A long, uninsulated steam line with a diameter of 100 mm and a surface emissivity of 0.8 transports steam at 150°C and is exposed to atmospheric air and large surroundings at an equivalent temperature of 20°C. (a) Calculate the rate of heat loss per unit length for a calm day. (b) Calculate the rate of heat loss on a breezy day when the wind speed is 8

Heat loss per unit length = 642.358 W/m

The heat loss per unit length on a breezy day during 8 m/s speed is = 1760.205 W/m

Explanation:

From the information given:

Diameter D

Surface emissivity ε = 0.8

Temperature of steam = 150° C = 423K

Atmospheric air temperature

Velocity of wind V = 8 m/s

To calculate average film temperature:

To calculate volume expansion coefficient

From the table of "Thermophysical properties of gases at atmospheric pressure" relating to 358 K of average film temperature; the following data are obtained;

Kinematic viscosity (v) = 21.7984 × 10⁻⁶ m²/s

Thermal conductivity k = 30.608 × 10⁻³ W/m.K

Thermal diffusivity ∝ = 31.244 × 10⁻⁶ m²/s

Prandtl no. Pr = 0.698

Rayleigh No. for the steam line is determined as follows:

The average Nusselt number is:

However, for the heat transfer coefficient; we have:

Hence, Stefan-Boltzmann constant

Now;

To determine the heat loss using the formula:

Now; here we need to determine the Reynold no and the average Nusselt number:

However, to determine the avg. Nusselt no by using Churchill-Bernstein correlation, we have;

SO, the heat transfer coefficient for forced convection is determined as follows afterward:

Finally; The heat loss per unit length on a breezy day during 8 m/s speed is:

pretty sure it's false

explanation:

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