Question Bank • QB_APE_EXP_018 NAT • 2 Marks

Section 6: Agricultural Process Engineering

Evaporation and Drying

Subtopic: Concentration and drying of liquid foods – evaporators, tray, drum and spray dryers

Question Statement:

In a single-effect evaporator, juice boils at $60^\circ\text{C}$ (latent heat of vaporization $\lambda_v = 2358\text{ kJ/kg}$) producing $2000\text{ kg/h}$ of vapor. Saturated steam is supplied at $100^\circ\text{C}$ (latent heat of condensation $\lambda_s = 2257\text{ kJ/kg}$). If the overall heat transfer coefficient $U = 1800\text{ W/(m}^2\cdot\text{K)}$, the required heating surface area $A$ in $\text{m}^2$ is ________ (round off to 1 decimal place).

Answer Key:

18.2

Step-by-Step Solution & Explanation

1. Heat transfer rate $q$: $q = \dot{m}_v \lambda_v = \frac{2000\text{ kg/h}}{3600\text{ s/h}} \times 2358 \times 10^3\text{ J/kg} = 0.5556 \times 2.358 \times 10^6 = 1.310 \times 10^6\text{ W}$ 2. Temperature driving force $\Delta T = T_{steam} - T_{boil} = 100 - 60 = 40^\circ\text{C} = 40\text{ K}$. 3. Required heat transfer area $A$: $A = \frac{q}{U \Delta T} = \frac{1.310 \times 10^6\text{ W}}{1800\text{ W/(m}^2\cdot\text{K)} \times 40\text{ K}} = \frac{1.310 \times 10^6}{72,000} \approx 18.194\text{ m}^2 \approx 18.2\text{ m}^2$
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