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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
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