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

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 W_u 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 Po P_o = 100 kPa, atmospheric temperature To T_o = 300 K.

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

Properties: The average temperature of the air is Tavg=(T1+T2)/2=(400 K+300 K)/2=350 K T_{avg} = (T_1 + T_2) / 2 = (400 \ K + 300 \ K) / 2 = 350 \ K , air has a gas constant of R = 0.2870 kJ / kgK (Appendix 1), and air at 350 K has specific heat at constant pressure cp c_p = 1.008 kJ / kgK (Appendix 4), specific heat of air at constant volume at 350 K cv c_v = 0.721 kJ / kgK (Appendix 4).

Governing equations:

Maximum useful work          Wu=−mϕ=−m[(u−uo)+Po(v−vo)–To(s−so)+V22+gz] 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]

Ideal gas equation                                              Pv = RT

Entropy change (ideal gas,                     Δs=cpln⁡T2T1−Rln⁡P2P1 \Delta s=c_p \ln \frac{T_2}{T_1} -R \ln\frac{P_2}{P_1}
constant specific heats)

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Specific internal energy contribution:

u–uo=cv(T – To)=0.721 kJ/kgK×(400 K – 300 K)=72.100 kJ/kg. u – u_o = c_v (T  –  T_o) = 0.721 \ kJ/kgK \times (400 \ K  –  300 \ K) = 72.100 \ kJ/kg.

Boundary work per unit mass contribution:

Po(v−vo)=PoR⟮TP–ToPo⟯=100 kPa×0.2870 kJ/kgK⟮400 K800 kPa–300 K100 kPa⟯=−71.750 kJ/kg P_o (v-v_o)=P_oR\left\lgroup\frac{T}{P} – \frac{T_o}{P_o} \right\rgroup =100 \ kPa \times 0.2870 \ kJ/kgK \left\lgroup\frac{400 \ K}{800 \ kPa} – \frac{300 \ K}{100 \ kPa} \right\rgroup = -71.750 \ kJ/kg

Specific entropy contribution: To(s−so)=To⟮cpln⁡TTo−Rln⁡PPo⟯ T_o (s-s_o)=T_o\left\lgroup c_p \ln \frac{T}{T_o}-R \ln \frac{P}{P_o} \right\rgroup

=300 K×⟮1.008 kJ/kgK×ln⁡400 K300 K−0.2870 kJ/kgK×ln⁡800 kPa100 kPa⟯=−92.045 kJ/kg.  = 300 \ K \times \left\lgroup1.008 \ kJ/kgK \times \ln \frac{400 \ K}{300 \ K} -0.2870 \ kJ/kgK \times \ln \frac{800 \ kPa}{100 \ kPa} \right\rgroup =-92.045 \ kJ/kg.  

Total useful work:

Wu=−mϕ=−500 kg[72.100 kJ/kg – 71.750 kJ/kg+92.045 kJ/kg]=−46198 kJ. W_u = -m\phi = -500 \ kg \left[72.100 \ kJ/kg  –  71.750 \ kJ/kg +92.045 \ kJ/kg\right] =-46198 \ kJ.

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

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