---
title: "Coolant heat rejection and radiator air flow"
subject: "Automotive Technology"
chapter: 4
key: "semester_5/automotive_technology:radiator-heat-load-numerical"
source: https://www.bctengineer.com/q/semester_5/automotive_technology/radiator-heat-load-numerical
site: BCT Engineering
---

# Coolant heat rejection and radiator air flow

Automotive Technology · Chapter 4: Engine Cooling and Lubrication

- Asked in: year not given, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/automotive_technology/radiator-heat-load-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_5/automotive_technology/chapter-4

## Question

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.

## 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.

## Answer

**Given:** $\dot m_c = 1.5$ kg/s, $c_{pc} = 3.6$ kJ/kg K, $\Delta T_c = 8$ K, $\rho_c = 1030$ kg/m³, $\Delta T_a = 25$ K, $c_{pa} = 1.005$ kJ/kg K.

#### (i) Heat rejected

$$
Q = \dot m_c c_{pc} \Delta T_c = 1.5 \times 3.6 \times 8 = 43.2\ \text{kW}
$$

#### (ii) Coolant volume flow (water pump capacity)

$$
\dot V = \frac{\dot m_c}{\rho_c} = \frac{1.5}{1030} = 1.456\times10^{-3}\ \text{m}^3/\text{s} = 1.456\ \text{L/s} = 87.4\ \text{L/min}
$$

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

$$
\dot m_a = \frac{Q}{c_{pa} \Delta T_a} = \frac{43.2}{1.005 \times 25} = 1.72\ \text{kg/s}
$$

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

#### Fuel energy rate

$$
\dot E_f = \frac{43.2}{0.28} = 154.3\ \text{kW}
$$

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

## Check yourself

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

1. Which expression gives the heat rejected to the coolant, and what is its value?
   A. m cp dT = 1.5 x 3.6 x 8 = 43.2 kW
   B. m dT/cp = 3.3 kW
   C. m cp = 5.4 kW
   D. m cp dT = 1.5 x 4.18 x 8 = 50.2 kW
   Answer: A. Q = m cp dT using the given cp of 3.6 kJ/kg K gives 43.2 kW. (Key point: Heat rejected)

2. How is volume flow found, and what is the result?
   A. mass flow x density = 1545 L/s
   B. mass flow / density = 87.4 L/min
   C. mass flow / density = 87.4 L/s
   D. mass flow x density = 87.4 L/min
   Answer: B. V = 1.5/1030 = 1.456 L/s, which is 87.4 L/min. (Key point: Coolant volume flow)

3. What principle gives the air mass flow?
   A. Heat lost by coolant equals heat gained by air: m_a = 43.2/(1.005 x 25) = 1.72 kg/s
   B. Air flow equals coolant flow
   C. Air flow = Q x cp
   D. Air gains twice the heat lost
   Answer: A. Energy balance across the radiator gives 1.72 kg/s of air. (Key point: Air flow by heat balance)

4. If 43.2 kW is 28 % of the fuel energy, what is the fuel energy rate?
   A. About 12.1 kW
   B. About 154 kW
   C. About 43.2 kW
   D. About 1.5 kW
   Answer: B. Fuel energy = 43.2/0.28 = 154.3 kW. (Key point: Fuel energy)

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