---
title: "Cooling water and radiator air flow calculation"
subject: "Internal Combustion Engines"
chapter: 7
key: "semester_6/internal_combustion_engines:engine-coolant-flow-radiator-numerical"
source: https://www.bctengineer.com/q/semester_6/internal_combustion_engines/engine-coolant-flow-radiator-numerical
site: BCT Engineering
---

# Cooling water and radiator air flow calculation

Internal Combustion Engines · Chapter 7: Engine cooling

- Asked in: year not given, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/internal_combustion_engines/engine-coolant-flow-radiator-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_6/internal_combustion_engines/chapter-7

## Question

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 $c_{pw} = 4.187$ kJ/kg K and $c_{pa} = 1.005$ kJ/kg K.

## 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.

## Answer

#### (a) Fuel energy and fuel consumption
$$
\begin{aligned}
Q_{fuel} &= \frac{BP}{\eta_{bth}} = \frac{60}{0.30} = 200\ \text{kW} \\
\dot{m}_f &= \frac{Q_{fuel}}{LCV} = \frac{200}{44\,000} = 4.545 \times 10^{-3}\ \text{kg/s} = 16.36\ \text{kg/h}
\end{aligned}
$$

#### (b) Heat to coolant and water flow
$$
\begin{aligned}
Q_w &= 0.30 \times 200 = 60\ \text{kW} \\
\dot{m}_w &= \frac{Q_w}{c_{pw}\,\Delta T_w} = \frac{60}{4.187 \times 10} = 1.433\ \text{kg/s} = 86\ \text{kg/min}
\end{aligned}
$$

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

$$
\dot{m}_a = \frac{Q_w}{c_{pa}\,\Delta T_a} = \frac{60}{1.005 \times 15} = 3.98\ \text{kg/s}
$$

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

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the fuel energy rate found from brake power and efficiency?
   A. Q = BP x efficiency = 18 kW
   B. Q = BP / brake thermal efficiency = 200 kW
   C. Q = BP + 30% = 78 kW
   D. Q = BP / 0.7 = 85.7 kW
   Answer: B. Q_fuel = 60/0.30 = 200 kW, so fuel flow = 200/44000 = 16.4 kg/h. (Key point: Fuel energy from efficiency)

2. How is the cooling water flow found, and what is it?
   A. 60/(4.187 x 10) = 1.43 kg/s
   B. 200/(4.187 x 10) = 4.78 kg/s
   C. 60 x 4.187 x 10 = 2512 kg/s
   D. 60/(1.005 x 10) = 5.97 kg/s
   Answer: A. Heat to coolant is 0.30 x 200 = 60 kW; mass flow = Q/(cp dT) = 60/41.87 = 1.43 kg/s. (Key point: Water flow)

3. How is the air flow through the radiator found?
   A. Heat rejected by the water divided by cpa x air temperature rise = 3.98 kg/s
   B. Brake power divided by cpa x 15
   C. Fuel energy divided by cpa x 15
   D. Water flow x 15
   Answer: A. In steady state the radiator passes all 60 kW to the air: m = 60/(1.005 x 15) = 3.98 kg/s. (Key point: Radiator air flow)

## Related in Internal Combustion Engines

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