A pipe #2

Classical Mechanics Level pending

What would need to be the temperature difference $\Delta t$ between the inner and the outer surface of a copper pipe with thermal conductivity $\lambda = 0.9\ \text{cal} \cdot \text{cm}^{-1} \cdot \text{s}^{-1} \cdot \text{grad}^{-1}$ , and with radii $r_1 = 5\ \text{cm}$ and $r_2 = 6\ \text{cm}$ , if through the outer surface $S = 5000\ \text{cm}^2$ of the pipe is supposed to radially flow an amount of heat $Q^* = 6000\ \text{cal} \cdot \text{s}^{-1}$ per second ?

Details and assumptions:

Give your answer to 2 decimal places.
Give your answer in $^\circ \text{C}$ .

The answer is 1.46.

1 solution

André Hucek
Oct 25, 2017

First we should define the length of the pipe $l$ , which gives the area surface $S = 2 \pi r_2 l$ , so

$\displaystyle l = \frac{S}{2 \pi r_2} = \frac{5000}{\pi \cdot 12} = 132.6\ \text{cm}$ .

The heat amount per second $\displaystyle Q^* = \frac{Q}{\tau} \hspace{4cm} \color{#3D99F6}(1)\ \small \color{#3D99F6} \text{from}\ q = \frac{Q}{l \cdot \tau} = - \lambda 2 \pi x \frac{dt}{dx}$

is equal to $Q^* = q \cdot l$ .

So according to $\color{#3D99F6}(1)$ , we get the expression $\displaystyle q = 2 \pi \lambda \frac{t_1 - t_2}{\text{lg} \frac{r_2}{r_1}}$ ,

$\implies \Delta t = \frac{Q^*}{2 \pi \lambda l} (\text{lg}\ r_2 - \text{lg}\ r_1) = \frac{6000}{2 \pi \cdot 0.9 \cdot 132.6} (4.094 - 3.912) = \boxed{1.46 ^\circ \text{C}}$ .

A pipe #2

The answer is 1.46.

1 solution

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