Question

For the inclined-tube manometer the pressure in pipe A is 0.8psi. The fluid in both pipes A and B is water, and the gage fluid in the manometer has a specific gravity of 2.6. What is the pressure in pipe B corresponding to the differential reading shown?

Answer 1

Calculate the specific weight of the gage fluid.

\(\gamma_{f}=(\mathrm{SG}) \gamma_{w}\)

Here, the specific weight of the fluid is \(\mathrm{SG}\) and the specific weight of water is \(\gamma_{w}\).

Substitute \(2.6\) for \(\mathrm{SG}\) and \(62.4 \mathrm{lbf} / \mathrm{ft}^{2} \gamma_{w}\)

\(\begin{aligned} \gamma_{f} &=2.6(62.4) \\ &=162.24 \mathrm{lbf} / \mathrm{ft}^{3} \end{aligned}\)

Convert the units of pressure from psi to psf.

\(p_{A}=0.6 \mathrm{psi}\)

\(=0.6 \frac{\mathrm{lbf}}{\mathrm{in}^{2}} \times\left(\frac{144 \mathrm{in}^{2}}{1 \mathrm{ft}^{2}}\right)\)

\(=86.4 \mathrm{lbf} / \mathrm{ft}^{2}\)

Apply the principal of manometry to the system.

\(p_{A}+\gamma_{w} h_{A}=\gamma_{f} l_{f} \sin \theta+\gamma_{w} h_{B}+p_{B}\)

Here, the pressures at \(A\) and \(B\) are \(p_{A}\) and \(p_{B}\), the length of the fluid column in \(l_{f}\), inclination of the tube is \(\theta\), the heights of the water columns as shown in the figure are \(h_{A}\) and \(h_{B}\).

Substitute \(86.4 \mathrm{lbf} / \mathrm{ft}^{2}\) for \(p_{A}, 62.4 \mathrm{lbf} / \mathrm{ft}^{2},\left(\frac{3}{12}\right) \mathrm{ft}\) for \(h_{A}, 162.24 \mathrm{lbf} / \mathrm{ft}^{3}\) for \(\gamma_{f},\left(\frac{8}{12}\right) \mathrm{ft}\)

for \(l_{f}, 30^{\circ}\) for \(\theta\), and \(\left(\frac{3}{12}\right) \mathrm{ft}\) for \(h_{B}\).

\(86.4+(62.4)\left(\frac{3}{12}\right)=(162.24)\left(\frac{8}{12}\right) \sin 30^{\circ}+(62.4)\left(\frac{3}{12}\right)+p_{B}\)

\(86.4+15.6=54.08+15.6+p_{B}\)

\(p_{B}=32.32 \frac{\mathrm{lb}}{\mathrm{ft}^{2}} \times\left(\frac{1 \mathrm{ft}^{2}}{144 \mathrm{in} .^{2}}\right)\)

\(p_{B}=0.224 \mathrm{psi}\)

Therefore, the pressure at \(B\) is \(0.224\) psi