An insulated pipe carries steam at 300°C. The pipe is made of stainless steel (with k = 15 W/mK), has an inner diameter is 4 cm, and has an outer diameter is 4.5 cm. Fiberglass insulation (with k= 0.038 W/mK and ε = 0.9) of 2.5 cm thickness covers the pipe. Consider the pipe to be inside a large room with air and wall temperatures of 20°C. If the convective heat transfer coefficient inside the pipe is 75 W/m² and the convective heat transfer coefficient outside the pipe is 10 W/m², find the rate of heat loss per-unit-length (W/m) from the pipe and the temperature of the outer surface of the insulation. Note 1: You may find helpful example problems in Incropera and DeWitt. Note 2: You will need to guess a temperature for the outside of the pipe to compute an estimated hrad. After solving for the temperature of the outside using your thermal circuit, evaluate how close your original estimate was. Then iterate (repeat the math) with the new surface temperature. Consider coding your computations in R or Python to make this go faster.

The answers to the question are

(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W

(ii) The temperature of the outer surface of the insulation is 49.89 °C

Explanation:

To solve the question, we note that the heat transferred is given by

Where

= Temperature at the inside of the pipe = 300 °C

= Temperature at the outside of the pipe = 20 °C

r₁ =internal  radius of pipe = 4.0 cm

r₂ = Outer radius of pipe = 4.5 cm

r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm

= 15 W/m·K

= 0.038 W/m·K

= 75 W/m²·K

= 10 W/m²·K

Plugging in the values in the above equation where for a unit length L = 1 m, we have

Q = 131.32 W

From which we have, for the film of air at the pipe outer boundary layer

Where for the air film on the pipe outer surface is given by

where A =area of the outside of the pipe

= = 0.227 K/W

Therefore

131.32 W = which gives

= 49.89 °C

Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)

Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)

T₁ = Surface temperature of the pipe = 49.89 °C and

T₂ = Temperature of the surrounding = 20.00 °C

Plugging in the values gives, q' = 0.307 W per m²

Total heat lost per unit length = 131.32 + 0.307 =131.62 W

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .