, $0.0022$ for $f$, $25\text{m}$ for $L$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $2\text{}\text{m}$ for $z_1$ in Equation (XIV).

  $\begin{array}{c}t=\frac{{\left(10\text{m}\right)}^2}{{\left(5\text{cm}\right)}^2}\sqrt{\frac{1+0.0022\frac{25\text{m}}{5\text{cm}}}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}}\left({\left(2\text{m}\right)}^{\frac{1}{2}}\right)\\ =\frac{{\left(10\text{m}\right)}^2}{{\left(5\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}\sqrt{\frac{1+0.0022\frac{25\text{m}}{5\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)}}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}}\left({\left(2\text{m}\right)}^{\frac{1}{2}}\right)\\ =88480.20\text{s}\left(\frac{1\text{h}}{3600\text{s}}\right)\\ =24.7\text{h}\end{array}$

Substitute $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $509.3\dot{\vee }$ for $V_{2,\left(t\right)}$, $5\text{cm}$ for $D$ and $1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$ for $\mu$ in Equation (XV).

  $\mathrm{Re}=\frac{\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(509.3\dot{\vee }\right)\left(5\text{cm}\right)}{\left(1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}$ (XVIII)

Substitute $3.55\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}$ for $\dot{\vee }$ in Equation (XVIII).

  $\begin{array}{c}\mathrm{Re}=\frac{\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(509.3\left(3.55\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\right)\right)\left(5\text{cm}\right)}{\left(1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\\ =\frac{\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(509.3\left(3.55\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\right)\right)\left(5\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}{\left(1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\\ =90.04\times {10}^3\end{array}$

Substitute $90.04\times {10}^3$ for $\mathrm{Re}$, $0$ for $\epsilon$ and $5\text{cm}$ for $D$ in Equation (XVI).

  $\begin{array}{c}\frac{1}{\sqrt{f}}=-2\log \left(\frac{\frac{0}{5\text{cm}}}{3.7}+\frac{2.5}{\left(90.04\times {10}^3\right)\sqrt{f}}\right)\\ \frac{1}{\sqrt{f}}=-2\log \left(\frac{2.787\times {10}^{-5}}{\sqrt{f}}\right)\\ f=0.0184\end{array}$

Conclusion:

The discharge rate of water via the pipe is $3.55\text{L}/\text{s}$.

The time taken to empty the swimming pool is $24.7\text{h}$.

The given value of the friction factor is not accurate and the reasonable value of friction factor is $0.0184$.