Question

A shell-and-tube heat exchanger with two shell passes and eight tube passes is used to heat ethyl alcohol \(\left(c_{p}=2670\right.\) $\mathrm{J} / \mathrm{kg} \cdot \mathrm{K})\( in the tubes from \)25^{\circ} \mathrm{C}\( to \)70^{\circ} \mathrm{C}\( at a rate of \)2.1 \mathrm{~kg} / \mathrm{s}$. The heating is to be done by water $\left(c_{p}=4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( that enters the shell side at \)95^{\circ} \mathrm{C}$ and leaves at \(45^{\circ} \mathrm{C}\). If the overall heat transfer coefficient is \(950 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the heat transfer surface area of the heat exchanger.

Short answer

Based on the given information and calculations, the heat transfer surface area of the shell-and-tube heat exchanger is approximately 11.97 m².

Step-by-step solution

Calculate the heat transfer rate for ethyl alcohol

We'll first find the heat transfer rate (Q) for ethyl alcohol using the given mass flow rate (m), specific heat capacity (c_p), and temperature difference (ΔT): Q = m * c_p * ΔT where m = 2.1 kg/s (mass flow rate) c_p = 2670 J/(kg·K) (specific heat capacity of ethyl alcohol) ΔT = T_final - T_initial = 70°C - 25°C = 45 K (temperature difference) Now, we can plug in the given values and calculate Q: Q = 2.1 kg/s * 2670 J/(kg·K) * 45 K ≈ 253215 J/s = 253.215 kW

Calculate the Log Mean Temperature Difference (LMTD)

Next, we will calculate the LMTD using the given temperatures of ethyl alcohol and water as follows: LMTD = \(\frac{\Delta T_{1}-\Delta T_{2}}{\ln \left(\Delta T_{1} / \Delta T_{2}\right)}\) where ΔT_1 = T_hot_in - T_cold_out = 95°C - 70°C = 25 K (temperature difference between the hot inlet and cold outlet) ΔT_2 = T_hot_out - T_cold_in = 45°C - 25°C = 20 K (temperature difference between the hot outlet and cold inlet) Now, we can calculate the LMTD: LMTD ≈ \(\frac{25 - 20}{\ln\left(\frac{25}{20}\right)}\) ≈ 22.37 K

Determine the heat transfer surface area

Using the overall heat transfer coefficient (U), heat transfer rate (Q), and LMTD, we can now determine the heat transfer surface area (A) using the following equation: Q = U * A * LMTD We need to solve for A: A = \(\frac{Q}{U \cdot \text{LMTD}}\) where U = 950 W/(m²·K) (overall heat transfer coefficient) Q = 253215 W (heat transfer rate) LMTD = 22.37 K Now, we can plug in the values and calculate the heat transfer surface area: A ≈ \(\frac{253215}{950 \cdot 22.37}\) ≈ 11.97 m² Therefore, the heat transfer surface area of the heat exchanger is approximately 11.97 m².