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Some Charge Air Cooling Thoughts

Using any type of compressor, whether turbocharger, belt driven centrifugal supercharger, roots compressor or lysholm, heats the air during compression. The amount of heat is a function of the pressure ratio of the output to input of the compressor and the efficiency of the compressor. Heating the air creates two side effects: reducing the density of the air for a given manifold pressure, resulting in less power and the need for reduced spark timing, over rich mixtures, or water or alcohol injection (which all cost power) to control detonation plus the threat of damage ranging from blown head gaskets to destroyed pistons, rods, bearings, etc.. By cooling the compressor outlet air, we regain the lost power due to all of these factors and reduce the risk of detonation. There are a number of ways to cool the air: 1) one of the simplest is just to inject water into the hot air stream and use the heat of vaporization to cool the air but at the cost of combustion heat being used to heat the water vapor; 2) use alcohol injection in the same manner as 1) but this has more benefits in that it raises the average octane level of the total fuel charge and it will take you right to the fuel class; 3) air to air intercooling which has plumbing and air flow issues, and 4) water to air intercooling which can lower the charge air temepratures even below ambient but with the penalty of adding weight to the car, as well as the plumbing issues aand failure modes of the plumbing and pumps. Because I am old and getting older every day (each day seems to advance my age by 2 days), I have taken on a consultant for this article: Phillip Guziec. He is an engineer I ran across on the Fordnatics list and is far handier on thermodynamics than I. He has assisted me in the analysis and proofing of this article: help that was needed and for which I am grateful.

I am going to try and provide some thoughts on the advantages and disadvantages of these types of charge air cooling and then to do some analyses to give some scope to the solutions involved. I am going to use my turbocharged motor for analysis. Most of the information provided will give you some background when you talk to tradespeople who know this stuff far better than I.

Some generic thoughts
Did you ever really think about your motors' needs? Let's think about my Sunbeam Alpine and my initial goals: to go faster than Sir Henry Segrave when he set the absolute world's land speed record in 1927. I chose the modified sports class for obvious reasons; Sir Henry's car was a Sunbeam and it is a sports car. After deciding what I wanted to do, I next determined how to do it. This resulted from using my spread sheet (see the download page) to determine how much horsepower I was going to require to just exceed the old 1927 record (thus becoming the owner of the world's fastest Sunbeam). I think I now have every bit of information to make a rational guess at the charge air cooling items listed. I am going to tackle number 4) first because it is easier to me. If you want to delve into any of the below methods for you car specifically, here is what you will need to know: the flywheel horsepower you need to meet your speed goal, the efficiency of your compressor if blown, the distance to run from start to final trap, aand the barometric pressure and ambient temperature at race time. Not much, huh?

Organic to all of the Analyses
Ok, here are the items I needed for analysis of cooling methods for my car: top speed of 204 mph, approximately 500 FWHP, blown, and at Bonneville on the long course. You will need to know a bit about how much boost is needed to get your horsepower. Keep in mind that you will need less boost the cooler your intake charge. In my case it 10 psig should do the trick. The reason you need this is to determine the amount of heat that your compressor puts into the air. Don't worry about the math as I am going to make a spread sheet to help you figure it all out.

Some Assumptions
I have never run my car so I had to figure out a way to determine how long it takes to make a run on the Bonneville Long course. The idea is to find out how long your gee whiz bang motor needs to be making power at full boogie. Yeah, I know that because of traction issues that the throttle may be feathered a tad during the run to top speed but if we analyse it at full song then some conservativism is built in. Ok? Here is how I did it...

I assumed that the speed at the end of the first mile was 80% of top speed, speed at the end of the second mile was 80% of the remaining speed, third mile 80% of the remaining speed and 4th and 5th miles at speed. You may choose to define it how ever you want, but the inportant aspect is to figure out how long your engine will be making the horsepower number. So...

a) Time to cover the first mile: 80% of my top speed of 204 mph is 163.2 mph. The average speed is (0 + 163.2) / 2 = 81.6 mph or 119.6 feet per second. Since a mile is 5280 feet then the time to cover the first mile is 5280 / 119.6 = 44 seconds.

b) Time to cover the second mile: 80% of the remaining speed is 0.8 times (204 mph - 163.2 mph) = 32.68 mph. Add this to the first mile exit speed to get 163.2 mph + 32.6 mph = 195.84 mph. The average speed for the second mile is (195.84 mph + 163.2 mph) / 2 = 179.5 mph = 263.3 feet per second. Then the time to cover the second ile is 5280 / 263.3 = 20.05 seconds.

c) Time to cover the third mile: 80% of the remaining speed is 0.8 times (204 mph - 195.84 mph) = 6.53 mph. Add this to the second mile exit speed to get 195.84 mph + 6.53 mph = 202.36 mph. The average speed for the third mile is (195.84 mph + 202.36 mph) / 2 = 199.1 mph = 292.02 feet per second. Then the time to cover the third mile is 5280 / 292.02 = 18.08 seconds.

d) The time to cover the fourth mile is found similarly by noting that we are at top speed here and skipping the 80% business. Top speed for me is 204 mph or 299.2 feet per second. This covers the 4th mile in 5280 / 299.2 = 17.65 seconds.

e) Time for the 5th mile, in my case is exactly the same as the 4th mile: 17.65 seconds.

The total time then is a) + b) + c) + d) + e) = 44 + 20.05 + 18.08 + 17.65 + 17.65 = 117.43 seconds. Lets just round this of to 120 seconds or a full 2 minutes. I plotted my data, see the figure below just to get a feel for the data. It looks sorta right although you may have a different scheme to get the total run time. At the shorter venues, like Maxton or El Mirage, you jest need to average the speed over the distance to the trap to get the full power run time. It will be conservative. Example: Maxton is a mile to the trap, so in my case, if I wanted to go 204 there then the average speed would be 102 mph or 149.6 mph = 219.4 feet per second. This means that the power would be on for 5280 / 219.4 = 24.06 seconds. Similarly for El Mirage, the distance to the trap is 1.3 miles (I think!). It would take me about 31 seconds to make a run. The importance is that you just need to know how long you will be at full power on any course. Do what ever you need to do to get that time. I have included the method I used for the long course in the spread sheet.

Now I need to determine how much air that my motor will use in the time to make the full run. Now you see why we needed time! And we have to make some more assumptions. I assume a Brake Specific Fuel Consumption of 0.55 pounds of fuel per hour per horsepower. This is pretty standard, and for you normally aspirated folk, this should be around 0.5. This is for GASOLINE only. If you are using alky or nitro then you need to know the Brake Specific Fuel Consumption,BSFC, and the air fuel ratio. For me, I chose an air fuel ratio of 12.5:1 as this is the best power ratio unless you need to use excess fuel to aviod detonation. So, at 500 hp my motor needs 500 hp times 0.55 lb fuel / hp hour or 275 pounds of fuel if it ran for a full hour. But, it is just going to run for 2 minutes so the amount of fuel consumed is (275 lbs/hour /(60 min/hour) * 2 minutes = 9.167 pounds of fuel in a full run; about a gallon and a half, give or take a tad. Now at a air fuel ratio of 12.5:1 the motor will need 12.5 * 9.167 = 114.6 pounds of air for the full run. It does not matter how you get this air, but you do need to know it's temperature.

So how to get the air temperature? Well, it is not too bad and I will include it in the spread sheet as well as show you how, here. There is a set of formulas that are used. First, I work in the Rankine temperature system That is my roots and that's just the way it is. So al temperatures are Rankine ( but a single degrees is the same as Fahrenheit).

T2 = T1 * (P2 / P1) 0.286
T2 is the compressor outlet temperature
T1 is the compressor inlet temperature which is the same as the ambient temperature (95o F
P2 is the boost pressure plus ambient pressure (10 psig)
P1 is the ambient pressure (12.6 psia for Bonneville, 14.7 at sea level)
0.286 is the ratio of specific heats-1 divided by that ratio is [gamma -1]/gamma
Remember the Rankine thing...


T2 = (460 + 95) * [(10 + 12.6)/12.6]0.286 = 655.9o R

For a compressor outlet air temperature of

T2 = 655.9o R - 460o = 195.94o F

Now I need to adjust the theoretical temperature rise to account for the compressor efficiency or lack of. In my case the T03 turbos have about 65% efficiency at the levels I want to play with.

Then the real temperature rise is...

Trise = [T2 - T1] / 0.65
Trise = [195.94o F - 95o F] / 0.65 = 155.3o F

So, the temperature of the charge air is

Tcharge air = Trise + Tambient
Tcharge air = 155.3o + 95o = 250.3o F

Now we know how hot the air is coming out of the compressor. I have chosen to reduce the charge air temperature to 70o F, to add power from air density, keep detonation at bay and even add timing back into the ignition system to get even more power. So the required charge air temperature drop is

Delta T = Tactual charge air - Tdesired charge air = 250.3 o F - 70o F = 180.3o F

Now for the amount of heat in the charge air. Remember we dermined the amount of air needed way back there? It was 114.6 lbm. The equation for the amount of BTUs is given as

q = cp * mair * Delta T

q is the amount of heat in the air and to be removed, in BTU
cp is the specific heat of air at a constant pressure BTU/lbm oF = 0.24
mair is the mass of air = 114.6 lbm
Delta T is what was determined above

q = 0.24 * 114.6 * 180.3 = 4959 BTU

The results of these calculations are used in all of the charge air cooling methods below. I point out to you that there is a down loadable excel file which takes care of a lot of the math on this. It is is the files section. Go look for it.

1) Water Injection Method
Water injection is a tried a true method of cooling the charge air. The problem has been for many of us knowing how much to inject. Well, hopefully, this will shed some light. The magic of water injection is the cooling power of the evaporation of water or "phase change".The thing that has to be recognized is that water undergoes a phase change when going from a liquid to a vapor or gas. This is called the Latent Heat of Vaporization for Water. There any number of sources on the web or even in old text books, which have the value for the latent heat. There appears to be some variation in the value ranging from around 970 to 1003 BTU/Lbm. This is considerably higher than that of the latent heat of fusion (#4 below) and stems from the fact that it is harder to make something boil than to freeze it. I am going to use the 970 BTU/lbm value because it is the lowest number and that builds in conservatism. It takes 970 BTU to turn a pound mass of water into vapor or steam (saturated liquid to a saturated vapor). We know how many BTUs there are to be removed, so it is relatively straight forward to find the amount of water needed.

Water required = BTU to be removed from the air / latent heat of vaporization of water


Water required = 4959 BTU / 970 BTU/lbm = 5.11 lbm

Remember that water weighs 8.34 lb per gallon, so this would only be 0.61 gallons of water.

Dividing this by the amount of time we will be running under full power, 2 inutes in my case gives us the flow rate for our water injection system. In my case:

5.11 lbm / 2 minutes = 2.56 lbm / min
2.56 lbm / 8.34 lbm / gal = 0.31 gal / min
1 / 0.31 = 3.26 minutes / gallon

Now having said that, it gets a bit more complicated because now you need to find an injector or spray nozzle that can spray about 2.5 lbs/minute for my case, yours might be different. A litle research on the web should find one pretty easily though. You can also just use a small hole in the end of a tube as an injector and, by spraying water into a gallon jug and timing, can calibrate the hole with a drill. The next complication is when to begin spraying the water. One school of thought is to use a hobbs switch to turn on the spray when boost reaches a set point level. In this case it would be either on or off. I like the idea of using satged switches, say maybe three of then with different set points so that at a low boost level, one comes on and as boost increasesadds the other 2 incrementally. This would really work pretty well for a supercharged application because boost is a function of rpm and full boost wont happen until the rev limiter is hit. On turbo applications, boost can come on full at low rpms so the full flow is needed right up front. Another thought is that a device like an FMU used with a high pressure pump would increase the line pressure and hence spray more in demand with the boost level.

An elegant solution is to build a pressurized water tank and plumb the boost pressure to the tank and the outlet of the water injection to the inlet of the turbo or compressor. This way, increasing boost will increase the flow rate of water. Placing the water tank slightly below the injector outlet will assure a small boost is required to initiate water flow. The injector can be sized using a pressure regulator to pressurize the tank to full boost and measuring the flow rate using the jug and stopwatch method. Be carefull using this, however, because if the water is in the form of droplets and they impinge on the impellor blades then serious damage to the compressor could occur. Also, the tank itself must be able to withstand the boost pressure, so good tank design is essential.

A really cool thought is that this could be used in conjunction with either #3 or # 4 below to reduce charge air temperature before it reached the intercooler by spraying into the air duct work before the intercooler. An argument exists that spraying the water after the intercooler could be more beneficial, however, I am skeptical at this point.

I am not a design organization so you are left to your own devices to figure out how to plumb and activate the spray system. Because water vaporizes better at low pressure and a high flow velocity promotes a low pressure, via Bernoulli effect, then the best place to locate the injector is in the smallest cross section flow area of the duct work. If I were going to do a dual system, like I mentioned above, I would use an intercooler circulation pump that could generate say about 30 psig and just spray the ice water from the coolant tank and add coolant tank capacity. Oh, and don't forget that you have to boost reference the spray system because you are spraying into a pressurized system (just like for an EFI fuel regulator). One final thought: you may need to reduce the amount of water because it quenches the fire in the combustion chamber. If your motor doesn't run so well, just dial back on the amount of water until it does. Doing this however lessens the cooling effect so be circumspect in this.

2) Alcohol Injection Method
The calculations are virtually the same as for water injection except that the latent heat of vaporization for methanol is lower, about 503 BTU/lbm.


Methanol required = 4959 BTU / 503 BTU/lbm = 9.86 lbm

Methanol is of course less dense than water, specific gravity is only 0.78 that of water: it will take approximately 1.52 gallons of methanol to make a run using my conditions.

So it takes more methanol, but methanol gives something back. It adds octane and is of course combustible, assuming you have sufficient air for combustion. That means when you spray it into the air duct and hence into the motor not only are you cooling the charge air but you are increasing the performance of the fuel! Neat, huh. Well it comes at a price. Remember you are putting in enough air to match the fuel air ratio of 12.5:1 as in my case. Methanol requires more fuel per amount of air for max power or complete combustion and makes max power at about 7:1 air fuel ratio. Decreasing the amount of gasoline to compensate for the added alcohol may have detrimental performance implications. Adding more fuel vial methanol injection, with out compensating means that the system will run richer than expected. So, you need to consider this right up front. Be careful, though, you don't want to detonate you prize baby as soon as you hit the boost by leaning it out. As before, you need to find some kind of nozzle that will spray about 0.76 lbm/min for my case. Yours will be different of course. As above, I would use some sort of staged spray system and or an FMU like device and boost reference the regulator or pump. One final problem with injection of methanol into the air duct: it makes quite a KABOOM if you get any kind of backfire. If you use it before the intercooler as indicate might be possible in #1 above then you can get quite a volume of air:fuel inside the duct work. I suspect it might be spectacular! So if you use alcohol injection, spray it after the intercooler and as close to the throttle body or carb as possible. But not so close that it doesn't have time to vaporize (jeeze....another analyses...no, no stop me, please, somebody....).

3) Air-To-Air Intercooling
This is the most common of intercooling methods because it is reasonably fool proof and does not have a fluid that has to be replaced. It has one side effect that hinders top speed in LSR because it requires some sort of air flow over the intercooler core.which adds aero drag from ducting the air to the intercooler and out. On streamliners, lakesters, altereds, or any of those classes where the front end can be closed up, it does not seem to be worth while. On cars which may have their radiator open to airflow, it might be the easiest way. It is of course subject to the ambient air conditions. It is not possible to get the charge air cooler than ambient air using this method without adding water spray or ice on the intercooler or by spraying nitrous over the intercooler. Right up front, I am telling you I don't know much about the design of heat exchangers, but I am willing to try and make a stab at it with some rationalizing assumptions...the bottom line however, is that I cannot make the required temperature drop using an air-to-air intercooler by itself (cannot get below ambient).

For my motor, the CFM is determined as follows:

CFM = [rpm * displ of one cylinder * Number of cylinders / 2] / 1728
CFM = [6500 * {4.032 * 3.1415 * 3.0 / 4} * 8 /2] /1728 = 575 CFM

Because there are obstructions in the air flow, such a runner surface friction, friction in the air itself, pumping losses due to friction, the volumetric efficiency is always than 100%; mine is around 93% for a normally aspirated motor. So my actual CFM is 575 * 0.93 or about 535. There is a mystique about CFM associated with the volume of air you can flow into an engine. The CFM calculated is ALL that you can get into the engine, volume flow wise. Notice that the equation to determine CFM has does not use pressure or temperature or density to calculate. CFM is based on mechanical conditions only. You can get more MASS by pressurizing the engine with boost but you cannot flow more than calculated with volumetric efficiency thrown in. You change the voumetric efficiency by porting, polishing, removing small imperfections in the runner walls, increasing valve size, etc. You may argue that ram tuning gets more air into the cylinder and it does simply because the intake valve is open when the standing wave in the air flow (due to wave tuning which has to do with the rpm the motor is operating at) is at a higher pressure. So it is like a mini supercharger and you can get virtual volumetric efficiencies higher than 100% with ram tuning on fully prepared race engines.

Designing an air to air itercooler is far beyond my capabilities, although we could all muddle through it. I suggest that if air to air is what you want and need then by all means go to the trades people such as Turbonetics and Spearco. I have their catalogs and the table below is mostly from these. What confounds me, as it might you, is the fact that there seems to be no rhyme or reason for the different levels of flow or horsepower capabilities. I tried to develop a "rule of thumb" or "figure of merit" so that it would be simple. Sorry. There are some thoughts though in that you want the slowest charge air flow as possible through the core which means a charger air face that is relatively large. A skinny, long cooler has higher charge air velocities which while resulting in the same relative cooling as a short fat one, has larger pressure drops due to the increased Reynolds Number. See? I knew I could work that in somehow. I think that the slection is skewed because of the limited sizes of cores available. So a core will support many diffeent flows and power levels. I would suggest following the example below and then contacting Spearco/Turbonetics with your rough information and let them complete the selection process.

In this example, I would like to keep the velocity of the charge air through the intercooler to the average charge air velocity, 115 ft/sec, through the core. The reason is to keep pressure losses to a minimum. And to eliminate any compressiblity effects. Note: if you want to know the speed of sound for any air temperature here is how you do it:

a (ft/sec) = 49.02 * Square Root(460 + Temp)

If the air flow is 115 ft/sec I can determine the area needed for the charge air face of the core. Now the plate and bar type core is the best but has an effective flow area of only 0.45 times the face area.

Charge Air Face Area (sq inches) = (CFM * 144) / (Velocity * 60 * 0.45)
CFM = cubic feet per minute (mechanical flow times volumetric efficiency of your engine)
144 = conversion between square inches and square feet
Velocity = 115 ft/sec assumed or desired charge air flow flow through the core
60 = conversion between minutes and seconds
0.45 = increases core area size because of the type of core (plate and bar - this is a Spearco number)

Plugging in my numbers...

Charge Air Face Area (sq inches) = (535 * 144) / (115 * 60 * 0.45) = 24.8 square inches needed

As can be seen in the table, there are a couple of thickness of core available from Spearco: 3.5 inches and 4.5 inches, with some custom "cost a lots" thrown in for good measure. If I elect to use a 3.5 inch thick core then the other dimension of the chage air face is easily found:

Heigth of Core = Charge Air Face Area / Core Thickness = 24.8 in2 / 3.5 inches = 7.089 = 7.1 inches

Now I have 2 of the3 dimensions of the core. Only the width is left. Here I tried several schemes to get someting that would work. I found the average volume of all the Spearco cores and the average horsepower supported by all the different cores and came up with a horsepower per cubic inch of core volume. Is this correct? I do not know, but for sure, when you chat with the experts you may have more knowledge than they!

So, if I get the average HP/ in3 number from the table, which is 1.51 hp / in3, then I can calculate the core volume needed.

Core Volume Required (in3) = Horse Power / (HP / in3)
Core Volume Required (in3) = 500 / 1.51= 331.12 in3

Now, if I divide the core volume by the charge air face dimensions I can get the core length needed.

Core Length Required (in) = Core Vol / Charge Air Face Dimensions = 331.12 in / 24.8 in = 13.35 in

You now know more than I do!

Spearco Data Table
Air to Air Intercooler
Spearco #s       My NumbersChg Air Flow AreaChg Air Vel. 
Core P/NCore AssyCFMHPD - inchH - inchW - inch Vol - in3in2ft/secHP/in3
2-1102-2205406303.509.246.00 194.0414.5589.052.78
2-1132-2223202153.506.8011.75 279.6510.7171.711.14
2-1142-29011007703.505.5032.00 616.008.66304.761.79
2-1151-22110807603.5018.486.00 388.0829.1189.052.78
2-1162-2236454303.5013.6011.75 559.3021.4272.271.15
2-1182-2286954852.2510.0018.50 416.2510.13164.741.67
2-1192-2277205003.5015.705.38 295.3624.7369.882.44
2-1202-2387405203.507.9024.00 663.6012.44142.741.12
2-1612-2257004703.507.9012.88 356.1312.44135.021.97
2-1672-22610507003.5015.8012.88 712.2624.89101.271.47
2-1722-24510106753.5010.4010.50 382.2016.38147.992.64
2-1812-246150010003.5020.8010.50 764.4032.76109.891.96
2-1732-2574603104.5010.5016.50 779.6321.2651.920.59
2-1742-23710507004.5016.408.75 645.7533.2175.881.63
2-175 9206154.5025.106.50 734.1850.8343.441.25
2-1762-23312008004.5026.2011.00 1296.9053.0654.280.93
2-1782-2477605103.507.8020.00 546.0012.29148.471.39
2-179 10006704.5019.7020.75 1839.4939.8960.160.54
2-1802-2357605103.5013.0017.25 784.8820.4889.080.97
2-1822-248150010003.5015.6020.00 1092.0024.57146.521.37
2-1922-2426304203.509.5012.40 412.3014.96101.051.53
2-1942-24312608403.5019.0012.40 824.6029.93101.051.53
2-1952-2495003353.509.408.38 275.5414.8181.051.81
2-2962-25010006703.5018.808.38 551.0829.6181.051.81
2-1972-23414009404.5010.9319.50 959.1122.13151.811.46
2-2032-2556404303.506.5224.00 547.6810.27149.581.17
2-2052-25612809003.5013.0424.00 1095.3620.54149.581.17
2-2072-26610106753.0014.2018.50 788.1019.17126.451.28
2-2082-2585253503.506.5017.25 392.4410.24123.081.34
2-2092-2626704503.007.1318.50 395.449.62167.171.69
2-2102-2676704506.005.9011.75 415.9515.93100.941.61
2-2112-26811807806.0011.8011.75 831.9031.8688.891.42
2-2122-2637905304.505.1218.50 426.2410.37182.871.85
2-2162-2727905303.5010.5028.00 1029.0016.54114.650.77
2-2172-27312808503.5021.0028.00 2058.0033.0892.880.62
2-2182-2747905304.505.8516.50 434.3611.85160.051.82

4) Water-To-Air Intercooling
Here is where Phil caught me. I had dumbly used the temperature of the water as 32o F and worked from there. This will work but only if you tow a tank car behind you. The trick is to use the Latent Heat of Fusion for Ice. This works because ice melts at a constant temperature of 32o F making this efficient and doable. Check the web for the value of the latent heat, nah, Phil already did that and, of course, I checked him. The latent heat for ice is approximately 143.2 BTU/lbm. That means it takes that much heat per pound of ice to melt it at 32o F. So...

Ice required = BTU to be removed from the air / latent heat of fusion of ice


Ice required = 4959 BTU / 143.2 BTU/lbm = 34.6 lbm

Remember that water weighs 8.34 lb per gallon, so it you froze water in gallon jugs, this would only be 4.15 gallons of ice.

Some closing thoughts here. Remember that the ice is in pounds and not gallons if you are using crushed ice. You need to make sure that you have the mass of ice. I would used crushed ice or raid the cooler at the motel. You need a tank that can be drained down after each run. The tank needs to be just large enough to hold the crushed ice mass and a supply of water to pump around the intercooler. In this case say around 5 or 6 gallons total because the water will get in between the ice pieces. Use a pump that will pump a fair amount of liquid, say 5 gpm( Spearco has a nice new ABS plastic one that wont rust!, p/n 2-1441 at 6 - 8 gpm). This way, the return water wont be heated too much. And above all, Insulate the tank the supply and return lines and the intercooler body itself!

As I did for he air to air intercooler selection above, I tried to do the same for water to air. However, the data is lacking. The data is also for much warmer water than 32o F, more like 70o. As you can see, the water to air, even when the water is at ambient temperature, the water to air is almost twice as efficient as the air to air. But it is deeper than that because the intercooler is much smaller volume wise to reach this point. So, if I had to recommend how you select a water to air core, I would say build the biggest you can find space for. That way, if the water gets warmer than desired due to a long wait in line, then it will still perform very well. In all cases, consult with the experts! Use this only as a guide in chatting with them.

Spearco Data Table
Water to Air Intercooler
Spearco #s       My NumbersChg Air Flow AreaChg Air Vel. 
Core P/NCore AssyCFMHPD - inchH - inchW - inch Vol - in3in2ft/secHP/in3
2-1702-2301500*4.510.1259 410.0620.50175.58*
2-1712-231700*4.510.1254.5 205.0320.5081.94*
2-2002-252450*3.6563.5 76.659.86109.59*
2-2012-264**3.6567 153.309.86**
2-2022-281**61213.51 972.7232.40**
2-2062-265*4002.254.657 73.244.71*5.46
2-2132-270*40034.258.63 110.035.74*3.64
2-2142-271**64.258.63 220.0711.48**
2-2152-269*6004.54.657 146.489.42*4.10
2-2192-280**121213.51 1945.4464.80**
*2-261*25009910.25 830.2536.45*3.01
*2-259*25009910.25 830.2536.45*3.01
*2-254*25004.5920.5 830.2518.23*3.01
*2-275*1000994.5 364.5036.45*2.74
*2-276*10008.510.636 542.1340.66*1.84
*2-271-A*6004.510.636 287.0121.53*2.09
*2-271-B*8004.2519.263 245.5736.83*3.26
*2-280*250013.51412 2268.0085.05*1.10
*missing data          

Please note that the example for my motor below uses already established components for the core and the analyses looks a bit odd. However, ig you download the Excel file, chgair.xls, from the download page, it has all the materials to do this from scratch. That spread sheet will provide you with dimensions similar to the Spearco numbers for core sizes.

So where do I fit in this design stage. Well, I have an old T-bird air to air core that I am converting to water to air. The Turbo book says when you do this reverse the flow paths such that the old air cooling path now becomes the charge air path and the old charge air path becomes the water coolant path. Why, I do not have the foggiest, but I believe the experts. That being the case, my core is about 3 inches 6 inches by 9 inches for the air path and with the 0.45 fudge factor thrown in I have about 24 square inches for the flow area. With my air flow of 535 CFM my velocity through the core is

Core Face Flow (ft/sec) = (CFM * 144 in2/ft2) / (Flow Face Area in2 * 60 sec/min)


Core Face Flow (ft/sec) = (535 ft3/min * 144 in2/ft2) / (24 in2 * 60 sec/min) = 53 ft/sec (36 mph)

Since my core is about three inches thick, then my core volume is 3 * 6 * 9 cubic inches or 162 cubic inches. At the average hp per cubic inch (remember this was for a water temp of 70 degrees F from the Turbonetics catalog) then my core should support 3.02 * 162 = 489 hp. While this is a bit lower than expected, a glance at the table above will reveal that there are 2 very large intercoolers which skew the hp/cubic inch average. Taking these out and I have an acceptable water to air intercooler. It compares well with the 700 hp version runing at only 70 degrees water temp. So I think I am still good to go. Remember, as in all things, YMMV.

Another option, if it can be packaged , is to run the intercooler in a tank of ice and water, especially if you can mount the intercooler horizontally so the water can run vertically through the core. The heating of the water will make the water circulate through thermosiphoning, keeping the whole thing at 32o degrees.

For the really adventourous, you can run a brine solution (salt) which will bring the temperature of the ice water down to 0o F. Sounds like magic but this is how ice cream was made back in the olden days. By adding salt, you melt some of the ice but the heat used in melting brings the temperature of the whole mixture down to 0o F. As long as there is extra salt in the solution, the temperature will stay at 0o F. The extra temperature drop equates to free power from the cooler air increasing the air density plus extra spark timing or margin from detonation. Just rerun the calculations above using 0o, throw in a little extra for the initial chilling of the mixture and be careful of corrosion. One final caution using this method. You can get the air too cold and that will in effect cause some of the fuel, when injeced or carb'd into the manifold to become droplets rather than vapor. Efficiency suffers and power may, in fact, even go down.

Copyright (C) 1998 - 2004, all dates inclusive, L.E. Mayfield - All Rights Reserved

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