Question

Steam at 120°C is continuously passed through a 50 cm long rubber tube of inner and outer radii 1.0 cm and 1.2 cm. The room temperature is 30°C. Calculate the rate of heat flow through the walls of the tube. Thermal conductivity of rubber = 0.15 J s⁻¹ m⁻¹°C⁻¹.

Answer 1

Inner radii = r = 1 cm = 10^–2 m
Outer radii = R = 1.2 cm = 1.2 × 10^–2 m
Length of the tube, l = 50 cm = 0.5 m
Thermal conductivity, k = 0.15 Js^–1 m^–1 °C^–1

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Let us consider a cylindrical shell of length l, radius x and thickness dx.
Rate of flow of heat q$=\frac{d\mathrm{Q}}{dt}$
$\frac{dQ}{dt}=\frac{-KA∆T}{dx}$

Here, the negative sign indicates that the rate of heat flow decreases as x increases.
$$\begin{gathered} q=-K\left(2\mathrm{\pi }xl\right).\frac{\mathrm{dT}}{\mathrm{d}x} \\ {\int }_r^R\frac{\mathrm{d}x}{x}=-\frac{2\mathrm{\pi }Kl}{q}{\int }_{{\mathrm{T}}_1}^{{\mathrm{T}}_2}dT \\ {\left[\ln \left(x\right)\right]}_r^R=-\frac{2\mathrm{\pi K}l}{q}\left({\mathrm{T}}_2-{\mathrm{T}}_1\right) \\ \Rightarrow q=\frac{2\mathrm{\pi }kl\left(T_1-T_2\right)}{\ln \left({\displaystyle \frac{R}{r}}\right)} \\ q=\frac{2\mathrm{\pi }\times 0.15\times 0.5\left(90\right)}{\ln \left({\displaystyle \frac{1.2\times {10}^{-2}}{1\times {10}^{-2}}}\right)} \\ q=262.9\mathrm{J}/\sec \end{gathered}$$