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Cooling water and radiator air flow calculation

Short answer

From brake power 60 kW and brake thermal efficiency 30%, fuel energy input is 200 kW, so fuel consumption = (200/44000)×3600 = 16.4 kg/h. Coolant heat load is 30% of 200 kW = 60 kW; water mass flow = 60/(4.187×10) = 1.43 kg/s. Air flow through radiator = 60/(1.005×15) = 3.98 kg/s. Results: (a) 16.4 kg/h fuel, (b) 1.43 kg/s water, (c) 3.98 kg/s air.

A four-stroke engine develops a brake power of 60 kW with a brake thermal efficiency of 30%. The coolant removes 30% of the fuel energy. If the temperature rise of the cooling water is limited to 10 K, find (a) the fuel consumption in kg/h for a lower calorific value of 44 000 kJ/kg, (b) the mass flow rate of the cooling water, and (c) the mass flow rate of air through the radiator if the air temperature rises by 15 K. Take cpw=4.187c_{pw} = 4.187 kJ/kg K and cpa=1.005c_{pa} = 1.005 kJ/kg K.

(a) Fuel energy and fuel consumption

Qfuel=BPηbth=600.30=200 kWm˙f=QfuelLCV=20044 000=4.545×10−3 kg/s=16.36 kg/h\begin{aligned} Q_{fuel} &= \frac{BP}{\eta_{bth}} = \frac{60}{0.30} = 200\ \text{kW} \\ \dot{m}_f &= \frac{Q_{fuel}}{LCV} = \frac{200}{44\,000} = 4.545 \times 10^{-3}\ \text{kg/s} = 16.36\ \text{kg/h} \end{aligned}

(b) Heat to coolant and water flow

Qw=0.30×200=60 kWm˙w=Qwcpw ΔTw=604.187×10=1.433 kg/s=86 kg/min\begin{aligned} Q_w &= 0.30 \times 200 = 60\ \text{kW} \\ \dot{m}_w &= \frac{Q_w}{c_{pw}\,\Delta T_w} = \frac{60}{4.187 \times 10} = 1.433\ \text{kg/s} = 86\ \text{kg/min} \end{aligned}

(c) Air through the radiator

The radiator rejects all of the heat taken by the water to the air (steady state):

m˙a=Qwcpa ΔTa=601.005×15=3.98 kg/s\dot{m}_a = \frac{Q_w}{c_{pa}\,\Delta T_a} = \frac{60}{1.005 \times 15} = 3.98\ \text{kg/s}

At an air density of about 1.15 kg/m3^3 this is about 3.46 m3^3/s.

Answer: (a) 16.4 kg/h of fuel; (b) 1.43 kg/s of cooling water; (c) 3.98 kg/s of air.

✓ Answer checked by Claude Opus 5.5 (high effort)

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