Problem

Water is flowing. Calculate the gage reading when the velocity in the 12-in.-diameter pipe...

Water is flowing. Calculate the gage reading when the velocity in the 12-in.-diameter pipe is 8 ft/s.

Step-by-Step Solution

Solution 1

Calculate the velocity in the 6 in diameter pipe using continuity equation.

$$ \begin{aligned} &A_{1} V_{1}=A_{2} V_{2} \\ &V_{1}=\frac{A_{2}}{A_{1}} V_{2} \end{aligned} $$

Here, \(A_{1}\) is the area of 12 in diameter pipe and \(A_{2}\) is the area of 6 in diameter pipe.

Substitute \(8 \mathrm{ft} / \mathrm{s}\) for \(V_{1},\left(\frac{\pi}{4} \times 12^{2}\right)\) in \(^{2}\) for \(A_{1},\left(\frac{\pi}{4} \times 6^{2}\right)\) in \(^{2}\) for \(A_{2}\).

$$ \begin{aligned} V_{2} &=\frac{\left(\frac{\pi}{4} \times 12^{2}\right)}{\left(\frac{\pi}{4} \times 6^{2}\right)} \times 8 \\ &=32 \mathrm{ft} / \mathrm{s} \end{aligned} $$

Calculate the head loss in the pipe with 12 in diameter.

$$ H_{a}=f_{1} \frac{L_{1}}{D_{1}} \frac{V_{1}^{2}}{2 g} $$

Here, \(L_{1}\) is the length of the pipe, \(g\) is the acceleration due to gravity, \(D_{1}\) is the diameter of the pipe and \(f_{1}\) is the friction factor.

Substitute \(0.02\) for \(f_{1}, 150 \mathrm{ft}\) for \(L_{1}, 32.2 \mathrm{ft} / \mathrm{s}^{2}\) for \(g,(12 / 12) \mathrm{ft}\) for \(D_{1}, 8 \mathrm{ft} / \mathrm{s}\) for \(V_{1}\).

$$ \begin{aligned} H_{n} &=0.02 \times \frac{150}{(12 / 12)} \times \frac{8^{2}}{2 \times 32.2} \\ &=2.98 \mathrm{ft} \end{aligned} $$

Calculate the head loss in the pipe with 6 in diameter.

$$ H_{/ 2}=f_{2} \frac{L_{2}}{D_{2}} \frac{V_{2}^{2}}{2 g} $$

Here, \(L_{2}\) is the length of the pipe, \(D_{2}\) is the diameter of the pipe, \(f_{2}\) is the friction factor.

Substitute \(0.02\) for \(f_{2}, 8 \mathrm{ft}\) for \(L_{2}, 32.2 \mathrm{ft} / \mathrm{s}^{2}\) for \(g,(6 / 12) \mathrm{ft}\) for \(D_{2}\) and \(32 \mathrm{ft} / \mathrm{s}\) for \(V_{2}\).

$$ \begin{aligned} H_{i 2} &=0.02 \times \frac{8}{(6 / 12)} \times \frac{32^{2}}{2 \times 32.2} \\ &=5.09 \mathrm{ft} \end{aligned} $$

Obtain the contraction loss coefficient from fig. \(8.15\),

For the area ratio of \(0.25\), contraction loss coefficient, \(K_{e}=0.35\).

Calculate the head loss due to sudden contraction in the pipe.

$$ H_{e}=K_{c} \frac{V_{2}^{2}}{2 g} $$

Here, \(K_{c}\) is the loss coefficient.

Substitute \(0.35\) for \(K_{c}, 32 \mathrm{ft} / \mathrm{s}\) for \(V_{2}\) and \(32.2 \mathrm{ft} / \mathrm{s}^{2}\) for \(g\).

$$ \begin{aligned} H_{e} &=0.35 \times \frac{32^{2}}{2 \times 32.2} \\ &=5.6 \mathrm{ft} \end{aligned} $$

Obtain the contraction loss coefficient from "Loss coefficients for flow through sudden area changes figure".

For the area ratio of \(0.25\), expansion loss coefficient, \(K_{e}=1\).

Calculate the head loss due to expansion in the pipe.

$$ H_{e}=K_{e} \frac{V_{2}^{2}}{2 g} $$

Here, \(K_{e}\) is the expansion loss coefficient.

Substitute 1 for \(K_{e}, 32 \mathrm{ft} / \mathrm{s}\) for \(V_{2}, 32.2 \mathrm{ft} / \mathrm{s}^{2}\) for \(g\).

$$ \begin{aligned} H_{e} &=1 \times \frac{32^{2}}{2 \times 32.2} \\ &=15.9 \mathrm{ft} \end{aligned} $$

Calculate the gage reading using energy equation.

$$ \begin{aligned} &\left(\frac{p_{1}}{\rho}+\frac{V_{1}^{2}}{2}+g z_{1}\right)-\left(\frac{p_{2}}{\rho}+\frac{V_{2}^{2}}{2}+g z_{2}\right)=H_{r} \\ &\left(\frac{p_{1}}{\rho g}+\frac{V_{1}^{2}}{2 g}+z_{1}\right)-\left(z_{2}\right)=\left(H_{n}+H_{12}+H_{c}+H_{e}\right) \\ &p_{1}=\rho g\left(H_{n}+H_{n 2}+H_{e}+H_{e}\right)-\rho g\left(z_{1}-z_{2}\right)-\frac{\rho V_{1}^{2}}{2} \end{aligned} $$

Here, \(V_{1}\) is the velocity at the entrance of the pipe, \(p_{1}\) is the pressure at the entrance of the pipe, \(\rho\) is the density of the fluid, \(g\) is the acceleration due to gravity, \(z_{2}\) is the head at location 2 , \(H_{e}\) is the expansion head loss, \(H_{i}\) is the head loss due to pipe friction.

Substitute \(62.4 \mathrm{lbf} / \mathrm{ft}^{3}\) for \(\rho g, 2.98 \mathrm{ft}\) for \(H_{n}, 5.09 \mathrm{ft}\) for \(H_{i 2}, 5.6 \mathrm{ft}\) for \(H_{e}, 15.9 \mathrm{ft}\) for \(H_{e}\), \(1.94 \mathrm{lbf} \cdot \mathrm{s}^{2} / \mathrm{ft}^{4}\) for \(\rho, 0\) for \(z_{1}, 30 \mathrm{ft}\) for \(z_{2}\), and \(8 \mathrm{ft} / \mathrm{s}\) for \(V_{\mathrm{t}}\).

$$ \begin{aligned} p_{1} &=62.4 \times(2.98+5.09+6.12+15.9)-62.4 \times(-30)-\frac{1.94 \times 8^{2}}{2} \\ &=3688 \mathrm{lbf} / \mathrm{ft}^{2} \\ &=25.6 \mathrm{psi} \end{aligned} $$

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