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Heat-to-Coolant Determination for Race Car

  1. Dec 31, 2009 #1
    Happy New Year 2010 everyone,

    I need your help with the determination of heat rejection to coolant from engine. I need to define this in order to finalise the geometry of a suitable radiator.

    I have an Excel spreadsheet from my team mate about the results obtained at the lab last year. I believe the lab was to determine the heat rejection from the engine to coolant by measuring in and out temperatures of coolant (100% water) and plug those in a well known expression

    Q = m X c X delta(T)

    Here is the description from the guy from last year.

    "The data for heat rejection calculations was acquired during the engine testing at Tickford. Temperature and pressure data of the coolant was collected against engine speed. A Grundfos water pump was used during testing, supplying coolant at a constant rate of 0.9 kg/s. The coolant is 100% water. This data was used to calculate heat rejection to the coolant against engine speed using Equation 7 1. The heat exchanger used during the test was a water to water module with flow rate controls on the cool water side. These were set such that the coolant into the engine after warm up was at 700C."

    I then plotted the graphs comparing the heat (from calculation) to the developed power by engine. A bizzarre result arisen - the heat to coolant is a lot greater than engine power, not equal to engine power according to rule of thumb saying 33% of total energy from fuel is equally converted to driving power and heat rejection to cooling system.

    Can anyone suggest what happens? Was there something wrong? What should I do next?

    Should I stick with a more good-looking, but unusual result from the experiment or rely on the less-academic conventional rule of thumb which take the heat as equal (or 70% suggested by some rad manufacturers) of the engine power?

    If I went for the experimental one, I would end up having so much bigger rad (around 80 kW from graph), opposing to everyone's expectation as they think the current rad is too big. We did have an overheating issue between the competition last year which I think was caused by too small rad. The current rad is design for dissipating only 37 kW of heat.

    Can anybody reckon me some other more accurate method on determining the desire heat rejection rate of a new radiator?

    Thank you very much in advance.
  2. jcsd
  3. Dec 31, 2009 #2


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    Water at 700C? That's a high pressure coolant system!

    Yes there's something wrong.

    Assuming you mean 70 degrees C, there are a few possibilities:

    1. Your brake power measurement is inaccurate
    2. Your heat to water measurement is inaccurate
    3. You've buggered up a calculation somewhere

    Show us your values and we'll have a look. Heat to jacket water is a notoriously difficult thing to measure accurately because you're dealing with such low temperature differentials, and mass flow is difficult to determine accurately (are you SURE the actual water flow stayed at 0.9kg/min? This is a lot for a small engine. Is that the flow corresponding to the temperature differential you're using, or for the primary side of the heat exchanger?).

    Don't stick with a 'rule of thumb', work out where you've gone wrong, you should be able to measure within a couple of percent of actual heat rejection with k-type thermocouples and a half-decent flow meter.

    Finally, while your engine will be rejecting most heat to coolant at full load (i.e. when pelting along) but your flow across the radiator will be high here too; don't forget to consider low speed/stationary use where the air flow will be low though you may still have bucket loads of heat to reject.
  4. Dec 31, 2009 #3


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    Not sure if it helps, but have you seen Ranger Mike's thread here in the ME forum on formula race car cooling?


  5. Jan 2, 2010 #4
    Happy New Year brewnog

    Thank you very much for your response. Sorry, there was something wrong with the water temp. I meant 70 C.

    Regarding the coolant flow, that is even higher as it was 0.9 kilogram per "second". Unfortunately, that is the only information I have from the lab. I agree with you that this is too high, and this is why the heat rejection is unrealistically high ???

    Do you think I should try to get a performance curve of the water pump at different engine speed? What I know is that the water pump used in the lab is not the same as the water pump used in the engine and during the real drive. Also, the real water pump is the original one that comes with the engine - Honda CBR600RR motorcycle engine.

    1. your suggestion about the measured power - I think this is quite correct, I will show you the data we have in a spreadsheet

    2. Yes, quite agree there was something wrong with the heat rejection calculation
  6. Jan 2, 2010 #5

    Here is what I have from last year;

    The first picture shows the logged engine power and torque, also fuel consumption rate [kg/hr] at different engine speed.

    The ones that we use are highlighted in pale blue, the first column shows the engine speeds. The second and third show engine torque and power espectively.

    The last column shows the fuel flow rate in kg/hr

    Attached Files:

  7. Jan 2, 2010 #6
    The second attachment is another data logged during the test.

    It shows two coolant temperatures going into and coming out from the "engine" highlighted in blue and red respectively.

    Again, at different engine speed

    Attached Files:

  8. Jan 2, 2010 #7
    Now, I will show you my calculation.

    First, I determined the potential power from combusting the fuel.

    Using >>> P [kW] = fuel mass flow rate [kg/s] X LCV

    I then obtained the second column (total power from buring fuel)

    The third column is the heat to coolant I would obtain if the 33% rule of thumb was applied.

    The forth column shows the actual percentage of the driving power from engine compared to the total power available from burning fuel.

    Attached Files:

  9. Jan 2, 2010 #8
    The last calculation was the determination of heat-to-coolant by calculation.

    I used the expression; Q [kW] = coolant mass flow rate [kg/s] X heat capacity (roughly 4.186) X temp. difference [in K or C]

    Attached Files:

    • heat.jpg
      File size:
      59.5 KB
  10. Jan 2, 2010 #9

    Sorry for not attaching the graphs last time. Here you go!

    First picture shows the torque and power curves of the current engine, would not be any greater after a few modifications I believe.

    Second picture shows the comparison of heat-to-coolant and the engine power - as I have mentioned, heat rejection to coolant is unrealistically greater than engine power !!! really strange.

    Third picture shows percentages of heat rejection to coolant and power output to the total available power from combusting the fuel. - again, the heat rejection was around 40% whereas the power output was only about 25% Is there any comment on this>?

    Finally, what would I do to finalise my target heat rejection of the cooling system, more specifically the radiator?

    Attached Files:

  11. Jan 3, 2010 #10


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    I'd definitely consider rechecking your water flow.

    Can you upload your spreadsheet somewhere?
  12. Jan 3, 2010 #11
  13. Jan 3, 2010 #12


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    Cheers, looks like that's a new version of Excel or something, I'll try and have a look at it tomorrow on my own computer.
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