Engine Coolant Heat Dissipation Calculations

[Jay_]

Okay, I created this topic because the other one seem to get no responses?!

I found this paper that was useful to me :

http://people.bath.ac.uk/enscjb/airtex.pdf

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:

http://www.engineeringtoolbox.com/ethylene-glycol-d_146.html

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?

Also,
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?

 

[Chestermiller]

Okay, I created this topic because the other one seem to get no responses?!

I found this paper that was useful to me :

http://people.bath.ac.uk/enscjb/airtex.pdf

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

It can't be the bhp. Otherwise the coolant flow would be huge. It must be the power to run the water pump.

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:

http://www.engineeringtoolbox.com/ethylene-glycol-d_146.html

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?

The latter.

Also,
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?

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.

Chet

 

[Jay_]

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. Let's consider RPM of 4000, I get a volumetric flow rate of 4.6 L/sec. Is that too high?

##$CALCULATIONS$##

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

$#END OF CALCULATIONS$#

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.

 

[Jay_]

Since I have the brake horsepower, I could use the graph on this thread :

https://www.physicsforums.com/threads/coolant-flow-rate.566125/

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?

 

[Chestermiller]

Since I have the brake horsepower, I could use the graph on this thread :

https://www.physicsforums.com/threads/coolant-flow-rate.566125/

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?

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).

Chet

 

[Diandrea]

exactly

 

[Chestermiller]

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. Let's consider RPM of 4000, I get a volumetric flow rate of 4.6 L/sec. Is that too high?

##$CALCULATIONS$##

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

$#END OF CALCULATIONS$#

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.

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.

Chet

 

[Jay_]

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?

 

[Chestermiller]

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?

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.

Chet

 

[billy_joule]

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.

 

[Jay_]

Chestermiller could you post me those curves please?

 

[Chestermiller]

Chestermiller could you post me those curves please?

http://files.engineering.com/getfil...34-bc19-41780227b743&file=cooling_montage.jpg

 

[Jay_]

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.

 

[Chestermiller]

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.

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.

Chet

 

[Jay_]

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.

 

[Jay_]

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?

 

[Chestermiller]

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?

Sure. It really doesn't matter much. Average temperature is best.

 

[Jay_]

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.

 

[Chestermiller]

So I could co with the V8 engine data?

That's what I would do. At this point, what other choice is there?

Would it be approximately the same for a water pump of today?

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.

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?

Probably, for the engine.

Chet

 

[Jay_]

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.

 

[Jay_]

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.

http://www.bimmerfest.com/forums/showthread.php?t=499204

 

[Chestermiller]

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.

Chet

 

[Chestermiller]

That 12 minute drive was when the engine was up to temperature, correct?
And, it was at about 50 mph and 1500 rpm?

Chet

 

[Jay_]

Quoting myself

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

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

 

[Jay_]

I used the rule of thumb in this thread (post 2).

http://www.eng-tips.com/viewthread.cfm?qid=161237

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? :(

 

[Chestermiller]

Quoting myself

Hi Jay. Let me say the following first:

The mass of fuel burnt looks about right. Based on that, a total energy of 17.6 kWh also looks right. Based on that, a total average rate of energy delivery of 88 kW looks OK. One of the references I sent you gives a rule of thumb of how this energy is distributed: 1/3 goes to powering the car, 1/3 goes out the exhaust, and 1/3 is removed by the radiator. This means that, on average, about 30 kW is the rate of cooling provided by the radiator (presumably under cruising conditions). For a 15 C temperature drop in the radiator, this would imply a coolant rate of 0.6 kg/sec = 0.6 L/sec = 11 gpm. If the average engine speed was 1350 rpm (presumably the time average), this would suggest that the coolant rate in gpm is on the order of 0.01 times the engine speed in rpm. You will recall that earlier, I had suggested a factor of 0.015.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 don't understand the basis of this methodology, so I can't comment on it.

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 suggested 60 gpm at 4000 rpm. At lower engine speeds, it would be lower. Later, I suggested rpm x 0.015

I may have just calculated the energies wrong. Does 96 kWh for the coolant seem too high?

Yes, way high. See above.

Or does 17.6 kWh for the fuel sound too low?

No. It sounds OK. See above.

You should try to do the initial experiments at constant cruising speed with constant engine speed, and with the engine at temperature.

Chet

 

[Jay_]

I don't have the car anymore to do the experiment.

Presently, I have two (sensible) things to go with for the coolant flow rate :
1. flow rate in gpm = 0.015 * rpm
2. flow rate in gpm = 0.2*bhp

I did some scouting calculations. I found the flow rate at rpm values of 1500, 2000, 3000, 3500 and 4500.

According to 1.
@1500 ----> 22.5 gpm
@2000 ----> 30 gpm
@3000 ----> 45 gpm
@3500 ----> 52.5 gpm
@4500 ----> 67.5 gpm

According to 2.
@1500 ---> 22.93 gpm
@2000 ---> 18.06 gpm (a dip)
@3000 ---> 29.71 gpm
@3500 ---> 35.86 gpm
@4500 ---> 45.78 gpm

They seem to give similar values, but it varies at the greater rpm values. Which seems more correct? In that link he said "Somewhere around 0.2 usgpm/bhp at max revs. Does he mean at the higher rpms? If that is the case, I could use the proportionality of 0.015 till 2000, and then go by the 0.2 times brake horsepower?

I just tried using this 0.2 times brake horsepower. Despite that I still get the total energy to be higher than the fuel burnt. So, where else am I messing it up?
Value --> min, mean, max

Mflow (kg/sec) = 1.1469 , 1.8181 , 3.1783
Vflow (m^3/sec) = 0.0011 , 0.0018 , 0.0031 (in gpm these are 18 , 28.5 , 49.6)
Density (kg/m^3) = 1008.7 , 1010.7 , 1015.3
CpCool (J/kg degC) = 3583.2 , 3594.3 , 3598.6
delT (deg C) = 0 , 7.6268 , 11

Each of these seem fine to me, does anything still seem off? But I still get the energy as 21.85 kWh, which is more than the fuel burnt, which is 18.10 kWh.

 

[Chestermiller]

I don't have the car anymore to do the experiment.

Presently, I have two (sensible) things to go with for the coolant flow rate :
1. flow rate in gpm = 0.015 * rpm
2. flow rate in gpm = 0.2*bhp

I did some scouting calculations. I found the flow rate at rpm values of 1500, 2000, 3000, 3500 and 4500.

According to 1.
@1500 ----> 22.5 gpm
@2000 ----> 30 gpm
@3000 ----> 45 gpm
@3500 ----> 52.5 gpm
@4500 ----> 67.5 gpm

According to 2.
@1500 ---> 22.93 gpm
@2000 ---> 18.06 gpm (a dip)
@3000 ---> 29.71 gpm
@3500 ---> 35.86 gpm
@4500 ---> 45.78 gpm

I very strongly recommend going with 3. flow rate in gpm = 0.01 * rpm. Look how this would compare with 2. It would be a great match at every engine speed except for the very lowest...and, at the lowest engine speed is where you are experiencing your troubles. This would clear up all those troubles. Also, since the engine is driving the water pump, the pump impeller rotational speed will be proportional to the engine speed. So, as long as there is not too big a pressure change across the radiator (i.e., too big a pressure drop for the coolant within the engine flow channels), the coolant flow rate should be proportional to the engine speed.

They seem to give similar values, but it varies at the greater rpm values. Which seems more correct?

Option 3.

In that link he said "Somewhere around 0.2 usgpm/bhp at max revs. Does he mean at the higher rpms? If that is the case, I could use the proportionality of 0.015 till 2000, and then go by the 0.2 times brake horsepower?

I would go with option 3.

I just tried using this 0.2 times brake horsepower. Despite that I still get the total energy to be higher than the fuel burnt. So, where else am I messing it up?
Value --> min, mean, max

Mflow (kg/sec) = 1.1469 , 1.8181 , 3.1783
Vflow (m^3/sec) = 0.0011 , 0.0018 , 0.0031 (in gpm these are 18 , 28.5 , 49.6)
Density (kg/m^3) = 1008.7 , 1010.7 , 1015.3
CpCool (J/kg degC) = 3583.2 , 3594.3 , 3598.6
delT (deg C) = 0 , 7.6268 , 11

Each of these seem fine to me, does anything still seem off? But I still get the energy as 21.85 kWh, which is more than the fuel burnt, which is 18.10 kWh.

Here is a sample calculation:

Engine speed = 1000 RPM
Volume flow rate coolant = 10 gpm = 1.34 cfm = 84.3 kg/min = 1.41 kg/sec
Cp=3600 J/kgC
del T = 10 C
Q = 50.8 kW
In 12 minutes, heat lost in radiator = 10.2 kWh
46% of the energy in the fuel predicted to be lost in radiator. This still seems a little high. I might go with a factor even lower than 0.01, say 0.008.

Chet

 

[Jay_]

What I have done as of now is calculate the instantaneous values of heat, added them all and compared it with the fuel burnt. I will get the average of the samples and compute the power, multiply it by time and get energy and see what that gives me.

This still seems a little high. I might go with a factor even lower than 0.01, say 0.008

But on what basis are we choosing any of these numbers? I wonder if any auto heads could help in giving an estimate here on the flow rate.

 

[Jay_]

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?

 

[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?