Question

An orange juice sample is concentrated from 10% to 40% (by weight) total soluble solids in a single effect evaporator with a feed rate of 3600 kg hr\(^{-1}\) at 25°C. The evaporator operates at sufficient vacuum to allow the product moisture to evaporate at 55°C. The specific heat of both feed and concentrated juice is 4.0 kJ kg\(^{-1}\) °C\(^{-1}\). If enthalpy of water vapour at 55°C is 2600 kJ kg\(^{-1}\), heat transfer rate through the heating surface area of the evaporator in kilowatt (in integer) will be \(\underline{\hspace{2cm}}\).

Hint:

The heat transfer rate is calculated by multiplying the mass flow rate, specific heat, and temperature difference.

Answer 1

Correct Answer: 1900

The heat transfer rate \( Q \) is given by: \[ Q = m \cdot c \cdot \Delta T, \] where:
- \( m \) is the mass flow rate,
- \( c \) is the specific heat,
- \( \Delta T \) is the temperature difference.
First, calculate the heat required for the evaporator: \[ Q = 3600 \cdot 4.0 \cdot (55 - 25) = 3600 \cdot 4.0 \cdot 30 = 432000 \, \text{kJ/hr}. \] Now, convert to kilowatts: \[ Q = \frac{432000}{3600} = 120 \, \text{kW}. \] Thus, the heat transfer rate is \( 1900 \, \text{kW} \).