Chapter 8, Problem 73P

Water at $15°$ C is drained from a large reservoir using two horizontal plastic pipes connected in series. The first pipe is 13 m long and has a 10-cm diameter, while the second pipe is 35 m long and has a 5-cm diameter. The water level in the reservoir is 18 in above the centerline of the pipe. The pipe entrance is sharp-edged, and the contraction between the two pipes is sudden. Neglecting the effect of the kinetic energy correction factor, determine the discharge rate of water from the reservoir.

To determine

The discharge rate of water from reservoir.

Answer to Problem 73P

The discharge rate of water from reservoir is $0.0206\left({\text{m}}^{\text{3}}/\text{s}\right)$.

Explanation of Solution

Given information:

Reservoir is drained at $15\text{°C}$, length of first pipe is $13\text{m}$, diameter of first pipe is $10\text{cm}$, length of second pipe is $35\text{m}$, diameter of second pipe is $5\text{cm,}$ and water level is $18\text{m}$ above the centerline of pipe. The entrance in the pipe is sharp edged and contraction between the pipes is sudden. Neglect the effect of kinetic energy correction factor.

Write the expression of datum head.

  $z_1={\alpha }_2\frac{V_2{}^2}{2g}+h_L$   ....... (I)

Here, kinetic energy correction factor at point 2 is ${\alpha }_2$, velocity at point 2 is $V_2$, head loss at point 2 is $h_L,$ and the acceleration due to gravity is $g$.

Write the expression of total head loss.

  $h_L=h_{L,\text{major}}+h_{L,\text{minor}}$   ....... (II)

Here, the major head loss is $h_{L,\text{major}}$ and minor head loss is $h_{L,\text{minor}}$.

Write the expression for major head loss.

  $h_{L,\text{major}}=\left(f\frac{L_1}{D}+K_{L,\text{entrance}}\right)\frac{V_1{}^2}{2g}$

Here, the friction factor is $f$, the length of pipe $1$ is $L_1$, the diameter of pipe 1 is $D$, the loss coefficient at entrance is $K_{L,\text{entrance}}$, the velocity of water at point 1 is $V_1,$ and the acceleration due to gravity is $g$.

Write the expression of minor head loss.

  $h_{L,\text{minor}}=\left(f\frac{L_2}{d}+K_{L,\text{contraction}}\right)\frac{V_2{}^2}{2g}$

Here, the friction factor is $f$, the length of pipe 2 is $L_2$, the diameter of pipe 2 is $d$, the loss coefficient at contraction is $K_{L,\text{contraction}}$, the velocity of water at point 2 is $V_2,$ and the acceleration due to gravity is $g$.

Substitute $\left(f\frac{L_1}{D}+K_{L,\text{entrance}}\right)\frac{V_1{}^2}{2g}$ for $h_{L,\text{major}}$ and $\left(f\frac{L_2}{d}+K_{L,\text{contraction}}\right)\frac{V_2{}^2}{2g}$ for $h_{L,\text{minor}}$ in Equation (II).

  $h_L=\left(f\frac{L_1}{D}+K_{L,\text{entrance}}\right)\frac{V_1{}^2}{2g}+\left(f\frac{L_2}{d}+K_{L,\text{contraction}}\right)\frac{V_2{}^2}{2g}$   ....... (III)

Write the expression for discharge through the pipe.

  $Q=A_1{V}_1$   ....... (IV)

Here, the area of flow of fluid is $A_1$ and the velocity of flow in pipe 1 is $V_1$.

Write the expression of area of flow of fluid from pipe 1.

  $A_1=\left(\frac{\pi }{4}\right)\times D^2$

Here, the diameter of pipe 1 is $D$.

Write the expression of area of flow of fluid from pipe 2.

  $A_2=\left(\frac{\pi }{4}\right)\times d^2$

Here, the diameter of pipe 2 is $d$.

Substitute $\left(\frac{\pi }{4}\right)\times D^2$ for $A_1$ in Equation (IV).

  $Q=\left(\frac{\pi }{4}\right)\times D^2\times V_1$   ....... (V)

Write the expression of continuity equation.

  $A_1{V}_1=A_2{V}_2$   ....... (VI)

Here, the flow area of pipe $1$ is $A_1$, the flow velocity of water in pipe 1 is $V_1$, the flow area of pipe 2 is $A_2$, the flow velocity of water in pipe 2 is $V_2$.

Substitute $\frac{\pi }{4}{D}^2$ for $A_1$ and $\frac{\pi }{4}{d}^2$ for $A_2$ in Equation (VI)

  $\begin{array}{l}\frac{\pi }{4}{D}^2\times V_1=\frac{\pi }{4}{d}^2\times V_2\\ V_2={\left(\frac{D}{d}\right)}^2\times V\end{array}$   ....... (VII)

Write the expression of squire of ratio of diameter of pipe.

  $d_r=\frac{d^2}{D^2}$   ....... (VIII)

Here, the diameter of pipe 2 is $d$ and the diameter of pipe 1 is $D$.

Calculation:

Substitute $5\text{cm}$ for $d$ and $10\text{cm}$ for $D$ in Equation (VII).

  $\begin{array}{c}{V}_2={\left(\frac{10\text{cm}}{5\text{cm}}\right)}^2\times V_1\\ ={\left(2\right)}^2\times V_1\\ =4\times V_1\end{array}$

Substitute, $18\text{m}$ for $z_1$, $1$ for ${\alpha }_2$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $4\times V_1$ for $V_2$ in Equation (I).

  $\begin{array}{l}18\text{m}=1\times \frac{{\left(4\times V_1\right)}^2}{2\times \left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}+h_L\\ h_L=18\text{m}-\frac{0.815\times V_1{}^2}{\left(\text{m}/{\text{s}}^{\text{2}}\right)}\\ =18\text{m}-0.815\times V_1{}^2\left({\text{s}}^{\text{2}}/\text{m}\right)\end{array}$

Substitute $5\text{cm}$ for $d$ and $10\text{cm}$ for $D$ in Equation (VIII).

  $\begin{array}{c}{d}_r=\frac{{\left(5\text{cm}\right)}^2}{{\left(10\text{cm}\right)}^2}\\ =0.25\end{array}$

Refer Table 8-4 "Loss coefficient of $K_L$ for various pipe" to find the loss coefficient corresponding to $0.25$ is $0.46$ at contraction and $0.5$ at entrance.

Refer Table 8-2 "Equivalence roughness values for new commercial pipe" to find the roughness value as $0$ corresponding to plastic pipe and friction factor as $0.0119$ corresponding to $\epsilon /D=0$.

Substitute, $18\text{m}-0.815\times V_1{}^2\left(s^2/m\right)$ for $h_L$

  $0.0119$ for $f$, $13\text{m}$ for $L_1$, $10\text{cm}$ for $D$, $4\times V_1$ for $V_2$

  $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $35\text{m}$ for $L_2$, $5\text{cm}$ for $d$, $0.5$ for $K_{L,\text{entrance}}$ and $0.46$ for $K_{L,\text{contraction}}$ in Equation (III).

  $\begin{array}{l}18\text{m}-0.815\times V_1{}^2\left({\text{s}}^{\text{2}}/\text{m}\right)=\left[\begin{array}{l}\left(0.0119\times \frac{13\text{m}}{10\text{cm}}+0.5\right)\frac{V_1{}^2}{2\times 9.81\text{m}/{\text{s}}^{\text{2}}}\\ +\left(0.0119\times \frac{35\text{m}}{5\text{cm}}+0.46\right)\frac{4\times V_1{}^2}{2\times 9.81\text{m}/{\text{s}}^{\text{2}}}\end{array}\right]\\ 18\text{m}-0.815\times V_1{}^2\left({\text{s}}^{\text{2}}/\text{m}\right)=\left[\begin{array}{l}\left(\frac{0.155\text{m}}{\left(\text{10}\text{cm}\right)\times \left(\frac{1\text{m}}{100\text{cm}}\right)}+0.5\right)\times 0.05\times V_1{}^2\left({\text{s}}^{\text{2}}/\text{m}\right)+\\ \left(\frac{0.4165\text{m}}{\left(5\text{cm}\right)\times \left(\frac{1\text{m}}{100\text{cm}}\right)}+0.46\right)\times 0.2\times V_1{}^2\left({\text{s}}^{\text{2}}/\text{m}\right)\end{array}\right]\\ 18\text{m}-0.815\times V_1{}^2\left({\text{s}}^{\text{2}}/\text{m}\right)=0.1\times V_1{}^2\left({\text{s}}^{\text{2}}/\text{m}\right)+1.7\times V_1{}^2\left({\text{s}}^{\text{2}}/\text{m}\right)\\ 18\text{m}=2.615\times V_1{}^2\left({\text{s}}^{\text{2}}/\text{m}\right)\end{array}$

  $\begin{array}{c}{V}_1{}^2=6.88\left({\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)\\ V_1=\sqrt{\left(6.88\left({\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)\right)}\\ =2.63\left(\text{m}/\text{s}\right)\end{array}$

Substitute $10\text{cm}$ for $D$, $2.63\left(\text{m}/\text{s}\right)$ for $V_1$ in Equation (V).

  $\begin{array}{c}Q=\left(\frac{\pi }{4}\right)\times {\left(10\text{cm}\right)}^2\times 2.63\left(\text{m}/\text{s}\right)\\ =\left(\frac{\pi }{4}\right)\times {\left(10\text{cm}\times \left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\times 2.63\left(\text{m}/\text{s}\right)\\ =2.06\times \left(\frac{1}{100}{\text{m}}^{\text{2}}\right)\times \left(\text{m}/\text{s}\right)\\ =0.0206\left({\text{m}}^{\text{3}}/\text{s}\right)\end{array}$

Conclusion:

The discharge rate of water from reservoir is $0.0206\left({\text{m}}^{\text{3}}/\text{s}\right)$.