Question

A well-insulated, thin-walled, double-pipe, counter-flow heat exchanger is to be used to cool oil \(\left(c_{p}=\right.\) \(\left.2.20 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) from \(150^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\) at a rate of \(2 \mathrm{kg} / \mathrm{s}\) by water \(\left(c_{p}=4.18 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) that enters at \(22^{\circ} \mathrm{C}\) at a rate of \(1.5 \mathrm{kg} / \mathrm{s}\) Determine \((a)\) the rate of heat transfer and \((b)\) the rate of entropy generation in the heat exchanger.

Short answer

Question: Calculate (a) the rate of heat transfer and (b) the rate of entropy generation in a well-insulated, thin-walled, double-pipe, counter-flow heat exchanger used to cool oil by water. Answer: (a) The rate of heat transfer is 484 kW, and (b) the rate of entropy generation is 0.4396 kW/K.

Step-by-step solution

Determine the temperature of the water when it leaves the heat exchanger

To find the final temperature of the water, we have to apply an energy balance equation. Since the heat exchanger is well-insulated, the rate of heat transfer from oil to water (Q) is equal to the difference between the energy leaving and entering the system. For the oil: \(Q = m_o \cdot c_{p_o} \cdot(T_{o, out} - T_{o, in})\) For the water: \(Q = m_w \cdot c_{p_w} \cdot(T_{w, out} - T_{w, in})\) As \(Q_oil = Q_water\), we have: \(m_o \cdot c_{p_o} \cdot(T_{o, out} - T_{o, in}) = m_w \cdot c_{p_w} \cdot(T_{w, out} - T_{w, in})\) Plug in the given values and solve for the unknown \(T_{w, out}\): \(2 \text{ kg/s} \cdot 2.20 \text{ kJ/kg°C} \cdot (40 - 150) = 1.5 \text{ kg/s} \cdot 4.18 \text{ kJ/kg°C} \cdot (T_{w, out} - 22)\) \(T_{w, out} = 59.13^{\circ}C\)

Calculate the rate of heat transfer

Now that we have the final temperature of the water when it leaves the heat exchanger, we can calculate the rate of heat transfer (Q). From the energy balance equation, we can use the oil side: \(Q = m_o \cdot c_{p_o} \cdot(T_{o, out} - T_{o, in})\) Plug in the given values and the calculated water outlet temperature: \(Q = 2 \text{ kg/s} \cdot 2.20 \text{ kJ/kg°C} \cdot (40 - 150)\) \(Q = -484 \text{ kW}\) (Negative sign denotes that the heat is transferred from the oil to the water)

Calculate the rate of entropy generation

The rate of entropy generation (S_gen) in the heat exchanger can be calculated using the following equation: \(S_{gen} = m_o \cdot c_{p_o} \cdot \ln\frac{T_{o, out}}{T_{o, in}} + m_w \cdot c_{p_w} \cdot \ln\frac{T_{w, out}}{T_{w, in}}\) Plug in the values and calculated temperatures: \(S_{gen} = 2 \text{ kg/s} \cdot 2.20 \text{ kJ/kg°C} \cdot \ln\frac{40}{150} + 1.5 \text{ kg/s} \cdot 4.18 \text{ kJ/kg°C} \cdot \ln\frac{59.13}{22}\) \(S_{gen} = -0.4396 \text{ kW/K}\) (Negative sign denotes entropy being generated within the system) The results are: (a) The rate of heat transfer is \(484 \text{ kW}\), and (b) the rate of entropy generation is \(0.4396 \text{ kW/K}\).