Question

A 10-m-long, 8-cm-outer-radius cylindrical steam pipe is covered with 3 -cm- thick cylindrical insulation with a thermal conductivity of $0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. If the rate of heat loss from the pipe is \(1000 \mathrm{~W}\), the temperature drop across the insulation is (a) \(58^{\circ} \mathrm{C}\) (b) \(101^{\circ} \mathrm{C}\) (c) \(143^{\circ} \mathrm{C}\) (d) \(282^{\circ} \mathrm{C}\) (e) \(600^{\circ} \mathrm{C}\)

Short answer

Answer: The temperature drop across the insulation of the cylindrical steam pipe is 143°C.

Step-by-step solution

Write down the given data.

The given data for this exercise are as follows: Length of the pipe (L): 10 m Outer radius of the cylindrical pipe (r1): 8 cm or 0.08 m in SI units Thickness of the cylindrical insulation (t): 3 cm or 0.03 m in SI units Thermal conductivity of the insulation (k): 0.05 W/(m·K) Rate of heat loss from the pipe (Q): 1000 W

Find the inner and outer radii of the insulation.

To find the temperature drop across the insulation, we first need to determine the inner and outer radii of the insulation. The inner radius is the same as the outer radius of the pipe, and the outer radius is the sum of the inner radius and the thickness of the insulation: Inner radius of the insulation (r2): r1 = 0.08 m Outer radius of the insulation (r3): r2 + t = 0.08 m + 0.03 m = 0.11 m

Use the formula for heat loss rate through a cylindrical surface.

The formula for the heat loss rate through a cylindrical surface is given by: \(Q = 2\pi k L \frac{\Delta T}{\ln(\frac{r3}{r2})}\) Our goal is to find the temperature drop across the insulation (\(\Delta T\)). Rearranging the formula to solve for \(\Delta T\), we get: \(\Delta T = \frac{Q \ln(\frac{r3}{r2})}{2\pi k L}\)

Insert the given data into the formula and solve for the temperature drop.

Now that we have all the data, we can insert it into the formula to calculate the temperature drop across the insulation: \(\Delta T = \frac{1000 \, W\, \ln(\frac{0.11 \, m}{0.08 \, m})}{2\pi(0.05 \, \frac{W}{m\cdot K})(10 \, m)}\) Calculating the value, we get: \(\Delta T = 143^{\circ} \mathrm{C}\) The correct answer is (c) \(143^{\circ} \mathrm{C}\).