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 rate of heat transfer, and temperature difference across the insulation layer for the given system. Answer: The thermal resistances are approximately \(R_{i} \approx \frac{1}{360\pi} \mathrm{K} / \mathrm{W}\), \(R_{ins} \approx \frac{0.05}{2.34\pi} \mathrm{K} / \mathrm{W}\), and \(R_{o} \approx \frac{1}{294.637\pi} \mathrm{K} / \mathrm{W}\). The steady rate of heat transfer is approximately \(\dot{Q} \approx 7462.56 \mathrm{~W}\). The temperature difference across the insulation layer is approximately \(\Delta T_{ins} \approx 15.98 \mathrm{~K}\).

Step-by-step solution

Calculate the thermal resistance of the convection at the inside surface of the vessel

The formula for calculating the thermal resistance of the inner surface (\(R_{i}\)) due to convection is: $$ R_{i} = \frac{1}{h_{i} A_{i}}, $$ where \(h_{i} = 40 \frac{\mathrm{W}}{\mathrm{m}^2 \cdot \mathrm{K}}\) is the inside heat transfer coefficient and \(A_{i}\) is the inside surface area of the vessel. Since the vessel is a sphere, we can calculate its surface area with: $$ A_i = 4 \pi r_i^2, $$ where \(r_{i} = 1.5 \mathrm{~m}\) is the radius of the sphere. Calculating the surface area, $$ A_i = 4 \pi (1.5)^2 = 9\pi \mathrm{~m}^2. $$ Now we can calculate the thermal resistance of the inside surface: $$ R_{i} = \frac{1}{40 (9\pi)} = \frac{1}{360\pi} \mathrm{K} / \mathrm{W}. $$

Calculate the thermal resistance of the insulation layer

The formula for calculating thermal resistance of the insulation (\(R_{ins}\)) is: $$ R_{ins} = \frac{r_o - r_i}{4 \pi k (r_{o}r_{i})}, $$ where \(r_{o} = r_{i} + t_{ins}\) is the outer radius of the insulation layer, \(t_{ins}= 0.05 \mathrm{~m}\) is the thickness of the insulation, and \(k = 0.20 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}}\) is the thermal conductivity of the insulation material. Calculating the outer radius: $$ r_o = 1.5 + 0.05 = 1.55 \mathrm{~m}. $$ Now, calculating the thermal resistance of the insulation layer: $$ R_{ins} = \frac{1.55 - 1.5}{4 \pi (0.20) (1.55)(1.5)} = \frac{0.05}{2.34\pi} \mathrm{K} / \mathrm{W}. $$

Calculate the thermal resistance of the convection at the outside surface of the insulation layer

The formula for calculating the thermal resistance of the outer surface (\(R_{o}\)) due to convection is: $$ R_{o} = \frac{1}{h_{o} A_{o}}, $$ where \(h_{o} = 10 \frac{\mathrm{W}}{\mathrm{m}^2 \cdot \mathrm{K}}\) is the outside heat transfer coefficient and \(A_{o}\) is the outside surface area of the insulation layer. Calculating the surface area \(A_{o}\): $$ A_o = 4 \pi (1.55)^2 = 9.604\pi \mathrm{~m}^2. $$ Now we can calculate the thermal resistance of the outside surface: $$ R_{o} = \frac{1}{10 (9.604\pi)} \approx \frac{1}{294.637\pi} \mathrm{K} / \mathrm{W}. $$

Calculate the overall thermal resistance and steady rate of heat transfer

The overall thermal resistance (\(R_{total}\)) is the sum of the resistances of the three layers: $$ R_{total} = R_{i} + R_{ins} + R_{o}. $$ Calculating \(R_{total}\): $$ R_{total} = \frac{1}{360\pi} + \frac{0.05}{2.34\pi} + \frac{1}{294.637\pi} \approx \frac{1}{106.455\pi} \mathrm{K} / \mathrm{W}. $$ Now we can find the steady rate of heat transfer (\(\dot{Q}\)) using the formula: $$ \dot{Q} = \frac{\Delta T}{R_{total}}, $$ where \(\Delta T = T_{air} - T_{fluid} = 22 - 0 = 22 \mathrm{~K}\) is the temperature difference between the surrounding air and the fluid. Calculating the steady rate of heat transfer: $$ \dot{Q} = \frac{22}{\frac{1}{106.455\pi}} = 22(106.455\pi) \approx 7462.56 \mathrm{~W}. $$

Calculate the temperature difference across the insulation layer

We can determine the temperature difference across the insulation layer, \(\Delta T_{ins}\), using the formula: $$ \Delta T_{ins} = R_{ins} \dot{Q}. $$ Calculating \(\Delta T_{ins}\): $$ \Delta T_{ins} = \frac{0.05}{2.34\pi} (7462.56) \approx 15.98 \mathrm{~K}. $$ #Summary#The thermal resistances of the inside surface, insulation layer, and outside surface are approximately: (a) \(R_{i} \approx \frac{1}{360\pi} \mathrm{K} / \mathrm{W}\), \(R_{ins} \approx \frac{0.05}{2.34\pi} \mathrm{K} / \mathrm{W}\), and \(R_{o} \approx \frac{1}{294.637\pi} \mathrm{K} / \mathrm{W}\). (b) The steady rate of heat transfer is approximately \(\dot{Q} \approx 7462.56 \mathrm{~W}\). (c) The temperature difference across the insulation layer is approximately \(\Delta T_{ins} \approx 15.98 \mathrm{~K}\).

Key concepts

Understanding Heat Transfer

Heat transfer is a fundamental concept in understanding how energy moves from one place to another. It is an essential process in various scientific and engineering disciplines, including thermodynamics and climate science. In the context of a spherical vessel with insulation, heat transfer primarily occurs by conduction through the materials and by convection at the surfaces.

When a fluid at a lower temperature is stored inside a vessel, heat will naturally transfer from the warmer surrounding air to the cooler fluid until equilibrium is reached. The rate of this heat transfer is influenced by factors such as the temperature difference between the environments, the thermal properties of the insulation material, and the heat transfer coefficients of the surfaces in contact with the fluid and the air. In practice, engineers aim to minimize heat transfer to maintain the fluid at a desired temperature by using insulation with low thermal conductivity and designing appropriate surface properties for the vessel.

Improving the comprehensibility of this concept, it's important to remember that thermal resistance is analogous to electrical resistance; it is a measure of how difficult it is for heat to flow through a material. The higher the thermal resistance, the better the insulation property of the material, and the slower the rate of heat transfer.

Conduction in Spherical Coordinates

Conduction in spherical coordinates can be visualized when examining heat flow through objects such as balls or planets, where the heat moves radially from the center to the outside or vice versa. For a uniform spherical shell like the insulation of the vessel, the heat will flow equally in all radial directions.

To calculate the thermal resistance due to conduction in a spherical coordinate system, it is vital to consider the geometry of the sphere. Unlike flat surfaces, where the area through which heat conducts is constant, in a spherical shell, the area changes as you move from the inner radius to the outer radius. The thermal resistance calculation takes into account the radii of the inner and outer surfaces as well as the thermal conductivity of the insulating material. This concept is crucial for understanding how different shapes affect the path of heat transfer and subsequently how to control or use this transfer in practical applications.

Why This Matters

Grasping the peculiarities of heat conduction in spherical coordinates will allow students to tackle problems involving spherical objects with more insight and precision. Understanding that the surface area increases as the radius becomes larger is important when calculating thermal resistance for design and analysis in engineering and environmental science.

Insulation Thermal Properties

The thermal properties of insulation are integral to its effectiveness in preventing heat transfer. The key property that determines insulation effectiveness is thermal conductivity, denoted by 'k,' which measures how readily a material can conduct heat. Lower thermal conductivity indicates better insulating properties since less heat is transferred through the material over time.

In addition to thermal conductivity, the thickness of the insulation also plays a significant role. Thicker layers of insulation generally provide greater resistance to heat flow, assuming the same material is used. In the exercise, the insulation's thickness and its material's thermal conductivity were used to calculate the insulation's thermal resistance. This resistance affects both the steady rate of heat transfer and the temperature gradient across the insulation layer.

Real-World Application

In real-world scenarios, such as building construction or refrigeration, selecting the right insulation material with appropriate thermal properties is crucial for energy efficiency. By using insulation with a low thermal conductivity and sufficient thickness, energy costs can be reduced, and thermal comfort can be improved. This concept helps students understand the practical decisions made in insulation material choice and its impact on performance.