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• Notes

\begin{array}{l}{\text { A gas is contained in a vertical, frictionless piston-cylinder device. The piston has a mass of } 3.2 \mathrm{kg}} \\ {\text { and across-sectional area of } 35 \mathrm{cm} 2 \text { . }} \\ {\text { A compressed spring above the piston exerts a force of } 150 \mathrm{N} \text { on the piston. }} \\ {\text { If the atmospheric pressure is } 95 \mathrm{kPa} \text { , determine the pressure inside the cylinder }}\end{array}

$P(A)-P_{atm }(A)$$-F_{sprng}- w= 0$

$P(A)=P_{atm }(A)$$+F_{sprng}+ w$

$÷(A)$

$P=P_{a t m}+$$\left(\frac{F_{sprng}}{A}+\frac{w}{A}\right)$

$P=P_{a t m}+\frac{F_{sprn}+w}{A}$

$=9 5+$$\frac{150*3.2(9.81)}{35 * 10^{-4}}$$\left(\frac{1}{1000}\right)={147 \mathrm{kPa}}$

$\begin{array}{l}{\text { Consider the system shown in Fig. Pl- } 86 . \text { If a change of } 0.7 \mathrm{kPa} \text { in the pressure }} \\ {\text { of air causes the brine-mercury interface in the right column to drop }} \\ {\text { by } 5 \mathrm{mm} \text { in the brine level in the right column while the pressure in }} \\ {\text { the brine pipe remains constant, determine the ratio of } \mathrm{A} 2 ! \mathrm{AX}}\end{array}$

$\text { Before } g-$                $P_{A_{1}}+\rho_{w} g h{w}+$$\rho_{\mathrm{Hy}}gh_{hg1}-\rho_{br}gh_{b.1}=P_B$

$\text { After g- }$                $P_{A{2}}+\rho_{w} g h{w}$$+\rho_{H y} g h_{H y 2}$$-\rho_{br} g h_{b r_{2}}=P_{B}$

(After - Before)

$P_{A 2}-P_{A 1}+\rho_{H y} g \Delta h_{Hy}$$- \rho_{b r} g \Delta h_{br}=0$

$\frac{P{A1}-P_{ A2}}{\rho_{ w g}}=$$S G_{H g} \Delta h_{H g}-S G_{b r} \Delta h_{b r}=0$

$P_{A 2}-P_{A 1}$$=-0.7 \mathrm{kPs}=-700 \mathrm{N} / \mathrm{m}^{2}=-700{\mathrm{kg}}/ \mathrm{ms}^{2}$

$\Delta h_{br }=0.005 \mathrm{m} \longrightarrow$$\Delta h_{H g}=\Delta h_{H g \ right}+\Delta h_{H \ le f}$

$=\Delta h_{b r}+\Delta h_{b r} (A2/A1)$

$=\Delta h_{b r}\left(1+\frac{A{2}}{A{1}}\right)$

$\frac{700}{1000 * 9.81}=(13.96*(0.005)(1+\frac{A1}{A2})-(1.1)(0.005))$

${A{2} / A{1}}=0.134$