I am given the temperature profile below and the two equations:

$$Q = (2\pi R_0)LU_0(T_{1\infty}-T_{2\infty})\tag1\label1$$ $$\frac{1}{U_0R_0} = \frac{1}{h_1R_0} +\frac{\ln(R_\mathrm a/R_0)}{k}+ \frac{1}{h_2R_\mathrm a}\tag2\label2$$

temperature profile

and I am asked

How does \eqref{1} simplify for the temperature profile shown, i.e. the case where the temperature in the wall is very close to $T_{1\infty}$}?

I am not sure if this means that I am supposed to "remove" the middle term in \eqref{2} since it will more or less be the same as the first term and thus the overall heat transfer coefficient would look like this:

$${U_0} = \left(\frac{1}{h_1} + \frac{R_0}{h_2R_\mathrm a}\right)^{-1}$$

What I don't understand is that I just simplified $\ce{Eq(2)}$ and not \eqref{1} which the question asks for. Or is it that by simplifying \eqref{2} I in turn simplify \eqref{1}, or have I completely misunderstood the question.

  • 1
    $\begingroup$ Actually, the correct answer should be $$U_0=\frac{h_2R_a}{R_0}$$. These are shown as two separate equations only for mathematical convenience. You could eliminate Uo by combining them into a single equation. What do you get if you substitute the approximation I wrote into Eqn. 1? $\endgroup$ Apr 24 '19 at 11:52
  • $\begingroup$ @ChetMiller Oh I forgot about the first term. Thank you! If I subtitute Your approximation into Eq.1 I get, $\ce{Q=(2πL)h_2R_a(T_{1∞}−T_{2∞})}$ $\endgroup$
    – lotte07
    Apr 24 '19 at 12:29

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