# Modelling heat loss to cylinder walls in a piston engine.

I am building a mathematical model of a piston engine which calculates the pressure and temperature of the cylinder gas every 1° of crank angle.

I'm having some difficulty with the heat transfer to and from the cylinder wall (to when the Tg > Tw and from when Tg < Tw).

I found the heat transfer equation on EngineeringToolbox.com

q = 2 π k (ti - to) / ln(ro / ri) (1)

where

q = heat transferred per unit time per unit length of cylinder or pipe (W/m, Btu/hr ft)

k = thermal conductivity of the material (W/m.K or W/m oC, Btu/(hr oF ft2/ft))

to = temperature outside pipe or cylinder (K or oC, oF)

ti = temperature inside pipe or cylinder (K or oC, oF)

ln = the natural logarithm

ro = cylinder or pipe outside radius (m, ft)

ri = cylinder or pipe inside radius(m, ft)

http://www.engineeringtoolbox.com/conductive-heat-loss-cylinder-pipe-d_1487.html

Which gives me an heat loss in W. How do I convert the loss in W to a reduction in temperature in °K?

I am building a mathematical model of a piston engine which calculates the pressure and temperature of the cylinder gas every 1° of crank angle.

I'm having some difficulty with the heat transfer to and from the cylinder wall (to when the Tg > Tw and from when Tg < Tw).

I found the heat transfer equation on EngineeringToolbox.com

q = 2 π k (ti - to) / ln(ro / ri) (1)

where

q = heat transferred per unit time per unit length of cylinder or pipe (W/m, Btu/hr ft)

k = thermal conductivity of the material (W/m.K or W/m oC, Btu/(hr oF ft2/ft))

to = temperature outside pipe or cylinder (K or oC, oF)

ti = temperature inside pipe or cylinder (K or oC, oF)

ln = the natural logarithm

ro = cylinder or pipe outside radius (m, ft)

ri = cylinder or pipe inside radius(m, ft)

http://www.engineeringtoolbox.com/conductive-heat-loss-cylinder-pipe-d_1487.html

Which gives me an heat loss in W. How do I convert the loss in W to a reduction in temperature in °K?

Whats missing is a loss to either air or water. Grab a heat/mass textbook.

Ranger Mike
Gold Member
one area you have to account for is heat carried away by motor oil...dont forget if the inside bottom of the piston was not getting sprayed with high pressure oil the piston would melt and the wrist pin would seize. Also the piston rings drag and cause friction as well as carry away significant heat. A more difficult to determine area to find temperature transfer is the valves. They are at the forefront of heat transfer interacting with cool intake mixture and hot exhaust gasses and finally oil at the valve spring keeper area..as a matter of fact the whole valve is exposed to heat transfer.

Mike. All relevant yes, but if I can't even get the loss to the walls sorted then the other losses are moot.

Hey kozy, not an engineer here but got a quick pertinent question -

When the ignition timing of a spark ignited gasoline engine is retarded from the optimal value the peak pressure and peak temperature decrease and I believe the fuel continues to burn for more crank angles so more of the cylinder walls are exposed to combustion heat / there is more heat rejection to the coolant.

That's the explanation of the relationship that I've mostly found that exists between ignition timing and heat transfer to the engine.

Not sure if that is accurate, I get the impression that there is a fixed amount of heat energy in the fuel, if less is converted to useful work because of low pressures on the piston, more of it ends up going into the engine as waste heat and out the exhaust.

I think I've also read some things about the idle speed of engines and fuel consumption being related because IIRC if the walls cool off they absorb more heat from the combustion reaction, that would have created useful work. The other thing related is the engine's need to maintain the catalyst at high temps. Though I don't know much about whether it can cool off enough to lose effectiveness.

I also remember reading that larger displacement engines have a proportional loss in fuel economy because of the larger surface area of the cylinder walls being like a large heatsink. The bigger the cylinder the more heat rejection, inevitably, it seems.

Anyway I'm not trying to help you just making some comments, would appreciate anyone who can clarify these things more.

And I am not sure if anyone mentioned it but the residual exhaust gas will absorb heat as well.

Hey kozy, not an engineer here but got a quick pertinent question -

When the ignition timing of a spark ignited gasoline engine is retarded from the optimal value the peak pressure and peak temperature decrease and I believe the fuel continues to burn for more crank angles so more of the cylinder walls are exposed to combustion heat / there is more heat rejection to the coolant.

That's the explanation of the relationship that I've mostly found that exists between ignition timing and heat transfer to the engine.

Not sure if that is accurate, I get the impression that there is a fixed amount of heat energy in the fuel, if less is converted to useful work because of low pressures on the piston, more of it ends up going into the engine as waste heat and out the exhaust.

I think you are right there, it's just figuring out how to model that heat loss at every ° interval that is stumping me a little.

Hi kozy

and I think you must consider the heat transfer by conduction through the lubricant and also the heat transfer by convection of the exhaust gasess.

THe heat transfer rate is congruent with values I have found, however I am unsure of how I apply this to my model.

I have a for loop, and for each ° crank angle, it calculates the either the pressure, and then the resulting temperature, or during combustion, the temperature and then the resulting pressure.

For each interval, I get a pressure value in KPa and a temperature in °K. I am wondering how to apply 2.5 MW/m2 heat transfer to these values...

Your second link looks like it might cover this, but I need to get my head around it!

Last edited:
Chestermiller
Mentor
You don't have to model the heat transfer to the wall over every time interval. Even though the heat flux to the wall is time dependent over each time interval, the rpms are so high that the temperature just a slight distance into the wall is going to be constant due to the thermal inertia of the metal. So all you need to do is solve the steady state heat conduction equation within the metal for heat transfer from the cylinder wall to the coolant. You use the time average heat flux at the cylinder wall as the boundary condition.

Chet