Question

A hot, freshly-sterilised fermentation medium is cooled in a double-pipe heat-exchanger. The medium enters the inner pipe of the exchanger at 95 \(^\circ C\) and leaves the exchanger at 40 \(^\circ C\). Cooling water, flowing counter-currently to the medium, enters the annulus of the exchanger at 15 \(^\circ C\) and leaves the exchanger at 45 \(^\circ C\). The overall heat transfer coefficient is 1350 W m\(^{-2}\) °C\(^{-1}\). The rate of heat transfer per unit area will be _________ W/m². (Round off to the nearest integer)

Hint:

In heat exchangers, the log mean temperature difference (LMTD) is crucial for calculating the heat transfer rate. Always use the appropriate temperature difference for counter-current flow.

Answer 1

Correct Answer: 48000

Step 1: Understand the heat transfer equation. The rate of heat transfer per unit area (\(q\)) in a heat exchanger is given by the equation: \[ q = U \cdot A \cdot \Delta T_m, \] where:
\(U\) is the overall heat transfer coefficient (1350 W/m²°C),
\(A\) is the area of heat transfer (which cancels out in this case, as we are finding per unit area),
\(\Delta T_m\) is the log mean temperature difference (LMTD).
Step 2: Calculate the log mean temperature difference (LMTD).
The LMTD for counter-current flow is given by:
\[ \Delta T_m = \frac{(\Delta T_1 - \Delta T_2)}{\ln\left(\frac{\Delta T_1}{\Delta T_2}\right)}, \] where:
\(\Delta T_1 = T_{{hot, in}} - T_{{cold, out}} = 95 - 45 = 50 \, ^\circ C\),
\(\Delta T_2 = T_{{hot, out}} - T_{{cold, in}} = 40 - 15 = 25 \, ^\circ C\).
Thus, \[ \Delta T_m = \frac{(50 - 25)}{\ln\left(\frac{50}{25}\right)} = \frac{25}{\ln(2)} \approx 25 / 0.693 = 36.07 \, ^\circ C. \] Step 3: Calculate the heat transfer rate per unit area.
Now we can calculate the rate of heat transfer per unit area: \[ q = U \cdot \Delta T_m = 1350 \times 36.07 \approx 48695.5 \, {W/m}^2. \] Thus, the rate of heat transfer per unit area is \( \boxed{48700} \, {W/m}^2 \).