Question

A single effect evaporator is used to concentrate orange fruit juice with 2.7:1 concentration ratio of product solid to feed solid. The juice is fed to the evaporator at \(100\ kg.h^{-1}\) and \(30\,^{\circ}\text{C}\), where the product moisture evaporates at \(60\,^{\circ}\text{C}\). The specific heat of juice is \(3.9\ kJ.kg^{-1}.^{\circ}\text{C}^{-1}\). The latent heat of vaporisation at 60 °C is \(2450\ kJ.kg^{-1}\). Heat to accomplish the evaporation, in \(MJ.h^{-1}\), is nearest to

• A. 165.96
• B. 117.06
• C. 154.26
• D. 142.56

Hint:

Use the solids balance to get the evaporated water mass, then add sensible heat to reach boiling and latent heat to vaporise it.

Answer 1

The Correct Option is A

Step 1: Find the product flow rate from the concentration ratio.
Solids are conserved: feed solids equal product solids, \( F x_f = P x_p \). The ratio \( x_p/x_f = 2.7 \) means \( P = F/2.7 \).
With \( F = 100\ kg/h \), \( P = 100/2.7 = 37.04\ kg/h \).

Step 2: Find the water evaporated.
Vapour removed \( V = F - P = 100 - 37.04 = 62.96\ kg/h \).

Step 3: Add the sensible heat to bring feed up to boiling.
Sensible heat \( Q_s = F c_p (T_2 - T_1) = 100 \times 3.9 \times (60-30) = 11700\ kJ/h \).

Step 4: Add the latent heat to vaporise the water removed.
\( Q_L = V \times \lambda = 62.96 \times 2450 = 154254\ kJ/h \).

Step 5: Total the heat load and convert units.
\( Q = Q_s + Q_L = 11700 + 154254 = 165954\ kJ/h \approx 165.96\ MJ/h \).

Final Answer:
The evaporator needs close to 165.96 MJ per hour, matching option (A). \[ \boxed{Q \approx 165.96\ MJ.h^{-1}} \]