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I'm sorry, Mark, I did not show that step.
You are correct the weight (mass) of water(or any other cooling medium) is
an important factor as is its specific heat.
In the example you used - where we have
a static 2 gallons capacity of water, It would actually only take 8*2 = 16 lbms
*10 = 160 BTU to raise the temp of the water 1 degree F. The difference is
in one case we are talking about raising the temperature of a fixed static
amount of water which can not readily get rid of the heat, in the other (our
radiator engine case) we are talking about how much heat the coolant can
transfer from engine to radiator. Here the flow rate is the key factor.
But lets take your typical 2 gallon cooling system
capacity and see what we can determine.
If we take our 2 gallons and start moving it from
engine to radiator and back we find that each times the 2 gallons circulates it
transfers 160 BTU (in our specific example!!). So at our flow rate of 30 gpm we
find it will move that 160 BTU 15 times/minute (at 30 gallons/minute the 2
gallons would be transferred 15 times). So taking our 160 BTU that it took
to raise the temp of our 2 gallons of static water 10F that we now have
being moved from engine to radiator 15 times a minute = 160*15 = 2400
BTU/Min. Amazing isn't it? So no magic, just math
{:>). So that is how our 2 gallons of water can transfer 2400 BTU/min
from engine to radiator. It also shows why the old wives tale about "slow
water" cooling better is just that (another story about how that got
started)
In the equation Q = W*deltaT*cp that
specifies how much heat is transferred ,we are not talking
about capacity such as 2 gallons capacity of a cooling
system but instead are talking about mass flow. As long
as we reach that flow rate 1 gallon at 30 gpm or 1/ gallon at 60 gpm or
1/4 gallon at 120 gpm all will remove the same amount of heat.
However if you keep increasing the flow rate and reducing the volume
you can run into other problems - like simply not enough water to keep your
coolant galleys filled {:>), so there are limits.
Our 2 gallon capacity is, of course,
simply recirculated at the rate of 30 gpm through our engine (picking up heat-
approx 2400 BTU/min in this specific example) and then through our radiator
(giving up heat of 2400 BTU/Min to the air flow through the
radiators) assuming everything works as planned. IF the coolant does
not give up as much heat in the radiators (to the air stream) as it picks up in
the engine then you will eventually (actually quite quickly) over heat your
engine.
The 240 lb figure I used in the previous example
comes from using 8 lb/gal (a common approximation, but not precise as you point
out) to calculate the mass flow.
The mass flow = mass of the medium (8 lbs/gallon
for water) * Flow rate(30 gpm) =240 lbs/min mass flow. Looking at the units we
have
(8 lbs/gallon)*(30 Gallon/minute) canceling out the
like units (gallons) leaves us with 240 lb/minute which is our mass flow in this
case.
Then using the definition of the BTU we have 240
lbs of water that must be raised 10F. Using our heat transfer equation
Q = W*deltaT*cp, we have Q = 240*10*1 = 2400
BTU/minute is required to increase the temperature of this mass flow by
10F
Using the more accurate weight of water we would
have 8.34*30 = 250.2 lbm/minute so the actual BTU required is
closer to 2502 BTU/min instead of my original 2400 BTU/Min, so there is apporx a
4% error in using 8 lbs/gallon. If we could ever get accurate enough where
this 4% was an appreciable part of the total errors in doing our back of the
envelope thermodynamics then it would pay to use 8.34 vice 8, but I don't think
we are there, yet {:>).
Now the same basic equation applies to the amount
of heat that the air transfers away from out radiators. But here the mass
of air is much lower than the mass of water so therefore it takes a much higher
flow rate to equal the same mass flow. What makes it even worse is
that the specific heat of air is only 0.25 compared to water's 1.0.
So a lb of air will only carry approx 25% the heat of a lb of water, so again
for this reason you need more air flow.
if 30 gpm of water will transfer 2400 bTu of engine
heat (using Tracy's fuel burn of 7 gallon/hour), how much air does it require to
remove that heat from the radiators?
Well again we turn to our equation and with a
little algebra we have W = Q/(DeltaT*Cp) = 2400/(10*1) = 240 lbm/min. Not a
surprise as that is what we started with.
But now taking the 240 lbm/min mass flow and
translating that into Cubic feet/minute of air flow. We know that
a cubic foot of air at sea level weighs approx 0.076 lbs. So 240
lbm/(0.076 lbm/Cubic foot) = 3157 cubic feet/min to equal the same mass as
the coolant. But since the specific heat of air is lower (0.25) that water, we
actually need 75% more air mass or 1.75 * 3157 = 5524.75 CFM air flow at sea
level. Now I know this sounds like a tremendous amount of air but stay with me
through the next step.
Taking two GM evaporator cores with a total frontal
area of 2*95 = 190 sq inches and turning that in to square feet = 1.32 sq ft we
take our
5524.75 cubic feet minute and divide by 1.32 sq ft
= 4185 ft/min for the required air velocity to move that much air volume through
our two evaporator cores. To get the air velocity in ft/sec
divide 4185/60 = 69.75 ft/sec airflow velocity through our radiators
or 47.56 Mph. Now that sounds more reasonable doesn't it??
Now all of this is simply a first order
estimate. There are lots of factors such as the density of the air which
unlike water changes with altitude, the temperature of the air, etc. that can
change the numbers a bit. But, then there is really not much point in
trying to be more accurate given the limitations of our experimentation accuracy
{:>).
Also do not confuse the BTUs required to raise the
temperature of 1 lb of water 1 degree F with that required to turn water in to
vapor - that requires orders of magnitude more BTU.
Hope this helped clarify the matter.
Ed
Ed Anderson
RV-6A N494BW Rotary Powered
Matthews, NC
----- Original Message -----
Sent: Friday, August 13, 2004 8:32
AM
Subject: [FlyRotary] Re: DeltaT Coolant
was : [FlyRotary] Re: coolant temps
Ed,
Please humor me (a non-engineer) while I
ask a dumb question. If it takes 1BTU to raise 1lb of water 1 degree,
and you factor in 30 gpm flow to come up with a 2400 BTU requirement for a 10
degree rise for 1 lb of water, where does the number of pounds of water figure
into the equation, or do we just ignore that issue? Water is 8.34
lbs/gal, and say you have 2 gallons of coolant, that would be 16.68 lbs.
Seems that we would need to multiply the 2400 figure by 16.68 to arrive at a
total system requirement of 40,032 BTU/min. What am I missing
here?
Mark S.
At 09:58 PM 8/12/2004
-0400, you wrote:
Right you
are, Dave
Below is one
semi-official definition of BTU in English units. 1 BTU is amount of
heat to raise 1 lb of water 1 degree Fahrenheit.
So with Tracy's 30 gpm flow of
water = 240 lbs/min. Since its temperature is raised 10 degree F we
have
BTU = 240 * 10 * 1 = 2400
BTU/min
I know I'm ancient
and I should move into the new metric world, but at least I didn't do
it in Stones and Furlongs {:>)
Ed
The Columbia Encyclopedia, Sixth
Edition. 2001.
British thermal
unit
abbr. Btu, unit for
measuring heat quantity in the customary system of English units of
measurement, equal to the amount of heat required to raise the
temperature of one pound of water at its maximum density [which occurs at a
temperature of 39.1 degrees Fahrenheit (°F) ] by 1°F. The Btu may also be
defined for the temperature difference between 59°F and 60°F. One Btu is
approximately equivalent to the following: 251.9 calories; 778.26
foot-pounds; 1055 joules; 107.5 kilogram-meters; 0.0002928 kilowatt-hours. A
pound (0.454 kilogram) of good coal when burned should yield 14,000 to
15,000 Btu; a pound of gasoline or other
.
Ed
Anderson
RV-6A N494BW Rotary Powered
Matthews, NC
- ----- Original Message -----
- From: DaveLeonard
- To: Rotary motors in
aircraft
- Sent: Thursday, August 12, 2004 8:12 PM
- Subject: [FlyRotary] Re: DeltaT Coolant was : [FlyRotary] Re:
coolant temps
- Ed, are those units right. I know that the specific heat of
water is 1.0 cal/(deg Celsius*gram). Does that also work out to 1.0
BTU/(deg. Farhengight * Lb.) ?
-
- Dave Leonard
- Tracy my calculations shows your coolant temp drop is where it should
be:
-
- My calculations show that at 7 gph fuel burn you need to get rid of
2369 BTU/Min through your coolant/radiators. I rounded it off to
2400 BTU/min.
-
- Q = W*DeltaT*Cp Basic Heat/Mass Flow equation With water
as the mass with a weight of 8 lbs/ gallon and a specific heat of 1.0
-
- Q = BTU/min of heat removed by coolant mass flow
-
- Assuming 30 GPM coolant flow = 30*8 = 240 lb/min mass flow.
specific heat of water Cp = 1.0
-
-
- Solving for DeltaT = Q/(W*Cp) = 2400/(240*1) =
2400/240 = 10 or your delta T for the parameters specified should be
around 10F
-
- Assuming a 50/50 coolant mix with a Cp of 0.7 you would have
approx 2400/(240 *0.7) = 2400/168 = 14.2F so I would say you do not fly
with
-
- a 50/50 coolant mix but something closer to pure water.
But in any case, certainly in the ball park.
-
- You reported 10-12F under those conditions, so I would say condition
is 4. Normal operation
-
- Ed
-
- Ed Anderson
- RV-6A N494BW Rotary Powered
- Matthews, NC
-
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