Question

Steam at \(280^{\circ} \mathrm{C}\) flows in a stainless steel pipe $(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$ whose inner and outer diameters are \(5 \mathrm{~cm}\) and \(5.5 \mathrm{~cm}\), respectively. The pipe is covered with \(3-\mathrm{cm}\)-thick glass wool insulation $(k=0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\(. Heat is lost to the surroundings at \)5^{\circ} \mathrm{C}$ by natural convection and radiation, with a combined natural convection and radiation heat transfer coefficient of $22 \mathrm{~W} / \mathrm{m}^{2}$. . Taking the heat transfer coefficient inside the pipe to be \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat loss from the steam per unit length of the pipe. Also determine the temperature drops across the pipe shell and the insulation.

Short answer

Answer: The rate of heat loss per unit length of the pipe is approximately 1091.8 W/m. The temperature drop across the pipe shell is 0.16°C, and the temperature drop across the insulation is 274.4°C.

Step-by-step solution

Calculate the resistance of each layer

For this problem, we need to find the thermal resistance of each layer, which can be represented as: \(R_\text{layer} = \frac{L_\text{layer}}{k_\text{layer}A_\text{layer}}\) Where: - \(R_\text{layer}\) is the thermal resistance of the layer - \(L_\text{layer}\) is the thickness of the layer - \(k_\text{layer}\) is the thermal conductivity of the material - \(A_\text{layer}\) is the surface area of the layer We will calculate the resistances for three layers: inside the pipe, pipe wall, and insulation.

Calculate resistance inside the pipe

Using the given parameters, we can find the resistance inside the pipe (\(R_\text{in}\)): \(R_\text{in} = \frac{1}{h_\text{in}A_\text{in}} = \frac{1}{80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \cdot \pi(0.05 \mathrm{~m})^2} \approx 0.0007958~ \mathrm{m}^{2} \cdot \mathrm{K} / \mathrm{W}\)

Calculate resistance of the pipe wall

Now, we can calculate the resistance of the pipe wall (\(R_\text{pipe}\)): \(R_\text{pipe} = \frac{\ln{(r_\text{out}/r_\text{in})}}{2\pi k_\text{pipe}L} = \frac{\ln{0.055/0.05}}{2\pi(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})(1 \mathrm{~m})} \approx 0.0001465~ \mathrm{m}^{2} \cdot \mathrm{K} / \mathrm{W}\)

Calculate resistance of the insulation

Finally, let's find the resistance of the insulation (\(R_\text{ins}\)): \(R_\text{ins} = \frac{\ln{(r_\text{out,ins}/r_\text{out})}}{2\pi k_\text{ins}L} = \frac{\ln{(0.055 + 0.03) / 0.055}}{2\pi(0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})(1 \mathrm{~m})} \approx 0.25113~ \mathrm{m}^{2} \cdot \mathrm{K} / \mathrm{W}\)

Calculate total resistance

Now, we can find the total resistance (\(R_\text{total}\)): \(R_\text{total} = R_\text{in} + R_\text{pipe} + R_\text{ins} = 0.0007958 + 0.0001465 + 0.25113 \approx 0.25207~ \mathrm{m}^{2} \cdot \mathrm{K} / \mathrm{W}\)

Calculate the rate of heat loss

Using the total resistance and the temperature difference between the steam and surroundings, we can calculate the rate of heat loss per unit length of the pipe: \(q = \frac{T_\text{steam} - T_\text{surroundings}}{R_\text{total}} = \frac{280^{\circ} \mathrm{C} - 5^{\circ} \mathrm{C}}{0.25207~ \mathrm{m}^{2} \cdot \mathrm{K} / \mathrm{W}} \approx 1091.8~\mathrm{W} / \mathrm{m}\)

Calculate temperature drops

Now, we can find the temperature drops across the pipe shell and insulation using their respective resistances: \(\Delta T_\text{pipe} = R_\text{pipe}q = 0.0001465 \times 1091.8 \approx 0.16^{\circ} \mathrm{C}\) \(\Delta T_\text{ins} = R_\text{ins}q = 0.25113 \times 1091.8 \approx 274.4^{\circ} \mathrm{C}\) Using these steps we now have both the rate of heat loss per unit length of the pipe and the temperature drops across the pipe shell and insulation. The heat loss is approximately \(1091.8~\mathrm{W}/\mathrm{m}\), while the temperature drop across the pipe shell is \(0.16^{\circ} \mathrm{C}\), and the temperature drop across the insulation is \(274.4^{\circ} \mathrm{C}\).