Refer to Example Problem 1.39, where for a 69-m^3 helium tank compressed under a pressure of 800 kN/m^2 abs and at a temperature of 10^◦ C, the density of the helium in the tank was determined to be 1.36 kg/m^3 , and the weight of the helium in the tank was determined to be 920.804 N. The helium in

Question

Refer to Example Problem 1.39, where for a 69- [latex]m^{3} [/latex] helium tank compressed under a pressure of 800 kN/[latex]m^{2} [/latex] abs and at a temperature of [latex]10^{\circ }C [/latex], the density of the helium in the tank was determined to be 1.36 kg/[latex]m^{3} [/latex], and the weight of the helium in the tank was determined to be 920.804 N. The helium in the tank is further compressed under isentropic (frictionless and no heat exchange with the surroundings) to a volume of 35 [latex]m^{3} [/latex]. (a) Determine the resulting density of the helium in the tank. (b) Determine the resulting pressure of the helium in the tank. (c) Determine the resulting bulk modulus of elasticity of the helium in the tank.

Accepted Answer

(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 [latex] ho =\frac{M}{V} [/latex] and thus the weight of the helium in Equation 1.44 [latex]\gamma =\frac{W}{V}=\frac{Mg}{V}= ho g[/latex] remains a constant as follows:

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

[latex]g:=9.81 \frac{m }{sec^{2}} [/latex] [latex] ho _{1} :=1.36\frac{kg}{m^{3}} [/latex] [latex]W_{1} :=920.804 N^{3} [/latex]

[latex]V_{2} :=35 m^{3} [/latex] [latex]W_{2}=W_{1} :=920.804 N^{3} [/latex] [latex]\gamma _{2} :=\frac{W_{2}}{V_{2}} =26.309\frac{N}{m^{3}} [/latex]
[latex] ho _{2} := \frac{\gamma _{2} }{g} =2.682\frac{kg}{m^{3}} [/latex]

(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 [latex]\frac{p}{ ho ^{k} } =C[/latex] as follows:

[latex]K:=1.66 [/latex] [latex]\frac{P_{1}}{(1.36)^{k}\frac{kg}{m^{3}} } =4.802 imes 10^{5} \frac{m^{2}}{s^{2}} [/latex]

Guess value: [latex]P_{2} =1\frac{N}{m^{2}} [/latex]

Given

[latex]\frac{P_{1}}{(1.36)^{k}\frac{kg}{m^{3}} }=\frac{P_{}}{(2.682)^{k}\frac{kg}{m^{3}} } [/latex]
[latex]P_{2}: =Find ( ho _{2} )=2.47 imes 10^{6} \frac{N}{m^{2}} [/latex]

(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 [latex]E_{\upsilon } =\frac{dp}{-\frac{dV}{V} } =\frac{dp}{\frac{d ho }{ ho } } = ho \frac{dp}{d ho } = ho \frac{d(C ho ^{k} )}{d ho } = ho KC ho ^{k-1} \frac{d ho }{d ho} =kC ho^{k} =kp[/latex]), as follows:

[latex] E_{v} :=kp_{2} =4.1 imes 10^{6} \frac{N}{m^{2}} [/latex]

Table A.5
Physical Properties for Some Common Gases at Standard Sea-Level Atmospheric Pressure at Room Temperature [latex](68^{\circ } or 20^{\circ }C )[/latex]
Gas at [latex]68^{\circ }[/latex]	Chemical Formula	Molar Mass (m) slug=slug- mol	Density (ρ) [latex]slug/ft^{3} [/latex]	Absolute (Dynamic) Viscosity (μ) [latex]10^{-6} Ib-sec/ft^{2} [/latex]	Gas Constant (R) [latex]ft-Ib/(slug-^{\circ }R )=ft^{2}/(sec^{2} -^{\circ }R )[/latex]	Specific Heat		Specific Heat Ratio, [latex]K=C_{ ho }/C_{\upsilon }[/latex]
						[latex]C_{ ho }[/latex]	[latex]C_{\upsilon } [/latex]
						[latex]ft-Ib/(slug-^{\circ }R )=ft^{2}/(sec^{2} -^{\circ }R )[/latex]
Air		28.960	0.002310	0.376	1715	6000	4285	1.40
Carbon dioxide	[latex]CO_{2} [/latex]	44.010	0.003540	0.310	1123	5132	4009	1.28
Carbon monoxide	CO	28.010	0.002260	0.380	1778	6218	4440	1.40
Helium	He	4.003	0.000323	0.411	12.420	31.230	31.230	1.66
Hydrogen	[latex]H_{2}[/latex]	2.016	0.000162	0.189	24.680	86.390	86.390	1.40
Methane	[latex]CH_{2}[/latex]	16.040	0.001290	0.280	3100	13.400	13.400	1.30
Nitrogen	[latex]N_{2}[/latex]	28.020	0.002260	0.368	1773	6210	4437	1.40
Oxygen	[latex]O_{2}[/latex]	32.000	0.002580	0.418	1554	5437	3883	1.40
Water vapor	[latex]H_{2}O[/latex]	18.020	0.001450	0.212	2760	11.110	8350	1.33
at [latex]20^{\circ } C[/latex]		kg/kg-mol	[latex]kg/m^{3} [/latex]	[latex]10^{-6} N-sec/m^{2} [/latex]	[latex]N-m/(kg-^{\circ}K )=m^{2} /(sec^{2}-^{\circ}K )[/latex]	[latex]N-m/(kg-^{\circ}K )=m^{2} /(sec^{2}-^{\circ}K )[/latex]
Air		28.960	1.2050	18.0	287	1003	716	1.40
Carbon dioxide	[latex]CO_{2} [/latex]	44.010	1.8400	14.8	188	858	670	1.28
Carbon monoxide	CO	28.010	1.1600	18.2	297	1040	743	1.40
Helium	He	4.003	0.1660	19.7	2077	5220	3143	1.66
Hydrogen	[latex]H_{2}[/latex]	2.016	0.0839	9.0	4120	14.450	10.330	1.40
Methane	[latex]CH_{2}[/latex]	16.040	0.6680	13.4	520	2250	1730	1.30
Nitrogen	[latex]N_{2}[/latex]	28.020	1.1600	17.6	297	1040	743	1.40
Oxygen	[latex]O_{2}[/latex]	32.000	1.3300	20.0	260	909	649	1.40
Water vapor	[latex]H_{2}O[/latex]	18.020	0.7470	10.1	462	1862	1400	1.33

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

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