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Homework Help: Thermal Conductivity of Steel experiment

  1. Oct 17, 2014 #1
    1) PROBLEM:
    Task is to find the thermal conductivity of steel experimentally.
    I have conducted relevant experiments and I am trying to solve for the thermal conductivity (k) itself. My final however is ten times too large in size, and I am having trouble identifying where the error is coming from. Nothing seems to be adding up and its become a complete nightmare.
    The process is as follows;

    a) Forbes Bar - long bar heated from one end in steady state. Temperature recorded at intervals with respect to distance from the source.
    b) Suspend heated sample in air and record temperature drop with respect to time. (This part is for heat transfer coefficient).
    c) Calorimeter to find specific heat capacity of sample of steel.

    3)OUTCOME: (and relevant values + equations)...

    Graph for part a) plotted using θ = θ0e-αx ⇒lnθ = lnθ0 - αx i.e. α = -gradient
    graph for b) similarly from θ = θ0e-βt i.e. β = -gradient
    The values I obtained are α=1.7, β=0.001

    My value for Specific heat capacity ≅475
    P=perimeter = 0.04m
    A = area = 1x10-4
    L = 0.02m

    β = [(P/A + 2/L)h]/ρS
    α = √(hP/KA)

    where ρ = density, h = heat transfer coefficient, K = thermal conductivity,

    Any help is greatly appreciated. I have the feeling my value for alpha may not be right but I can't be sure and need some advice.

    Many thanks!
  2. jcsd
  3. Oct 17, 2014 #2


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    Your values should have units. Ignoring them leads to many common sources of errors.

    What is the shape of your bar and what is L?
  4. Oct 17, 2014 #3
    Shape is a rectangle and length is 0.02m as I mentioned above. i.e. a 2x1x1(cm)^3 steel rectangle
    Should h be somewhere between 1 and 10?I've been consistent in SI throughout sorry for not adding in here.
  5. Oct 17, 2014 #4


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    Ah, length is just for (b) and (c), okay.

    I guess "ρS" and "KA" (kA?) are both in the denominator here.

    You have two equations with two unknowns (h and k), it should be possible to solve for both. That is the whole idea of measurement (b).
  6. Oct 17, 2014 #5
    What makes you think that h for experiment b is going to be the same as for experiment a? I would expect that h in experiment a would be lower than in experiment b. What specific values (including units) did you get for h and k?

  7. Oct 18, 2014 #6
    I am aware, and I have done so, but my final value of K(steel) is around 600, when an textbook puts the actual value around 50.
    My graphs retrieved α=1.7 and β=0.001.

    h is not solved in either experiment, it is deduced from the results of part b.And I see no reason why it should differ seeing as it is the heat transfer coefficient between the steel and air, which doesn't change in part a or b.

    I get h=4.6 and K approx 640 if i remember correctly.

    The units are SI. h = W/m2K K = W/m.K
  8. Oct 18, 2014 #7
    The geometry in part a is completely different from that in part b. If the bar in part a is vertical and the heat transfer coefficient is the result of natural convection, the thermal boundary layer thickness in part b is going to grow much thicker than that in part a, and the average heat transfer coefficient is going to be much lower. Is the bar horizontal or vertical?
    If the heat transfer coefficient were lower in part a than in part b, K would be lower.

  9. Oct 19, 2014 #8
    Hi Chet,

    In part a) Bar is horizontal with heater on the left end (heat transfer tending to the right end). (Heats end)
    In part b) bar is horizontal on top of a heating plate. (Heats side face)

    From what you have said previously, should I take my obtained value of hB and find a new value hA from a relationship between the sizes of the bars?

    Many thanks!
  10. Oct 19, 2014 #9


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    I don't think that is possible.
    It would have been useful to make (b) with the same bar as (a), I think.
  11. Oct 19, 2014 #10
    This was not possible, but shouldn't be necessary.
    Part a used a forbes bar (between 0.5 and 1metre) - we don't take readings of temperature all the way along, just take a range of values because its in steady-state. (did from 0.013m to 0.213m in equal spaces of 0.02m). The cross area in part a was 0.009m x 0.009m square (8.1x10-5m2)

    Part b used a small sample as I've said. Cross-area of this was 1x10-4m2 (square of 1cm sides)

    I shouldn't need to know the length in part a. Knowing the rate of convective losses in b should be enough to know for all samples of the material in question since the air around it is (for sake of this experiment) the same.
  12. Oct 19, 2014 #11


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    The air is the same, but convection will look different if your sample has a different shape.
  13. Oct 19, 2014 #12
    I agree with this. Also, Salford, did you really get straight lines on your two plots? I would have thought with natural convection that the semilog graphs would not have been straight lines. Look up natural convection to see how the heat transfer coefficient is determined for this situation.

  14. Oct 22, 2014 #13
    I got straight line graphs because I manipulated the exponential expression to obtain a straight line gradient for the values α and β.

    Θ = Θ0exp(-αx) ⇒ lnΘ = lnΘ0 + (-α)x ... WHICH is of the form y=mx + c (straight line equation). Equation for beta analogous with β replacing α and time replacing position (x)

    After some thought, I have eventually obtained a value just over 40 for thermal conductivity of steel. Not great, but I didn't do enough runs through experiment to judge data in truth. Im happy with it in the end and quite glad to put it to bed!
  15. Oct 22, 2014 #14
    Excellent. Incidentally, just for clarification, I wasn't asking how you obtained a straight line analytically. I was asking whether the actual data fell close a straight line on your semi-log plot. No need to respond.

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