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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
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