Question

Cow milk is pasteurized at a flow rate of 1 kg.s$^{-1}$ in a counter-current heat exchanger using hot water as the heating medium. The milk enters at 15°C and exits at 50°C. The specific heat of milk is 3.5 kJ.kg$^{-1}$.°C$^{-1}$. Hot water enters at 75°C and exits at 60°C. Overall heat transfer coefficient $U = 1800$ W.m$^{-2}$.°C$^{-1}$. Find the heat transfer surface area in m$^2$ (rounded off to 2 decimal places). Assume steady-state.

Hint:

In counter-current heat exchangers, always calculate LMTD (log mean temperature difference) carefully—it balances inlet and outlet conditions.

Answer 1

Correct Answer: 1.8

Step 1: Heat gained by milk.
\[ Q = \dot{m} \, c_p \, \Delta T \] \[ Q = 1 \times 3.5 \, \text{kJ/kg.°C} \times (50 - 15) \, °C \] \[ Q = 3.5 \times 35 = 122.5 \, \text{kJ/s} = 122.5 \, \text{kW} \]
Step 2: Apply heat exchanger equation.
\[ Q = U A \Delta T_{lm} \]
Step 3: Calculate $\Delta T_{lm$ (log mean temperature difference).}
For counter-current flow: \[ \Delta T_1 = T_{h,in} - T_{c,out} = 75 - 50 = 25^\circ C \] \[ \Delta T_2 = T_{h,out} - T_{c,in} = 60 - 15 = 45^\circ C \] \[ \Delta T_{lm} = \frac{\Delta T_2 - \Delta T_1}{\ln \left( \frac{\Delta T_2}{\Delta T_1} \right)} \] \[ \Delta T_{lm} = \frac{45 - 25}{\ln (45/25)} = \frac{20}{\ln (1.8)} = \frac{20}{0.588} = 34.01^\circ C \]
Step 4: Solve for $A$.
\[ Q = U A \Delta T_{lm} \] \[ 122500 \, W = 1800 \times A \times 34.01 \] \[ A = \frac{122500}{1800 \times 34.01} = \frac{122500}{61218} = 2.00 \, m^2 \] Final Answer: \[ \boxed{2.00 \, m^2} \]