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Chapter 7 · 6 hours

Engine cooling

Practice questions

Practice questions and answers

4 exam-style questions on this chapter, written for this site from the official syllabus. We haven’t found past IOE papers for this subject yet; if you have some, share them in the community.

  • Practice · 3+4 marks

Why is cooling of an IC engine necessary? Describe the air cooling system and differentiate between air-cooled and water-cooled engines.

Answer

Need for cooling

About 25-35% of the fuel energy goes to the cylinder walls. Gas temperatures reach 2000 degrees C or more, so without cooling:

  • The lubricating oil film burns off and the piston and rings seize.
  • The metal loses strength; valves and piston crowns burn or crack because of thermal stress.
  • Knock and pre-ignition become severe in SI engines, and volumetric efficiency drops since the charge is heated.

The cooling system must remove only the necessary heat. Overcooling lowers thermal efficiency and causes acid corrosion and oil dilution. The wall temperature is kept at about 150-200 degrees C (liner surface), with the oil film below 160 degrees C.

Air cooling system

Heat is taken directly from the cylinder and head to the atmosphere by fins cast on their outer surfaces. The fins increase the surface area. Air flows over them due to the vehicle motion (motorcycles) or a fan and shrouds (tractors, small stationary engines). The cylinders are spaced and finned, and baffles direct the air. The fin area, air speed and material (aluminium has high conductivity) set the cooling.

Air-cooled versus water-cooled

BasisAir cooledWater cooled
MediumAir through finsWater in jackets, radiator
ComponentsFins, fan, shroudPump, radiator, thermostat, hoses, fan
Weight, costLighter, cheaperHeavier, costly
MaintenanceEasy, no leakage or freezingLeaks, scale, freezing need antifreeze
CoolingLess uniform, less effectiveUniform and effective
Warm-upFastSlower, thermostat controlled
NoiseMoreLess (water jacket damps noise)
Power per sizeLower; high compression ratio limitedHigher
UseMotorcycles, small engines, aircraftCars, trucks, large engines
  • Practice · 8 marks

Describe the thermosyphon and forced-circulation (pump) water cooling systems with sketches. Explain the function of the main components of a water cooling system.

Answer

Thermosyphon system

        radiator (top tank)
        ^                 |
  hot water rises     cooled water falls
        |                 v
   cylinder jacket <-- radiator bottom tank

Water heated in the jacket becomes lighter and rises to the top of the radiator. It is cooled by air as it flows down through the tubes, and the denser cold water returns by gravity to the bottom of the jacket. No pump is needed, but the radiator must be higher than the engine and the flow is slow and depends on temperature. It is used on old and small engines.

Pump (forced circulation) system

 Jacket --> [Thermostat] --> Radiator top tank
   ^                              |
   |                          tubes + fan
   +---- [Water pump] <--- bottom tank
         (bypass when thermostat closed)

A centrifugal pump driven by the crankshaft belt drives the water through the jackets and radiator. The flow rate is fast and controlled, giving more uniform cooling. The radiator may be mounted at any level. Used on nearly all automobile engines.

Components

ComponentFunction
Water jacketsPassages around cylinders and head that carry heat to the water
Water pumpCentrifugal pump; circulates the coolant
RadiatorTop tank, honeycomb of tubes with fins, bottom tank; transfers heat from water to air
FanDraws air through the radiator at low vehicle speed
ThermostatWax or bellows valve; closes below about 80-85 degrees C so that the engine warms up quickly, then opens
Radiator pressure capHolds 0.5-1.0 bar, raising the boiling point
Hoses and drain cockConnections and draining
Temperature gauge/sensorMonitor the coolant
Antifreeze (ethylene glycol)Prevents freezing and raises boiling point
  • Practice · 4 marks

Write short notes on the variation of gas temperature in the cylinder during the cycle and the flow of heat from the gas to the coolant.

Answer

Variation of gas temperature

The gas temperature changes strongly during the cycle.

 T (K)
 2500|        /\
     |       /  \
 1500|      /    \.
     |     /       ..
  700| ___/            ..____
  300|---------------------------> crank angle
     suction comp. comb. exp. exhaust
  • During suction and compression the charge temperature rises from about 300-350 K to 700-900 K.
  • After ignition it rises quickly to a peak of about 2200-2800 K in an SI engine (about 2000 K in CI), a little after TDC.
  • During expansion it falls to about 1200-1500 K at exhaust valve opening and then to about 700-900 K in the exhaust.
  • The peak lasts only a very short time. The mean temperature over the cycle is about 1000 K, while the metal wall stays at 400-500 K.

Heat flow

Heat flows from the hot gas to the wall by convection and radiation (radiation matters mainly in CI engines because of soot), then through the wall by conduction, and finally from the outer surface to the coolant by convection:

Q=Tg−Tc1hgA+tkA+1hcAQ = \frac{T_g - T_c}{\dfrac{1}{h_g A} + \dfrac{t}{kA} + \dfrac{1}{h_c A}}

The wall temperature varies only slightly through the cycle because the metal thermal inertia averages the gas temperature fluctuations, so heat flow is based on the mean gas temperature. The heat flux is highest at the piston crown, exhaust valve and cylinder head, and depends on gas velocity, turbulence, load and speed. About 25-35% of fuel energy goes to the coolant.

  • Practice · 6 marks

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.

Answer

(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.

Written from the official syllabus. Questions and answers are written for this site; check them against your class notes.

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