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

The rate of heat flow through the walls of the tube is $q \approx 233 \mathrm{J} / \mathrm{s}$

Explanation:

Step 1:

Let,

Inner radius $r_{1}=1 \mathrm{cm}$

Outer radius $r_{2}=1.2 \mathrm{cm}$

Steam temp. $\mathrm{T}_{1}=120 \bullet \mathrm{c}$

Room temp. $\mathrm{T}_{2}=30 \bullet \mathrm{c}$

Length of tube $\mathrm{I}=50 \mathrm{cm}$

Thermal conductivity $\mathrm{K}=0.15 \mathrm{JS}^{-1} \mathrm{m}^{-1} \bullet \mathrm{c}^{-1}$

Step 2:

Rate of heat flow $=\frac{d Q}{d t}$

$q=\frac{d Q}{d t}=-\frac{K A \Delta T}{d x}=-\frac{K(2 \pi x) t \Delta T}{d x}$

$\frac{d x}{x}=-\frac{2 \pi K l d T}{q}$

Step 3:

Integrating both sides

$\int_{r_{1}}^{r_{2}} \frac{d x}{x}=-\frac{2 \pi K \imath \int_{T_{1}}^{T_{2}} d T}{q}$

$q=-\frac{2 \pi K l\left(T_{2}-T_{1}\right)}{\ln \frac{r_{2}}{r_{1}}}$

By substituting the values in above equation

$q \approx 233 \mathrm{J} / \mathrm{s}$