E5.2 Composite Cylinder

Example 5.2: A 2 " steel pipe carries steam at 300 °F. It is lagged with 0.5 " of rock wool. The surrounding air is at 70 °F. What will be the heat loss per linear foot? The following data are applicable.

Inside film coefficient, h_i = 1500 Btu/(hr·ft²·°F);

Outside film coefficient, h_o = 2.23 Btu/(hr·ft²·°F);

Thermal conductivity of steel, k₁ = 28 Btu/(hr·ft·°F);

Thermal conductivity of rock wool, k₂ = 0.033 Btu/(hr·ft·°F)

Solution:

Steam temperature at the center of the pipe, t₁, is 300 °F. The bulk temperature of the air, t₅, is 70 °F. Surface temperatures (t₂, and t₄) are not known. The temperature at the steel-wool interface (t₃) is also unknown. The inside diameter of a 2" pipe having standard thickness (represented by Schedule 40) is 2.067". Outside diameter of the pipe can be found to be 2.374". Now we can find the appropriate radii. These are

The insulation is 0.5" thick, so r₃ = 0.0989 + 0.5/12 = 0.1406 ft. Thickness of the cylindrical sections can be found as follows

This gives

The length of the pipe is 1 foot. If R₁ and R₄ are equivalent resistances for film coefficients, then their values can be found as

However, R₂ and R₃ are conductive resistances. Their values can be found as

These values are found to be R₂ = 7.985´10^-4 (hr·°F)/Btu, and R₃ = 1.695 (hr·°F)/Btu. And the overall resistance is given as:

The heat flow can be found by dividing overall temperature difference by the resistance.

Using basic concept, driving force is a product of flow and resistance.

Δt₁ = 0.13 °F, Δt₂ = 0.08 °F, Δt₃ = 176.83 °F, Δt₄ = 52.96 °F

And interface temperatures are t₂ = 299.87 °F, t₃ = 299.79 °F, t₄ = 122.96 °F

The bulk temperature can be verified to be 122.56 - 52.96 = 70 °F.