Q. 195.0( 1 Vote )

# Steam at 120°C is

Answer :

**Given:**Steam temperature: T

_{1}= 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: T

_{2}= 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: T

_{1}to T

_{2}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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