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*
*
* WARNING, Danger, Disclaimer
*
* The following is an attempt by me to understand the sizing requirements
for an *
* auxilliary water tank for cooling in my car. I am not a Thermodynamicist
and *
* have only passing acquaintance with the subject. If someone has a better
*
* analysis, publish it to the list so we can all benefit. If there are
assumptions *
* made which can be bettered then let me know! I'd like to update my
analysis. *
*
*
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I am at the same place in determining cooling system minimum size, so this
is a good
opportunity to validate L.Kvach Butters' cooling system analysis. I spent
the morning running around to get the Circle Track issue that Kvach
mentioned and when I did find it, I could not findthe info he stated. So I
will take it on faith....
The results from using a simple analysis like Butter's confirms his
numbers, however, this is way too conservative and results in a way larger
than needed tank (~ 35 - 40 gal). If radiation and convection are included
then the tank size can be reduced to around 15 gallons (if my assumptions
are anywhere close).
I caution those who would mount the tank in the rear. Why? Well when you
put that much mass in the rear the cg moves backward towards the center of
pressure. When the car is balance for and aft and it looses traction, the
car is very apt to swap ends with the heavy end. End of $0.02.
I will be further analysing my combination to determine if the inclusion of
a radiator (vw rabbit size with elec fans) can keep me under boiling temp
without the extra tank. I may still have to use a small tank.
Analysis
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************Some givens and assumptions are required to establish the
playing field for the analysis:
1) A gallon of gasoline weighs 6.22 pounds.
2) A gallon of water weighs 8.432 pounds
3) Heat value of a pound of gasoline is 19,384 BTU
4) 1 BTU is the heat required to raise 1 pound of water 1 degree F
5) Normally Aspirated Motor requires 0.5 lb fuel per HP
6) Boosted engines require 0.55 lb fuel per HP
7) Cooling system is without a radiator
8) Starting water temperatures are a) 40, b) 50, c) 60 degrees F
9) 5 mile runup
10) Pareto's law for speed buildup - 80% of speed is obtained in first
mile, remaining 20% of the speed is made during last four miles.
Rule of thumb: heat loss through the cooling system is approximately 33% of
that available.
Assume Horse Power Output at the flywheel is 650. Assume max speed is 250 mph.
How big does a closed system water tank need to be to keep the motor below
boiling?
Quick Analysis:
First guess is to say that right out of the gate, the motor is WOT and
making its horsepower.
Speed is down because the cars mass must be accelerated from 0.
The average time for the first mile is:
250 mph x 0.8 = 200mph (at end of 1st mile)
200mph = 3.33 miles per minute
Average speed for first mile = 0 - 3.33/2 = 1.67 miles per minute
Average time for first mile = 1 min/ 1.67 = .6 min = 36 seconds to cover
first mile.
Similarly the times for the reamining miles are:
Mile 2 = 16.36 sec
Mile 3 = 14.75 sec
Mile 4 = 14.52 sec
Mile 5 = 14.46 sec
Total time at WOT = 96 seconds
= 1.6 minutes
= 0.02669 hours
So how much fuel is burned in that amount of time?
650 hp x 0.55 lb fuel/hp/hr = 357.5 lbs fuel per hour
357.5 lbs fuel/hr x 0.02669 hours = 9.542 pounds of fuel (about a gallon
and a half)
So how much heat is liberated during this process?
19384 BTU/lb of Gasoline x 9.542 pounds = 184,962 BTU
So how much heat goes into cooling water?
Rule of thumb says 33%
Therefore 0.33 x 184,962 BTU = 61,038 BTU
So how many pounds of water is required, if heated to 220F?
a) if water is initially at 40F?
Delta T = 180 F
61,038 BTU /180 F = 339.1 lbs
= 40.7 gallons
Using the assumptions for my car, accounting for the differences, the
numbers that Kvach
developed are generally correct.
The big thing missing in this analysis is the mass of the engine. If a
motor weighs approximately 450 pounds and it consists of "cold" metal then
it will take some time to bring it to temperature.
Also, the oil carries away quite a bit of heat. But the engine is the
biggest factor so lets see if this can be calculated to reduce the tank
size. Time also becomes a factor because of the lag associated with heating
the block. In other words, the engine takes a while to get hot which
reduces the size of the water tank.
Now for my engine with 650 HP, the heat flux available to heat the
block/heads and the cooling water can be determined:
650HP x 0.55 lbs fuel/hp/hr = 357.5 lbs fuel burned per hour
If 1/3 is lost to heating the block and coolant water then available heat
is 357.5/3 = 119.17 lb/hr
The heat flux is 119.17 lb/hr x 19384 BTU/lb = 2,309,926.7 BTU/hr
= 641.65 BTU/sec
This is the heat flux available to heat the block and coolant as the motor
is running at WOT.
Here it gets a little tricky. At the same time the engine is being heated,
it is being cooled by the coolant flow.
Qtot = Specific heat of iron times mass of iron times delta T (F) +
specific heat of water times mass of water times delta T (F)
Qtot = 641.56 BTU/sec time time for run = 641.56 BTU/Sec x 96 sec = 61,589 BTU
Therefore 61589 BTU = 0.1 BTU/lb x 450 lb x 180 F + 1 BTU/lb x BTUh20 x 180F
Solving for BTUh20 = 61589 - .1 x 450 x 180 = 53849 BTU that the water must
absorb.
Back solving for the lbs of water needed
53849 = 1 x M x 180
M = 299 lbs
Since a gallon of water weighs 8.432 lbs then it takes 35.5 gallons.
So the numbers are validated one more time 35.5 vs 40 gallons.
But radiation and conduction were not considered when heating the block and
water. A lot of heat is lost in the plumbing between the motor and tank
through conduction.
Lets assume that 1/6 of the available heat is lost through radiation (bolck
to surrounding structure, sheet metal, etc.) at the block area, 1/8 is lost
through conduction (block to air) in the engine compartment, 1/6 is lost at
the tank through radiation and 1/10 through conduction. Are these good
assumptions? Not a clue, just a starting point.
These losses can account for about 55% % of the available heat.
Qtot then = 641.56 x .45 = 288.7 BTU/sec
Running back through the numbers then gets to a tank size of 13.1 gallons.
This does not account for any kind of radiator and electric fan in the system.
L.E. Mayfield
124 Maximillion Drive
Madison, Al. 35758-8171
1-256-837-1051
old >>>>>>>> http://www.hsv.tis.net/~mayfield
new >>>>>>>> http://home.hiwaay.net/~lemayf
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