Question Bank • QB_DFE_PF_018 NAT • 2 Marks

Section 7: Dairy and Food Engineering

Preservation of Food

Subtopic: Preservation of food by cooling and freezing

Question Statement:

An industrial blast freezing tunnel freezes $1200\text{ kg/h}$ of green peas from $20.0^\circ\text{C}$ to $-18.0^\circ\text{C}$. The initial freezing point is $T_f = -1.0^\circ\text{C}$. Specific heat of unfrozen peas is $C_{pu} = 3.60\text{ kJ/(kg}\cdot\text{K)}$, specific heat of frozen peas is $C_{pf} = 1.90\text{ kJ/(kg}\cdot\text{K)}$, and latent heat of fusion is $\lambda_f = 260.0\text{ kJ/kg}$. The total refrigeration load required to freeze the product in $\text{kW}$ is ________ (round off to one decimal place).

Answer Key:

122.6

Step-by-Step Solution & Explanation

Multi-Approach Solution: **Method 1: Three-Stage Enthalpy Removal** 1. Sensible cooling of unfrozen peas ($20.0^\circ\text{C} \to -1.0^\circ\text{C}$, $\Delta T_1 = 21.0\text{ K}$): $q_1 = 1200\text{ kg/h} \times 3.60\text{ kJ/(kg}\cdot\text{K)} \times 21.0\text{ K} = 90720\text{ kJ/h}$ 2. Latent heat of freezing: $q_2 = 1200\text{ kg/h} \times 260.0\text{ kJ/kg} = 312000\text{ kJ/h}$ 3. Sensible cooling of frozen peas ($-1.0^\circ\text{C} \to -18.0^\circ\text{C}$, $\Delta T_2 = 17.0\text{ K}$): $q_3 = 1200\text{ kg/h} \times 1.90\text{ kJ/(kg}\cdot\text{K)} \times 17.0\text{ K} = 38760\text{ kJ/h}$ 4. Total heat removal rate: $q_{\text{total}} = 90720 + 312000 + 38760 = 441480\text{ kJ/h}$ 5. Refrigeration capacity in $\text{kW}$: $P = \frac{441480\text{ kJ/h}}{3600\text{ s/h}} = 122.633\text{ kW} \approx 122.6\text{ kW}$ **Method 2: Specific Enthalpy Sum** $\Delta h = (3.60 \times 21) + 260.0 + (1.90 \times 17) = 75.6 + 260.0 + 32.3 = 367.9\text{ kJ/kg}$ $P = \frac{1200}{3600} \times 367.9 = \frac{1}{3} \times 367.9 = 122.63\text{ kW}$
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