too much oil pressure?, high OT on the freeway Options

[VaccaRabite]

My 2056 is getting elevated oil temps on the freeway. I am using a high volume pump, and seeing about 40 pounds of pressure on the freeway when my oil is up to temp.

What I think is happening is that there is too much pressure, and the oil is not being pumped through the oil cooler due to the check valve closing the passage. However, this is just a guess, as I don't have any way to test it. All I know is that when I am driving at revs, oil temp goes up and if I keep driving at higher revs the oil temps stay up). if I drop revs, oil temps fall back into place, but I can't do that on the freeway without building a lot of head heat.

Before I go through the trouble and expense of adding an oil cooler, I want to make sure that I have checked out the simple fixes first.

What do I do to make sure oil if going through the oil cooler? What is the point where pressure cuts off passage through the cooler?

All tin is in place, and fan housing flaps divert air over the oil cooler 100% of the time (no thermostat in place, so the flaps are always set to cool).

Zach

[70_914]

Sorry to hijack the thread discussing basic thermodynamics but I will make a couple real world examples- since now I have agreement that fluid speed is the only variable in a closed system with a fixed cooler.

Example A: You are a college student. You are going to WWU taking plastics engineering with a double minor in physics and chemistry. You want to throw a killer party, and decide that running copper coils throgh a bucket of ice water is a good way to cool your beer as it exits a keg. The ice water is a constant temperature, the keg beer is also a constant temperature. You have a pressure valve to control flow through the copper coils. Your beer won't be the same cold temperature if you crank the pressure up to 50 psi and blast it through the coils, it will be colder if it goes through the coils a little slower and allows the ice water to remove more of the heat from the beer.

Example B: You are a fry cook and want to leave work early. You have a large griddle that needs to be cold before you can go home. You also have a piece of ice that is the exact width of this griddle. You can take the ice and rush it across the surface of the griddle at 2 inches per second and have over half the ice cube left at the other end- inefficient because the ice was not able to remove all the heat from the griddle and did not do as much work as it could of, or you could move the ice across the griddle at 1 inch per second and at the very last moment the ice is completely melted- very efficient because the ice was able to do all the work possible.

Both examples are abstract, neither example shows specifically a type IV engine with a pass through cooler, both examples show thermodynamics. Laws of physics don't change on a whim.

The overall heat transfer coefficient for a wall or heat exchanger can be calculated as:

1 / U A = 1 / h1 A1 + dxw / k A + 1 / h2 A2 (1)

where

U = the overall heat transfer coefficient (W/m2K)

A = the contact area for each fluid side (m2)

k = the thermal conductivity of the material (W/mK)

h = the individual convection heat transfer coefficient for each fluid (W/m2K)

dxw = the wall thickness (m)

The thermal conductivity - k - for some typical materials:

•Polypropylene PP - 0.12 W/mK
•Stainless steel - 21 W/mK
•Aluminum - 221 W/mK

The convection heat transfer coefficient - h - depends on

•the type of fluid - gas or liquid
the flow properties such as velocity
•other flow and temperature dependent properties

Heat transfer coefficient for some common fluids:

•Air - 10 to 100 W/m2K
•Water - 500 to 10 000 W/m2K