Question

In a one-shell and two-tube heat exchanger, cold water with inlet temperature of \(20^{\circ} \mathrm{C}\) is heated by hot water supplied at the inlet at \(80^{\circ} \mathrm{C}\). The cold and hot water flow rates are $5000 \mathrm{~kg} / \mathrm{h}\( and \)10,000 \mathrm{~kg} / \mathrm{h}$, respectively. If the shell-andtube heat exchanger has a \(U A_{s}\) value of \(11,600 \mathrm{~W} / \mathrm{K}\), determine the cold water and hot water outlet temperatures. Assume $c_{p c}=4178 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\( and \)c_{p t}=4188 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}$.

Short answer

Based on the given data and calculations, the cold water outlet temperature in the one-shell and two-tube heat exchanger is 53.33°C, and the hot water outlet temperature is 64.67°C.

Step-by-step solution

Calculate the heat capacities of cold and hot streams

Calculate the heat capacities for both cold and hot streams using their mass flow rates and specific heats: For the cold stream: \(m_c c_{p c} = 5000 \frac{\text{kg}}{\text{h}} * 4178 \frac{\text{J}}{\text{kg} \cdot \text{K}} = 20,890,000 \frac{\text{J}}{\text{h} \cdot \text{K}}\) For the hot stream: \(m_t c_{p t} = 10000 \frac{\text{kg}}{\text{h}} * 4188 \frac{\text{J}}{\text{kg} \cdot \text{K}} = 41,880,000 \frac{\text{J}}{\text{h} \cdot \text{K}}\)

Determine the Maximum Possible Heat Transfer Rate

Choose the minimum value of heat capacities and the maximum value of temperature difference: Minimum heat capacity: min(20,890,000 ; 41,880,000) = 20,890,000 J/(h·K) Maximum temperature difference: ΔT = |80 - 20| = 60 K Maximum Possible Heat Transfer Rate = (Minimum heat capacity) * (Maximum temperature difference) = 20,890,000 J/(h·K) * 60 K = 1,253,400,000 J/h

Calculate the Heat Transfer Rate

Given the heat exchanger UA_s value of 11,600 W/K, we can calculate the heat transfer rate as follows: Effectiveness = \(\frac{Q}{Q_{\text{max}}}\) Rearranging the formula for Q, we find Q = Effectiveness * Q_max (where Q_max = 1,253,400,000 J/h) Next, we find the effectiveness of the heat exchanger using the given UA_s value: Effectiveness = \(\frac{\text{UA}_s}{\text{min}(m_c c_{p c},m_t c_{p t})}\) Effectiveness = \(\frac{11,600\frac{\text{W}}{\text{K}}}{20,890,000\frac{\text{J}}{\text{h} \cdot \text{K}}}\) = 0.55509 (unitless) Now we can calculate the heat transfer rate Q: Q = 0.55509 * 1,253,400,000 J/h = 696,373,237 J/h

Determine the Cold and Hot Water Outlet Temperatures

Now we have the required heat transfer rate Q, and we can determine the outlet temperatures for the hot and cold streams. For the cold stream: \(Q = m_c c_{p c} (T_{c \text{ out}} - T_{c \text{ in}})\) Rearranging to solve for \(T_{c \text{ out}}\): \(T_{c \text{ out}} = T_{c \text{ in}} + \frac{Q}{m_c c_{p c}} = 20^{\circ}\mathrm{C} + \frac{696,373,237\frac{\text{J}}{\text{h}}}{20,890,000\frac{\text{J}}{\text{h} \cdot \text{K}}}\) \(T_{c \text{ out}} = 53.33^{\circ}\mathrm{C}\) For the hot stream: \(Q = m_t c_{p t} (T_{t \text{ in}} - T_{t \text{ out}})\) Rearranging to solve for \(T_{t \text{ out}}\): \(T_{t \text{ out}} = T_{t \text{ in}} - \frac{Q}{m_t c_{p t}} = 80^{\circ}\mathrm{C} - \frac{696,373,237\frac{\text{J}}{\text{h}}}{41,880,000\frac{\text{J}}{\text{h} \cdot \text{K}}}\) \(T_{t \text{ out}} = 64.67^{\circ}\mathrm{C}\) Thus, the cold water outlet temperature is \(53.33^{\circ}\mathrm{C}\), and the hot water outlet temperature is \(64.67^{\circ}\mathrm{C}\).