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