Entropy Change in Heat Exchange
An ideal gas CP = 7R/2 is heated in a steady flow heat exchanger from 70oC to 190oC, by another stream of the same ideal gas entering at 320oC. The flow rates of the two streams are the same.
(i) Calculate DS of the two gas streams for countercurrent flow
(ii) What is DStotal?
(iii) If heating stream enters at 200oC, what is DStotal?
Basis: 1 mole of gas
Entropy change for the constant pressure heating/cooling process:
DS = nCP ln (T2/T1)
Where T1 and T2 are respectively initial and final temperatures.
(i) Entropy change of the cold gas stream:
T1 = 273 + 70 = 343 K
T2 = 273 + 190 = 463 K
n = 1
Therefore, DS = (7R/2) x ln (463/343) = 1.05 R
Entropy change of the Hot gas stream:
T1 = 273 + 320 = 593 K
Since the flow rates of the two streams are same,
DThotstream = DTcoldstream = 190 - 70 = 120 K
T2 = T1 - 120 = 593 - 120 = 473 K
n = 1
Therefore, DS = (7R/2) x ln (473/593) = -0.79 R
(ii) DStotal = 1.05 R + (-0.79R) = 0.26 R
(iii) Heating stream inlet temperature = 200oC
Therefore, outlet temperature of heating stream = 200 - 120 = 80oC
DS for this stream = (7R/2) x ln (353/473) = -1.024 R
DStotal = 1.05 R + (-1.024 R) = 0.026 R