#### 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}.

#### Solution

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}

Top View

Let us consider a cylindrical shell of length *l*,

radius *x* and thickness *dx*.

Rate of flow of heat `q =( dQ)/dt`

`(dQ)/dt = -(KADeltaT)/dx`

Here , the negative sign indicates that the rate of heat flow decreases as *x* increases.

`q=-K(2pixl).(dT)/(dx)`

`int_r^R dx/x = -(2piKL)/q int_{T_1}^{T_2}dT`

`["ln" (x)]_r^R = (2piKL)/q (T_2 -T_1)`

`⇒ q = (2piKL(T_1 -T_2))/( "in" (R/r)`

`q = (2pi xx 0.15 xx 0.5xx(90)}/{"ln"((1.2xx10^-2)/(1xx10^-2))`

`q = 262.9 ` J/s