feel like a total idiot - temp rise in heated pipe

[rhirsch]

I've searched for this throughout the forums. Its seems so simple, but I am simply not getting it:

I have a pipe with gas flowing in it. I know the input temp and output temp I want. I know how much power I am going to put to the heater I know the diameter. I know the flow rate. I think I know everything to be able to discern the required length of the tube to achieve the desired output temp. But I am not seeing how to do it.

Here is the data:
Tin=5 deg C
Tout (desired) = 120 deg C
Mass flow rate, mdot= 3.6 kg/day
Heat input, qconv = 300W
Inner Tube diameter Di= .5 in
Outder Tube diameter Do = .5125 in (sorry about units switch)
h = 25 W/Km2

The problem I am having is that I think I can do the problem if I know the heat flux, but this flux is dependent on the length of the tube (because the surface area increases with length).

so I know
qconv=qdot*Vtube and
qdot=mdot * Cp *(To-Ti)

assuming its insulated and all the heat goes from the heater, through the pipe and into the gas. Then:

qconv=mdot * Cp * (To-Ti) * A * L

and then

L = qconv/(mdot * Cp * (To-Ti) * A)

but this means that the length gets longer if I increase the power for the same temp differential. This makes no sense, and why don't I need the convection coefficient?

I haven't done this stuff in years (decades) and I am obviously doing something wrong. This should be simple....


argh

 

[Compositepro]

If you are using electric heat you have direct control over power input and heat transfer coefficients are irrelevant.
The pipe has to be long enough to contain the heater, which can be designed to be less than an inch. The smaller you make the heater the more likely it is to burn-up if you lose gas flow.

Look at the design of Sylvania compressed air heaters.

http://www.sylvania.com/content/display.scfx?id=003684141

 

[rhirsch]

mintjulep

I dont get that. the only power input to the system is to the heater, this heat must be shared between the loss into the flowing gas stream and the loss through the insulation to the ambient air. So why would you assign the entire 300W to q2?

Compositepro
I have control over the power input but I am fixing it to 300W. The way the model works is that the entire aluminum block will be generating 300W because I will just presume that the block spreads the heat evenly. At this point I dont care about the actual heater implementation. I just want to spend 300 watts heating a block and find out how long the tube in the block needs to be in order to get my gas temp up to 120C

 

[Unotec]

I think you are in the right track. Just remember that your focus is not how much heat you're generating but how much the gas it taking away.
You have to calculate the heat removal rate per unit of length. It's been a while since the books, but I think chapter 3 on the fundamentals of heat and mass transfer by Frank Incropera has the exact same scenario you. Basically you have a radial system with energy generation.
It is a simple process but with many equations.

<<A good friend will bail you out of jail, but a true friend
will be sitting beside you saying ” Damn that was fun!” - Unknown>>

 

[MintJulep]

You have already determined that you need 300 W INTO THE GAS.

If you don't put 300 watts into the gas, you will NEVER get the temperature out you need.

So the heater needs to provide those 300 watts + whatever is lost through the outer insulation.

 

[rhirsch]

MintJulep,

Ah, OK it was just a misunderstanding. I have a 300W heater. I dont know how much will go into the gas (yet).

But yeah, for the purposes of this analysis I think either one is fine.

unotec,
I'll take a gander hopefully it is there.

thanks.

 

[rhirsch]

Still being dense and still not getting my answer:

Ok, I looked at incropera book again. I see the example you are talking about but it is not answering what I need. As per MintJulep, I know the amount of heat going into my gas stream. I need to know how long a tube needs to be to get my temp to rise by 115 degrees.

Ok, so lets choose a length 100 CM. and I have 300 watts. I'll even choose a diameter 1.25 cm.

so now I know my heat input per unitlength:

q'=300/(2*pi*r)

ok, so now how do I figure out the temp rise associated with this heat input?

i'm totally missing something I feel like im right back where I started
and perhaps I am, just, in this case I have picked a length along with the power.

q"=Qconv/(Pi*D*L)=7600 W/m2

tout=Tin+q"*2*pi*r/(mdot*Cp)*L

whoops! and there goes the involvement of the L which means I am back to where I started.

This very frustrating. If 300 watts goes into my flowing gas stream, there must be a maximum temp that is gets to for a fixed length, flow, and Cp. Forget the implementation, I can keep increasing the heater power until 300W goes into the gas stream, but the freakin length must matter somehow!

This seems so damn basic and I simply can not find it in the book! So, who can show me how stupid I am and how easy this is. All the information should be here. I'll list it again:

Q watts goes into the flowing gas
the gas flows at mdot Kg/s
The temp going into the pipe is T1
The temp coming out of the pipe is T2
The heat capacity of the gas is Cp
The Inner Diameter of the pipe is Di
The outer Diameter of the pipe is Do
The length of the pipe is L
the convection coefficient is h

you get the gist.

Im just looking for the relationship to find T if I have everything else, or to find L if I have everything else. All of these terms must matter.

every time I do it, the damn length term keeps dropping out.
If I increase the flow rate, the output temperature must drop so the flow rate must matter.
based on a flow rate, there must be a minimum length for enough heat to get into the gas stream to raise the temp to T2.
I realize the pipe temp will change with these scenarios, so the pipe properties must matter.


-help....

 

[rhirsch]

you know what? never mind

http://www.watlow.com/literature/specsheets/files/heaters/batcx20000605.pdf

 

[prex]

Please do read also the advice that is given. Go back and see what was said about wall temperature: this datum is lacking in your list, you need it to solve your problem.
And if you don't know it, make a guess, use the set temp for the thermostat in the heater or whatever.

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

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http://www.levitans.com

: Air bearing pads

 

[rhirsch]

I think if you go back to my 11:35 posat on oct 22, you'll see that I understand about the temperatures and picking htem. Which is fine. I was having trouble regardless of what they were, that is why I kept everything as variables.

Here is what I ended up doing (for those that may search for a similar thing).

I presumes that I could insulate the system well enough that most of the of the heat (an order of magnitude) went into the gas as opposed to leaving through the insulation.

Then I presumed a pipe size and that 300 watts went into the gas.

then I got a volumentric heat generation rate of a specific pipe size of A*L. qdot=Q/(A*L), where A is the crossectional area of the pipe. I presumes all the heat is flowing into the pipe and it is acting like a generator.

Then using hte heat generation rate, i found Tout based on various lengths of tubing. As the tubing gets longer, the tout rises as does the power. using:

Tout=(qdot*A*L+mdot*Cp*Tin)/(mdot*Cp)

Once I have Tout, then I know delta T and

q=qdot*A*L

noting that in each equation I am changing only L.

This gives me reasonable numbers and tell me that if I have a 30cm tube of 1/2" I will get the temperature I need with 120 watt of heat. I will use a 300 watt heater, and hope for the best.

 

[prex]

So your conclusion is that q=qdot*A*L=Q : in other words you confirm that what you assumed as an input is the result.
The point you miss is that the outlet temperature does not depend on length: if you need 120 W to heat up your fluid flow rate, this will be true for a heating length of 30 or 300 cm.
If your heater delivers 300 W and is 30 cm long, well that should be fine with your numbers. Go on with testing, you'll probably get a higher outlet temperature and regulating the heater you'll come to the result you need.

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes for fun rides

http://www.levitans.com

: Air bearing pads

 

[IRstuff]

see:

http://www.inhalation.net/turbulent_mixing.htm

which describes how long a pipe you need for mixing length

TTFN

FAQ731-376

 

[rhirsch]

Prex,
Yeah, but the way that I set it up is that the longer the tube the more power goes into it. I just picked a length a found a qdot that I used for whatever length. so I'm not sure I missed the point, I just used a guess at a particular instance and extrapolated form there.

I know its not the most accurate way to do it. The correct answer requires h and a good understanding of how much heat leaves the system and how hot the pipe gets (never mind that its not really a pipe, but a block with holes in it).

Thats why I doubled it, and will let my controller do the work.

 

[IRstuff]

You need to get your story straight. Earlier, you claimed it was open loop, so there is no control, and now, there is. Which is it?

And, you keep refering to "more power." That's a math error. Your input power is divided aross the length you design for. There is never more power demand than what you put in.

TTFN

FAQ731-376