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