Question

In a double pipe cocurrent heat exchanger, operating at steady state, oil (specific heat capacity = \(2100\ \mathrm{J\,kg^{-1}\,^\circ C^{-1}}\)) entering at \(7\ \mathrm{kg\,s^{-1}}\) at \(100^\circ\mathrm{C}\) is used to heat water (specific heat capacity = \(4200\ \mathrm{J\,kg^{-1}\,^\circ C^{-1}}\)) flowing at \(3.5\ \mathrm{kg\,s^{-1}}\) from \(20^\circ\mathrm{C}\) to \(50^\circ\mathrm{C}\). If the overall heat transfer coefficient is \(291\ \mathrm{W\,m^{-2}\,^\circ C^{-1}}\), the required heat transfer area (in \(\mathrm{m^2}\)) is ______ (rounded off to one decimal place).

Hint:

First find the oil outlet temperature from an energy balance, then compute the cocurrent LMTD from the inlet-end and outlet-end temperature differences, and use \(Q=UA\Delta T_{lm}\).

Answer 1

Correct Answer: 35

Step 1: Heat duty from water.

\[ Q = (3.5)(4200)(30) = 441000\ \mathrm{W} \]

Step 2: Oil outlet temperature.

\[ 441000 = (7)(2100)(100-T_{o,out}) \Rightarrow T_{o,out}=70^\circ\mathrm{C} \]

Step 3: Terminal temperature differences.

\[ \Delta T_1=80,\ \Delta T_2=20 \]

Step 4: LMTD.

\[ \Delta T_{lm} = \frac{60}{\ln 4} = 43.281^\circ\mathrm{C} \]

Step 5: Area.

\[ A = \frac{441000}{(291)(43.281)} = 35.02\ \mathrm{m^2} \]

\[ \boxed{A \approx 35.0\ \mathrm{m^2}} \]