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Thanks, Dave.
You are right, IF you are able to dump more heat into the air per cubic
foot of flow then you require less flow. I am not certain about 50F for
the DeltaT, from what Todd and others reported for GM cores it may be closer
to 30F. But, a different core can certainly provide a different deltaT.
Ed
Ed Anderson
RV-6A N494BW Rotary Powered
Matthews, NC
----- Original Message -----
From: <daveleonard@cox.net>
To: "Rotary motors in aircraft" <flyrotary@lancaironline.net>
Sent: Friday, August 13, 2004 5:34 PM
Subject: [FlyRotary] Re: Answer to when is 2 gallons enough?Re: DeltaT
Coolant was : [FlyRotary] Re: coolant temps
> Ed, great discussion. You forgot to account for the fact that the delta T
of the air will be much higher than that of the water. My guess is about 50
deg., this would lower our required mass of air down even lower.
>
> Dave Leonard
>
> >
> > 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<mailto:msteitle@mail.utexas.edu>
> > To: Rotary motors in aircraft<mailto:flyrotary@lancaironline.net>
> > 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<http://www.bartleby.com/65/en/Englsh-u.html>, 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<mailto:daveleonard@cox.net>
> > To: Rotary motors in
aircraft<mailto:flyrotary@lancaironline.net>
> > 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
> >
>
>
> >> Homepage: http://www.flyrotary.com/
> >> Archive: http://lancaironline.net/lists/flyrotary/List.html
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