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Coolant heat rejection and radiator air flow

Short answer

Heat rejected to coolant: 43.2 kW. Coolant volume flow rate: 87.4 L/min. Required air mass flow: 1.72 kg/s. Fuel energy rate: 154 kW.

A water-cooled engine circulates 1.5 kg/s of coolant (50/50 water-glycol, c_p = 3.6 kJ/kg K, density 1030 kg/m³). The coolant is cooled by 8 K in the radiator. Air passes through the radiator and its temperature rises by 25 K (c_p of air = 1.005 kJ/kg K). Calculate (i) the heat rejected to the coolant, (ii) the volume flow rate of the coolant in L/min, and (iii) the mass flow rate of air required. If the heat rejected is 28 % of the fuel energy, find the fuel energy rate.

Given: m˙c=1.5\dot m_c = 1.5 kg/s, cpc=3.6c_{pc} = 3.6 kJ/kg K, ΔTc=8\Delta T_c = 8 K, ρc=1030\rho_c = 1030 kg/m³, ΔTa=25\Delta T_a = 25 K, cpa=1.005c_{pa} = 1.005 kJ/kg K.

(i) Heat rejected

Q=m˙ccpcΔTc=1.5×3.6×8=43.2 kWQ = \dot m_c c_{pc} \Delta T_c = 1.5 \times 3.6 \times 8 = 43.2\ \text{kW}

(ii) Coolant volume flow (water pump capacity)

V˙=m˙cρc=1.51030=1.456×10−3 m3/s=1.456 L/s=87.4 L/min\dot V = \frac{\dot m_c}{\rho_c} = \frac{1.5}{1030} = 1.456\times10^{-3}\ \text{m}^3/\text{s} = 1.456\ \text{L/s} = 87.4\ \text{L/min}

(iii) Air flow (heat balance: heat lost by coolant = heat gained by air)

m˙a=QcpaΔTa=43.21.005×25=1.72 kg/s\dot m_a = \frac{Q}{c_{pa} \Delta T_a} = \frac{43.2}{1.005 \times 25} = 1.72\ \text{kg/s}

For air density 1.1 kg/m³ at the radiator, this is 1.72/1.1=1.561.72/1.1 = 1.56 m³/s.

Fuel energy rate

E˙f=43.20.28=154.3 kW\dot E_f = \frac{43.2}{0.28} = 154.3\ \text{kW}

Answer: Q=43.2Q = 43.2 kW; coolant flow =87.4= 87.4 L/min; air flow =1.72= 1.72 kg/s; fuel energy =154= 154 kW.

✓ Answer checked by Claude Opus 5.5 (high effort)

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