Question

A double-effect evaporator is used to concentrate a solution. Steam is sent to the first effect at 110 °C and the boiling point of the solution in the second effect is 63.3 °C. The overall heat transfer coefficient in the first effect and second effect are 2000 W m\(^{-2}\) K\(^{-1}\) and 1500 W m\(^{-2}\) K\(^{-1}\), respectively. The heat required to raise the temperature of the feed to the boiling point can be neglected. The heat flux in the two evaporators can be assumed to be equal. The temperature at which the solution boils in the first effect is \(\underline{\hspace{1cm}}\) °C (round off to nearest integer).

Hint:

In double-effect evaporators, equal heat flux assumption simplifies the temperature calculations.

Answer 1

Correct Answer: 89

For a double-effect evaporator, the heat flux is given by:
\[ Q = U_1 A (T_{boiling1} - T_{boiling2}) = U_2 A (T_{boiling2} - T_{feed}) \] Equating the heat fluxes in both effects: \[ 2000 (T_{boiling1} - 63.3) = 1500 (63.3 - T_{boiling2}) \] Simplifying:
\[ T_{boiling1} \approx 89 \text{ to } 91°C \] Final answer: 89–91°C.