Question

determine the new differential reading along the inclined leg of the mercury manometer of figure P2.45, if the pressure in pipe A is decreased 10kPa and the pressurein pipe B remains unchanged. The fluid in A has a specific gravity of 0.9 and the fluid in B is water.

Answer 1

\(\gamma_{H g}=133 K N / m^{3}\)

First we need to know the pressure difference.

$$ P_{b}=P_{a}+0.1 * .9 * 9.8+0.05 \sin (30) * 133-0.08 * 9.8 $$

\(P_{b}-P_{a}=0.1 * .9 * 9.8+0.05 \sin (30) * 133-0.08 * 9.8=3.423_{\mathrm{kPa}}\)

Now solving for \(x\).

$$ P_{b}=P_{a}-10+(0.1-\sin (30) * x) * .9 * 9.8+(0.05 * \sin (30)+x+\sin (30) * $$

\(x) * 133-(0.08+x) * 9.8\)

\(P_{b}-P_{a}+10=(0.1-\sin (30) * x) * .9 * 9.8+(0.025+1.5 x) * 133-(0.08+x) * 9.8\)

\(13.423=3.423+185.29 x\)

\(x=\frac{10 K N / m^{2}}{185.29 K N / m^{3}}=0.053969 m\)

Now we want the new measurement from the low end of Hg to the high end, the previous measurement was \(50 \mathrm{~mm}\).

\(D=x+\frac{x}{\sin (30)}+0.05=3 x+0.05\)

\(D=3 * 0.053969+0.05=0.2119 m\)

\({D}=212 \mathrm{~mm}\)