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Ed Anderson wrote:
Heat transfer equation Q =W*DeltaT*Cp, with W =
mass flow down by 30%. So to get rid of the same Q of heat (and since Cp doesn't
change that much) it would appear that means the delta T term would need to
increase by 30% for Q quantity to remain the same. But, I don't know exactly how
a 79Deg colder incoming air would affect the Delta T term.
BIll??
Ed, the equation above represents the heat *picked
up* by the air, and is directly proportional to the Delta T that you can achieve
in the air stream. So, if mass flow is down 30%, for equal heat rejection, you
would need the air to change temperature by an offsetting amount. Easier to use
actual numbers rather than %.
For example, if W = 100, DeltaT = 30 F, and Cp =
0.25, then Q ~ 750.
Reduce W to 70, and to get Q ~ 750 you would need a
deltaT of 42.8 F. There fore lowering the inlet temperature can increase the
temperature carrying capability of the air flowing through the
radiator.
However, there is another complicating factor, and
that is the transfer of the heat from the water to the air. Here you need to use
the log-mean delta T, to see if you can get the heat *into* the
air.
| Calculate Log-Mean
Temp Difference for 90 F day, 180 water inlet, 140 air outlet
(assumed) |
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| Input |
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| Th,in |
180 |
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| Th,o |
150 |
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| Tc,in |
90 |
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| Tc,o |
140 |
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| LMTD |
49.3 |
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| R factor |
0.6 |
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| P factor |
0.555555556 |
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| F factor from chart |
0.91 |
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| DelT_lm |
44.9 |
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| Shows a LMTD of 44.9 F for heat transfer |
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Assuming we are now flying at 0 F altitude, keeping all else the
same.
| Calculate Log-Mean
Temp Difference |
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| Input |
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| Th,in |
180 |
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| Th,o |
150 |
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| Tc,in |
0 |
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| Tc,o |
140 |
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| LMTD |
83.2 |
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| R factor |
0.214285714 |
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| P factor |
0.777777778 |
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| F factor from chart |
0.94 |
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| DelT_lm |
78.2 |
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we get a 78 F LMTD.
So the heat transfer from the exchanger surface to the air
Q = h*A*LMTD will increase, a 90F drop in air temperature gives a 33F
increase in driving force for getting the heat from the water to the air.
Bill Schertz
KIS Cruiser # 4045
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