Designing a coil for heat exchanging

[KFCeng]

Hello,

I'm trying to calculate the heat exchange area of a coil needed to maintain an oven at 80 °C, but I'm finding difficulties that you may help me with.

The process is as follow: there is a cubic oven that needs to be kept at 80 °C, so for this I'm designing coils to be set on its 3 sidewalls (the 4th sidewall is the oven's door). The fluid inside the coils enters as vapor at X kg/h and condenses throughout the process, exiting the coils as a liquid.

To calculate the heat exchange area needed I started with the energy balance:
Qresult = Qcoil - Qlosses
Qresult = m*Δh - U*A*ΔT (mass flow of the vapor inside the coil and its enthalpy, for the first term, and heat transfer that leaves the oven through its walls, for the second term)

The problem is the mass flow of the vapor that enters the coil is very high (already established) resulting in a really high Qcoil when compared to Qlosses. That gives us way more heat then we need to just keep the oven at 80 °C, let's put it this way.

My question is, first of all, is that approach correct? If so, is that a way to decrease the heat entering the system without changing the mass flow of the vapor?

And also (a very fundamental question, sorry) how can we find the area from this?

 

[Snickster]

I believe that the correct way to look at it is that when the heat input to the oven from the coil equals the heat loss from the oven to the environment at 80C then the temperature will reach equillibrium temperature of 80C desired. In other words:

Qcoil/oven = Qlosses to environment at 80C equillibrium temperature
or
m*Cp*dTa = Ub*Ab*dTb = Uc*Ac*dTc Note that the left side is the heat loss from the fluid just base on energy balance dH, the middle is the heat crossing from the fluid thru tubes to the oven - this actual sets the left side energy loss from the fluid due to dH, and on the right is heat crossing from oven to the environment - all must be equal at equillibrium temperature of 80C

Where m = mass flow of gas
Cp = specific heat of gas
dTa = change in temperature of gas
Ub = overal heat transfer coefficient between inside of tube/gas flowing and outside of tube/oven air (U = 1/Sum of Resistances of inside tube film, tube wall and outside tube film; 1/U = 1/(hi +Rtube + ho) see attached manual) Note the film resistance are based on many factors most important being fluid heat transfer properties and fluid velocities.
Ab = heat transfer surface area of tube
dTb = difference between temperature inside tubes and oven (note for a coil heat transfer dT is actually the LMTD - log mean temperature difference)
Uc = overal heat transfer coefficient between inside of oven and outside of oven (may be inside and outside air film plus wall construction could include air gaps or insulation layers).
Ac = heat transfer surface area oven
dTc = temperature difference between air inside and outside of oven.

So you see it get pretty complicated for find all these values to plug into the equations

As you flow gas through the tubes the heat transfer to the oven will be greater at first due to the high dT between gas temperature in tube and oven temperature, at the same time heat loss will be lowest since oven is at lowest temperaure difference to the outside of oven or basically dTc = 0. So the oven gradually heats up since the heat input from the coil is less than the heat loss from the oven to the environment until 80C design temperature is reached and equillibrium between heat input and loss occurs if you designed the system correctly.

So basically all you have to do is size the area Ab of the coil such that the heat input to the oven from the coil equals the heat loss at 80C design temperature from the oven to the environment. The m*Cp*dTa side of the equation has to always equal the equilibrium heat transfer rate (the dTa will self-adjust automatically). Attached is a heat exchanger manual that goes into factors involved for shell and tube heat exchangers design and in calculation of tube area and heat transfer coefficients, etc., which can be applied to your situation also.

 

[IRstuff]

My question is, first of all, is that approach correct? If so, is that a way to decrease the heat entering the system without changing the mass flow of the vapor?

I think you need to step back for about 10 seconds and realize that the heat exchange is a function of the area of the piping exposed within the chamber, right? If you had no exposed piping inside the chamber, there would be almost no heat transfer. If you don't want to do calculations, you might consider putting shutters over the coils and changing the amount of piping exposed that way

TTFN (ta ta for now)
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[georgeverghese]

Can you adjust vapor temp, if you cannot adjust vapor flowrate?
If your vapor temp is somewhat high, then there is a radiation component also in addition to natural convection. If so, radiation view factors and surface emissivities come into play, and things get complicated. Read up DQ Kern-Process Heat Transfer or Perry Chem Engg Handbook chapter on Radiation. This configuration is similar to that in industrial furnaces.

 

[MatthewL]

KFC, try controlling the pressure on the vapor to make the boiling point 80C. As long as your oven is below 80C, the vapor will condense, releasing heat into the oven. Once the temp reaches 80C, no further condensation will occur, therefore, no further heating. The mass flow of vapor doesn't matter, as long as you can recirculate it back to the heat source. Just make sure you bring the vapor in the top, and condensed liquid/excess vapor out the bottom, to prevent liquid hammer in the lines.

Matt

Quality, quantity, cost. Pick two.

 

[LittleInch]

Or just change the length of the coil?

Or wrap insulation around it?

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