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Ed, I really like your explanations - the math and attention to explaining
the units. Good work.
Now, about "slower water cooling better". I'd like to hear "the rest of the
story". Here's why I ask: I went into a car "window tint" shop 3 weeks
ago to shop for tint on 3 windows on south side of my house. Took care of
that busines - and noticed some nice after-market anodized blue aluminum
housings and stuff hanging on a wall display. I asked if this was an
electric water pump and asked a question about it. The "tint" guy said,
"You'll have to ask ____, the speed shop guy, who shares the 4 shop bays
with me."
I went out and visited with the racing guy - yep, used electric water pumps
and can fix me up with an in-line pump for my rotary engine on RV-6 since
aluminum housings are only for specific V-8s. He was working on his drag
racer and pointed to the thermosat housing and some large washers. "Then
thar washers are different sizes for restricting and adjusting the flow rate
through the radiator. Makes it cool better." So, "thar you have it".
Doesn't make sense to me
David
----- Original Message -----
From: "Ed Anderson" <eanderson@carolina.rr.com>
To: "Rotary motors in aircraft" <flyrotary@lancaironline.net>
Sent: Friday, August 13, 2004 9:14 AM
Subject: [FlyRotary] Answer to when is 2 gallons enough?Re: DeltaT Coolant
was : [FlyRotary] Re: coolant temps
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 -----
From: Mark Steitle
To: Rotary motors in aircraft
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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