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 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( in the tubes from \)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)\( that enters the shell at \)95^{\circ} \mathrm{C}$ and leaves at \(60^{\circ} \mathrm{C}\). If the overall heat transfer coefficient is $800 \mathrm{~W} / \mathrm{m}^{2} . \mathrm{K}$, determine the heat transfer surface area of the heat exchanger using \((a)\) the LMTD method and \((b)\) the \(\varepsilon-\mathrm{NTU}\) method. Answer: \((a) 11.4 \mathrm{~m}^{2}\)

Short answer

(b) Effectiveness-NTU Method: #tag_title# 4. Calculate Effectiveness and NTU #tag_content# In this method, we first calculate the effectiveness (\(\epsilon\)) of the heat exchanger: $$\epsilon = \frac{Q_{actual}}{Q_{max}} = \frac{mc_p\Delta T}{(mc_p)_{min}(\Delta T)_{max}}$$ Since the mass flow rate and specific heat capacity of the cold fluid (ethyl alcohol) are smaller than those of the hot fluid (water), the cold fluid will control the effectiveness: $$(mc_p)_{min} = m_{c}c_{p_c} = 2.1\;\text{kg/s} \cdot 2670\;\text{J/kg K} = 5610\;\text{W/K}$$ $$(\Delta T)_{max} = T_{h_{in}} - T_{c_{in}} = 95^{\circ}\text{C} - 25^{\circ}\text{C} = 70^{\circ}\text{C}$$ Substituting the values, we get: $$\epsilon = \frac{253935\;\text{W}}{5610\;\text{W/K} \cdot 70\;\text{K}} \approx 0.65$$ Next, we calculate the Number of Transfer Units (NTU) using the formula: $$\text{NTU} = \frac{Q_{actual}}{(mc_p)_{min} \cdot (1 - \epsilon)}$$ Substituting the values, we get: $$\text{NTU} = \frac{253935\;\text{W}}{5610\;\text{W/K} \cdot (1 - 0.65)} \approx 2.44$$ #tag_title# 5. Calculate Surface Area using Effectiveness-NTU Method #tag_content# Now, we can calculate the heat transfer surface area (\(A\)) using the effectiveness-NTU method with the formula: $$A = \frac{\text{NTU}\cdot(mc_p)_{min}}{U}$$ Substituting the values, we get: $$A = \frac{2.44 \cdot 5610\;\text{W/K}}{800\;\text{W/m}^2\text{K}} \approx 17.2\;\text{m}^2$$ The heat transfer surface area of the heat exchanger using the effectiveness-NTU method is approximately \(17.2\;\text{m}^2\). In conclusion, using the LMTD and effectiveness-NTU methods, the heat transfer surface area of the shell-and-tube heat exchanger is approximately \(11.4\;\text{m}^2\) and \(17.2\;\text{m}^2\), respectively.

Step-by-step solution

1. Calculate Heat Transfer Rate for Ethyl Alcohol

Using the given specific heat capacity, mass flow rate, and temperature change of the ethyl alcohol, we can calculate the heat transfer rate \(Q\) (in Watts) using the formula: $$Q = mc_p\Delta T$$ where \(m = 2.1\;\text{kg/s}\), \(c_p = 2670\;\text{J/kg K}\), \(\Delta T = T_{out} - T_{in} = 70^{\circ}\text{C} - 25^{\circ}\text{C} = 45^{\circ}\text{C}\). \(Q = (2.1\;\text{kg/s})(2670\;\text{J/kg K})(45\;\text{K}) = 253935\;\text{W}\) (a) LMTD Method:

2. Calculate Log Mean Temperature Difference (LMTD)

In this method, we use the temperature profiles of both fluids. The LMTD is calculated using the expression: $$\text{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1/\Delta T_2)}$$ where \(\Delta T_1 = T_{h_{in}} - T_{c_{out}} = 95^{\circ}\text{C} - 70^{\circ}\text{C} = 25^{\circ}\text{C}\), and \(\Delta T_2 = T_{h_{out}} - T_{c_{in}} = 60^{\circ}\text{C} - 25^{\circ}\text{C} = 35^{\circ}\text{C}\). Calculating LMTD, we get: $$\text{LMTD} = \frac{25 - 35}{\ln(25/35)} \approx 29.51\;\text{K}$$

3. Calculate Surface Area using LMTD Method

The formula to calculate the heat transfer surface area (\(A\)) using the LMTD method is: $$A = \frac{Q}{U\cdot \text{LMTD}}$$ where \(U = 800\;\text{W/m}^2\text{K}\) is the overall heat transfer coefficient. \(A = \frac{253935\;\text{W}}{800\;\text{W/m}^2\text{K} \cdot 29.51\;\text{K}} \approx 11.4\;\text{m}^2\) The heat transfer surface area of the heat exchanger using the LMTD method is approximately \(11.4\;\text{m}^2\).