: Heat transfer - temperature gradient value at a certain point

[libelec]

URGENT: Heat transfer - temperature gradient value at a certain point

Homework Statement

Given a very long cylinder of inner R₁ = 0,01 m, outer radius R₂ = 0,1 m, that transports water at 150ºC, and surrounded by air at 25ºC, find the temperature gradient value at R = 0,07m. \lambda = 500 kcal/mhºC

The Attempt at a Solution

My question is more theoretical than anything. I'm assuming estationary regime, and that the temperatures of the water and the air are uniform. Then, shouldn't the temperature gradient value be the same at any point? How am I supposed to calculate it for a certain point?

I know that for a cylinder, the heat transfer Q = -\lambda*grad(T) = -\lambda*2\pi*R*L*(T_inside - T₁)/(ln (R/R₁)). But I just don't see it.

Thanks.

 

[JaWiB]

This is just a guess, but I take it to mean the radial distance from the center. That is, the pipe is made up of a wall .09m thick and .01m radius for the space in which the water flows (otherwise, what does inner and outer radius mean?) The gradient is in the pipe material, otherwise you'd have a jump from 150 C to 25 C where the gradient is undefined

 

[libelec]

This is just a guess, but I take it to mean the radial distance from the center. That is, the pipe is made up of a wall .09m thick and .01m radius for the space in which the water flows (otherwise, what does inner and outer radius mean?) The gradient is in the pipe material, otherwise you'd have a jump from 150 C to 25 C where the gradient is undefined

Yes, but shouldn't the gradient in the pipe be the same for every R?

It's a hollow cylinder, that's what I meant with "inner radius".

 

[JaWiB]

Maybe since the area increases as you move from the inner wall to the outer wall, the gradient must decrease so that the rate of heat flow is constant.

Oh, and shouldn't the equation be Q = -lambda*A*grad(T) rather than Q= -lambda*grad(T)?

 

[libelec]

Isn't that one the equation for power?

 

[JaWiB]

I think you're right, it should be dQ/dt in my equation. But I still think what I said makes sense since it looks like your original equation is the heat flux according to wikipedia:

http://en.wikipedia.org/wiki/Conduction_(heat )

I'm not really sure what the answer to this is, though; I'm just trying to figure it out too :)

 

[ideasrule]

You can divide the hollow cylinder into infinitely many hollow cylinders with thickness of dr. Since dr is infinitesimal, the surface area inside and outside the cylinder are both 2*pi*r*L. The heat conduction equation gives Q=-lambda*A*dT/dr, where Q is constant. Integration will easily give you T as a function of r.

 

[libelec]

You can divide the hollow cylinder into infinitely many hollow cylinders with thickness of dr. Since dr is infinitesimal, the surface area inside and outside the cylinder are both 2*pi*r*L. The heat conduction equation gives Q=-lambda*A*dT/dr, where Q is constant. Integration will easily give you T as a function of r.

Isn't that the equation for the power conducted? I thought Q was measured in J/m².s, and, given the units or lambda, I would get J/s, which is power.

Anyway, supposing you're right, let me see if I understood well.

First I integrate to get the expression of that Q in terms of r. I get Q = -\lambda2*\pi*L/(ln(R/R₁))*(T - T₁), being R₁ the distance between the center of the hollow cylinder and the inner wall, and T₁ the temperature inside. Then, given my R = 0,07 m, I can calculate T in R.

Then how do I replace this information so that I get the gradient of temperature in R? Do I have to assume grad(T) = (T - T₁)/(R)?

Thanks

 

[ehild]

Q is the heat transferred in unit time from the hot inner part to the cold outer part of the cylinder. You can get it from the equation you derived by replacing R2 for R and T2 for T.

Q = -\lambda *2\pi *L/(ln(R2/R1))*(T2 - T1)

You can assume stationary state, that is, the temperature does not change with time and depends only on R. In this case, the transferred heat is the same across any cylindrical surface with radius R. You derived the formula for Q as function of R and T. You know Q, so you can find the T(R) function, and its gradient, too.

ehild

 

[libelec]

OK, thanks.

I have another question, regarding a non-stationary state problem:

A body of mass M = 1kg, specific heat capacity c = 0,84 J/g°C, temperature T₀ = 40°C and surface S = 100 cm² is dropped into a fluid of temperature \theta = 20°C. It receives net heat per unit of time dQ/dt = 58 J/s. Assuming it has a conductivity \lambda = infinite and emisivity \epsilon = 0, find the expression of its temperature as function of time. If now \epsilon isn't 0, find the differential equation for the body's heating.

My question is, this is a Newton's Law of Cooling problem, right? Since \lambda = infinite, Biot's number is 0 (much smaller than 1). The thing is that I thought that the equation for the heat transfer was based on convection. That is:

dQ/dt = h.S.(\theta - T(t)).

But I am not given h. How am I supposed to do this?

Thanks

 

[ideasrule]

This has nothing to do with convection, Newton's laws of cooling, or Biot's number. Since Q=mc*delta-T, dQ/dt = mc*(delta-T/dt), so temperature rises linearly with time. The purpose of setting lambda=infinity is so the entire body of mass 1 kg has uniform temperature throughout.

 

[libelec]

OK, but how does it rise linearly with time, if it submerged in a fluid colder than the body itself? By definition, heat should be pouring out of the body and into the fluid, right?

 

[ideasrule]

My bad, it cools linearly with time.

 

[libelec]

But there's a problem with the solution you gave: if it goes as dQ/dt = m.c.dT/dt, then dT grows linearly, since dQ/dt > 0, and forever: then there's no T_f. And then I'm not using the \theta temperature at all. I'm not considering the environment.

What do I do there?

 

[ehild]

The body gains heat at the rate of 58 J/s, and losses heat because of heat transfer to the environment.

dQ/dt=58-h*A(T-\theta) 1.)

A is the surface area of the body and h is the heat transfer coefficient, its value is missing, but you can see the form of solution without knowing its numerical value.

Because of the net heat gained or lost, the temperature of the body changes at a rate dT/dt:

dQ/dt=c*m*dT/dt 2.)

Comparing 1.) and 2.), you get a differential equation for T(t):

<br /> c*m*dT/dt=58-h*A(T-\theta)<br />



This is an inhomogeneous linear first-order equation, easy to solve with the starting condition T(0)=40 oC.

ehild

 

[libelec]

Got it. Thanks.