1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Internship Question - Heat Transfer!

  1. Jan 5, 2012 #1
    Hello, I'm not actually asking a homework question. I'm working at an internship and have been asked a practical question about how long it will take a steel rod to change temperatures.

    1. The problem statement, all variables and given/known data
    Because it's not a problem from a textbook, I may not have all the required data, but I can certainly try to find it.

    We have a steel rod at 41°F. It is in a room that is 68°F The rod is 3/4" in diameter, and 6" long. How long will it take for the pipe to reach 68°F?

    I've been using these numbers:
    The Specific Heat (Cρ) of Carbon Steel is: 0.49 (kJ/kg K), 0.12 (kcal/kg°C), or 0.12 (Btu/lbm°F)
    I got those numbers from this website.
    I also calculated the area of the rod to be A = .84375∏ inches cubed.

    2. Relevant equations
    I haven't taken a Heat Transfer course yet, but I've spent the past several hours reading about it on the internet (haha). I've been plugging in numbers into all sorts of equations, trying to find an answer. These are the equations I've been using are:

    Rate = k•A•(T1 - T2)/d
    Q = h*A*(Ts-T∞) (where T∞ is the temperature of the air)
    (Q_dot) = (m_dot)*(c_p)(T1-T2)

    3. The attempt at a solution
    Because I haven't taken a Heat Transfer course yet, and I honestly have never seen these equations before, I don't think I did anything right. I think I am very close, but need a little guidance. I know that Q is going to be power, and I think that I can use Watt*second for that, but I'm still having a hard time relating these equations to time.

    Can someone give me a hand?
     
  2. jcsd
  3. Jan 5, 2012 #2
    Finding heat transfer coefficients is very difficult, usually such data is gathered experimentally and then estimates can be made based on that.
     
  4. Jan 6, 2012 #3
    Do you know what numerical integration is? I can get you through this problem if you consider this a lumped mass system. This would be a good assumption because the boundary conditions are weak.
     
  5. Jan 9, 2012 #4
    Yep, I've take Calculus 1 and 2. If you're willing to explain, I should be able to follow along.
     
  6. Jan 9, 2012 #5
    OK, here goes. At the temperatures provided heat is conducted radially through the solid object faster than it can pass through the surface. There is a ratio called the Biot number which is h*L/k, where h is the overall heat transfer coefficient, L is a thickness of the object (radius in this case), and k is the thermal conductivity. If this number is < 1 you can consider this a lumped mass system meaning you can assume the object is at uniform temperature as it changes temperature with time. Given these temperatures, Bi is small. You can compute this for yourself. If you could not assume this, you would be solving a PDE in cylindrical coordinates. We'll also assume this is a one dimensional problem which is reasonable due to the length versus radius of the bar.

    Heat will get into the object 2 ways: natural convection and radiation.

    If you cannot calculate them, I'll help you with that later.

    The way to go about solving this is write a differential expression for the change in internal energy of the bar and relate it to the boundary conditions. In other words, in a small amount of time, dtheta, a small amount of heat will enter the object. The small amount of heat that is dependent on the object's temperature will raise the object's temperature by an amount dT in time dtheta.

    An expression for the change in internal energy would be: rho * C * dm * dT/dtheta where rho is density, C is specific heat, dm is the mass of a coin shaped section of the rod, and theta is time. Dm will be the differential volume of a slice of the rod and contain a dL term (length of cylinder).
    But the boundary condition applies to the surface and it also contains the dL term so it drops out. Surface area would be perimeter times dL.

    Equate to the boundary conditions to the differential change in internal energy. To solve it, assume a small time step and advance the equation one step at a time re-evaluating the boundary condition at each time step.

    Let me know where the confusion points are.
     
  7. Jan 9, 2012 #6
    Ok, I'm fairly sure I understand the process, but the specifics are harder to understand. After filling a few sticky pages up with notes, I cleaned them up into a few images for simplicity.

    Even though you implied that the Biot number wasn't necessary, I tried to solve for it anyway. I don't know how to find h (the overall heat transfer coefficient), and I just used a stock number from this website to get a number for k (thermal conductivity). Here's how I have it laid out:
    ppn1B.png
    So I'm stuck there.

    Next I moved on to try solving for the change in internal energy dE. I got the density ρ from wikipedia, the specific heat C from this website, and all that was left was dm and (dT/d∅).
    3ksGv.png

    I wasn't sure what to do for (dT/d∅), so I tried dm:
    2516F.png

    The only one I feel somewhat confident in is what I have for dm. But I'm still not exactly sure about everything, and I'm not sure what to do next.

    The roadblock I've encountered is that I haven't taken Differential Equations, so I don't know how to solve an equation like this.
     
  8. Jan 9, 2012 #7
    For these temperatures and the fact the rod is in air, the overall heat transfer coefficient is on the order of 1 BTU/hr-ft^2-R. You can change the units. We'll get to the boundary conditions once some things get cleared up. But the above will let you determine the Biot number when put in appropriate units.

    Your dm is correct.

    Your energy change is correct. Once you have the boundary conditions (heat added per unit time), equate them to dT/dtheta multiplied by rho, C, dm.

    The boundary conditions are:

    Natural convection:

    h = 0.19 * (T - Tinf)^.333

    The energy input per unit time-area from natural convection is

    Ec = h * (T - Tinf) = 0.19 * (T - Tinf)^1.333

    Units are BTU/hr-ft^2, T is absolute temperature (see below).

    Above is straight from JP Holeman's Heat Transfer textbook. It is one of many correlations for natural convection.

    The energy input per unit time-area from radiation is:

    Er = sigma * episilon * (T^4 - Tinf^4)

    where sigma is Stephan Boltzmann constant and episilon is emissivity which you could assume to be near unity. T is temperature of rod and Tinf is ambient temperature in degrees Rankine, a constant.

    Set the expression for the change in internal energy to the sum of these two boundary conditions multiplied by the bit of surface area. The dL's will drop out. Clear the coefficient on the time derivative so you are left with something that figuratively looks like this:

    dT/dtheta = A (fc(T) + fr(T))

    where A is a constant made up of rho, C, etc. fc(T) is natural convection BC. fr(T) is radiation BC. Make certain your units are uniform. Unit uniformity is crucial.

    To solve it write it as:

    dT = (A (fc(T) + fr(T))) * dtheta

    or

    delta(T) = (A (fc(T) + fr(T))) * delta theta

    Insert the initial temperature and a time step, say 1/4 hour. Evaluate RHS which will give you the delta(T). Add the delta(T) to the T from previous time step and re-evaluate the RHS. Repeat this process. You could write a small program to do this with smaller time steps to get more accurate results. This is known as explicit integration. It is the easiest to set up but is unstable if too large a time step is chosen.

    If the problem had linear boundary conditions, you could do it by solving the ODE in closed form.

    I have a finite element program so I can easily check your results with these boundary conditions.

    Good work.
     
  9. Jan 9, 2012 #8
    Vzh4T.png

    I don't understand what "1 BTU/hr-ft^2-R" is. Do you mean some crazy hyphenated unit-grouping like this? It looks like gibberish to me.
    LX0wX.png

    Boundary Conditions:
    I have Ec (energy from natural convection) and Er (energy from radiation).

    What do you mean by "The dL's will drop out"? In the Ec and Er equations you provided, there were no dL variables. Also, why did you go from Ec and Er to fc(T) and fr(T)? I know you say that they are natural convection BC and radiation BC, but I don't know what the BC means, and how to relate the energy (Ec and Er) to the...whateveryoucallthems (fcT and frT).

    This looks very clear and broken-down (THANKS!) but I am not able to interpret the variables, unfortunately.

    It's all greek to me...


    I believe I have disappointed you.
     
  10. Jan 9, 2012 #9
    Overall heat transfer coefficient, when multiplied by area and temperature difference gives heat transfer in BTU/hr. Hence, the units BTU/hr-ft^2-degree. Hyphens are standard nomenclature for multiplication.

    heat transfer = U * A * (T - Tinf)

    heat transfer = BTU/hr

    "What do you mean by "The dL's will drop out"? In the Ec and Er equations you provided, there were no dL variables."

    dm will contain a dL which is height of coin shaped element. Surface area also contains a dL and surface area is multiplied by both radiation and natural convection conditions. Thus dL appears in every term of differential equation.

    A(fc(T) + fr(T)) is (Ec + Er)*surface area*(delta theta)/(rho*c*dm)

    surface area =2*pi*r*dL
    dm = rho*pi*r^2*dL
    delta theta is time step for moving solution forward
    c is specific heat

    dL is length of coin shaped element

    Does this help? I realize you are getting into something for which you have no background.
     
  11. Jan 11, 2012 #10

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Unless thsi is some sort of trick quantum physics question, the answer is obviously ∞. The rod never quite reaches ambient!

    Now, if you had asked for 67F, that would be a different story ...
     
  12. Jan 11, 2012 #11
    I assumed the intern was trying to learn something about actually carrying out an elementary numerical integration. It gets to within 3 degrees of the ambient temperature in about 3.5 hours. Mathematically speaking, it never gets there.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Internship Question - Heat Transfer!
Loading...