HX coolant too fast to cool?

[frankiee]

Is it possible to have coolant go thru an engine too fast and have the coolant not pick up enough heat so the engine overheats?
I think it would be like a simple heat exchanger.
Is it possible to have the coolant go so fast that the heat exchange is lessened?

 

[TBP]

I doubt that the flow being too great is your problem. If it really is that high, I'd be worried about erosion damage.

I'd check that the water pump is actually moving coolant through the engine block. The pump itself? Stuck thermostat? Fouled radiator? Might even be a head gasket leak into one or more of the cylinders. That would give you a low coolant level, with no sign of an external leak.

 

[ione]

The heat removal (heat load) from the engine establishes the coolant flow rate. The geometry (sections) and the flow rate determine the fluid velocity. Increasing the coolant velocity (i.e. making the flow more turbulent) just promotes the heat transfer process (with all the remaining boundary conditions kept consistent). The problem could be just the opposite of too high coolant velocities. What I mean is the presence of stagnation areas in the circuit where the coolant velocity are too low. A poor design (geometry) could led to this scenario which implies a worse performance of the cooling system.

 

[zekeman]

The problem the OP posted is theoretical and does not speak to the practicality of requiring a larger radiator to transfer the heat since the coolant temperatures exiting the jacket would be lower.
So, sorry to advise the OP, he loses to engineer B.

 

[eadwine]

It's all about mass flow rate and temp. rise/fall between inlet and outlet. Essentially, higher flow, less temp rise/fall, lower flow more temp. rise/fall. The objective becomes to optimize the heat exchange rate and the size of the heat exchanger to accomplish the req'd. heat removal or addition to the system.

Remember Q=UAT, but this is all covered in the Crane Tech. Notes or the Compressed Air Data Book.

 

[redpicker]

One issue absent from this disucssion is that if rubber hoses are used to connect the radiator to the pump, high flow rates can create a vacuum at the pump inlet high enough to cause the hose to colapse, decreasing the flow.

While not an example of a high flow lowering the cooling capacity, it is an example of how attempting a high flow can decrease cooling capacity.

rp

 

[IRstuff]

That would only occur if the pump wasn't at, or near, the lowest gravitional point in the system, and if the fluid was't glycol/water and could vapor lock. And the typical water pump is designed and placed so something like that can't happen.

As near as I can tell, the OP was speaking to a hypothetical, yet physically realizable scenario.

TTFN

FAQ731-376

 

[MrBTU]

You guys.....
Higher flow rate means higher LMTD and higher mass flow rate, resulting in higher Q, since Q = m cp LMTD. And, since Q = U A LMTD and LMTD and U are both increasing with increasing flow rate, then your engine block is cooler, not hotter. Stop dwelling on residence time....

 

[eadwine]

MrBTU, If we hold "A" constant, "U" constant and increase flow so that "T" or "lmtd" decreases, then "Q" must decrease. Right?

 

[MrBTU]

But U increases with increased flow rate....

 

[zekeman]

The problem as stated could be rephrased to say;
If the engine overheats due to fouling, then does increasing the flow ALWAYS result in lowering the jacket temperature.If you prove that proposition then you proved that the heat transfer rate increases.

So without any heavy math, and with the better basic assumption that the engine puts out heat flux, say q BtU/HR-FT^2 and the heat is removed in a simple heat exchanger, a tube with area A and perimeter l, length, L
Writing 2 equations
1) rho*V*c*dT/dx=q*l whose solution is
T=T0+q*l*L/(rho*V*c)

2)h*(Tw-T)=q

Solving 2) for Tw after substitution for T, we get

3) Tw=q/h+T0+q*l*L/(rho*V*c)
where
T0= entrance water temperature
Tw =jacket temperature
rho= density
c= specific heat
V = velocity
Since h has been shown to be proportional to V^.8 (see my posts above)then 3) looks like
4) Tw=q/aV^.8+q*l*L/(rho*V*c)+T0
a= proportionality constant
Now looking at equation 4),it is clearly seen that increasing V results in a decrease of the first 2 terms and the last term, a constant, is the input temperature from the radiator.
Therefore Tw must be lowered by increasing V.

 

[eadwine]

Perhaps, we need to run a lab bench test and check our assumptions...

 

[electricpete]

zekeman - I suspect your conclusion is correct. However, you are modeling the engine-to-coolant heat transfer only... under assumption that the inlet temperature is constant, correct?

Can you add a coolant-to-air heat exchange into your model?

=====================================
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[zekeman]

Electricpete,

Good question.
My answer responds to the OP question of an "open loop" heat exchanger, but if you want to make it a real life situation then you close the loop with a radiator that is removing the same amount of total engine heat. In that case, as you know, a thermostat maintains control of the temperature entering the radiator, say 180 F, so the flow to the radiator and its temperature profile inside the radiator is independent of the fouling problem at the engine,which only causes the jacket Temperature to rise to accommodate the fouling. If you now increase the flow into the jacket, then the amount of increased flow rate would ends up bypassing the radiator,without affecting the flow through the radiator
So bottom line-- the exit temperature at the engine is always 180 and , of little consequence, but worth noting, owing to the increased flow, the entrance temperature to the jacket is somewhat increased because the delta T across the jacket is reduced.
Since.the maximum walll temperature is at the exit jacket wall where the water temperature is maximum.and at that point, you have q, the same flux I used previously, we can write
q=(Tw-180)/h
[BTW, we could have arrived at this without all the discussion following the the fsct that the water exits the engine at 180 F]

where h is proportional to V^.8 ; this leads to the obvious conclusion that the maximum wall temperature is reduced.

Seems that the actual closed loop case is easier to prove than the open loop.

 

[cyclops2]

This question was raised with EARLY Ford V8 passenger cars.
Stated. "If you remove the thermostat in the engine, the system WILL overheat."
As I was 15 at the time I remembered the statement well.
Did the Fords have radiators & fanblades maxed out??
Cocktail engineering of the overheating was. "Water did not stay in contact with the radiator long enough."

Rich

 

[Compositepro]

Again, the answer to the very simple original question is simply "no".

Water is in contact with the heat exchange surface 100% of the time. If it flows at a high rate a given unit of water will not heat-up as much as slower water so the water temperature will be lower. This increases the heat transfer rate per unit area of heat transfer surface. So per unit time the amount of heat transferred goes-up.

 

[Bribyk]

Some industrial nat gas engines have a minimum recommended external restriction to prevent "coolant flow rates [from becoming] excessive." The consequences of "excessive" flow are not given though.

Brian Bobyk - Hoerbiger Canada

 

[MrBTU]

In my 25 years with heat exchangers, the implications of too high coolant flow are:
1. tube and tubesheet erosion.
2. excessive cooling of the oil side which could bring oil temp too low and increase viscosity to a point where pumping loses were excessive.

We once had an oil cooler on the North Slope of Alaska that was acting up. The customer called and said that his oil side pressure drop was too high. We told him to pinch off his cooling water, hence raising the oil out temp (but still acceptable temp), decreasing the viscosity, and reducing his pressure drop.