I don't think the biggest problem at Bonneville for most is water
capacity. In our roadster, we used to run a 20 gallon tank with a stock
water pump and had to restrict the flow to a 3/8" discharge to get it
warm enough.
The volume of water circulation is most likely the biggest culprit. In
our present car, we have a four-core radiator with electric fan and
overflow tank with a 15 lb. cap. We don't run a stock pump, though, and
do have a bit of problem keeping cool enough on gasoline. Our pump is a
Jabsco centrifugal. I'm, in the process of trying to find a stock type
electric pump that will fit my application. (we drive an injector pump
off the cam). Any suggestions?
Tom
John Beckett wrote:
>
> 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
> >Sunbeam Alpine Bonneville Land Speed Racer,
> >'66 Hydroplane Drag Boat (390 FE)
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