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Engine Coolant Heat Dissipation Calculations

  1. Sep 24, 2014 #1
    Okay, I created this topic because the other one seem to get no responses?!

    I found this paper that was useful to me :


    It mentions the figure 2.0 - 2.6 L/min/kW. Is this kW, the bhp?

    Question 1:For a Nissan Maxima (3.5 L engine), could I take this number as 2.4?

    To get the mass flow rate I would have to multiply by the density of the coolant. Now I couldn't find the density of the coolant as a function of temperature. But I did find the specific gravity as a function of temperature:


    Question 2: For density, do I multiply the specific gravity with the density of water at 4 degrees C or do I multiply it with the density of water at the given temperature?

    Question 3: Since there is a temperature gradient across the radiator, which temperature do I consider while calculating the fluid properties hot_side temperature, or cold_side temperature or the mean?
  2. jcsd
  3. Sep 24, 2014 #2
    It can't be the bhp. Otherwise the coolant flow would be huge. It must be the power to run the water pump.
    The latter.
    Are you trying to calculate the heat load by determining the heat transfer coefficient inside the tubes, and on the air side of the tubes? The heat transfer coefficient on the air side is going to be difficult because you need to know the air flow rate across the tubes and also the enhancement effect of the fins.

  4. Sep 24, 2014 #3
    Hi Chet,

    No I am not calculating the heat transfer coefficients. I already have the coolant temperatures directly. I also have the specific heat capacity.

    What I need now is the mass flow rate.

    I have the density.

    I have the bhp. I thought the 2.4 L/min/kW was a constant related to the brake horsepower. Lets consider RPM of 4000, I get a volumetric flow rate of 4.6 L/sec. Is that too high?


    From graph 4000 RPM corresponds to 126.5 Wheel horsepower. I times that with 1.22 to get the brake horsepower because typical losses are 20-25% (I took 22%). I times that with 0.7457 to get 115.084 kW

    I multiply that with 2.4 to get 276.2 L/min or 4.6 L/sec


    I have few days Chet, let me know what would be the best way to go. I have tried everything - looking for equations, interpolations based on graphs, there just isn't enough things to go with. And I don't even have the car with me anymore, its not like I can experiment on it again.
    Last edited: Sep 24, 2014
  5. Sep 24, 2014 #4
    Since I have the brake horsepower, I could use the graph on this thread :


    The ordinate seems to give me a ratio (lets call it R). What are the units for 'R' it can't be dimensionless because its (coolant flow rate)/(brake power).

    Question : What I really don't understand is how do I go from RPM and Brake horsepower to Coolant flow rate

    Show me, consider the above case where RPM = 4000 corresponds to 115.084 kW brake horsepower. So, in that case what would my flow rate be?
  6. Sep 24, 2014 #5
    It's not the coolant flow rate. That would not make R dimensionless. It is the radiator heat load (which is the thing you are trying to measure). That 115 brake horsepower seems very low. The usual rule of thumb is 1 hp ~ 1kW (not exactly, but close).

  7. Sep 24, 2014 #6
  8. Sep 24, 2014 #7
    4.6 L/sec is about 75 gpm, which sounds like it could be in the ballpark. One of the graphs you had showed about 60 gpm through the core at 4000 rpm (fully open thermostat valve). So I would go with (rpm x 0.015) for the coolant flow rate through the core.

    You're not going to do much better unless you can get info from somewhere else, such as the pump manufacturer, the car manufacturer, the shop manual, or a good solid book on automotive engineering. Apparently, the references available on the internet are not doing the trick. Your library has auto engineering books as well as shop manuals.

  9. Sep 24, 2014 #8
    I got the value of 115 kW from a graph of the Honda 6th generation vehicle. I forgot which graph showed 60 gpm. Was it in the other thread?
  10. Sep 24, 2014 #9
    Yes. It was a figure on some 1950's coolant systems. Two of the curves showed the flow rate through the core. They were roughly linear with respect to the engine speed.

  11. Sep 24, 2014 #10


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    Regarding the heat load of a radiator:

    Don't forget typical car engines are around 20% efficient. In other words a car with 100kW power output at the crank will consume 500kW of petrol.
    The difference, 400kW is mostly heat, most of which the radiator has to get rid off. ie the heat load is four time greater than engine crank power.
  12. Sep 24, 2014 #11
    Chestermiller could you post me those curves please?
  13. Sep 24, 2014 #12
  14. Sep 24, 2014 #13
    The reason I didn't use that is because its for a V8. Could I really assume the same characteristics for a Honda 2007 V6?

    I feel we are going in circles. Could you lay out what I could do from start to end?

    I can interpolate equations, I have everything except the flow rate.
  15. Sep 25, 2014 #14
    I guess the reason that way you feel the way you do is that you just don't have enough information to do it right. You quoted a figure of 4.6 L/sec. That's about 75 gpm. So, if 60 gpm is too much for a 6 cylinder engine, 75 gpm is even worse. Please understand that you are not going to get the answer exact, and, at this point, you need to make a judgement call. That's why I'm suggesting going with 60 gpm at 4000 rpm. This part of the calculation is not going to be nearly as uncertain as the tailpipe calculation.

    Your professor is handcuffing you by not providing the resources to get what is needed.

  16. Sep 25, 2014 #15
    Okay. So I could just go with the V8 engine data then? You give me a green signal on that, I am going ahead sir.
  17. Sep 25, 2014 #16
    Also, let me know at which temperature should I consider the properties of the ethylene glycol-water mixture?

    Would it be at the mean of the T_cold and T_hot temperatures?
  18. Sep 25, 2014 #17
    Sure. It really doesn't matter much. Average temperature is best.
  19. Sep 25, 2014 #18
    So I could co with the V8 engine data? Would it be approximately the same for a water pump of today?

    I called AC Delco, they said the water pump operates between 250 to 290 bhp. But the lady in the line herself seemed confused and didn't seem to have a technical understanding of things. But would that range seem right?

    If anyone has worked with automotive parts please give some inputs I have about 5 more days to model this and the flow rate is the only thing I don't have.
  20. Sep 25, 2014 #19
    That's what I would do. At this point, what other choice is there?
    I don't know. I would have to spend some time studying a modern auto engineering book to research this. That's why I was suggesting that you get an auto engg book.
    Probably, for the engine.

  21. Oct 2, 2014 #20
    Hey I have something to work with now.

    But which of these data seem incorrect (I have minimum value to maximum value). This is for a 12 mile drive in the city :

    1. Engine coolant mass flow rate --> values from 2.18 kg/sec to 6.05 kg/sec
    2. Engine coolant volumetric flow rate --> 2.1 L/sec to 5.9 L/sec
    3. Difference in temperature between cool side and hot side in radiator --> 10 deg C to 21 deg C
    4. Mass of fuel burnt --> 1.3778 kg
    5. Energy of heat through Coolant --> 96 kWh
    6. Specific Heat of Coolant --> 3404.3 (S.I., at 300 K)
    7. Energy of fuel burnt based on calorific value of 46000 kJ/kg --> 17.6 kWh
    8. Heat rate (power) through coolant --> 122.5 kW to 431 kW

    The problem is of course, heat given from the coolant can't be less than the fuel energy burnt!

    My immediate guess is that the error is in 5. and 8.. But I get 8. through m*c*delT, and I can't say which is the one which is not the typical values.
  22. Oct 2, 2014 #21
    What would be a good estimate for the low side temperature of the radiator on an average? I relied on this graph (below link), and took it roughly as 70 degC for the city drive, and a lower value of 42 degC for the highway drive.

  23. Oct 3, 2014 #22
    It seems to me that the coolant flow rates must be high (by maybe a factor of about 4 or 5). The heat capacity looks a little high. For a 50:50 mixture of eg and water, the reference I saw gives 3140.

  24. Oct 3, 2014 #23
    That 12 minute drive was when the engine was up to temperature, correct?
    And, it was at about 50 mph and 1500 rpm?

  25. Oct 3, 2014 #24
    Quoting myself
    Is that the specific heat value at 300 K? I interpolated these :
    http://homepage.usask.ca/~llr130/physics/HeatCapcityOfAntiFreeze.html. At 27 deg C (300 K) the value is close to what I have in standard units.

    Regarding the flow rate, could someone give me a rule of thumb on this?

    What I have done as of now, is interpolate the :
    1. Interpolate the BHP as a function of RPM
    2. Considered 80% efficiency from the BHP to the water horsepower, and then used the SAE constant of 2.4 L/min/kW to find the volumetric flow rate. Then I multiplied it by density to find the mass flow rate.

    I went ahead with it because it seems to be in the right magnitude. You mentioned in post # 14 to go with 60 gpm, which is 1 gallon/sec or 3.8 L/sec, my figure is very much within that, I would think.

    I may have just calculated the energies wrong. Does 96 kWh for the coolant seem too high? Or does 17.6 kWh for the fuel sound too low? Other parameters
    Average speed : 12.2159, min = 0, max = 23.3 m/s
    Average Engine Speed: 1350.7, min = 700.2, max = 2137.8
  26. Oct 3, 2014 #25
    I used the rule of thumb in this thread (post 2).


    Using that, my flow rate shows from 1.1 L/sec to 3.1 L/sec which is better? I also made my cold side temperature greater (to 80 deg C). But I still get the coolant heat energy to be greater than the energy of the fuel burnt - which is impossible!

    This is how I computed the fuel burnt. I have the per second log of the airflow in kg/sec. I divide that with 14.7 (assuming A-F ratio of 14.7) and get the fuel flow in kg/sec. I add all the values and get the total kg of fuel burnt. I times this with 46000 (calorific value) to get the energy.

    What am I doing wrong? :confused::(
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