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