Fruit juice entering at 30 °C is being pasteurized in a double pipe counter flow heat exchanger using hot water entering at 97 °C and leaving at 65 °C. Specific heat of juice and water are 3.8 kJ·kg⁻¹·K⁻¹ and 4.2 kJ·kg⁻¹·K⁻¹, respectively. Flow rate of juice is 0.3 kg·s⁻¹ and that of water is 0.4 kg·s⁻¹. Considering an overall heat transfer coefficient of 1 kW·m⁻²·K⁻¹, the area of heat exchanger required, in m², is __________. (Rounded off to 3 decimal places)
Official Answer Key:
1.900 to 2.200
Step-by-Step Solution & Explanation
Step 1 — heat duty from the hot stream:
Q = ṁ_w c_w ΔT_w = 0.4 × 4.2 × (97 − 65) = 1.68 × 32 = 53.76 kW
Step 2 — juice outlet temperature from the energy balance:
Q = ṁ_j c_j (T_out − 30)
53.76 = 0.3 × 3.8 × (T_out − 30) = 1.14 (T_out − 30)
T_out − 30 = 47.16 ⇒ T_out = 77.16 °C
Step 3 — terminal temperature differences (counter-flow).
Hot fluid in (97 °C) meets juice out (77.16 °C): ΔT₁ = 19.84 °C
Hot fluid out (65 °C) meets juice in (30 °C): ΔT₂ = 35.00 °C
Step 4 — log mean temperature difference:
LMTD = (ΔT₂ − ΔT₁)/ln(ΔT₂/ΔT₁) = (35.00 − 19.84)/ln(35.00/19.84)
= 15.16/ln(1.7641) = 15.16/0.56765 = 26.71 °C
Step 5 — heat transfer area:
A = Q/(U × LMTD) = 53.76/(1 × 26.71) = 2.013 m²
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