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Date: Fri, 13 Aug 2004 15:31:51 -0400
From: "ericruttan@chartermi.net" <ericruttan@chartermi.net>
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To: Rotary motors in aircraft <flyrotary@lancaironline.net>
Subject: Re: [FlyRotary] Re: Answer to when is 2 gallons enough?Re: DeltaT
 Coolant was : [FlyRotary] Re:  coolant temps
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Let me add a trifle.

Car racers I ran with as a kid with less creativity and more "results=20
oriented" said the same thing, not because they had any fancy thermometer=
s,=20
but because they raced with the thermostats out, and boiled over.

The reasoning was hard to argue with.  Put the thermostat (read restricti=
on)=20
in and the car ran fine.  Take it out and it would boil over.

How can a student of physics argue with that!

Years later I found out that the water pump was cavitating due to the=20
reduced pressure.  With the restriction the pressure was higher and the=20
cavitating less likely.

With the addition of a better water pump vane design the boil overs stopp=
ed=20
when the thermostats where removed, and the cars ran cooler, and faster.


Ed Anderson wrote:
> Ok, David, since you asked.
>=20
> In the early days of car racing (and still) some folks decided to do so=
me
> data collection about the effects of different factors.
>=20
> Being aware of the value of data they placed "Mom's" cooking  thermomet=
ers
> 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.
>=20
> Here is what let to the mistaken belief that "Slow Water cools better".=

>=20
> Folks observed that if they measure the entry and exit temperature of w=
ater
> as it passed through a radiator an interesting thing.  The slower the w=
ater
> flowed the greater the temperature difference between the water enterin=
g and
> the water exiting the radiator.  AND  that observation is absolutely
> correct.  The reason of course is the longer the water stays in the rad=
iator
> 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.
>=20
> The assumed the greater temperature drop of the slower water was good a=
s 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!!!
>=20
> Where they failed is in considering the effects of the "slower" water o=
n the
> ENTIRE system  - not just the radiator, but the engine which this is al=
l
> 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 mo=
ving
> water will as the recent examples show.
> However, there are folks to this day will swear by "Slow" water.  I arg=
ued
> with old man Lou Ross for 45 minutes on the phone one time trying to re=
ason
> 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.
>=20
> Now, I have read that in some cases restrictors have helped cooling - b=
ut
> 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 rem=
oved
> from the engine (always assuming you get rid of the heat through a radi=
ator
> before the coolant returns to the engine).
>=20
> There David, hope that answers your question.
>=20
> Ed
>=20
> Ed Anderson
> RV-6A N494BW Rotary Powered
> Matthews, NC
> ----- Original Message -----=20
> 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
>=20
>=20
>=20
>>Ed, I really like your explanations - the math and attention to explain=
ing
>>the units.  Good work.
>>
>>Now, about "slower water cooling better".  I'd like to hear "the rest o=
f
>=20
> the
>=20
>>story".   Here's why I ask:  I went into a car "window tint" shop 3 wee=
ks
>>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 aluminu=
m
>>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 bay=
s
>>with me."
>>
>>I went out and visited with the racing guy - yep, used electric water
>=20
> pumps
>=20
>>and can fix me up with an in-line pump for my rotary engine on RV-6 sin=
ce
>>aluminum housings are only for specific V-8s.  He was working on his dr=
ag
>>racer and pointed to the thermosat housing and some large washers.  "Th=
en
>>thar washers are different sizes for restricting and adjusting the flow=

>=20
> rate
>=20
>>through the radiator.  Makes it cool better."  So, "thar you have it".
>>Doesn't make sense to me
>>
>>David
>>
>>----- Original Message -----=20
>>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 Cool=
ant
>>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 o=
f
>>water, It would actually only take 8*2 =3D 16 lbms *10 =3D 160 BTU to r=
aise
>=20
> the
>=20
>>temp of the water 1 degree F.  The difference is in one case we are
>=20
> talking
>=20
>>about raising the temperature of a fixed static amount of water which c=
an
>>not readily get rid of the heat, in the other (our radiator engine case=
)
>=20
> we
>=20
>>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 wha=
t
>=20
> we
>=20
>>can determine.
>>
>>If we take our 2 gallons and start moving it from engine to radiator an=
d
>>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 w=
ill
>>move that 160 BTU 15 times/minute (at 30 gallons/minute the 2 gallons
>=20
> would
>=20
>>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
>=20
> from
>=20
>>engine to radiator 15 times a minute =3D 160*15 =3D 2400 BTU/Min. Amazi=
ng
>=20
> isn't
>=20
>>it?   So no magic, just math {:>).  So that is how our 2 gallons of wat=
er
>>can transfer 2400 BTU/min from engine to radiator.  It also shows why t=
he
>>old wives tale about "slow water" cooling better is just that (another
>=20
> story
>=20
>>about how that got started)
>>
>>
>>In the  equation Q =3D W*deltaT*cp that specifies how much heat is
>=20
> transferred
>=20
>>,we are not talking about capacity such as 2 gallons capacity of a cool=
ing
>>system but instead are talking about mass flow.  As long as we reach th=
at
>>flow rate  1 gallon at 30 gpm or 1/ gallon at 60 gpm or 1/4 gallon at 1=
20
>>gpm all will remove the same amount of heat.  However if you keep
>=20
> increasing
>=20
>>the flow rate and reducing the volume you can run into other problems -=

>=20
> like
>=20
>>simply not enough water to keep your coolant galleys filled {:>), so th=
ere
>>are limits.
>>
>>Our  2 gallon capacity is, of course, simply recirculated at the rate o=
f
>=20
> 30
>=20
>>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 wor=
ks
>=20
> as
>=20
>>planned.  IF  the coolant does not give up as much heat in the radiator=
s
>=20
> (to
>=20
>>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=

>=20
> the
>=20
>>mass flow.
>>
>>The mass flow =3D mass of the medium (8 lbs/gallon for water) * Flow ra=
te(30
>>gpm) =3D240 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 =3D W*deltaT*cp, we have Q =3D 240*10*1 =3D 2400 BTU/minute is requir=
ed to
>>increase the temperature of this mass flow by 10F
>>
>>Using the more accurate weight of water we would have  8.34*30 =3D  250=
=2E2
>>lbm/minute  so the actual BTU required is closer to 2502 BTU/min instea=
d
>=20
> of
>=20
>>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 lo=
wer
>>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 specifi=
c
>>heat of air is only 0.25 compared to water's 1.0.  So a lb of air will
>=20
> only
>=20
>>carry approx 25% the heat of a lb of water, so again for this reason yo=
u
>>need more air flow.
>>
>>if 30 gpm of water will transfer 2400 bTu of engine heat (using Tracy's=

>=20
> fuel
>=20
>>burn of 7 gallon/hour), how much air does it require to remove that hea=
t
>>from the radiators?
>>
>>Well  again we turn to our equation and with a little algebra we have W=
 =3D
>>Q/(DeltaT*Cp) =3D 2400/(10*1) =3D 240 lbm/min. Not a surprise as that i=
s what
>=20
> we
>=20
>>started with.
>>
>>But now taking the 240 lbm/min mass flow and translating that into Cubi=
c
>>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) =3D 3157 cu=
bic
>>feet/min to equal  the same mass as the coolant. But since the specific=

>=20
> heat
>=20
>>of air is lower (0.25) that water, we actually need 75% more air mass o=
r
>>1.75 * 3157 =3D 5524.75 CFM air flow at sea level. Now I know this soun=
ds
>=20
> like
>=20
>>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 =3D 19=
0 sq
>>inches and turning that in to square feet =3D 1.32 sq ft we take our
>>5524.75 cubic feet minute and divide by 1.32 sq ft =3D 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 =3D=

>=20
> 69.75
>=20
>>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
>=20
> factors
>=20
>>such as the density of the air which unlike water changes with altitude=
,
>=20
> the
>=20
>>temperature of the air, etc. that can change the numbers a bit.  But, t=
hen
>>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 -----=20
>>  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
>=20
> takes
>=20
>>1BTU to raise 1lb of water 1 degree, and you factor in 30 gpm flow to c=
ome
>>up with a 2400 BTU requirement for a 10 degree rise for 1 lb of water,
>=20
> where
>=20
>>does the number of pounds of water figure into the equation, or do we j=
ust
>>ignore that issue?  Water is 8.34 lbs/gal, and say you have 2 gallons o=
f
>>coolant, that would be 16.68 lbs.  Seems that we would need to multiply=

>=20
> the
>=20
>>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 =3D 240 lbs/min.  Since its
>>temperature is raised 10 degree F we have
>>
>>    BTU =3D 240 * 10 * 1 =3D 2400 BTU/min
>>
>>    I know I'm ancient and  I should move into the new metric world, bu=
t
>=20
> at
>=20
>>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
>=20
> raise
>=20
>>the temperature of one pound of water at its maximum density [which occ=
urs
>>at a temperature of 39.1 degrees Fahrenheit (=B0F) ] by 1=B0F. The Btu =
may
>=20
> also
>=20
>>be defined for the temperature difference between 59=B0F and 60=B0F. On=
e Btu
>=20
> is
>=20
>>approximately equivalent to the following: 251.9 calories; 778.26
>>foot-pounds; 1055 joules; 107.5 kilogram-meters; 0.0002928 kilowatt-hou=
rs.
>=20
> A
>=20
>>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 -----=20
>>      From: DaveLeonard
>>      To: Rotary motors in aircraft
>>      Sent: Thursday, August 12, 2004 8:12 PM
>>      Subject: [FlyRotary] Re: DeltaT Coolant was : [FlyRotary] Re:
>=20
> coolant
>=20
>>temps
>>
>>
>>      Ed, are those units right.  I know that the specific heat of wate=
r
>=20
> is
>=20
>>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
>=20
> should
>=20
>>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 =3D W*DeltaT*Cp  Basic Heat/Mass Flow equation  With water as t=
he
>=20
> mass
>=20
>>with a weight of 8 lbs/ gallon and a specific heat of 1.0
>>
>>      Q =3D BTU/min of heat removed by coolant mass flow
>>
>>       Assuming 30 GPM coolant flow =3D 30*8 =3D 240 lb/min mass flow.
>=20
> specific
>=20
>>heat of water  Cp =3D 1.0
>>
>>
>>       Solving for DeltaT =3D Q/(W*Cp) =3D 2400/(240*1)  =3D  2400/240 =
=3D 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 app=
rox
>>2400/(240 *0.7) =3D 2400/168 =3D 14.2F so I would say you do not fly wi=
th
>>
>>       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 condit=
ion
>>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
>=20
>=20
>=20
>=20
>>> Homepage:  http://www.flyrotary.com/
>>> Archive:   http://lancaironline.net/lists/flyrotary/List.html
>=20
>=20