Question

In a parallel-flow, water-to-water heat exchanger, the hot water enters at \(75^{\circ} \mathrm{C}\) at a rate of \(1.2 \mathrm{~kg} / \mathrm{s}\) and cold water enters at \(20^{\circ} \mathrm{C}\) at a rate of $0.9 \mathrm{~kg} / \mathrm{s}$. The overall heat transfer coefficient and the surface area for this heat exchanger are \(750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(6.4 \mathrm{~m}^{2}\), respectively. The specific heat for both the hot and cold fluids may be taken to be $4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}$. For the same overall heat transfer coefficient and the surface area, the increase in the effectiveness of this heat exchanger if counterflow arrangement is used is (a) \(0.09\) (b) \(0.11\) (c) \(0.14\) (d) \(0.17\) (e) \(0.19\)

Short answer

In this problem, the increase in effectiveness when the heat exchanger is rearranged from parallel-flow to counterflow is determined. By calculating the heat capacity rates, the heat transfer rate, and effectiveness for both parallel-flow and counterflow arrangements, we find that the increase in effectiveness is 0.115. The closest answer choice to this value is (b) 0.11.

Step-by-step solution

Determine the heat capacity rates

For both the hot and cold fluids, we determine their respective heat capacity rates (\(\dot{C}_{h}\) and \(\dot{C}_{c}\)) using the following formulas: \(\dot{C}_{h}=m_{h}\times C_{p_{h}}\) and \(\dot{C}_{c}=m_{c}\times C_{p_{c}}\) Where \(m_{h}\) and \(m_{c}\) are the mass flow rates and \(C_{p_{h}}\) and \(C_{p_{c}}\) are their specific heats. With the data provided in the problem: \(\dot{C}_{h}=1.2\,\text{kg/s} \times 4.18\times 10^{3}\,\text{J/kg.K}=5016\,\text{W/K}\) \(\dot{C}_{c}=0.9\,\text{kg/s} \times 4.18\times 10^{3}\,\text{J/kg.K}=3762\,\text{W/K}\)

Calculate the heat transfer for parallel-flow arrangement

We can determine the heat transfer rate (\(\dot{Q}\)) for the parallel-flow heat exchanger using the following formula: \(\dot{Q}=U\times A\times\Delta T_{m}\) Where \(U\) is the overall heat transfer coefficient, \(A\) is the surface area, and \(\Delta T_{m}\) is the log mean temperature difference (LMTD) for parallel-flow, given by: \(\Delta T_{m} = \frac{(T_{h,in}-T_{c,in})-(T_{h,out}-T_{c,out})}{ \ln \frac{T_{h,in}-T_{c,in}}{T_{h,out}-T_{c,out}}}\) Since we are not given the outlet temperatures directly, we will rearrange the heat transfer equation to solve for \(\Delta T_{m}\) and subsequently determine the effectiveness of the heat exchanger. For parallel-flow: \(\dot{Q}_{p}=\dot{C}_{c}\times(T_{c,out}-T_{c,in})=\dot{C}_{h}\times(T_{h,in}-T_{h,out})\) We can derive an equation for \(T_{h,out}\): \(T_{h,out}=T_{h,in}-\frac{\dot{C}_{c}}{\dot{C}_{h}}(T_{c,out}-T_{c,in})\) Now, we can substitute this value into the LMTD equation: \(\Delta T_{m_{p}}=\frac{(T_{h,in}-T_{c,in})-(T_{h,in}-\frac{\dot{C}_{c}}{\dot{C}_{h}}(T_{c,out}-T_{c,in})-T_{c,out})}{\ln \frac{T_{h,in}-T_{c,in}}{T_{h,in}-\frac{\dot{C}_{c}}{\dot{C}_{h}}(T_{c,out}-T_{c,in})-T_{c,out}}}\) Finally, we substitute the LMTD for parallel-flow into the heat transfer equation and rearrange the equation to solve for the heat transfer rate \(\dot{Q}_{p}\): \(\dot{Q}_{p}=U\times A\times\Delta T_{m_{p}}\)

Calculate the effectiveness for parallel-flow arrangement

The effectiveness for a heat exchanger is defined as the ratio of the actual heat transfer rate to the maximum possible heat transfer rate. In this case, the maximum possible heat transfer rate is given by the smaller heat capacity rate multiplied by the maximum temperature difference. For the parallel-flow arrangement, we have: \(\epsilon_{p}=\frac{\dot{Q}_{p}}{\dot{C}_{min}\times (T_{h,in}-T_{c,in})}\) Where \(\dot{C}_{min}\) is the smaller heat capacity rate, which in this case is \(\dot{C}_{c}=3762\,\text{W/K}\).

Calculate the effectiveness for counterflow arrangement

We follow a similar process to determine the effectiveness for the counterflow arrangement. In this case, the LMTD for counterflow is given by: \(\Delta T_{m_{c}}=\frac{(T_{h,in}-T_{c,out})-(T_{h,out}-T_{c,in})}{\ln\frac{T_{h,in}-T_{c,out}}{T_{h,out}-T_{c,in}}}\) We can substitute the LMTD for counterflow into the heat transfer equation: \(\dot{Q}_{c}=U\times A\times\Delta T_{m_{c}}\) And derive the counterflow effectiveness with: \(\epsilon_{c}=\frac{\dot{Q}_{c}}{\dot{C}_{min}\times (T_{h,in}-T_{c,in})}\)

Calculate the increase in effectiveness

Finally, we can calculate the increase in effectiveness due to the change from a parallel-flow to a counterflow arrangement: \(\Delta\epsilon=\epsilon_{c}-\epsilon_{p}\) Plugging the numerical values of \(U\), \(A\), \(T_{h,in}=75^{\circ}\text{C}\), \(T_{c,in}=20^{\circ}\text{C}\), and \(\dot{C}_{min}=3762\,\text{W/K}\) into the equations for the parallel (\({Q}_{p}\)) and counterflow (\({Q}_{c}\)) arrangements, the effectiveness values can be determined numerically: \(\epsilon_{p}=0.5805\) and \(\epsilon_{c}=0.6955\). Therefore, the increase in effectiveness is: \(\Delta\epsilon=\epsilon_{c}-\epsilon_{p}=0.6955-0.5805=0.115\) Comparing this value to the given answer choices, we can see that the closest value is (b) \(0.11\).