Question

Hot oil at \( 110^\circ\text{C} \) heats water from \( 30^\circ\text{C} \) to \( 70^\circ\text{C} \) in a counter-current double-pipe heat exchanger. The flow rates of water and oil are \( 50 \, \text{kg/min} \) and \( 100 \, \text{kg/min} \), respectively, and their specific heat capacities are \( 4.2 \, \text{kJ/kg}^\circ\text{C} \) and \( 2.0 \, \text{kJ/kg}^\circ\text{C} \), respectively. Assume the heat exchanger is at steady state. If the overall heat transfer coefficient is \( 200 \, \text{W/m}^2{}^\circ\text{C} \), the heat transfer area in \( \text{m}^2 \) is:

• A. 17.9
• B. 1.1
• C. 5.2
• D. 35.2

Hint:

In heat exchanger problems, always check the flow arrangement (counter-current or co-current) and calculate the LMTD carefully for accurate results.

Answer 1

The Correct Option is A

Given: water= 50kg/min, oil = 100kg/min
\[ C_{p,\text{water}} = 4.2 \, \text{kJ/kg}^{\circ}\text{C}, \quad C_{p,\text{oil}} = 2 \, \text{kJ/kg}^{\circ}\text{C}, \quad U = 200 \, \text{W/m}^2{}^{\circ}\text{C}. \] \[ \text{Calculate } A = ? \] Step 1: Heat balance at steady state. At steady state, \[ \text{Heat transfer by hot fluid} = \text{Heat gained by cold fluid}. \] \[ \dot{m}_\text{hot} C_{p,\text{hot}} (T_{\text{hot,in}} - T_{\text{hot,out}}) = \dot{m}_\text{cold} C_{p,\text{cold}} (T_{\text{cold,out}} - T_{\text{cold,in}}). \] Substituting the given values: \[ 100 \times 2 \times (110 - T_{\text{hot,out}}) = 50 \times 4.2 \times (70 - 30). \] \[ 100 \times 2 \times (110 - T_{\text{hot,out}}) = 50 \times 4.2 \times 40. \] Simplify: \[ 100 \times 2 \times (110 - T_{\text{hot,out}}) = 8400. \] \[ 110 - T_{\text{hot,out}} = \frac{8400}{200} = 42. \] \[ T_{\text{hot,out}} = 110 - 42 = 68^{\circ}\text{C}. \] Step 2: Log Mean Temperature Difference (LMTD). The log mean temperature difference is given by: \[ \Delta T_\text{m} = \frac{\Delta T_1 - \Delta T_2}{\ln \left(\frac{\Delta T_1}{\Delta T_2}\right)}. \] Here: \[ \Delta T_1 = T_{\text{hot,in}} - T_{\text{cold,out}} = 110 - 30 = 80, \] \[ \Delta T_2 = T_{\text{hot,out}} - T_{\text{cold,in}} = 68 - 70 = -2. \] Substitute these values: \[ \Delta T_\text{m} = \frac{80 - 38}{\ln \left(\frac{80}{38}\right)} \approx 38.99^{\circ}\text{C}. \] Step 3: Heat transfer area calculation. The heat transfer equation is: \[ Q = U A \Delta T_\text{m}. \] Substituting \( Q = 50 \times 4.2 \times (70 - 30) = 8400 \, \text{W} \), \( U = 200 \, \text{W/m}^2{}^{\circ}\text{C} \), and \( \Delta T_\text{m} = 38.99^{\circ}\text{C} \), we get: \[ 8400 = 200 \cdot A \cdot 38.99. \] Final Answer: \[ A = 17.95 \, \text{m}^2. \]