Question

A spherical vessel, \(3.0 \mathrm{~m}\) in diameter (and negligible wall thickness), is used for storing a fluid at a temperature of $0^{\circ} \mathrm{C}\(. The vessel is covered with a \)5.0$-cm-thick layer of an insulation \((k=0.20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The surrounding air is at \(22^{\circ} \mathrm{C}\). The inside and outside heat transfer coefficients are 40 and 10 $\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\(, respectively. Calculate \)(a)$ all thermal resistances, in \(\mathrm{K} / \mathrm{W},(b)\) the steady rate of heat transfer, and \((c)\) the temperature difference across the insulation layer.

Short answer

Question: Calculate the thermal resistances, steady heat transfer rate, and temperature difference across the insulation layer of the spherical vessel. Answer: To find the thermal resistances, steady heat transfer rate, and temperature difference across the insulation layer, follow these steps: 1. Calculate the inner and outer radius of the vessel: \(R_1 = 1.5\,\text{m}\) and \(R_2 = 1.55\,\text{m}\). 2. Calculate the individual thermal resistances: \(R_{cond}\), \(R_{conv_1}\), and \(R_{conv_2}\). 3. Calculate the total thermal resistance: \(R_{tot} = R_{conv_1} + R_{cond} + R_{conv_2}\). 4. Calculate the steady rate of heat transfer: \(q = \frac{\Delta T}{R_{tot}}\), with \(\Delta T = T_{inside} - T_{outside}\). 5. Calculate the temperature difference across the insulation layer: \(\Delta T_{insulation} = q \times R_{cond}\). Using the given data, you will be able to calculate the thermal resistances, steady heat transfer rate, and temperature difference across the insulation layer for the given spherical vessel.

Step-by-step solution

Calculate the inner and outer radius of the vessel

We are given that the diameter of the spherical vessel is \(3.0\,\text{m}\). Considering negligible wall thickness, the radius of the inside sphere will be: \(R_1 = \frac{\text{diameter}}{2} = \frac{3.0}{2} = 1.5\,\text{m}\) The insulation layer is \(5.0\,\text{cm}\) thick. To find the radius of the outside sphere, we add this thickness to the inner radius: \(R_2 = R_1 + 5.0\,\text{cm} \times \frac{1\,\text{m}}{100\,\text{cm}} = 1.5\,\text{m} + 0.05\,\text{m} = 1.55\,\text{m}\)

Calculate the thermal resistances

There are three modes of heat transfer happening: conduction through the insulation, convection inside the vessel, and convection outside the vessel. The thermal resistance for conduction through the insulation, \(R_{cond}\), is given by: \(R_{cond} = \frac{1}{4 \pi k} \left(\frac{R_2^3 - R_1^3}{R_2 - R_1}\right)\) with \(k=0.20\,\frac{\text{W}}{\text{m}\cdot\text{K}}\), and \(R_1 = 1.5\,\text{m}\) and \(R_2 = 1.55\,\text{m}\) The thermal resistance for convection inside the vessel, \(R_{conv_1}\), is given by: \(R_{conv_1} = \frac{1}{h_1 A_1}\) with \(h_1=40\,\frac{\text{W}}{\text{m}^2\cdot\text{K}}\), and \(A_1 = 4 \pi R_1^2\) The thermal resistance for convection outside the vessel, \(R_{conv_2}\), is given by: \(R_{conv_2} = \frac{1}{h_2 A_2}\) with \(h_2=10\,\frac{\text{W}}{\text{m}^2\cdot\text{K}}\), and \(A_2 = 4 \pi R_2^2\)

Calculate the total thermal resistance

The total thermal resistance \(R_{tot}\) is given by the sum of the individual resistances: \(R_{tot} = R_{conv_1} + R_{cond} + R_{conv_2}\)

Calculate the steady rate of heat transfer

The formula for steady heat transfer through a system with the given total thermal resistance is: \(q = \frac{\Delta T}{R_{tot}}\) with the temperature difference \(\Delta T = T_{inside} - T_{outside}\), with \(T_{inside}=0\,\text{C}\) and \(T_{outside}=22\,\text{C}\).

Calculate the temperature difference across the insulation layer

The temperature difference across the insulation layer, \(\Delta T_{insulation}\), is given by: \(\Delta T_{insulation} = q \times R_{cond}\) Now that we have outlined the steps, we can use the given data to find the required thermal resistances, steady heat transfer rate, and temperature difference across the insulation layer.