Engine cooling systems and radiator heat transfer

Why it matters

Roughly a quarter to a third of the fuel energy in a typical IC engine must be carried away by the cooling system; the exact split depends on the engine and load, so take it from test data. If the coolant cannot reject that heat, cylinder heads crack, gaskets fail and oil breaks down; if it rejects too much, the engine runs cold, wears faster and emits more. Sizing the radiator, pump and fan is a direct application of the first law, convection and heat-exchanger theory, and the same methods now apply to EV battery and motor cooling loops.

Key ideas

Engine energy balance. Fuel energy splits into brake work, exhaust enthalpy, heat to coolant, heat to oil and radiation and convection from the block. The coolant heat must be removed continuously at full load, in slow traffic and on hot days, which together set the design point.

Liquid cooling circuit.

• Water pump (belt-driven or electric) circulates coolant through the block and head water jackets.
• Thermostat (wax-element valve, typically opening around 80–95 °C) keeps coolant in a bypass loop until the engine is warm, then opens the radiator path. It shortens warm-up and holds a stable operating temperature.
• Radiator: a compact cross-flow heat exchanger, flat aluminium tubes with louvred fins, both fluids unmixed. Coolant-side h is high; air-side h is low, so air-side area (fins) dominates the design.
• Pressure cap: pressurises the circuit (often around 100–140 kPa gauge; check the specification) to raise the coolant boiling point well above 100 °C, increasing the temperature difference to the air and preventing local boiling in the head. An expansion tank takes up volume changes.
• Fan: ram air is enough at road speed; at low speed or idle a mechanical (viscous-coupled) or electric fan provides the airflow.
• Heater core: a small radiator that uses engine heat to warm the cabin.

Coolant. Water has excellent specific heat and conductivity. Ethylene glycol mixtures (often about 50/50) lower the freezing point, raise the boiling point and carry corrosion inhibitors, at the cost of a lower specific heat (about 3.4–3.6 kJ/kg·K for 50/50) and higher viscosity, so more flow is needed for the same duty.

Heat transfer in the radiator. Coolant → tube wall by forced convection; through the wall by conduction; fins and tubes → air by forced convection, with fin efficiency reducing the effective area. Radiation is a small part despite the name. Overall resistance is dominated by the air side, so 1/(UA) ≈ 1/(η_o h_air A_air) plus smaller terms.

Sizing approaches.

• Coolant-side energy balance: Q = ṁ_c c_p,c ΔT_c gives the coolant flow for a chosen temperature drop (typically 5–10 K across the radiator).
• Air-side energy balance gives the air mass flow needed for an acceptable air temperature rise.
• ε–NTU with the cross-flow, both-unmixed relation (or chart) rates a given radiator; the relevant temperature difference is the initial temperature difference ITD = T_coolant,in − T_air,in.

Air-cooled engines (many two-wheelers) use finned cylinders and heads; the fin analysis of a separate lesson applies directly. They are simpler and lighter but run hotter and noisier and have poorer temperature control.

Faults that show up as overheating: low coolant, stuck-closed thermostat, failed fan or pump, a cap that does not hold pressure, and external fouling of radiator fins with dust and insects (an extra resistance on the critical air side) or internal scale.

Formulas

Q̇_c = ṁ_c c_p,c (T_c,in − T_c,out) — heat rejected by coolant, W; ṁ in kg/s; c_p in J/kg·K.

Q̇ = ṁ_a c_p,a (T_a,out − T_a,in) — heat gained by air.

Q̇ = U A F ΔT_lm,CF — LMTD form for cross-flow with correction factor F (from chart).

ITD = T_c,in − T_a,in; Q̇ = ε C_min ITD — ε–NTU form.

NTU = UA / C_min, C_r = C_min/C_max.

ε ≈ 1 − exp[(NTU^0.22 / C_r)(exp(−C_r NTU^0.78) − 1)] — cross-flow, both fluids unmixed (approximate relation; charts give the same).

1/(UA) = 1/(h_c A_c) + R_wall + 1/(η_o h_a A_a) — η_o: overall finned-surface efficiency on the air side.

V̇ = ṁ / ρ — coolant volume flow, m³/s (multiply by 60 000 for L/min).

Worked examples

Example 1 (standard, coolant flow). An engine produces 60 kW of brake power at 30% brake thermal efficiency, and 30% of the fuel energy goes to the coolant. The coolant is a 50/50 glycol–water mix (c_p = 3.5 kJ/kg·K, ρ = 1060 kg/m³), and the design temperature drop across the radiator is 8 K. Find the coolant heat load and the required coolant flow.

1. Fuel energy rate = 60/0.30 = 200 kW.
2. Heat to coolant: Q̇_c = 0.30 × 200 = 60 kW.
3. ṁ_c = Q̇_c/(c_p ΔT) = 60/(3.5 × 8) = 2.14 kg/s.
4. Volume flow = 2.14/1060 = 2.02 × 10⁻³ m³/s ≈ 121 L/min.
5. With plain water (c_p ≈ 4.18 kJ/kg·K) the same duty would need only 1.79 kg/s; glycol costs flow.

Example 2 (GATE level, rating the radiator). The coolant from Example 1 (C_c = 2.14 × 3500 = 7500 W/K) enters the radiator at 95 °C. Air at 35 °C passes through at 2.0 kg/s (c_p = 1005 J/kg·K). The radiator has UA = 1500 W/K, cross-flow with both fluids unmixed. Can it reject 60 kW?

1. C_a = 2.0 × 1005 = 2010 W/K = C_min; C_r = 2010/7500 = 0.268.
2. NTU = UA/C_min = 1500/2010 = 0.746.
3. NTU^0.22 = 0.938; NTU^0.78 = 0.796; exp(−0.268 × 0.796) = exp(−0.2133) = 0.808.
4. ε = 1 − exp[(0.938/0.268)(0.808 − 1)] = 1 − exp(−0.672) = 1 − 0.511 = 0.489.
5. Q̇ = ε C_min ITD = 0.489 × 2010 × (95 − 35) = 59.0 kW.
6. Outlets: coolant 95 − 59 000/7500 = 87.1 °C; air 35 + 59 000/2010 = 64.4 °C.

The radiator falls about 1 kW short at this airflow on a 35 °C day, so the coolant inlet temperature would rise until balance is reached. Remedies: more airflow (fan), more air-side area (UA) or a higher-pressure cap allowing a hotter coolant (bigger ITD).

Common mistakes

• Taking the temperature difference for radiator rating as coolant-out minus air-in; the ε–NTU method uses the inlet-to-inlet difference (ITD).
• Assuming coolant has C_min. Usually air has the smaller capacity rate.
• Using water's specific heat for a glycol mixture.
• Thinking a thermostat controls the fan; it controls coolant flow to the radiator (a separate switch or ECU controls the fan).
• Removing the thermostat to "cure" overheating; the engine then runs cold and the root cause remains.
• Adding radiator area on the coolant side; the controlling resistance is on the air side.

For GATE ME

Questions from this topic usually test energy balances on coolant and air (flow rate, temperature change, heat rejected), LMTD and ε–NTU applied to a radiator or oil cooler, and fin and convection ideas in an engine context. For automotive-flavoured questions, practise reading a short description, identifying C_min and writing Q̇ = ṁ c_p ΔT on both sides before using any heat-exchanger relation.

Quick check

1. A radiator rejects 5000 J in 10 s. What is the heat transfer rate?
2. Coolant flows at 0.1 kg/s with c_p = 4.18 kJ/kg·K, and 1000 W is removed. What is its temperature drop?
3. Why does a pressure cap help cooling?
4. In a car radiator, which fluid usually has C_min?

Answers: 1. 500 W. 2. 1000/(0.1 × 4180) = 2.39 K. 3. It raises the coolant boiling point, preventing boiling and allowing a larger temperature difference to the air. 4. The air.