Question 6.4: Problem: A cylinder contains 200 l of air at 150 kPa and 32 ...
Problem: A cylinder contains 200 l of air at 150 kPa and 32 °C. The pressure of the gas is constant while 70 kJ of heat are added to the gas. Find its entropy change (a) assuming constant specific heats and (b) using air tables.
Find: Entropy increase of gas ∆S due to heating.
Known: Initial volume V_1 = 200 l = 0.2 m³ , initial pressure P_1 = 150 kPa, initial temperature T_1 = 32 °C = 305.15 K, final pressure P_2 = P_1 , heat added Q_{12} = 70 kJ.
Assumptions: Air is an ideal gas with constant specific heats.
Governing Equations:
First law (constant pressure) Q_{12}=m(h_2 -h_1)
Entropy change (ideal gas, constant specific heats) \Delta s =s_2 – s_1= c_p \ln\frac{T_2}{T_1} -R \ln\frac{P_2}{P_1}
Properties: Air has gas constant R = 0.287 kJ / kgK (Appendix 1) and at 305 K (interpolation) air specific heat c_p = 1.0053 kJ / kgK (Appendix 4).
Mass of gas,
m=\frac{P_1V_1}{RT_1} =\frac{150 \ kPa \times 0.2 \ m^3}{0.2870 \ kJ.kgK \times 305.15 \ K} =0.34255 \ kg.
(a) Assuming constant specific heat:
Q_{12}=m(h_2 – h_1)= m c_p(T_2 – T_1)
70 \ kJ = 0.34255 \ kg \times 1.0053 \ kJ/kgK(T_2 – 305.15 \ K)
T_2= 508.69 \ K
The change in entropy is
\Delta S =m \Delta s= m \left\lgroup c_p \ln\frac{T_2}{T_1} -R \ln\frac{P_2}{\underbrace{P_1}_{=0} }\right\rgroup =m c_p \ln \frac{T_2}{T_1} .
Since P_2 = P_1 , the second term is zero, then
\Delta S =m \Delta s= 0.34255 \ kg \left\lgroup1.0053 \ kJ/kgK \times \ln \frac{508.69 \ K}{305.15 \ K} \right\rgroup =0.17598 \ kJ/K.
(b) Using gas tables:
From the tables, specific enthalpy h_1 (305 K) = 305.22 kJ / kg, specific entropy s_1 (305 K) = 1.71865 kJ / kgK. Using the first law,
h_2=\frac{Q}{m} +h_1=509.57 \ kJ/kg.
Looking at the air tables (Appendix 7) the value of h_2 lies between h(500 K) and h(510 K). The value of s_2 must also lie in this temperature range.
| T(K) | h (kJ/kg) | p_r | u (kJ/kg) | v_r | s^\circ (kJ/kgK) |
| 500 | 503.02 | 8.411 | 359.49 | 170.6 | 2.21952 |
| T_2 | 509.57 | s_2 | |||
| 510 | 513.32 | 9.031 | 366.92 | 162.1 | 2.23993 |
We can find the exact value of s_2 by linear interpolation (see Figure E6.4):
\frac{s_2 – 2.21952}{2.23993 – 2.21952} =\frac{509.57 – 503.20}{513.32 – 503.20} ,
s_2 = 2.2324 \ kJ/kgK.
\Delta S =m(s_2 – s_1)=0.34255 \ kg \times (2.2324 \ kJ/kgK – 1.71865 \ kJ/kgK)=0.17597 \ kJ/K.
Answer: The increase in entropy (a) assuming constant specific heat is 0.178 kJ / K and (b) using air tables is 0.176 kJ / K.
