Subtopic: Transient heat transfer in simple geometry
Question Statement:
A steam jacketed kettle is filled with a 250 kg liquid food with a specific heat of 3.1 kJ·kg⁻¹·K⁻¹. Steam used at 121 °C has a heat transfer coefficient of 2 kW·m⁻²·K⁻¹. The area of heat transfer is 0.15 m². Using lumped capacity method, the time required to heat the product from 30 °C to 95 °C, in minutes, is __________. (Rounded off to 2 decimal places)
Official Answer Key:
52.00 to 56.00
Step-by-Step Solution & Explanation
In the lumped capacity model the product is at a uniform temperature T(t), heated by a constant-temperature medium T_s:
m c_p (dT/dt) = h A (T_s − T)
Integrating between the initial and final temperatures:
t = (m c_p)/(h A) × ln[(T_s − T₀)/(T_s − T)]
Step 1 — time constant:
(m c_p)/(h A) = (250 × 3.1)/(2 × 0.15) = 775/0.3 = 2583.3 s
Step 2 — temperature ratio:
(T_s − T₀)/(T_s − T) = (121 − 30)/(121 − 95) = 91/26 = 3.5
Step 3 — time:
t = 2583.3 × ln(3.5) = 2583.3 × 1.25276 = 3236.3 s
Step 4 — convert to minutes:
t = 3236.3/60 = 53.94 minutes
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