Question 5.9: The two basic types of steady-flow heat exchanger are charac...
The two basic types of steady-flow heat exchanger are characterized by their flow patterns: cocurrent and countercurrent. Temperature profiles for the two types are indicated in Fig. 5.6. In cocurrent flow, heat is transferred from a hot stream, flowing from left to right, to a cold stream flowing in the same direction, as indicated by arrows. In countercurrent flow, the cold stream, again flowing from left to right, receives heat from the hot stream flowing in the opposite direction.
The lines relate the temperatures of the hot and cold streams, T_H \text { and } T_C respectively, to \dot{Q}_C , the accumulated rate of heat addition to the cold stream as it progresses through the exchanger from the left end to an arbitrary downstream location. The following specifications apply to both cases:
\begin{array}{llll} T_{H_1}=400 K & T_{H_2}=350 K & T_{C_1}=300 K & \dot{n}_H=1 mol \cdot s ^{-1} \end{array}
The minimum temperature difference between the flowing streams is 10 K. Assume the ideal-gas state for both streams with C_P = (7/2)R . Find the lost work for both cases. Take T_σ = 300 K.
With the assumption of negligible kinetic- and potential-energy changes, with
\dot{W}_s=0 \text {, and with } \dot{Q}=0 (there is no heat exchange with the surroundings) the energy balance [Eq. (2.29)] can be written:
\Delta\left[\left(H+\frac{1}{2} u^2+z g\right) \dot{m}\right]_{ fs }=\dot{Q}+\dot{W}_s (2.29)
\dot{n}_H\left(\Delta H^{i g}\right)_H+\dot{n}_C\left(\Delta H^{i g}\right)_C=0
With constant molar heat capacity, this becomes:
\dot{n}_H C_P^{i g}\left(T_{H_2} – T_{H_1}\right)+\dot{n}_C C_P^{i g}\left(T_{C_2} – T_{C_1}\right)=0 (A)
The total rate of entropy change for the flowing streams is:
\Delta\left(S^{i g} \dot{n}\right)_{ fs }=\dot{n}_H\left(\Delta S^{i g}\right)_H+\dot{n}_C\left(\Delta S^{i g}\right)_C
By Eq. (5.10), with the assumption of negligible pressure change in the flowing streams,
\frac{\Delta S^{i g}}{R}=\int_{T_0}^T \frac{C_P^{i g}}{R} \frac{d T}{T}-\ln \frac{P}{P_0} (5.10)
\Delta\left(S^{i g} \dot{n}\right)_{ fs }=\dot{n}_H C_P^{i g}\left(\ln \frac{T_{H_2}}{T_{H_1}}+\frac{\dot{n}_C}{\dot{n}_H} \ln \frac{T_{C_2}}{T_{C_1}}\right) (B)
Finally, by Eq. (5.27), with \dot{Q}=0 ,
\dot{W}_{\text {lost }}=T_\sigma \Delta(S \dot{m})_{ fs } – \dot{Q} (5.27)
\dot{W}_{\text {lost }}=T_\sigma \Delta(S \dot{m})_{ fs } (C)
These equations apply to both cases.
- Case I: Cocurrent flow. By Eqs. (A), (B), and (C), respectively:
\frac{\dot{n}_C}{\dot{n}_H}=\frac{400-350}{340 – 300}=1.25
\Delta\left(S^{i g} \dot{n}\right)_{ fs }=(1)(7 / 2)(8.314)\left(\ln \frac{350}{400}+1.25 \ln \frac{340}{300}\right)=0.667 J \cdot K ^{-1} \cdot s ^{-1}
\dot{W}_{\text {lost }}=(300)(0.667)=200.1 J \cdot s ^{-1}
- Case II: Countercurrent flow. By Eqs. (A), (B), and (C), respectively:
\frac{\dot{n}_C}{\dot{n}_H}=\frac{400 – 350}{390 – 300}=0.5556
\Delta\left(S^{i g} \dot{n}\right)_{ fs }=(1)(7 / 2)(8.314)\left(\ln \frac{350}{400}+0.5556 \ln \frac{390}{300}\right)=0.356 J \cdot K ^{-1} \cdot s ^{-1}
\dot{W}_{\text {lost }}=(300)(0.356)=106.7 J \cdot s ^{-1}
Although the total rate of heat transfer is the same for both exchangers, the temperature rise of the cold stream in countercurrent flow is more than twice that for cocurrent flow. Thus, its entropy increase per unit mass is larger than for the cocurrent case. However, its flow rate is less than half that of the cold stream in cocurrent flow, so that the total entropy increase of the cold stream is less for countercurrent flow. From a thermodynamic point of view, the countercurrent case is much more efficient. Because \Delta\left(S^{i g} \dot{n}\right)_{ fs }=\dot{S}_G , both the rate of entropy generation and the lost work for the countercurrent case are only about half the value for the cocurrent case. Greater efficiency in the countercurrent case would be anticipated based on Fig. 5.6, which shows that heat is transferred across a smaller temperature difference (less irreversibly) in the countercurrent case.