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 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 P_o = 100 kPa, atmospheric temperature T_o = 300 K.
Assumptions: Air is an ideal gas with constant specific heats.
Properties: The average temperature of the air is 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 c_p = 1.008 kJ / kgK (Appendix 4), specific heat of air at constant volume at 350 K c_v = 0.721 kJ / kgK (Appendix 4).
Governing equations:
Ideal gas equation Pv = RT
Entropy change (ideal gas, \Delta s=c_p \ln \frac{T_2}{T_1} -R \ln\frac{P_2}{P_1}
constant specific heats)
Specific internal energy contribution:
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:
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/kgSpecific entropy contribution: 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 \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:
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.