So from the other day...and I didn't want to clutter up the other guys thread any further.

And the Ground rules with me are, I am just talking and thinking out loud. I DO NOT KNOW IT ALL and even though I come off that way sometimes.....anyone is free to tell me I am wrong and please please please show me how I am wrong. This way I learn and that is what I am really all about....learning and getting to the truth of things that I don't quite understand.

Thermal conductance

Q. The story goes like this. if you are sitting in traffic. and somehow you increase the flow of coolant into your radiator, will you see a drop in temps exiting the bottom or outlet of the radiator.

Guys,

Gotta get an answer here. I am having to step up my knowledge on Thermal Conductance and Thermal properties for a new product line we've taken on.

And of course, none of the "homework" is fun unless it is car related.

Start thinking about your radiator.

Here is the question.

If you do nothing else to your cooling system....meaning no other variables enter the equation, save one.

The story goes like this. if you are sitting in traffic. and somehow you increase the flow of coolant into your radiator, will you see a drop in temps exiting the bottom or outlet of the radiator.

The discussion came from someone using an electric water pump vs. a belt driven water pump.

From what I can find, the Temperature Difference (or Delta T) will not change. The radiator can only dissipate "X" amount of heat so lets say the Delta T is 30° F

If you increase the amount of coolant entering the radiator all you are going to do is raise the over all temps of the Delta T.....not INCREASE the amount of Delta T.

So if the water coming in is "X" gpm(gallons per minute) 200° and leaving at 170°. increasing the GPM is only going to increase the........


And that is where I stop and scratch my beard and check the point on top of my head....look in the mirror at my pencil neck.......and also wring my hands.....but only after I straighten my pocket protector.

If the radiator is already cooling to its capacity, adding volume of coolant for it to cool is only going to increase the load which means the Delta T would be affected by the increased load.

Understand, we are not factoring in increased windflow on the OUTSIDE of the radiator.

A. Energy removed by the rad is:

Q=Cp(specific heat) [kJ/kg*C]*mass flowrate [kg/s] *delta T [C]

So with nothing else changing, an increase in flowrate will just mean a lower temperature difference between the inlet/outlet of radiator. This is the heat that is removed by the radiator. Since you are not changing waste heat of the car (still idle or whatever), the energy needed to remove from the engine to keep it cool stays the same. Presumably, you have a T-stat regulating the temperature at the inlet of the radiator (say 180*). Say you need a 30* delta T to start out with (outlet temp of 150*). Now, let's say you double the flow rate. Now, your outlet temp will be 165* (15* delta T).

Now as flow increases, there are some other advantages. The radiator will be more efficient (up to a point), because of better convective heat transfer. Similar for the engine. The faster flow would allow a smaller radiator, less air flow through the rad, or be able to handle a higher load (more power, work better in summer). There is a limit. At a certain point, no matter how good your convective heat transfer is you are still limited on how fast the heat can transfer through the radiator fins or through the block of the engine. This is referred to as being conduction limited.

Q. OK so that is all real good.

So lets say you are at the upper echelon of heat conduction.....then just increase the flow of water. this is where the T-outlet increases?

A. That equation always holds true for single phase (liquid only) heat transfer. If you have two phase (boiling), then you have an additional term due to the latent heat of vaporization (energy removed by boiling). That equation is the heat removed by the radiator (or the heat gained by the engine). These will always balance (at least try to) due to the radiator and the rest of the cooling system. There will only be an imbalance if the radiator is maxed out and cannot dissipate the heat that the engine is putting out. At this point, the coolant temperature will increase until overheating occurs.

Now, let's say you are maxed out and you kicked on additional flow. This is what would happen:
- temperature difference across the radiator will still decrease (above equation)
- the pump does adds some heat to the coolant (waste heat by the pump, extra load on the alternator running the pump). The amount is small, though.
- as flow continues to increase, the delta T will continue to decrease to a point where the delta T cannot be measured. The energy is still being removed at the same rate (perhaps enhanced some from the increased flow)

Basically, as the flow increases towards infinity, the temperature everywhere in the system becomes equal (say 180* everywhere). Really, the ability of the coolant system is determined by the radiator:
- air flow (convection) through the radiator
- surface area of radiator
- material of rad (high conductive material like aluminum can enhance heat transfer)
- average temperature of the coolant system
- ambient temperature

The max energy removed by the rad is:

Q= h (convective heat transfer coefficient) * A (area) * (Tcool-Tamb)

h- depends on fin design, air flow through rad
A - surface area of all the fins
Tcool - average coolant temp
Tamb - ambient temperature

If Qengine > Q rad, the coolant temp will increase using the above equation until there is a balance. If you didn't have to worry about localized boiling and such, T coolant will just keep increasing with engine load to make the heat balance. If the rad is larger, better flow through the rad, or lower ambient temp, the radiator can remove the same amount of energy with less increase of Tcoolant.

I am going to add some though, but little math.

My 92 uses a significantly different cooling methodology.

First off, the heater core is a constant flow unit. There is no shut off valve for it. That means it is always hot and always acting as a small radiator for engine temp control.

A little back story: My 92 has been losing coolant. It's also shown a tendency to run a little hot (not overheating). It would normally operate at the 195~205 temp range. After the heater core was bypassed, and I was back to surface street driving (never an issue at highway speeds) the temps started to rise... and the short drive to and from home became an exercise in watching the temps rise into the upper 220s.

The other reality - with the heater core by passed, the C68 climate control can't manage temps correctly. Remember, the heater core is constant flow, there are temp sensors inside the car that are constantly testing the air temp (or, in the case of the solar sensor, sun intensity). Without the heater core, the AC has nothing to fight with. You see, the AC Evap and the heater core ultimately share the same common air space. They are divided by the mixing door... more cool, less heat allowed in and visa-versa.

Ok... now why was my engine running hot? I am sure you have all added it up, but let me end the suspense: C68 fires the fans every time the AC compressor kicks on. the heater core adds about 2 quarts to the system, which has to have an effect on thermal efficiency. Now that I have heat, I am more prone to turning on the climate controls. Without heat, all I got was a whole lot colder! So with the additional coolant and the fans actually being used, the engine temps are staying about where they started - 195~205.

Ain't that fascinating?