A very thin and long pipeline, used for the transport of crude oil, is buried in

Question

A very thin and long pipeline, used for the transport of crude oil, is buried

in the earth such that its centerline is a distance of 1.5 m below the surface. The pipe has

an outer diameter of 0.5 m and is insulated with a layer of cellular glass 100 mm thick.

What is the heat loss per unit length of pipe under conditions for which heated oil at

120%uF0B0C flows through the pipe and the surface of the earth is at a temperature of 0%uF0B0C?

What is the temperature at the outer surface of the cellular glass insulation? Draw

schematically the isotherms inside the soil and the heat flux plot. Assume that thermal

resistance due to convection in the oil is neglected, and temperature is uniform on the

surfaces of the pipe and glass fiber.

Explanation / Answer

I can only help till finding the heat loss per unit length and the temp. of outer surface of cellular glass insulation..

Here we go ..

Thermal resistance for a cylindrical shell R = ln ( r2 / r1 ) / ( 2 *pi * k * L )

k = thermal conductivity

L = length of cylinder

r1 = inner radius

r2 = outer radius

here inner radius = outer radius of sylinder = 0.25 m

outer radius = 0.25 + insulaiton thickness = 0.25 + 0.1 = 0.35 m

Assumption : Temperature of outer surface of pipe = temperature of liquid = 120 celcius

so R1 = ln ( 0.35 / 0.25 ) / ( 2 *pi * k1 * L )

k1 = conductivity of glass

now for earth

layer of earth above the glass insulation = 1.5 - 0.35 = 1.15 m

so r2 = 1.15

r1 = 0.35

so R2 = ln ( 1.15 / 0.35 ) / ( 2 *pi * k2 * L )

k2 = thermal conductivity of earth

so heat transfer per unit length = ( 120 - 0 ) / [ R1 + R2 ]

Q = 120 * ( 2*pi * L ) / [ (ln ( 0.35 / 0.25 ) / k1 ) + ( ln ( 1.15 / 0.35 ) / k2 ) ]

so, Q / L = 753.98 / [ 0.3365 / k1 +   1.1896 / k2 ]

so heat transfer per unit length = Q/ L = 753.98 / [ 0.3365 / k1 +   1.1896 / k2 ]

you haven't given the value of k1 & k2 ... substitute them here in SI units and get the answer

Now for the glass temperature

Let T = temperature of outer surface of glass

now Q = ( 120 - T ) / R1

so,   753.98 / [ 0.3365 / k1 +   1.1896 / k2 ] = ( 120 - T ) * 2*pi * k1 /    0.3365

so, T = 120 - (120 * 0.3365 / k1 ) / [ 0.3365 / k1 +   1.1896 / k2 ]

again substitute the value of k1 and k2 ... \and please do rate .. it was a loong quesiton

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