Calculate the heat loss from 1 m of a 210 x 5 mm pipeline made of dry concrete covered with 50 mm
Calculate the heat loss from 1 m of a 210 x 5 mm pipeline made of dry concrete covered with 50 mm slag wool insulation. Temperature of the inner surface of the pipeline t1 = 180 C. Temperature of the outer surface of the insulation t3 = 50 C. Determine the temperature at the boundary of the wall and insulation.
l = 1m – pipeline length;
d1 is the inner diameter of the pipeline;
δd1 = 5mm = 0.005m – pipeline wall thickness;
d2 = 210mm = 0.21m – outer diameter of the pipeline;
δd2 = 50mm = 0.05m – insulation wall thickness;
d3 is the outer diameter of the insulation;
t1 = 180 ° C, t2, t3 = 50 ° C – temperature on the corresponding surfaces;
λ1 = 1.00 W / (m * ° C) – specific thermal conductivity of dry concrete;
λ2 = 0.05 W / (m * ° C) – specific thermal conductivity of slag wool;
R1 – thermal resistance of the pipeline;
R2 is the thermal resistance of the insulation;
P – heat loss.
Let’s calculate d1 and d3:
d1 = d2 – 2 * δd1 = 0.2 (m);
d3 = d2 + 2 * δd2 = 0.31 (m).
Formula for thermal resistance of a pipe
Consider an infinitely thin layer of a pipe surface with a radius r and a thickness dr. For the thermal resistance of this layer, we get:
dR = dr / λS;
dR = dr / (2πrlλ);
dR = 1 / (2πlλ) * dr / r.
By integrating the equation from r1 to r2, we obtain the formula for the thermal resistance of the pipe:
R = 1 / (2πlλ) * ln (r2 / r1), or
R = 1 / (2πlλ) * ln (d2 / d1),
where r1, d1 and r2, d2 are the radii and diameters of the pipe surfaces.
Thermal resistance of the pipeline and insulation
R1 = 1 / (2πlλ1) * ln (d2 / d1);
R2 = 1 / (2πlλ2) * ln (d3 / d2).
Substituting the numerical values, we get:
R1 = 0.0078 (° C / W).
R2 = 1.2397 (° C / W).
Heat flow formula
For areas of concrete and insulation, we apply the formula for the heat flow:
{t1-t2 = P * R1
{t2-t3 = P * R2
Having solved the system of equations, we get:
P = (t1-t3) / (R1 + R2) = 104.2 (W);
t2 = t1 – P * R1 = 179.2 (° C).
Answer: P = 104.2W, t2 = 179.2 ° C.
