Problem: A compressed air tank contains 500 kg of air at 800 kPa and 400 K. How much work can be obtained from this if the atmosphere is at 100 kPa and 300 K? Find: Maximum work Wu that can be obtained from the compressed air. Known: Mass of air m = 500 kg, air pressure P = 800 kPa, air

Question

Problem: A compressed air tank contains 500 kg of air at 800 kPa and 400 K. How much work can be obtained from this if the atmosphere is at 100 kPa and 300 K?

Find: Maximum work [latex] W_u [/latex] that can be obtained from the compressed air.

Known: Mass of air m = 500 kg, air pressure P = 800 kPa, air temperature T = 400 K, atmospheric pressure [latex] P_o [/latex] = 100 kPa, atmospheric temperature [latex] T_o [/latex] = 300 K.

Assumptions: Air is an ideal gas with constant specific heats.

Properties: The average temperature of the air is [latex] T_{avg} = (T_1 + T_2) / 2 = (400 \ K + 300 \ K) / 2 = 350 \ K [/latex], air has a gas constant of R = 0.2870 kJ / kgK (Appendix 1), and air at 350 K has specific heat at constant pressure [latex] c_p [/latex] = 1.008 kJ / kgK (Appendix 4), specific heat of air at constant volume at 350 K [latex] c_v [/latex] = 0.721 kJ / kgK (Appendix 4).

Governing equations:

Maximum useful work [latex] W_u=-m\phi =-m\left[(u-u_o) + P_o(v-v_o) - T_o(s-s_o) + \frac{\pmb{V}^2}{2}+gz \right] [/latex]

Ideal gas equation Pv = RT

Entropy change (ideal gas, [latex] \Delta s=c_p \ln \frac{T_2}{T_1} -R \ln\frac{P_2}{P_1} [/latex]
constant specific heats)

Accepted Answer

Specific internal energy contribution:

[latex] u - u_o = c_v (T - T_o) = 0.721 \ kJ/kgK imes (400 \ K - 300 \ K) = 72.100 \ kJ/kg. [/latex]

Boundary work per unit mass contribution:

[latex] P_o (v-v_o)=P_oR\left\lgroup\frac{T}{P} - \frac{T_o}{P_o} ight group =100 \ kPa imes 0.2870 \ kJ/kgK \left\lgroup\frac{400 \ K}{800 \ kPa} - \frac{300 \ K}{100 \ kPa} ight group = -71.750 \ kJ/kg [/latex]

Specific entropy contribution: [latex] T_o (s-s_o)=T_o\left\lgroup c_p \ln \frac{T}{T_o}-R \ln \frac{P}{P_o} ight group [/latex]

[latex] = 300 \ K imes \left\lgroup1.008 \ kJ/kgK imes \ln \frac{400 \ K}{300 \ K} -0.2870 \ kJ/kgK imes \ln \frac{800 \ kPa}{100 \ kPa} ight group =-92.045 \ kJ/kg. [/latex]

Total useful work:

[latex] W_u = -m\phi = -500 \ kg \left[72.100 \ kJ/kg - 71.750 \ kJ/kg +92.045 \ kJ/kg ight] =-46198 \ kJ. [/latex]

Answer: The maximum amount of useful work that can be obtained from the compressed air is 46.2 MJ.

Question 6.15: Problem: A compressed air tank contains 500 kg of air at 800...

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