Question

A well-insulated heat exchanger is to heat water \(\left(c_{p}=4.18 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) from 25 to \(60^{\circ} \mathrm{C}\) at a rate of \(0.50 \mathrm{kg} / \mathrm{s}\) The heating is to be accomplished by geothermal water \(\left(c_{p}=4.31 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) available at \(140^{\circ} \mathrm{C}\) at a mass flow rate of \(0.75 \mathrm{kg} / \mathrm{s} .\) Determine \((a)\) the rate of heat transfer and \((b)\) the rate of entropy generation in the heat exchanger.

Short answer

The rate of heat transfer is approximately 73.15 kJ/s, and the rate of entropy generation is approximately -0.137 kJ/s · K.

Step-by-step solution

Calculate the temperature change for both fluids

First, let's find the temperature changes for both fluids. For water, the initial temperature is 25°C and the final temperature is 60°C, so the temperature change is: \(\Delta T_{water} = T_{2_{water}} - T_{1_{water}} = 60 - 25 = 35^{\circ} C\) For geothermal water, the initial temperature is 140°C, and we need to find the final temperature (\(T_{2_{geo}}\)) by using the heat transfer equation, which we will do in the next step.

Calculate the rate of heat transfer

Now let's find the rate of heat transfer (\(\dot{Q}\)) using the mass flow rate, specific heat, and temperature change of both fluids: \(\dot{Q} = m_{water} \cdot c_{p_{water}} \cdot \Delta T_{water} = m_{geo} \cdot c_{p_{geo}} \cdot \Delta T_{geo} \) Substitute the given values: \(\dot{Q} = (0.50 \mathrm{kg/s}) \cdot (4.18 \mathrm{kJ/kg \cdot ^{\circ}C}) \cdot (35 ^{\circ}C) = (0.75 \mathrm{kg/s}) \cdot (4.31 \mathrm{kJ/kg \cdot ^{\circ}C}) \cdot (140 - T_{2_{geo}})\) Solve for \(T_{2_{geo}}\): \(73.15 = 107.25 - 3.2375 \cdot T_{2_{geo}}\) \(T_{2_{geo}} \approx 10.56^{\circ} C\) Now calculate the rate of heat transfer: \(\dot{Q} \approx (0.50 \mathrm{kg/s}) \cdot (4.18 \mathrm{kJ/kg \cdot^{\circ} C}) \cdot (35 ^{\circ}C) \approx 73.15\ \mathrm{kJ/s}\) So, the rate of heat transfer is \(\approx 73.15\ \mathrm{kJ/s}\).

Calculate the rate of entropy generation

First, let's find the entropy change for water and geothermal water: \(\Delta S_{water} = m_{water} \cdot c_{p_{water}} \cdot ln\left(\frac{T_{2_{water}}}{T_{1_{water}}}\right) \approx (0.50 \mathrm{kg/s}) \cdot (4.18 \mathrm{kJ/kg \cdot ^{\circ}C}) \cdot ln\left(\frac{60}{25}\right)\) \(\Delta S_{geo} = m_{geo} \cdot c_{p_{geo}} \cdot ln\left(\frac{T_{2_{geo}}}{T_{1_{geo}}}\right) \approx (0.75 \mathrm{kg/s}) \cdot (4.31 \mathrm{kJ/kg \cdot^{\circ} C}) \cdot ln\left(\frac{10.56}{140}\right)\) Now, compute the rate of entropy generation: \(\dot{S}_g = \Delta S_{water} + \Delta S_{geo} \approx (0.50 \mathrm{kg/s}) \cdot (4.18 \mathrm{kJ/kg \cdot^{\circ} C}) \cdot ln\left(\frac{60}{25}\right) + (0.75 \mathrm{kg/s}) \cdot (4.31 \mathrm{kJ/kg \cdot^{\circ} C}) \cdot ln\left(\frac{10.56}{140}\right)\) \(\dot{S}_g \approx -0.137\ \mathrm{kJ/s \cdot K}\) Therefore, the rate of entropy generation is approximately \(-0.137\ \mathrm{kJ/s \cdot K}\).

Conclusion

The rate of heat transfer in the well-insulated heat exchanger is approximately \(73.15\ \mathrm{kJ/s}\), and the rate of entropy generation is approximately \(-0.137\ \mathrm{kJ/s \cdot K}\).