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.

 

 

Thanks, Bill

 

    Its nice to have someone with some knowledge of thermodynamics to keep me straight.

K&W also points out the log curve in the temperature gradient in the core from the walls to the center of the air passages through the core.

 

ED