No guessing here. In my /CC I run an 18 gal. tank. Have used up to 700 HP so
far with no cooling system problems through the five mile at Bonneville.
Also know of guys running tanks as small as 15 - 16 gal. successfully.
Ran the Chevette last year (approximately 450 HP) with a data logger for the
first time. When I played it back after the first run, with a 160 degree
thermostat, it ran the full five miles at the required 160 degrees.
I believe this system would run 1000 HP and not go over 180 degrees.
For you radiator guys try Keith Turks NASCAR qualifying radiator with built
in tank, it worked great as I never heard him even mention a cooling problem
this year at Bonneville or hot laps at Maxton.
Hope this helps.
John Beckett
ECTA & BNI record holder
-----Original Message-----
From: Lawrence E. & Cathy R. Mayfield <lemay@hiwaay.net>
To: land-speed@autox.team.net <land-speed@autox.team.net>
Date: Wednesday, November 17, 1999 12:43 PM
Subject: Cooling and Water Tanks
>***************************************************************************
*
>********
>*
> *
>* 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. *
>*
> *
>***************************************************************************
*
>********
>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
>***************************************************************************
*
>************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
>
>DrMayf@AOL.com
>lemay@hiwaay.net <<<<preferred
>
>Sunbeam Tiger, B9471136
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