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David,
Thanks for the explanation. I had thought of using an airspeed
indicator, but wasn't sure that it would give me an accurate reading
without using a pitot tube. Does the tube size need to be the same
as the pitot inlet hole?
If I used a water manometer instead, how do you convert inches to
mph? It seems too that it could make a difference as to where on
the surface of the radiator you take the reading.
One other thing came to mind after I posted my message. I'm not
sure the 3-rotor pumps 50% more coolant than the 2-rotor. While the
20B produces 50% more heat, I don't know if it pumps 50% more
coolant. I do know that the impeller is larger on the 20B.
Has anyone measured the flow of the 20B water pump?
Mark
(So close, yet so far)
At 10:44 AM 8/13/2004 -0500, you wrote:
"How does one determine the
airspeed across the radiator?"
Put a valve that functions like a "selectable" T fitting in
your airspeed
pitot line. (same as an "alternate static pressure valve/switch like
in a
Cessna?) Run the "new/additional" pitot line to a
spot behind your
radiator. Go fly at a specific speed, then switch your airspeed
indicator
from "regular pitot line" to the "radiator airspeed pitot
line" and write
down the results. Repeat at different speeds (climb, economy
cruise, high
cruise, top or race speed).
Or, can plumb in a simple water manometer. Kevin Horton has posted
lots of
stuff on that. I have the e-mails and websites saved if you want to
go that
route.
David
----- Original Message -----
From: "Mark Steitle" <msteitle@mail.utexas.edu>
To: "Rotary motors in aircraft"
<flyrotary@lancaironline.net>
Sent: Friday, August 13, 2004 10:02 AM
Subject: [FlyRotary] Re: Answer to when is 2 gallons enough?Re:
DeltaT
Coolant was : [FlyRotary] Re: coolant temps
Ed,
Wow, that was a great explanation. I think I even understood most
of
it. ;-) I'm assuming that for a 3-rotor I would just
multiply the 2400
BTU figure by 1.5 to arrive at a BTU requirement of 3600 for the
coolant? Same for the air requirement (240 * 1.5 = 360 lbs, or
5524.75 *
1.5 = 8287 CFM). My single radiator core is approx. 18" x
20" or 360 sq.
in. or 2.5 cubic feet. Then 8287/2.5 = 3314.8 ft/min (55.25
ft/sec) air
velocity, which equals 37.67 mph.
Does this look right? I can cool my engine at 37 mph airflow across
the
core? How does one determine the airspeed across the
radiator?
Mark
(See what you started???)
At 10:14 AM 8/13/2004 -0400, you wrote:
>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:
<mailto:msteitle@mail.utexas.edu>Mark
Steitle
>To:
<mailto:flyrotary@lancaironline.net>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
>><http://www.bartleby.com/65/en/Englsh-u.html>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:
<mailto:daveleonard@cox.net>DaveLeonard
>>To:
<mailto:flyrotary@lancaironline.net>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
>> Homepage:
http://www.flyrotary.com/
>> Archive:
http://lancaironline.net/lists/flyrotary/List.html
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