Engine Coolant Heat Dissipation Calculations

[Chestermiller]

Okay sir, the drive cycle is not 12 minutes. Its 1597 seconds, or 26 mins. and 37 secs. Under these circumstances, would you say 18 kWh of fuel burnt is too less? Because my coolant calculation average power comes to 49 kW, which is close to your 50.8 kW.

Total energy in 1597 seconds (0.443611 hours) =(49*0.443611) = 21.74 kWh

I calculate the fuel consumption as follows:
mean air flow = 0.0127 kg/sec (from OBD data)
considering air-fuel ratio of 14.7
mean fuel flow = 0.0127/14.7 = 8.6395 x 10^-4 kg/sec
In 1597 seconds, total fuel = 1.3797 kg

Calorific value of 47300 kJ/kg , means I burnt 65260.81 kJ = 18.13 kWh

Which part of my calculations is wrong?

How many miles were covered in this 26 minutes, and what was the gas mileage?

Was the engine at temperature during this test, or was it cold to start the test? If the engine wasn't at temperature, then the system was not operating at steady state and all bets are off. Also, of course, when the engine is cold, most of the coolant flow bypasses the radiator.

In my engineering judgement, the calculation of the maximum possible energy consumption, 18 kWh was correct, and it was the radiator calculation that was inaccurate. There are only 3 numbers that go into the radiator calculation: the temperature rise, the heat capacity, and the flow rate. Which of these three do you think is the inaccurate number? Everything points to overestimating the coolant flow rate through the radiator.

Chet

 

[Chestermiller]

BTW, please see the following: https://www.google.com/search?q=car radiator design&oq=car radiator de&aqs=chrome.1.69i57j0l3.7506j0j4&sourceid=chrome&ie=UTF-8

First reference: Designing a more effective car radiator.

Chet

 

[Jay_]

How many miles were covered in this 26 minutes, and what was the gas mileage?

Given that the average velocity was 12.2159 m/s, in 1597 seconds, we went 12.2159*1597 = 19508.7923 meters or 12.12 miles.

Fuel calculation method 1 (based on mileage)

Assuming city mileage of 19mpg, I burnt (12.216/19) = 0.64 gallons. Given the calorific value based on volume of 33.3 kWh/USgal (from wikipedia). I used 21.26 kWh.

Fuel calculation method 2 (based on fuel flow)

Given the mean air flow = 0.0127 kg/sec and the air-fuel ratio is 14.7, I get a mean fuel flow of 0.0127/14.7 = 8.6395 x 10-4 kg/sec.

In 1597 seconds, total fuel = 1.3797 kg

Taking even the higher calorific value of 47300 kJ/kg , means I burnt 65260.81 kJ = 18.13 kWh

Regarding the flow rate and the coolant heat dissipation

In page 2 of the link, they mention their radiator dissipates 70729 J/s or 70.73 kW, this is more than my value of 49 kW or the sample calculation you showed too.

Also, in page 6 of the link you provided they have the volumetric flow rate to be 30 gpm. Mine is even lesser than this at 28.5 gpm - it certainly seems to be in the range of typical values. I just can't see why it doesn't make sense when I compare them.

 

[Chestermiller]

How many miles were covered in this 26 minutes, and what was the gas mileage?

Given that the average velocity was 12.2159 m/s, in 1597 seconds, we went 12.2159*1597 = 19508.7923 meters or 12.12 miles.

Fuel calculation method 1 (based on mileage)

Assuming city mileage of 19mpg, I burnt (12.216/19) = 0.64 gallons. Given the calorific value based on volume of 33.3 kWh/USgal (from wikipedia). I used 21.26 kWh.

Fuel calculation method 2 (based on fuel flow)

Given the mean air flow = 0.0127 kg/sec and the air-fuel ratio is 14.7, I get a mean fuel flow of 0.0127/14.7 = 8.6395 x 10-4 kg/sec.

In 1597 seconds, total fuel = 1.3797 kg

Taking even the higher calorific value of 47300 kJ/kg , means I burnt 65260.81 kJ = 18.13 kWh

The above results all seem to be consistent with one another. This is what I calculate also. The radiator design paper indicates that only 1/3 of the heat of combustion is removed by the radiator (see Intro). This is consistent with what we have seen elsewhere. The rest of the combustion energy is supposed to drive the car mechanically and to go out the exhaust.

Regarding the flow rate and the coolant heat dissipation

In page 2 of the link, they mention their radiator dissipates 70729 J/s or 70.73 kW, this is more than my value of 49 kW or the sample calculation you showed too.

Also, in page 6 of the link you provided they have the volumetric flow rate to be 30 gpm. Mine is even lesser than this at 28.5 gpm - it certainly seems to be in the range of typical values. I just can't see why it doesn't make sense when I compare them.

It isn't clear what they did in their tests of the radiator. They may have done the tests in the laboratory without a car attached. But I'm not sure about this. If the tests were done in the laboratory, then they could have used much higher heat loads then in an actual car. They did show values of car speeds, but these were very low, and it isn't clear how these relate.

Chet

 

[Jay_]

Okay. So that means its the coolant energy that is higher than the actual number. And since we have everything that is makes sense, we are down to changing the coolant flow rate?

 

[Chestermiller]

Okay. So that means its the coolant energy that is higher than the actual number. And since we have everything that is makes sense, we are down to changing the coolant flow rate?

Yes. The heat of combustion is an absolute indicator of the maximum amount of energy available, so, to me, that means that we are down to changing the coolant flow rate.

The maximum horsepower that the car engine/gasoline is capable of delivering is measured on a dynomometer with the engine red lined in low gear and a huge torque applied to the wheels. Under these conditions, we are expecting a much much greater consumption of gasoline and a much greater measured mechanical power, on the order of 100's of kW. And the combustion energy will have to be substantially greater than this. And the energy removed by the radiator will also have to be high (if the car is run under these conditions for any significant interval of time). So 70 kW for the radiator cooling is not unreasonable under extreme conditions. But, under ordinary cruising conditions, it must at least be much less than the total energy delivered by the gasoline.

Chet

 

[Jay_]

Okay, I got a reference for the flow rate and it works. I get 31.2938% of the total fuel as the coolant heat energy, which I think is consistent with literature. From this link : http://talk.newagtalk.com/forums/thread-view.asp?tid=349481&DisplayType=flat&setCookie=1

The post by user Gerald J. I also got a pdf copy of the book Mark's Standard Handbook for Mechanical Engineers, so I will cite that in the report.

He mentioned it says 2 gallons/bhp-hour for large engines, to 4 gallons/bhp-hour for small engines. I took the middle number, and flow rate as 3 gallons/bhp-hour. This equals 0.05 gallons/minute/bhp, which is much like the constant we were trying to adjust earlier.

 

[Jay_]

However, he posted saying this is for a 90 degF (or 32 degC) rise. Should I change the value based on delT?

 

[Chestermiller]

To orient yourself on this (i.e., get yourself in the ballpark), you need to do a calculation where the temperature rise is about 15C, and the rate of heat removal is about 1/3 the rate of energy delivery by the gasoline. In your test, the average rate of energy delivery by the gasoline was about 41 kW. So the rate of heat removal by the radiator was about 13.5 kW. This would break down to about 0.25 kg/s through the radiator (0.25 L/s). This translates into about 4 gpm of coolant flow through the radiator (or 6 gpm with a 10 C temperature rise).

 

[Jay_]

As of now, in my city drive cycle the calculations seem correct. I get average coolant power = 12.3115 kW, fuel_power = 39.3417 kW, so its 31.29%

However, for my highway driving, I get coolant power = 11.78 kW, fuel_power = 64.70 kW, so its just 18.61% Does this make sense?

I had to estimate the radiator cool side temperature, and I put it as 80 deg C for both cases. Should I reduce it in the highway drive cycle?

Because this link seems to indicate that. See the temperature graph: http://www.bimmerfest.com/forums/showthread.php?t=499204 In the "drive on highway" part the radiator cool side temperature is much lower. So question is, should I lower it in my code as well. Maybe take it as 70 degC, instead of 80 degC like in the city drive?

Here are the percentages for various values of Tcold in the highway drive:
Deg C Coolant heat as percent of total fuel burnt

80 -- 18.16%
75 -- 22.60%
70 -- 27.05%
65 -- 31.49%
60 -- 35.94%

 

[Chestermiller]

Hey Jay. This is starting to shape up now. Good going.

I thought you were measuring the coolant temperatures at the inlet and outlet of the radiator. Is this not the case? If not, you now have two highly uncertain parameters involved in the heat load calculation, namely, the coolant flow rate and the coolant temperature out of the radiator. Please tell me you can measure the coolant temperature coming out of the radiator.

Chet

 

[Jay_]

I can't. But I have that graph and I am getting it from literature. I showed it to my professor, he said its okay since we are also out of time. Now, what I would like to know is based on automotive engineering:

1. Does that graph in the previous link make sense? Specifically, is the temperature of the cool side less in a highway drive as shown in the link above?
2. If not, what values would be good for the cool side temperatures? 80 deg C works well for the city drive.

 

[Chestermiller]

I can't. But I have that graph and I am getting it from literature. I showed it to my professor, he said its okay since we are also out of time. Now, what I would like to know is based on automotive engineering:

1. Does that graph in the previous link make sense? Specifically, is the temperature of the cool side less in a highway drive as shown in the link above?
2. If not, what values would be good for the cool side temperatures? 80 deg C works well for the city drive.

That graph in the previous link makes some sense. The lower the coolant flow rate through the radiator, the lower the outlet temperature is going to be (compared to the inlet temperature). Do you expect the coolant flow rate to be lower during highway driving, and do you expect the inlet temperature to be about the same as for city driving?

Chet

 

[Jay_]

My professor said I needn't assume any difference in the outlet temperatures, so I went with that. I also assumed the coolant flow to be based on the bhp, which depends on the rpm. So depending on that I don't get much of a difference.

 

[Chestermiller]

Okay. He's the boss.

Chet