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$$
\text { The water in a tank is pressurized by air, and the pressure is measured by a multifluid manometer }
$$

$$
\text { as shown in Fig. Pl-53. Determine the gage pressure of air in the tank if } l_{i,}^{\prime}=0.2 \mathrm{m}, \mathrm{h} 2=0.3 \mathrm{m}, \text { and }
$$

$$
h 3=0.46 \mathrm{m} . \text { Take the densities of water, oil, and mercury to be } 1000 \mathrm{kg} / \mathrm{m} 3,850 \mathrm{kg} / \mathrm{m} 3, \text { and } 13,600 \mathrm{kg} / \mathrm{m} 3, \text { respectively. }
$$

$$
P_{1}+\rho_{w} g h_{1}+\rho_{0} g h_{2}
$$$$
-\rho g h_{3}=P_{atm}
$$

$$
P_{1}-P_{a t m}=
$$$$
\rho g h_{3}-\rho_{w} g h_{1}
$$$$
-\rho_0 \mathrm{gh}_{2}
$$

$$
P_{1} - P_{a t m}
$$$$
=g\left(\rho_{m} h_{3}-\rho_{w} h_{1}-\rho_{0} h_{2}\right)
$$

$$
\text{P}_ \text{gage }=P_{1}-P_{\text { atm }}
$$
\(∴ \quad \text P_\text { 1gage }=g \left(\rho_m h_{3}-\rho_{w} h_{1}-\rho_{0} h_{2}\right) \)

$$
=9.81(13600 *(0.46)-1000(0.2)-850(0.3))
$$

$$
=56.9 \mathrm{kPa}
$$

$$
\text { Consider a U-tube whose arms are open to the atmosphere. Now water is poured into }
$$

$$
\text { the U-tube from one arm, and light oil }(p=790 \mathrm{kg} / \mathrm{m} 3) \text { from the other. One arm }
$$

$$
\text { contains } 70 \text { -cm-high water, while the other arm contains both fluids with an }
$$

$$
\text { oil-to-water height ratio of } 4 . \text { Determine the height of each fluid in that arm. }
$$

$$
\rho_{\text{ oil }}=790 \mathrm{kg} / \mathrm{m}^{3}
$$

$$
\rho_{w}=1000 \mathrm{kg} / \mathrm{m}^{3}
$$

$$
h_{a}=4 h_{w 2}=4(0.168)\left(\begin{array}{c}{0.673} {m}\end{array}\right)
$$

$$
P_{\text { bottom }}=P_\text { at m}+\rho_{w} g h w_{1}
$$

$$
P_\text {bottom = }
$$$$
P_\text { atm }
$$$$
+\rho_\text {w} ghw_ 2
$$$$
+\rho_{0} g h_{a}
$$

$$
\rho_ { w} ghw_1
$$$$
=\rho_{w} ghw_{2}+\rho_{a} g h a
$$

$$
\rho_{w} hw_{1}=\rho_{w} hw_{2}+\rho_{0} h_{a}
$$

\(÷ \rho_w\rightarrow h w_{1}= \frac{\rho_{w} hw_{2}}{\rho_w}+\frac{\rho_{0} h a}{\rho_{w}}\quad\longrightarrow h{w_{1}}=h w_{2}+\frac{\rho_{0} h_{4}}{\rho_{w}} \)

$$
\longrightarrow \quad 0.7=h w_{2}+
$$$$
\frac{790 \mathrm{ha}}{1000}
$$

$$
\longrightarrow \quad 0.7=h w_{2}+\frac{790hw_2 (4)}{1000}
$$

$$
\text { hw} _2=0.168
$$

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