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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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Find the temperature as a function of radius r in case of a cylindrical shell whose inner surface temperature and outer surface temperature are $ \theta_1$ and $ \theta_2$. The inner and outer radii of the cylindrical shell are $r_1$ and $r_2$.

$\begin {array} {1 1} (A)\;\theta_1 - (\theta_1 + \theta_2)\: ln\: (r_1/r_2) / ln(r_2/r_1) & \quad (B)\;\theta_1 + (\theta_1 - \theta_2)\: ln\: (r/r_1) / ln(r_2/r_1) \\ (C)\;\theta_1 - (\theta_1 - \theta_2)\: ln\: (r/r_1) / ln(r_1/r_2) & \quad (D)\;\theta_1 - (\theta_1 - \theta_2)\: ln\: (r/r_1) / ln(r_2/r_1) \end {array}$

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Consider a cylindrical shell of length $l$, internal and external radii $r_1$ and $r_2$, respectively. Let its inner surface be maintained at a steady temperature $ \theta_1$ and the outer surface at a steady temperature $ \theta_2 \: (\theta_2 < \theta_1).$
Consider an elementary cylindrical shell of thickness $dr$, at temperature difference $d \theta$
Considering temperature of the layer $ \theta$ at a distance $r$ from the axis,
$ \int_{r_1}^r \large\frac{dr}{r}$$ = - K \large\frac{2 \pi l}{t1}$$ \int_{\theta_1}^{\theta} \: d\theta$
$ \Rightarrow ln \bigg( \large\frac{r}{r_1} \bigg)$$ = K2 \pi l(\theta_1-\theta) / H$………….(i)
and $ \int_{r_1}^r \large\frac{dr}{r}$$ = - K \large\frac{2 \pi l}{H}$$ \int_{\theta_1}^{\theta_2} \: d\theta$
$ \Rightarrow ln \bigg(\large\frac{r_2}{r_1} \bigg)$$ = K2 \pi l(\theta_1-\theta_2)/H$………(ii)
Dividing (i) by (ii), $\large\frac{ln \bigg( \large\frac{r}{r_1} \bigg)}{ ln \bigg( \large\frac{r_2}{r_1} \bigg) }$$ = \large\frac{(\theta_1-\theta) }{ (\theta_1 – \theta_2)}$
$ \Rightarrow \theta = \theta_1 – (\theta_1 – \theta_2) \large\frac{ln \bigg( \large\frac{r}{r_1} \bigg)}{ ln\bigg( \large\frac{r_2}{r_1} \bigg)}$
Ans : (D)
answered Mar 19, 2014 by thanvigandhi_1
 
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