Question 1.43: Refer to Example Problem 1.39, where for a 69-m^3 helium tan...

(a) To determine the resulting density of the helium in the tank, one must note that the mass of the helium in Equation 1.33 \rho =\frac{M}{V} and thus the weight of the helium in Equation 1.44 \gamma =\frac{W}{V}=\frac{Mg}{V}=\rho g remains a constant as follows:

N:=kg\frac{m }{sec^{2}}                P_{1} :=800\frac{kn}{m^{2}}                V_{1} :=69m^{3}

g:=9.81 \frac{m }{sec^{2}}                \rho _{1} :=1.36\frac{kg}{m^{3}}                W_{1} :=920.804 N^{3}

V_{2} :=35 m^{3}                W_{2}=W_{1} :=920.804 N^{3}                \gamma _{2} :=\frac{W_{2}}{V_{2}} =26.309\frac{N}{m^{3}}
\rho _{2} := \frac{\gamma _{2} }{g} =2.682\frac{kg}{m^{3}}

(b) Given compression under isentropic (frictionless and no heat exchange with the surroundings), to determine the resulting pressure of the helium in the tank, one may apply Equation 1.156 \frac{p}{\rho ^{k} } =C as follows:

K:=1.66                            \frac{P_{1}}{(1.36)^{k}\frac{kg}{m^{3}} } =4.802\times 10^{5} \frac{m^{2}}{s^{2}}

Guess value:                    P_{2} =1\frac{N}{m^{2}}

Given

\frac{P_{1}}{(1.36)^{k}\frac{kg}{m^{3}} }=\frac{P_{}}{(2.682)^{k}\frac{kg}{m^{3}} }
P_{2}: =Find (\rho _{2} )=2.47\times 10^{6} \frac{N}{m^{2}}

(c) Given compression under isentropic (friction less and no heat exchange with the surroundings), the resulting bulk modulus of elasticity of the helium in the tank is equal to the product of the absolute pressure of the helium and the specific heat ratio, which is read from Table A.5 in Appendix A (Equation 1.158 E_{\upsilon } =\frac{dp}{-\frac{dV}{V} } =\frac{dp}{\frac{d\rho }{\rho } } =\rho \frac{dp}{d\rho } =\rho \frac{d(C\rho ^{k} )}{d\rho } =\rho KC\rho ^{k-1} \frac{d\rho }{d\rho} =kC\rho^{k} =kp), as follows: