Cooling system specifications

[lsrjunkie]

Hey all, I'm looking for information in calculating cooling system flow and capacity for a given engine. A little vague, I know, but I dont want to just get the biggest and baddest of everything and hope it will work. There has to be a way to figure flow rate, volume, etc.

[Harold Bettes]

Howdy to All that like this kind of stuff,

H2O flow required to carry away heat  from any device or component in a system can be calculated.

 GPM = [C (Hp)] /ΔT  Where GPM = gal/min, Hp = horsepower,   ΔT = (Tend - Tstart), as subscripted, ºF
This is using water as the liquid, so use a value for C as 5.1.

IF you want to use a simple calculation on how long in time it takes to heat up a given mass of water:
How long does it take to heat up a volume of Water?  With water volume re-circulated (closed system) and no additional water added to the system. 

Time = [(MH20 / HpBTU) (Tend - Tstart)]   where Time = minutes, MH2O = weight of water (lbs), 
HpBTU = Horsepower (BTU),  T = temperatures as subscripted, ºF
Plan on using the weight of water as about 8.3lbs/gal so if you have 20 gallons that would be 20 x 8.3 = 166 lbs. Use 42.44BTU/min/Hp, or 2546BTU/hr, so 100Hp = 4244BTU/min

The cooling load of an internal combustion engine can be varied, but easy calcualtions can use somewhere around 33% to 40% of flywheel horsepower.

SO, rock on with your $9 calculator and enjoy the process, but be realistic with the details.

Regards to All,
HB2

[John Burk]

Harold is your engine heat to cooling water = 33% to 40% of engine hp just throttle down time or does it factor in staging lane warmup ?

[lsrjunkie]

I'm not entirely sue what happened here. I guess a guy should be careful what he asks for. Thank you very much Harold but...

WOW! Apparently somebody should have paid way more attention in math class. I figured there was a way to calculate my requirements, but that is just a little over my head.

Help.    Please.

I may be way off, By using the first formula the way I think it's supposed to work...

For a 400bhp engine to maintain 185 degrees you need a flow of 11gpm, (I think).

[lsrjunkie]

And what exactly do you mean by the line,

"The cooling load of an internal combustion engine can be varied, but easy calcualtions can use somewhere around 33% to 40% of flywheel horsepower."

Is that meaning that heat going into the cooling system is lost HP? I know that the heat going into the cooling system is definitely wasted energy.

So much information... I love it, now to figure out how to obtain the correct info, decipher it, and put it to use.

[hotrod]

I think what Harold means is as follows.

An internal combustion engine loses about 33% of the thermal energy of the fuel it burns as heat to the cooling system.
So if it makes 400 flywheel horse power at full throttle, then it will be dumping the thermal equivalent of about 132 hp to the cooling system during that time

400 x .33 = 132

If you put that into the formula you get:

GPM = [C (Hp)] /ΔT
GPM = [5.1 (132)] / (change in temperature of the water in degrees F)

GPM = 673.2 / allowable change in water temp

So to get the final total needed water flow you need to decide what is acceptable temperature rise in your water reservoir.
If you start with a tank of tap water at 60 deg F and are willing to let the water rise to 180 deg F you have a change in temp of the water of 120 degrees.
Putting that into the formula you end up with:

GPM = 673.2 / 120
GPM = 5.61 gpm for that allowable heat rise.

If you only want to allow a 60 degree temperature rise for the water from say 100 degree starting temp to 160 degrees you get:

GPM = 673.2 / 60  =  11.22 GPM

[lsrjunkie]

That makes a little more sense. Thanks hotrod.

[hotrod]

John that fraction would be split out according to how much power the engine was making.

If the engine generates 10 hp at idle on the line it would be dumping only about 3-4 thermal hp to the water during warm up.

That heat loss should be proportional to the power actually being developed by the engine at any given time (ie how much fuel it is burning per minute).

[Sumner]

I'm kind of fond of math, but there are so many variables...

-- using only a cooling tank.

-- using a cooling tank and a rad-in-a-box.

-- preheating the water to X degress (whole cooling tank or just the rad-in-a-box??).

-- long course or short course car.

-- time at full throttle in each gear.

-- time at max HP.

-- time of the run.

-- on and on.

Might be best to find racers with a combo close to yours and find out what they are using and if it is working or what they would do different.

Sparky put us on to the rad-in-a-box and I think that it has so many positives and so few negatives that in addition to using it in Hooley's Stude...



http://purplesagetradingpost.com/sumner/Hooley%202013/13%20-%20hooley-construction-2013-13.html

.... I'll use the same setup basically in the lakester.  Benefits are:

-- you don't pre-heat your main cooling water.

-- the engine side of the system can now run under pressure as it was designed to do which helps to eliminate steam pockets.

-- the engine now uses its water pump for cooling and if the pump was to fail between the main cooling tank and the rad-in-a-box you have some safety built into the system heat wise and the pumps most of us use from the main tank aren't the most reliable even though they should be for the cost.

-- the flow to the rad-in-a-box from the main cooling tank isn't as critical as going from there to the engine as the water surrounding the rad-in-a-box and the water in the radiator itself can buffer short term heat spikes better than relying only on the water coming from the main tank via a pump.

Good luck,

Sum

[Rex Schimmer]

I think that you are missing something in Harold's equations. The delta T (the difference between the water temperatures) is the difference between the water temp going into the engine and the temp of the water coming out of the engine. Not the temp of the total water in the system. If you look at it this way then you get very different numbers. Using the example of the 400 hp engine with 132hp going to heat you probably really only wanting a delta T of say 15 degrees which would mean a flow rate of around 45 gpm,which goes along with the rule of thumb that you should have 10 gpm of water flow per 100 horse power(this is per my good friend Stewart Van Dyne, Van Dyne Engineering). You need high flows to maintain turbulent water flow through the engine for maximum cooling efficiency and if you are using a radiator you definitely want high flow rates to maintain turbulent flow.
Rex

[Sumner]

...which goes along with the rule of thumb that you should have 10 gpm of water flow per 100 horse power(this is per my good friend Stewart Van Dyne, Van Dyne Engineering). ....Rex

That is probably a good starting point and also points at the problem in trying to cool the engines with an inline pump between the cooling tank and the engine.  Meziere pumps are popular and what we use, but the largest is only 55 gpm (rated, actual??). 

Saying that we did use one (I think we have the 35 gpm one) with the B motor in the Stude and it was probably an 800-850 HP motor and we use it now to pump water to the rad-in-a-box but I'd like more as we are probably only circulating the 30 gallons of main tank water through the rad-in-a-box 1-2 times on a run.

Many people have run with this setup (inline pump/main tank) over the years with success though so it can work and does.  We found the best way to cool the motor is to make sure you turn the cooling pump on before or during a run  , otherwise you will be ....



http://purplesagetradingpost.com/sumner/hooley/hooley-2007-Story.html

...tearing things down  ,

Sum

[Stan Back]

Our formula . . .

Let's put a big water tank over the right rear tire.  Make it outta 1/8" steel with a 1/4" bottom.  Fill it up.  The little Focus pump starts with the engine -- "the Belt tells me so".  Highest temp after the five -- 185º.  Can't run back-to- back hardly ever (never have, probably won't).  Big 1-1/4"(?) gate valve can choke it down if we ever want to.  How many HP to drive it?  Probably less than the 100-150 more we need to recapture the record for the 5th time.

Sure is simple (just like the owner).

Stan

[lsrjunkie]

Damn Stan, That's why I like you!

[1leg]

...which goes along with the rule of thumb that you should have 10 gpm of water flow per 100 horse power(this is per my good friend Stewart Van Dyne, Van Dyne Engineering). ....Rex

That is probably a good starting point and also points at the problem in trying to cool the engines with an inline pump between the cooling tank and the engine.  Meziere pumps are popular and what we use, but the largest is only 55 gpm (rated, actual??). 

Saying that we did use one (I think we have the 35 gpm one) with the B motor in the Stude and it was probably an 800-850 HP motor and we use it now to pump water to the rad-in-a-box but I'd like more as we are probably only circulating the 30 gallons of main tank water through the rad-in-a-box 1-2 times on a run.

Many people have run with this setup (inline pump/main tank) over the years with success though so it can work and does.  We found the best way to cool the motor is to make sure you turn the cooling pump on before or during a run  , otherwise you will be ....

...tearing things down  ,

Sum

The pump in the picture is a WP136s 20gpm water pump.
All water pumps are rated without restriction(free flowing), The 55gpm will likely flow around 20-25 gpm after going thru the block and radiator. The thing often over looked is pump pressure. Most electric water pump made for racing engine will only create around 5-6 psi when a good belt driven pump will get around 20-40 psi.
In blown and fuel application you should really be running a belt driven pump, I would also recommend one for NA gas engines over 500 hp.

The other thing often over looked is hose size on the inlet side of water pump I would concider anything under 1.5" to small for a LS racing cooling system.

[Harold Bettes]

Howdy to All who like this kind of stuff,

First a kudo plus to hotrod as he did the math correctly for the example given for the flow rate.

I apologize for not doing the work and showing a step by step example and follow the process as hotrod did. In my mind, I had reduced it to a problem already solved because the OP had a question and stated he wanted to do some calcs to get to a solution for cooling. I supplied two methods to do so.

One was the flow rate required and the other was how much mass of water and a time calc. Nobody took a shot at that one which surprised me as there are so many water pots (volume types) used at Bonneville. The either low pressure or no pressure (local atmospheric pressure reference) systems are very common. Only caveat there is that at 4200'MSL, water will boil at approximately 204degF, so a guy wants to stay shy of that number with a non-pressureized system. I suggest that 190F is a max target for those type systems. 

So, if a guy has 20 gallons of H2O available (166lbs) and the starting temp is 80F and you are going to dump 132Hp in it (as in hotrod's example) with a final temp of 190F, how long does it take to get to the end point (temperature) - ? Time (minutes) = (166/5602) x (190F - 80F) = 3.3 minutes :cry:Hot stuff

A combination of the two methods of calculation will get the job done. 

Hope that makes some sense. IF not, as somebod else suggested, one can always copy a system that works with a similar package. Perhaps that is a good reason to always have a camera handy. OR at least keep a calculator and something to write with.

Regards to All,
HB2