(a) The temperature distributions along the heat exchanger, for the hot and cold streams, are shown in Figure (a). The results are for the quantitative analysis performed below. The thermal circuit diagram is shown in Figure (b).
(b) The definition of NTU for the two-stream heat exchange is given by NTU\equiv \frac{1}{R_{\sum}(\dot{M}c_{p})_{min}} , 0\leq NTU\lt \infty , i.e.,
NTU\equiv \frac{1}{R_{\sum}(\dot{M}c_{p})_{min}}
To determine (\dot{M}c_{p})_{min} , we need to compare the two streams, i.e.,
(\dot{M}c_{p})_{c}=\dot{M}_{c}c_{p,c} liquid water
(\dot{M}c_{p})_{h}=\dot{M}_{c}c_{p,h} liquid Dowtherm J.
From Table, we have
liquid water c_{p,c} = 4,220 J/kg-^{\circ }C Table
liquid Dowtherm J c_{p,h} = 2,400 J/kg-^{\circ }C Table .
Then
(\dot{M}c_{p})_{c}=0.5(kg/s) × 4,220(J/kg-^{\circ }C) = 2,110 W/^{\circ }C
(\dot{M}c_{p})_{h}=0.25(kg/s) × 2,400(J/kg-^{\circ }C) = 600 W/^{\circ }C.
Then
(\dot{M}c_{p})_{min}=(\dot{M}c_{p})_{h}= 600 W/^{\circ }C.
Next
NTU=\frac{1}{0.001(^{\circ }C/W) × 600(W/^{\circ }C)} = 1.667.
(c) The ratio of the heat capacitances C_{r} is defined by C_{r}\equiv \frac{(\dot{M}c_{p})_{min}}{(\dot{M}c_{p})_{max}} , 0 \leq C_{r} \leq 1, i.e.,
C_{r}\equiv \frac{(\dot{M}c_{p})_{min}}{(\dot{M}c_{p})_{max}} =\frac{(\dot{M}c_{p})_{h}}{(\dot{M}c_{p})_{c}} =\frac{600(W/^{\circ }C)}{2,110(W/^{\circ }C)} = 0.2844.
(d) The \epsilon_{he}-NTU relation for the counterflow arrangement is given in Table,i.e.,
\epsilon _{he}=\frac{1-e^{-NTU(1-C_{r})}}{1-C_{r}e^{-NTU(1-C_{r})}} -\frac{1 − e^{−1.667}(1 − 0.2844)}{ − 0.2844e^{−1.667(1−0.2844)}}
=\frac{1 − e^{−1.193}}{1 − 0.2844e^{−1.193}} = 0.7625
(e) The hot-stream exit temperature \left\langle T_{f,h}\right\rangle _{0} is determined from the definition of \epsilon _{he}, for the counterflow arrangement, given by \epsilon _{he}\equiv \frac{\Delta \left\langle T_{f}\right\rangle\mid_{(\dot{M}c_{p})_{min}} }{\Delta T_{max}}
\epsilon _{he}=\frac{\left\langle T_{f,h}\right\rangle_{L}-\left\langle T_{f,h}\right\rangle_{0} }{\left\langle T_{f,h}\right\rangle_{L}-\left\langle T_{f,c}\right\rangle_{0}}
Solving this for \left\langle T_{f,h}\right\rangle_{0}, we have
\left\langle T_{f,h}\right\rangle_{0}=\left\langle T_{f,h}\right\rangle_{L}-\epsilon _{he}(\left\langle T_{f,h}\right\rangle_{L}-\left\langle T_{f,c}\right\rangle_{0})
= 120(^{\circ }C) − 0.7625(120 − 20)(^{\circ }C) = 43.75^{\circ }C.
(f) The heat transfer rate \left\langle Q_{u}\right\rangle _{L-0} is determined from \left\langle Q_{u}\right\rangle _{L-0}=(\dot{M}c_{p})_{h}(\left\langle T_{f,h}\right\rangle_{0}-\left\langle T_{f,h}\right\rangle_{L} )=-(\dot{M}c_{p})_{h}\Delta T_{h}, i.e.,
\left\langle Q_{u}\right\rangle _{L-0}=-(\dot{M}c_{p})_{h}(\left\langle T_{f,h}\right\rangle_{L}-\left\langle T_{f,h}\right\rangle_{0} )=(\dot{M}c_{p})_{c}(\left\langle T_{f,c}\right\rangle_{L}-\left\langle T_{f,c}\right\rangle_{0})
Solving for \left\langle T_{f,c}\right\rangle_{L} , we have
\left\langle T_{f,c}\right\rangle_{L}=\left\langle T_{f,c}\right\rangle_{0}-\frac{(\dot{M}c_{p})_{h}(\left\langle T_{f,h}\right\rangle_{L}-\left\langle T_{f,h}\right\rangle_{0})}{(\dot{M}c_{p})_{c}} =\left\langle T_{f,c}\right\rangle_{0}-\frac{\left\langle Q_{u}\right\rangle_{L-0} }{(\dot{M}c_{p})_{c}}
=20(^{\circ }C) −\frac{600(W/^{\circ }C) × 76.25(^{\circ }C)}{2,110(W/^{\circ }C)} =20(^{\circ }C) −\frac{4.575 × 10^{4}(W)}{2,110(W/^{\circ }C)}
= 41.79^{\circ }C.
The inlet and exit temperatures are shown in Figure (a)
(g) The average heat exchanger convection resistance \left\langle R_{u}\right\rangle _{L} is given by \left\langle R_{u}\right\rangle _{L}\equiv \frac{1}{(\dot{M}c_{p})_{min}\epsilon _{he}} ,\epsilon _{he}=\epsilon _{he}(NTU,C_{r},heat exchanger type),
i.e.,
\left\langle R_{u}\right\rangle _{L}= \frac{1}{(\dot{M}c_{p})_{min}\epsilon _{he}}
=\frac{1}{600(W/^{\circ }C) × 0.7625} = 0.002186^{\circ }C/W
