ME 353 Heat Transfer 1

M.M. Yovanovich

P3MT91.MS

Problem 3 of Midterm Exam, October 1991.

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Pipline buried a distance of 1.5 m below the surface of the earth.

The outer diameter of the pipe is Do = 0.5 m. The pipe is insulated

with a layer of cellular glass of thickness t = 100 mm.

The conductivity of the soil is ksoil = 0.52 W/m K, and the

thermal conductivity of the insulation is kins = 0.069 W/m K.

The temperature of the oil is Toil = 120 deg C, and the temperature

of the earth surface is Tsurface = 0 deg C.

The thickness and the thermal conductivity of the pipe wall

are not given. Its thermal resistance is negligible relative to the

thermal resistances of the insulation and the soil.

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> restart:

System parameters.

> syspar:=
(Do = 0.5, t = 100/1000, z = 1.5, L = 1, kins = 0.069,
ksoil = 0.52, Toil = 120, Tsurface = 0);

a) Determine the shape factor per unit length of pipe and the

thermal resistance of the pipe/insulation/earth system.

From the attached Table of shape factors, the shape factor of the

pipe/insulation is known.

> Rsys:= Rins + Rsoil;

> Rins:= ln(D1/Do)/(2*Pi*kins*L);

> Rsoil:= 1/(ksoil*S);

> S:= 2*Pi*L/arccosh(2*z/D1);

> D1:= Do + 2*t;

b) Heat loss rate per unit length of pipe.

> q:= (Toil - Tsurface)/Rsys;

Computations of resistances and heat loss per unit length of pipe.

> S:= evalf(subs(syspar, S), 5);

> Rins:= evalf(subs(syspar, Rins), 5);

> Rsoil:= evalf(subs(syspar, Rsoil), 5);

> Rsys:= evalf(subs(syspar, Rsys), 5);

> q:= evalf(subs(syspar, q), 5);

The insulation and soil resistances are comparable.