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'sway it looks to me. Pass the antifreeze! Hic
Ed
Anderson
RV-6A N494BW Rotary Powered
Matthews, NC
----- Original Message -----
From: "David Carter" <dcarter@datarecall.net>
To: "Rotary motors in aircraft" <flyrotary@lancaironline.net>
Sent: Friday, August 13, 2004 6:09 PM
Subject: [FlyRotary] Re: Answer to when is 2 gallons enough?Re: DeltaT
Coolant was : [FlyRotary] Re: coolant temps
> So, the moonshiners slowed the flow and correctly and logically noted an
> increase in delta T from inlet to outlet of radiator, then focused on
> "longer time in radiator lets liquid cool more" - then failed to continue
> their test - to observe what would happen if they doubled the flow rate.
> If they got half the delta T in half the time (or slightly more than half
of
> delta T), then they could have, with an adequate "mountain education in
> thermodynamics", seen that they were extracting the same amount of heat -
> or slightly more - per unit of time, right? "Hey, Bill (hic), what the
> h_ll's he talking about? (Hic) What's the difference in temperature and
> heat? Ain't it the same thing? Mom's thermometer days 'temperature'?
What
> else we need to know?"
>
> David
>
> ----- Original Message -----
> From: "Ed Anderson" <eanderson@carolina.rr.com>
> To: "Rotary motors in aircraft" <flyrotary@lancaironline.net>
> Sent: Friday, August 13, 2004 1:13 PM
> Subject: [FlyRotary] Re: Answer to when is 2 gallons enough?Re: DeltaT
> Coolant was : [FlyRotary] Re: coolant temps
>
>
> > Ok, David, since you asked.
> >
> > In the early days of car racing (and still) some folks decided to do
some
> > data collection about the effects of different factors.
> >
> > Being aware of the value of data they placed "Mom's" cooking
thermometers
> > at the entrance and exit of a radiator mounted next to the still. With
a
> > fire burning under the still they proceeded to pump hot water .... OK,
> Ok,
> > getting serious.
> >
> > Here is what let to the mistaken belief that "Slow Water cools better".
> >
> > Folks observed that if they measure the entry and exit temperature of
> water
> > as it passed through a radiator an interesting thing. The slower the
> water
> > flowed the greater the temperature difference between the water entering
> and
> > the water exiting the radiator. AND that observation is absolutely
> > correct. The reason of course is the longer the water stays in the
> radiator
> > the more heat is removed from the water before it exits the radiator.
So
> > far they were on solid ground in their observations - however their
> > interpretation of the data is where they went afoul.
> >
> > The assumed the greater temperature drop of the slower water was good as
> it
> > showed more heat went out of the water - I mean how can you argue with
> > that?? Therefore slow water must be better at cooling - right? Well,
> > actually NOT!!!
> >
> > Where they failed is in considering the effects of the "slower" water on
> the
> > ENTIRE system - not just the radiator, but the engine which this is all
> > about in the first place. In reality lessening the flow of the coolant
> > (while it will cause a great deltaT of the coolant flowing through the
> > radiator) will fail to remove heat from the engine as fast as faster
> moving
> > water will as the recent examples show.
> > However, there are folks to this day will swear by "Slow" water. I
argued
> > with old man Lou Ross for 45 minutes on the phone one time trying to
> reason
> > it through with him but to no avail. It only stopped when I got a bit
> > frustrated and stated "Well, Lou, if slow water cools better --- then
> > Stopped water must cool best!!"" He of course knew that wasn't the case,
> but
> > still clung to that belief.
> >
> > Now, I have read that in some cases restrictors have helped cooling -
but
> > not by slowing the water flow. In some cases, it supposedly reduced
> > cavitation of the water pump, kept head pressure in the block higher,
> > promoted nucleated boiling and a number of things that I never bothered
to
> > look into. But all things equal more coolant flow equals more heat
> removed
> > from the engine (always assuming you get rid of the heat through a
> radiator
> > before the coolant returns to the engine).
> >
> > There David, hope that answers your question.
> >
> > Ed
> >
> > Ed Anderson
> > RV-6A N494BW Rotary Powered
> > Matthews, NC
> > ----- Original Message -----
> > From: "David Carter" <dcarter@datarecall.net>
> > To: "Rotary motors in aircraft" <flyrotary@lancaironline.net>
> > Sent: Friday, August 13, 2004 11:38 AM
> > Subject: [FlyRotary] Re: Answer to when is 2 gallons enough?Re: DeltaT
> > Coolant was : [FlyRotary] Re: coolant temps
> >
> >
> > > 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
> > >
> > >
> > >
> > > >> Homepage: http://www.flyrotary.com/
> > > >> Archive: http://lancaironline.net/lists/flyrotary/List.html
> >
> >
> >
> > >> Homepage: http://www.flyrotary.com/
> > >> Archive: http://lancaironline.net/lists/flyrotary/List.html
> >
> >
>
>
>
> >> Homepage: http://www.flyrotary.com/
> >> Archive: http://lancaironline.net/lists/flyrotary/List.html
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