Q. 195.0( 1 Vote )

# Steam at 120°C is

Given:
Steam temperature: T1 = 120 °C
Length of the tube : l = 50 cm = 0.5 m
Inner radii of the tube: r = 1 cm = 0.01 m
Outer radii of the tube : R = 1.2 cm = 0.012 m
Room temperature: T2 = 30 °C
Thermal conductivity of rubber: K = 0.15 J s–1 m–1 °C–1 Formula used:
Rate of amount of heat flowing or heat current is given as: Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness or length of the material.
Consider an element dx at a distance of x from the center between r and R. We will integrate this element dx to find total heat transferred from the tube.
Heat flow can be given as q = Δθ /Δt
In differential form: Here we used negative sign because the heat flow decreases with increase in thickness dx
Also, Area of the tube formed due to element dx can be given as
A = 2πxl
Here, x is the radius of the tube due to dx and l is the length of the tube.  Now we integrate both the sides, taking temperature from tube to surrounding: T1 to T2 and radii from r to R      Hence, the rate of heat flow through the walls of the tube is 232.50 J/s.

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