Forced Convection passed Heating Coil - A question of Understanding

[HughRob]

My question which seems embarrassing resolves around the final temperature of the downstream air flow.

If we have a duct and inside the duct is a heating coil.

Lets say the heating coil only covers 50% of the cross sectional area for arguments sake, it is obvious that only a proportion of the air will be in contact with the heater.

By the usual method we can calculate the convection transfer coefficient and therefore calculate the heat input that will occur.

To calculate the actual final mixed air stream temperature do I need to use the enthalpy change divided by the specific heat to calculate the temperature delta?

 

[speco]

HughRob,

The heat input is a function of the heat transfer coefficient and the LMTD. You need to calculate the air outlet temperature as well as the hot fluit outlet temperatur to get an LMTD.
Remember, the heat input is the same as the heat transferred.

So, it's a re-iterative process. Start with the mass flow of air and the hot fluid. I'm assuming that the hot fluid is non-isothermal (that is, not condensing in this case).

The total heat gained by the air is the same as the heat lost by the hot fluid. So, for each side, Q=mass * Cp * (t2-t1) (or T1-T2)

That is Q = heat transferred

Cp is specific heat of the fluid

T1 & T2 refer to the hot fluid inlet and outlet temperatures.

t1 & t2 refer to the cold fluid inlet and outlet temperatures.

At the same time...

Q = U * A * LMTD

where U is the overall heat transfer coefficient,

A is the surface area, and

LMTD is the effective temperature difference between the two fluids (Log-mean temperature difference) This calculation is available in any textbook on heat transfer.

So all three values of Q must be identical.

If you start with some real numbers for the mass flows, you will quickly be able to zero in on the value of Q that makes everything in balance.

Regards,

Speco (

http://www.stoneprocess.com