Main reason I'm asking is:

Not long ago I bought an '87 IROC 5.7 TPI L98, finally found one, been
looking for about two years and figured this is my window of opportunity
because now is probably the time to find them at reasonable prices....
another 5 to 10 years they will be pretty high priced I'd think......
anyway, these cars always seem to run really warm. A had a '94 Z-28 LT1 that
also seemed to run on the warm side also. And my buddy's '93 LT1 is the
same.

The IROC mostly sits in garage and I only drive it for fun in the warmer
months (Btw, I'm located in Virginia) and really could go without a heater
if it would help keep the engine temp down some..... Or I guess I could just
install a 160 or 180-90 stat instead... What do you guys think?
Who all here owns a L98 or LB9 or even an LT1? Do the LS1s also run fairly
hot? I figure they probably stay a bit cooler being an all aluminum
motor......
Bigjfig, Charles what do you think? I know Bigjfig will probably tell me to
leave it alone because thats the way it came from the factory.....

Another thing, I've owned a few 5.0 GT Mustangs, two 5.0 H.O (L69) Z-28s, an
LT1 Z-28 and I gotta say that this L98 IROC is the most fun vehicle to drive
out of all of them..... even if it is an automatic. My LT1 had alot more
top-end HP but the L98 has "loads" of low-end torque... I actually raced my
pals '93 LT1 and spanked him out of the hole, by the 8th mile mark we were
dead even and then his LT1 just walked me towards the end... man, you can
really tach up those LT1s... The L98 just runs out of breath around 4,500
RPM...... I think I'll add some shorty headers and SLP
runners..............fun stuff.

~John

John, that's a very interesting question that comes up from time to time so
let's go back to the basics.

The basic heat transfer formula for two convective cooling mediums in a heat
exchanger is:

Mc^T = Mc^T

where:

M = mass flow rate
c is a heat transfer constant we will ignore since it doesn't change in this
case
^T is the differential temperature between the heat exchanger inlet and
outlet

Reducing the equation gives us:

M^T = M^T

Which we refine to a radiator in a car as:

M^Twater = M^Tair

So the only variables we are looking at are mass flow rate of water through
the radiator and air flow through the radiator.

If the radiator has heat capacity greater than the heat generation of the
engine then an increase in the mass flowrate of the water (by removing the
obstruction to flow of the thermostat) with no change in air flow rate, with
a constant air inlet and outlet temperature will result in a decrease of
differential water temperature across the radiator. In the real world air
outlet temperature will increase but let's ignore that for now since it will
drive ^Twater up a bit. This gives us

^Twater = ^Tair/Mwater

Since Mwater is increasing ^Twater is decreasing.

A casual inspection of the above would lead to the conclusion the water is
travelling too fast through the radiator and the engine coolant temperature
will increase in the engine. But we aren't only dealing with water flow but
mass flowrate through the engine. The greater mass of water flowing through
the engine will have a higher heat capacity and will remove more heat from
the engine which will drive down exit water temperature from the engine. If
the radiator heat rejection rate exceeds the heat generation of the engine
due to combustion, the engine coolant temperature will decrease at all
points in the system, although the delta temperature across the radiator
decreases.

When engineers design heat exchangers, the flowrate of the liquid (or
liquids) must be taken into account but exceeding the design flowrate
usually decreases heat transfer efficiency by a couple of percent so it's
fairly negligible.

So based on the above we see that an increase in flowrate will remove more
heat from the engine but that comes with a few caveats as follows:

* The radiator must be in good shape and must be large enough for the engine
in the car.
* No external heat sources are added to the coolant such as a blown head
gasket.
* The water pump capacity must be adequate to prevent departure from
nucleate boiling.
* The increased flowrate must not suck the lower radiator hose flat and lead
to cavitation.
* The radiator fans (and air flow at speed) must be able to pass enough air
through the
  radiator to prevent overheating at idle.
* Warmup time to operating temperature will increase (running the A/C helps
in a warmup)
  and oil pressure may be unacceptably high (in the case of a high volume
oil pump).

Another aspect is cylinder wear versus engine temperature. Running the
engine below it's design temperature means the bores are tighter and the
rate of ring and cylinder wall wear will increase. This is a factor since
the engine will take longer to heat up to whatever stable temperature the
engine coolant ends up at and that temperature may be below optimum.

So as we following this path to it's bitter end, it's intuitively obvious to
the most casual observer that removing a thermostat to bandaid a problem in
the cooling system will have a negligible effect but if you are in the
borderline case that increasing the horsepower of your engine has strained
the radiator to it's limit, removing the thermostat might prevent incurring
the cost of a very expensive aluminum radiator.

Keep in mind that air flow and engine heat output vary with speed and engine
load but if the radiator can remove more heat that the engine can produce at
all times the engine coolant temperature can still be controlled.

I took this option in my wife's vette back in 92 and have had no problems
since but I do make sure I have time to warm the engine up before driving
it.

Test it and see what happens. An o-ring style water neck is a great help
when swapping to different (or no) thermostats.

Dave

PS - It's late, I'm tired so take all this with a grain of salt. ;^)