silly question for novice 1

[JohnDillon]

I am working on my first heat transfer design and I have all of my data, but I do not know what to do with it.

I know the following for the material on each side of a wall:
specific heat
viscosity
thermal conductivity
density
velocity

for the wall in between them I know the following:
thickness
mean surface area
thermal conductivity

i understand all about Reynalds, Nusselts, and Prandtl number as well.

i also know the temperature that both substances will initially be.

i am trying to figure out how long in time it will take to get one substance to drop from 350 F to 100 F. How do I apply that I know "substance A" is initially 350 F, but I need it to lower to 100 F through heat transfer.

i do not understand how to apply all of my knowns.

thanx all.

 

[dcasto]

All those unitless numbers won't mean much if this is a static heat transfeer.

Q = UA deltaTlnmean is your friend.

 

[Zapster]

Sounds like you have a transient/unsteady heat transfer problem. Often these types of problems are troublesome to solve unless you make some assumptions. Start with calculating the Biot numbers. If the Biot number is small then use a lumped-mass approximation method. Dig out your heat and mass transfer text book or go to the library. The one I have from school is “Heat and Mass Transfer,” by Frank M. White. Keep in mind that this type of calculation will yield an answer that may have more uncertainty than you would like.

This link will explain a little about Biot numbers:

http://en.wikipedia.org/wiki/Biot_number

 

[25362]

Yes, this is a typical unsteady state heat transfer. One procedure that can be used for heating or cooling of a large slab is the Gurney-Lurie diagram.

Probable sources:

Fluid flow and heat transfer by professor Lydersen - Wiley.
Conduction of Heat in solids by Carlswaw & Jaeger - Clarendon Press, Oxford.

 

[corus]

If it's a case of two walls with a fluid in between them then unless one wall radiates to another then you can treat them separately with heat loss to the fluid (say if it was water). Unless it's a lumped mass problem (as others have said) then it's a 1D problem where the boundary condition on one face is forced convection, and from the other is... maybe insulated? There's an analytical solution here

http://en.wikipedia.org/wiki/Heat_equation.

If there is radiation then you'd be better using a FE program or finite differences due to the non-linear terms of radiation.

corus

 

[Flareman]

JohnDillon

In broad brush terms, this is an Arrhenius type problem and will finish up in a form
Q1 = Qo * exp(-x/t) or Qo * (1 - exp(-x/t))
If you plot these curves with generic values you will easily see how they work. (use x/t between 0 and 10)

Q1 is the quantity of heat transferred at time x
Qo is the quantity of heat you need to transfer to get from the initial state to a final state of equilibrium.
Your problem is one where you have to find x for a particular value of Q1.
You can determine the final equilibrium state by examination of your problem and calculate a value for Qo.

If we differentiate the equation we get
dQ/dt = Qo * (-1/t)*exp(-x/t)
When x = 0, dQ/dt = -(1/t)
you have the starting conditions with the biggest dT
and you can calculate the rate of heat transfer (= dQ/dt)
so you evaluate t (= the time constant)
then you can pick off x for any other intermediate Q1.

Bear in mind that dQ/dt may have component values which, themselves, are temperature dependent so write the equations out in algebra first, rather than jumping in to a quick and dirty numeric answer.

David

 

[JohnDillon1]

for some reason my old name no longer works so I had to modify it, hmmmm. i emailed tech support.

update, after looking at your helpful hints I think the following best applies to my situation. the lump mass approx method looks perfect however to solve for what i need, i need to know the temp at either side of the stainless steel tubing.

my system has cool water at around 40 F on the outside. a piece of stainless steel tube in the middle. and 250 F nylon on the inside. i am having tons of issues with the nylon. the nylon is a solid at this temp, and it will actually be in pellets. the nylon will be moved via an auger at a very slow rate (say 1.524 m/hr). the nylon will eventually leave this system due to the auger at hopefully 100 F.

so how do i solve for the heat transfer coefficient of the nylon given the following: this will have the system as laminar flow. this assumes convection heat transfer.


Cp specific heat capacity 1670 J/kg-K
k thermal conductivity 0.25 W/m-K
v velocity 1.524 m/hr
p density 560.645 kg/m^3
R resistance 102317.0565
u viscosity ?? it is around for 1E+12 kg/m-s for solids but this gives me a "h" that doesnt make sense when i solve for it using the Nusselt number.

so that didnt seem to work so instead i assume it is conductive heat transfer and use the following:

k thermal conductivity 0.25 W/m-K
x thickness 0.889 m
A Surface Area 17.01674016 m^2
R resistance 0.208970694

but again this does not make sense cause i end up getting a temperature on the S.S. wall of less than 41 F when it is next to 350 F, and the water temp on the other side of the wall is 40 F.

what gives?

and thanx.

 

[25362]

It seems the thickness should be expressed in mm not m.

 

[JohnDillon1]

that is irrelvenent whether it is in m, mm, in, just how i have it in my excel formulas. i just seek how to solve for my unknowns, so i can finish my design.

when i did have the thickness in bottom part of my previous email, in reality, it is the diameter of the area where the nylon will be. not sure if i can do this.

i just cant get an answer that sounds right.

 

[IRstuff]

Your modelling approach seems to be off. Your nylon pellets are not the equivalent of a free-flowing liquid, unless the pellets are VERY small compared to the dimensions of the pipe. The fact that you've got a silly number for viscosity should be a clue. The pellets aren't liquid, so viscosity is not the right approach for describing how the material flows.

Moreover, "pellets" would not have complete physical contact with each other, so using the thermal conductivity nylon might be pointless, since the thermal conductivity is strictly limited by the contact area between pellets, again, unless the pellets are VERY tiny.

Your late introduction of the auger further complicates the problem. How do you know if the auger is contributing to a forward transmission of heat through its shaft?

I suggest you start over and simply, and clearly, define the problem, rather than piecemealing your solution, for which you haven't yet given all the parameters.

As a minimum, the most likely approach is to treat this as a solid conduction problem, BUT, with the assumption that the solid is more like a foam or aerogel. You need to come up with an effective thermal conductivity of a mass of pellets, which is not the conductivity of the solid material.

TTFN

FAQ731-376

 

[JohnDillon1]

i am trying to get a good approx, not an exact number. i can add an error factor to it later.

the pellets are around 3-5 mm. while the pipe diameter is about 1 meter. i am not really worried about the auger right now. the auger will move at such a slow rate that the heat caused by friction i will not care about. and the auger will actually be a hollow tube with the cold water flowing through it to aide in the heat transfer from the nylon, but that is something i can figure out myself, if somebody can help with my question.

how to calculate the temp on the either side of the wall given my givens since the nylon is a solid. and i have no idea how to figure out the heat loss of a solid being guided through a cooling tunnel.

 

[IRstuff]

It wasn't a question about friction, it was more related to the information you only now revealed, i.e., that the auger also provides cooling, and that the auger would actually have a higher impact on the cooling, given its substantially larger cooling surface area, potentially.

You need to derate the thermal conductivity to account for the fact that the beads do not form a continuous solid. Your thickness was too thick, it should only be midpoint between two cooling surfaces.

TTFN

FAQ731-376

 

[JohnDillon1]

the thickness i used was the inner diameter of the area where the nylon will be. i did this just to get an amount of nylon.

like i said, i could care less about the auger now, other than the fact it is what is transporting the nylon.

i do not understand what to do next.

 

[vpl]

John

There seem to be a number of changes in your system and what you're trying to solve for betweenn your original post and your latest one. Maybe there really aren't, but for those who are reading and replying to this problem, it has made it rather confusing.

Remember you're asking for free consulting and the answer you get is only going to be as good as the information provided. Assuming you can ignore parts of the problem and add an error factor is not a good solution (remember garbage in = garbage out).

May I suggest that you consider providing a drawing showing your system. This would allow you to better define your problem. Define your wall thickness, and specify your overall system length. Also, provide some additional information about the nylon pellets. How big are they? Is the density that you gave the density for solid nylon, or is it the actual density for your pellets entering the system. The size and density of the pellets would affect whether they can be treated as a fluid, in which case you might want to rethink your viscosity.

I would also strongly suggest that you go talk to whomever assigned you this as your first heat transfer design and get more information from them. They may have some additional information for you.

Patricia Lougheed

Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of the Eng-Tips Forums.

 

[MJCronin]

My two cents...

Using a liquid to cool a stream of solid pellets contained within a tube doesn't work well.

I think that you should consider another type of cooler....either using cool air in a fluidized bed configuration (like everybody else does) or consider a "vertical bin" cooler with the cooling media contained in plates.

Both have a downside... fluidized beds cost a lot in compressed air. The vertical bin cooler has a residence time that must be repected.

See:

http://www.idswater.com/water/us/Bulkflow_Technologies/Heat_Exchanger_Plates/202_0/g_supplier_3.html

Let us know what you decide.... ( a star would be nice..)

MJC

 

[chicopee]

The SS tube temp of 41F may make perfect sense since the temp of water is at 40F and has more intimate contact with the tube than the Nylon pellets which in themselves have low thermal conductivity therefore a poor ability to cool down rapidly. Also the temp profile thru the tube thickness is probably uniform depending in the value of the Boit Modulus as pointed out above by Zapster.

Without going thru the calculations, always draw a sketch of the problem, then assign equal incremental lengths starting from the hot zone to the cool zone. Assign temps of both medium at start and end of tube; may have to assume ( makes an ass of you and me) thorough mixing of pellets during travel time, therefore can deal with bulk temperature at each stage during travel time. Also assume negligible heat absorption by tube and heat loss to surrounding from pellets and uniform temp thru tube thickness.

Along each incremental length starting from the hot zone:

Heat loss from pellets @ 350F (Mn*Cpn*(350-Tni))=

Heat transfer thru tube wall=
(Tnbi-Twbi)/((ln (ri/ro/2*pie*k*l))
Tnbi(bulk temp of nylon@ position i)=(350+Tni)/2
Twbi(bulk temp of water@ position i)=(40+ Twi)/2
ri, ro= inside and outside radius of tube

Heat absorbed by water @ 40F (Mw*Cpw*(Twi-40)).

You'll need a spreadsheet to work this problem out; also you'll have to rearrange the relationship between Tnbi,Twbi,Tni and Twi.

 

[unclesyd]

There are 4 of these beast on our site and my only advice is "my father who_____".
I never been involved in the actual calculations but have have been around numerous discussion about same. I have never seen any actual calculations for any of the coolers/heaters. Everything I've been involved with has been the "Edisonian Variety of Investigation"

All coolers and one heater required extensive testing and modification before they could met the specified heat transfer numbers based on the core fluids.

Part of the reason was initially that the nylon pellets were showing slug flow all the way through the machine. Extensive modifications had to made to the screws to where the pellets would tumble/mix as they traveled through the machine. This included roughening the surface of the flights and adding blades to insure mixing as traveled through the machine. This was a trial and error operation. The rotors actually had to be removed for these modifications.

As you know the heat transfer for nylon is very poor and we actually use N2 to help with the heat transfer in a couple of the machines. During one startup we had the 1/8" pellets actually stack up and try to act as a solid mass.

I checked to see how the new machines were specified and was informed that they sized strictly by surface area based on the original machines, so the heat duty was essentially the same.
All our machines are inclined though one was supposed to work in the horizontal, nope.

If you don't have the machine installed make sure you can pull the rotors.

If I remember correctly that the heat transfer is zilch away from metal contact and poor even there. \

There is a step change in the heat transfer that accompanies any change in feed rate. This is due to the apparent density change as product travel through the machine.

Here is one type machine we have. There is some data on the screw flight areas that will come close to other screws.

You will have to copy this to address bar.

http://www.metsominerals.com/inetMi...FEC1256BD60046EF65/$File/TS_Holo-Flite-en.pdf