Question 4.15: Problem: Air is heated from 300 to 500 K. Find the change in...
Problem: Air is heated from 300 to 500 K. Find the change in specific enthalpy using (a) c_p evaluated at 25 °C, (b) c_p evaluated at T_{avg} , (c) a function c_p (T), (d) ideal gas tables.
Find: Change in specific enthalpy ∆h in four ways.
Known: Initial temperature T_1 = 300 K, final temperature T_2 = 500 K.
(a) For air at 25 °C, c_p = 1.004 kJ / kgK (Appendix 1):
\Delta h=c_p(T_2-T_1)=1.004 \ kJ/kgK \times (500 \ K-300 \ K)=200.800 \ kJ/kg
(b) The average temperature during the heating is T_{avg}=\frac{(T_2+T_1)}{2} =\frac{500 \ K + 300 \ K}{2} =400 \ K.
For air at 400 K, c_p = 1.013 kJ / kgK (Appendix 4):
\Delta h=c_p(T_2-T_1)=1.013 \ kJ/kgK (500 \ K-300 \ K)=202.600 \ kJ/kg
(c) From Appendix 5, for air
\overline{c}_p =Mc_p=28.11 + 0.1967 \times 10^{-2} \ T+0.4802 \times 10^{-5} \ T^2-1.966 \times 10^{-9} \ T^3:
\Delta \overline{h} =\overline{h} _2-\overline{h} _1=\int\limits_{T_1}^{T_2}{} \overline{c} _p(T)dT=\int\limits_{300 \ K}^{500 \ K}{} (28.11 + 0.1967 \times 10^{-2} \ T + 0.4802 \times 10^{-5} \ T^2-1.966 \times 10^{-9} \ T^3)dT
\Delta \overline{h} =\left[28.11 \ T + 0.9835 \times 10^{-3} \ T^2 + 0.1601 \times 10^{-5} T^3-0.4915 \times 10^{-9} \ T^4\right]_{300 \ K}^{500 \ K} \\ =5933.58 \ kJ/kmolK
\Delta h = \frac{\Delta\overline{h} }{M} =\frac{5933.58 \ kJ/kmolK}{28.97 \ kg/kmol} =204.818 \ kJ/kgK
(d) From Appendix 7, h(300 K) = 300.19 kJ / kgK and h(500 K) = 503.02 kJ / kgK:
∆h = h ( 500 K ) − h( 300 K )= 503.02 kJ / kgK – 300.19 kJ / kgK = 202.830 kJ / kgK
Answer: The changes in specific enthalpy are: (a) 200.8 kJ / kg, (b) 202.6 kJ / kg, (c) 204.8 kJ / kg, (d) 202.8 kJ / kg. The difference in the specific enthalpy change calculated using these different methods is small. Often, assuming constant specific heat is a reasonable assumption. If ideal gas tables are available they are the easiest way of calculating enthalpy changes.