Subject: Re: [FlyRotary] Re: Air Density at altitude Date: Tue, 9 Dec 2003 12:28:10 -0800 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_84815C5ABAF209EF376268C8" ------=_NextPart_84815C5ABAF209EF376268C8 Content-type: text/html; charset=US-ASCII Content-Transfer-Encoding: quoted-printable =

Ed Anderson wrote:

Heat transfer equation Q =3DW*DeltaT*Cp, wi= th W =3D=20 mass flow down by 30%=2E So to get rid of the same Q of heat (and since Cp d= oesn't=20 change that much) it would appear that means the delta T term would need to=20= increase by 30% for Q quantity to remain the same=2E But, I don't know exact= ly how=20 a 79Deg colder incoming air would affect the Delta T term=2E

BIll??

Ed, the equation above represents the heat = *picked=20 up* by the air, and is directly proportional to the Delta T that you can ach= ieve=20 in the air stream=2E So, if mass flow is down 30%, for equal heat rejection,= you=20 would need the air to change temperature by an offsetting amount=2E Easier t= o use=20 actual numbers rather than %=2E

For example, if W =3D 100, DeltaT =3D 30 F,= and Cp =3D=20 0=2E25, then Q ~ 750=2E

Reduce W to 70, and to get Q ~ 750 you woul= d need a=20 deltaT of 42=2E8 F=2E There fore lowering the inlet temperature can increase= the=20 temperature carrying capability of the air flowing through the=20 radiator=2E

However, there is another complicating fact= or, and=20 that is the transfer of the heat from the water to the air=2E Here you need = to use=20 the log-mean delta T, to see if you can get the heat *into* the=20 air=2E

Calculat= e Log-Mean=20 Temp Difference for 90 F day, 180 water inlet, 140 air outlet=20 (assumed) 180 = =

Input	<= FONT=20 face=3DArial size=3D2>	<= FONT=20 face=3DArial size=3D2>	<= FONT=20 face=3DArial size=3D2>	<= FONT=20 face=3DArial size=3D2>	<= FONT=20 face=3DArial size=3D2>	<= FONT=20 face=3DArial size=3D2>	<= FONT=20 face=3DArial size=3D2>	<= FONT=20 face=3DArial size=3D2>	<= FONT=20 face=3DArial size=3D2>	<= FONT=20 face=3DArial size=3D2>
Th,in

=20

<= FONT=20 face=3DArial size=3D2> Th,o 150 <= FONT=20 face=3DArial size=3D2> Tc,in 90 <= FONT=20 face=3DArial size=3D2> Tc,o 140 <= FONT=20 face=3DArial size=3D2> <= FONT=20 face=3DArial size=3D2> LMTD 49=2E3 <= FONT=20 face=3DArial size=3D2> <= FONT=20 face=3DArial size=3D2> R factor 0=2E6 <= FONT=20 face=3DArial size=3D2> P factor 0=2E555555556 <= FONT=20 face=3DArial size=3D2> F factor from chart 0=2E91 <= FONT=20 face=3DArial size=3D2> <= FONT=20 face=3DArial size=3D2> DelT_lm 44=2E9 <= FONT=20 face=3DArial size=3D2> <= FONT=20 face=3DArial size=3D2> <= FONT=20 face=3DArial size=3D2> <= FONT=20 face=3DArial size=3D2> <= FONT=20 face=3DArial size=3D2> <= FONT=20 face=3DArial size=3D2> <= FONT=20 face=3DArial size=3D2> <= FONT=20 face=3DArial size=3D2> <= FONT=20 face=3DArial size=3D2> <= FONT=20 face=3DArial size=3D2> <= FONT=20 face=3DArial size=3D2> <= FONT=20 face=3DArial size=3D2> <= FONT=20 face=3DArial size=3D2> <= FONT=20 face=3DArial size=3D2> Shows a LMTD of 44=2E9 F for heat transfer =

Assuming we are now flying at 0 F altitude, keeping all else the=20 same=2E

=20 Calculat= e Log-Mean=20 Temp Difference 180 150 0 140 0=2E214285714 0=2E777777778 0=2E94

Input	<= FONT=20 face=3DArial size=3D2>	<= FONT=20 face=3DArial size=3D2>
Th,in
Th,o
Tc,in
Tc,o

LMTD	83=2E2

R factor
P factor
F factor from chart

DelT_lm	78=2E2



	<= FONT=20 face=3DArial size=3D2>	<= FONT=20 face=3DArial size=3D2>

we get a 78 F LMTD=2E

So the heat transfer from the exchanger surface to the air

Q =3D h*A*LMTD will increase, a 90F drop in air temperature gives a 33F=20= increase in driving force for getting the heat from the water to the air=2E<= BR>

Bill Schertz
KIS Cruiser # 4045

= ------=_NextPart_84815C5ABAF209EF376268C8 Content-type: text/plain; charset=US-ASCII Content-Transfer-Encoding: quoted-printable Ed Anderson wrote:=20 Heat transfer equation Q =3DW*DeltaT*Cp, with W =3D mass flow down by 30%=2E= So to get rid of the same Q of heat (and since Cp doesn't change that much) it would a= ppear that means the delta T term would need to increase by 30% for Q quantity to = remain the same=2E But, I don't know exactly how a 79Deg colder incoming air would = affect the Delta T term=2E=20 BIll??=20 Ed, the equation above represents the heat *picked up* by the air, and is di= rectly proportional to the Delta T that you can achieve in the air stream=2E So, if= mass flow is down 30%, for equal heat rejection, you would need the air to change= temperature by an offsetting amount=2E Easier to use actual numbers rather than %=2E=20 For example, if W =3D 100, DeltaT =3D 30 F, and Cp =3D 0=2E25, then Q ~ 750=2E=20= Reduce W to 70, and to get Q ~ 750 you would need a deltaT of 42=2E8 F=2E Th= ere fore lowering the inlet temperature can increase the temperature carrying capabil= ity of the air flowing through the radiator=2E=20 However, there is another complicating factor, and that is the transfer of t= he heat from the water to the air=2E Here you need to use the log-mean delta T, to s= ee if you can get the heat *into* the air=2E=20 Calculate Log-Mean Temp Difference for 90 F day, 180 water inlet, 140 air ou= tlet (assumed)=20 Input =20 Th,in 180 =20 =20 Th,o 150 =20 Tc,in 90 =20 Tc,o 140 =20 =20 LMTD 49=2E3 =20 =20 R factor 0=2E6 =20 P factor 0=2E555555556 =20 F factor from chart 0=2E91 =20 =20 DelT_lm 44=2E9 =20 =20 =20 =20 =20 Shows a LMTD of 44=2E9 F for heat transfer=20 Assuming we are now flying at 0 F altitude, keeping all else the same=2E Cal= culate Log-Mean Temp Difference=20 Input =20 Th,in 180=20 Th,o 150=20 Tc,in 0=20 Tc,o 140=20 =20 LMTD 83=2E2=20 =20 R factor 0=2E214285714=20 P factor 0=2E777777778=20 F factor from chart 0=2E94=20 =20 DelT_lm 78=2E2=20 =20 =20 =20 =20 we get a 78 F LMTD=2E So the heat transfer from the exchanger surface to th= e air Q =3D h*A*LMTD will increase, a 90F drop in air temperature gives a 33F incr= ease in driving force for getting the heat from the water to the air=2E=0D=0A=20 Bill Schertz KIS Cruiser # 4045=20 ------=_NextPart_84815C5ABAF209EF376268C8--