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Question:

A water cooler of storage capacity 120 litres can cool water at a constant rate of P in a closed circulation system. The water from the cooler is used to cool an external device that generates constantly 3 kW of heat (thermal load). The temperature of water fed into the device cannot exceed 30°C and the entire stored 120 litres of water is initially cooled to 10°C. The entire system is thermally insulated. The minimum value of P (in watts) for which the device can be operated for 3 hours is: (Specific heat of water is 4.2 kJkg⁻¹K⁻¹ and density of water is 1000 kgm⁻³)

1600

3933

2067

2533

Solution:

Heat generated in 3 hrs = 3(3600) × 3 × 10³ = 324 × 10⁵ J
Heat used by water heater = msΔT = 120 × 1 × 4200 × 20 = 100.8 × 10⁵ J
Heat absorbed by coolant = Pt = 324 × 10⁵ − 100.8 × 10⁵ = 223.2 × 10⁵
P = 223.2 × 10⁵ / 3600 = 2067 W