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Why not to do the throttle body coolant bypass

76375 Views 82 Replies 40 Participants Last post by  jman
I've been lurking around here for a while, so here goes:

This is why it is a bad idea to bypass your throttle body coolant line:

First, we need to make some assumptions. The first set of these assumptions deals with the operating condition of the engine. Let’s assume that we are running the engine at full throttle at 4,000 RPM. We have a 3.0L engine that is efficient and has about an 85% volumetric efficiency. Our effective engine volume is 0.85*3.0L or 2.55L. Since we have a four stroke motor, we are pulling in 2.55L of air 2000 times a minute. Therefore we are ingesting 3.00 cubic feet of air per second, after unit conversions. Our throttle body’s inside diameter is 2.5 inches (D) and its total length is 2 inches (L).

The second set of assumptions deals with the air and coolant flowing through our throttle body. Let’s assume that we are pulling in air from outside the engine bay on a warm day. Our intake air is 80 degrees, Fahrenheit. At 80F and 1 atmosphere, air has the following properties:

Density (p): 0.0735 lbm/ft^3
Thermal conductivity (k): 0.01516 BTU/hr*ft*F
Kinematic Viscosity (v): 16.88*(10^-5) ft^2/s
Specific heat (Cp): 0.24 BTU/lbm*F
Prandtl Number (Pr): 0.708 Unitless

Let us also assume that our coolant is 280F and that our throttle body is not cooled significantly by the incoming air. That is, the surface temperature of the inside of the throttle body is always 280F.

Now we will figure out how much and how fast the air is entering the engine. Through simple calculations, knowing the throttle body dimensions and volumetric flow rate and the density of the air, we can find out the mass flow rate and velocity of the air entering the engine. These values are found to be 794 lbm/hr (m) and 88 ft/s (V).

We now need to know if the flow of the air is turbulent or laminar. This will allow us to determine what appropriate equations to use later. First, we need to find the Reynolds Number (a unitless number that allows one to know if the flow is turbulent or not). This is found by the equation:

Re = V*D/v

We find our Reynolds Number to be 110,000. This is definitely turbulent flow! (Anything over 10,000 is defined as fully turbulent flow)

We need to find our entry length, or the length of tubing needed for the flow to become fully turbulent. This value is defined as Lh = 10*D. This value is found to be 2.08 ft. This is acceptable, since there is, most likely, two feet of piping between the throttle body and the air filter.

Since our entry length is less than our actual piping length, we can use Dittus-Boulter equation to determine the Nusselt Number (Nu):. (Sorry about all of this name dropping)

Nu = 0.023*Re^.8*Pr^.4 = h*D/k

We find our Nusselt Number to be 214. The ‘h’ value above is the average heat transfer coefficient. Now, we can actually find the temperature of the air coming out of our throttle body. Solving for h in the above equation yields h equal to 15.58 BTU/hr*ft^2*F.

By using Newton’s law of cooling, where the rate of heat transfer (Q) is determined to be:

Q=h*(area of heat transfer)*(Surface temperature-Medium Temperature)

By using differential equations, natural logs and some other hocus pocus, we get the following equation:


Finally, by using the above equation, the outlet temperature can be determined to be 81.3 degrees Fahrenheit.

Now, using the SAE J1349 correction factor, you lose ~1% of your total power for each 10 degree increase in inlet air temperature. With this 1.3 degree increase, due to the throttle body coolant, you are losing 0.13% of your power. Or, on a 200hp car, you are losing 0.26hp. By overriding the coolant flowing through your throttle body, you are risking having your throttle body freeze open in cold weather (the whole purpose of running coolant through the throttle body in the first place). Hope this clears up any confusion.
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Replying to Topic 'Why not to do the throttle body coolant bypass'

Area of heat transfer = only the inside of the TB

I was ignoring any heat transfer from underhood heat, so

Replying to Topic 'Why not to do the throttle body coolant bypass'

Originally posted by dawei213Now if the temperature of the TB is like 1500 degrees (won't happen but I'm saying IF), then it might affect it.
Well, most aluminum melts at about 1220F, so I would be more concerned with injesting molten aluminum than inlet air temps if my throttle body was that hot!
Replying to Topic 'Why not to do the throttle body coolant bypass'

Yes I agree with what you are saying, I was just giving you a hard time. :)
Replying to Topic 'Why not to do the throttle body coolant bypass'

Originally posted by mmarfan

PS Doctor, no engine runs at near 100% efficancy.
I said volumetric efficiency, not thermal efficiency. 85% volumetric efficiency is not unheard of in a modern, normally aspirated engine.
Replying to Topic 'Why not to do the throttle body coolant bypass'

Originally posted by mmarfan

A reason to do it while drag racing:

HEAT! My Contour (Duratec) ran 15 flat after a 1 hour cool off time and it did not have coolent running through the TB. It ran high 15's when I took it right off the highway (like 15.8), and dropped 2 MPH <- more important. HP = Speed at the end of the track. So in turn if you let you car cool down, bypass the coolent, preventing the engine from sucking hot air into it, you should run better times. BUT LET IT COOL OFF! After you leave the track, re-attach it, well unless you have someone talking crap... :D

PS Doctor, no engine runs at near 100% efficancy.
Heat IS the enemy when it comes to drag racing, but I still would not bypass the coolant in my throttle body or worry about heat transfer through my inlet piping or manifold. From my earlier post, our inlet air is moving at 88ft/s. In a three foot long section of pipe, each molecule of air is only in there for 3/10 of a second. Hardly enough time to heat up significantly. If you have ever passed your hand through the flame of a candle (2000F), you know how important time is to heat transfer (well, maybe not if you burned yourself).

I would suspect that the reason your car ran faster when cool is due to coolant and inlet air temps. Many modern engine ECUs rely heavily on coolant temps when they select fuel and timing maps. Hondas are especially sensitive to this. When the car is running hot, the ECU retards timing and richens the mixture. This is because hot spots on engine sleeves can cause preignition. Also IATs play a major role. ECUs will advance timing with cooler air, yielding more horsepower. Hot inlet air tends to detonate more easily than cold air. This is the second benefit to a cold air intake (the first being the higher oxygen content of the cooler air).

Let's see how significant my 0.25 horsepower gain is. Let us also use a 'horsepower calculator' to help:

If I have a Mazda6s at 3400 lbs (159lb driver) from:

and I pull a 16.07 second 1/4 mile from:

I have 210.51 hp at the flywheel, close to 220hp. Now, if I modify my quarter mile times that will reflect a 0.25 hp gain, I arrive at a 16.0635 second quarter mile, or a gain of 0.0065 seconds. I know that this is not a very scientific method but 1/4 mile times have to be the worst judge of minor performance gains. Humans cannot even sign their name the same way twice, much less launch, shift and steer the car the same way down a 1320ft long track more than once. Let's not even go into environmental factors!
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