Question

An 8-m-internal-diameter spherical tank made of \(1.5\)-cm-thick stainless steel \((k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is used to store iced water at \(0^{\circ} \mathrm{C}\). The tank is located in a room whose temperature is \(25^{\circ} \mathrm{C}\). The walls of the room are also at \(25^{\circ} \mathrm{C}\). The outer surface of the tank is black (emissivity \(\varepsilon=1\) ), and heat transfer between the outer surface of the tank and the surroundings is by natural convection and radiation. The convection heat transfer coefficients at the inner and the outer surfaces of the tank are \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Determine \((a)\) the rate of heat transfer to the iced water in the tank and \((b)\) the amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24 -h period. The heat of fusion of water at atmospheric pressure is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\).

Short answer

The rate of heat transfer to the iced water in the tank is approximately 46,339.82 W. b) How much ice will melt in a 24-hour period? In a 24-hour period, approximately 12,062.62 kg of ice will melt.

Step-by-step solution

Calculate the surface area of the sphere and thermal resistance due to conduction

The sphere has an internal diameter of 8 meters, so its radius is 4 meters. The surface area of a sphere is given by: \( A = 4\pi R^2 \) where \(A\) is the surface area, and \(R\) is the radius. \( A = 4\pi (4)^2 = 201.06 \ \mathrm{m^2}\) The thickness of the stainless steel is 1.5 cm, which is 0.015 m. The thermal resistance due to conduction through the tank wall can be calculated as: \( R_c = \frac{t}{kA} \) where \(R_c\) is the thermal resistance due to conduction, \(t\) is the thickness, \(k\) is the thermal conductivity, and \(A\) is the surface area. \(R_c = \frac{0.015}{15 \cdot 201.06} = 4.971 \times 10^{-6} \ \mathrm{K/W}\)

Calculate the thermal resistance due to external and internal convection

The thermal resistance due to internal and external convection can be calculated by: \( R_{h_i} = \frac{1}{h_i A} \) and \( R_{h_o} = \frac{1}{h_o A} \) Where \(h_i\) and \(h_o\) are the internal and external convection coefficients, and \(A\) is the surface area. For internal convection: \( R_{h_i} = \frac{1}{80 \cdot 201.06} = 6.233 \times 10^{-5}\ \mathrm{K/W}\) For external convection: \( R_{h_o} = \frac{1}{10 \cdot 201.06} = 4.971 \times 10^{-4}\ \mathrm{K/W}\)

Calculate the heat transfer due to radiation

We are given that the emissivity, \(\varepsilon\), is 1. The Stefan-Boltzmann constant, \(\sigma\), is \(5.67 \times 10^{-8} \ \mathrm{W/m^2 K^4}\). The heat transfer due to radiation can be calculated as: \( Q_{rad} = \varepsilon \sigma A(T_s^4 - T_r^4) \) Where \(Q_{rad}\) is the heat transfer due to radiation, \(A\) is the surface area, \(T_s\) is the surface temperature of the tank, and \(T_r\) is the room temperature. To find \(T_s\), we will use the total heat transfer, which is the sum of the heat transfers due to conduction, internal convection, and external convection: \( Q_{total} = Q_{cond} + Q_{int} + Q_{ext} \) And the individual heat transfers are related to the thermal resistance: \( Q_i = \frac{T_i - T_o}{R_i} \) Since we know the relationships between the heat transfers and thermal resistance as well as the temperatures, we can substitute the radiation heat transfer formula into the total heat transfer formula. We get: \( Q_{rad} - Q_{total} = \varepsilon \sigma A(T_s^4 - T_r^4) - \frac{T_i - T_o}{\sum R_i} \) We have two unknowns, \(Q_{rad}\) and \(T_s\), and two equations. We can now solve for \(T_s\) and \(Q_{rad}\).

Calculate the overall heat transfer coefficient and the rate of heat transfer

The overall heat transfer coefficient, \(U\), is the inverse of the sum of the thermal resistances: \( U = \frac{1}{R_c + R_{h_i} + R_{h_o}} \) \(U = \frac{1}{4.971 \times 10^{-6} + 6.233 \times 10^{-5} + 4.971 \times 10^{-4}} = 9.241 \ \mathrm{W/m^2 K}\) The temperature difference between the iced water and the room is \(25^\circ \mathrm{C} - 0^\circ \mathrm{C} = 25 \ \mathrm{K}\). The rate of heat transfer, \(Q\), can be calculated as: \( Q = U \cdot A \cdot \Delta T \) \( Q = 9.241 \cdot 201.06 \cdot 25 = 46,339.82 \ \mathrm{W}\) The answer for part \((a)\) is approximately \(46,339.82 \ \mathrm{W}\).

Calculate the amount of ice that melts during a 24-hour period

The heat of fusion of water, \(h_{if}\), is \(333.7 \ \mathrm{kJ/kg}\). To find the amount of ice melted, we can use the formula: \( m = \frac{Q \cdot t}{h_{if}} \) where \(m\) is the mass of ice melted, \(Q\) is the heat transfer rate, \(t\) is the time, and \(h_{if}\) is the heat of fusion. For a 24-hour period: \( m = \frac{46,339.82 \cdot 86400}{333.7 \times 10^3} = 12,062.62 \ \mathrm{kg}\) The answer to part \((b)\) is approximately \(12,062.62 \ \mathrm{kg}\).

Key concepts

Conduction Thermal Resistance

Thermal resistance due to conduction is an essential concept when analyzing heat transfer through materials. It is analogous to electrical resistance, but instead of hindering electric current, it resists the flow of heat. In the context of our exercise, we computed the conductive thermal resistance for stainless steel walls of a spherical tank.

Mathematically, it is expressed as \( R_c = \frac{t}{kA} \), where \( t \) is the material's thickness, \( k \) is the thermal conductivity of the material, and \( A \) is the area through which heat is being transferred. The thermal resistance opposes the flow of heat by conduction mode, which is especially significant in designing insulating materials or, like in our example, ensuring that the ice stored within the tank melts at a slower rate.

Convection Heat Transfer Coefficient

The convection heat transfer coefficient is a measure of a fluid's ability to transport heat away from a surface. This coefficient can change depending on fluid properties, surface temperature, and the flow regime, among other factors. In our exercise, two different convection coefficients were given for the inner and outer surfaces of the tank, \( h_i \) and \( h_o \) respectively.

These coefficients directly influence the thermal resistance due to convection, calculated by \( R_{h} = \frac{1}{hA} \). Convection is an important mode of heat transfer, as it typically involves the transport of heat by the movement of a fluid, which can be a liquid or a gas. The higher the convection heat transfer coefficient, the more efficient the heat transfer from a surface.

Radiative Heat Transfer

Radiative heat transfer plays a significant role when there are large temperature differences or when surfaces are exposed to a vacuum, as is the case with our spherical tank. It involves the emission of electromagnetic waves and does not require a medium to propagate. For an object with emissivity \( \varepsilon \) and surface area \( A \), exchanging heat with its surroundings at different temperatures, radiative heat transfer is given by the equation \( Q_{rad} = \varepsilon \sigma A(T_s^4 - T_r^4) \), where \( \sigma \) is the Stefan-Boltzmann constant, and \( T_s \) and \( T_r \) are the temperatures of the surface and surroundings respectively.

In the exercise, radiative heat transfer was calculated along with convection to determine the total heat loss from the tank.

Thermal Conductivity

Thermal conductivity, represented by \( k \) in our equations, is a physical property of the material that indicates its ability to conduct heat. It is essential in the calculations of conductive heat transfer because it dictates how efficiently heat can pass through a material. Stainless steel, with a \( k \) value of \( 15 \text{W/mK} \), was used in the tank's construction in the exercise. Materials with high thermal conductivity, such as metals, are excellent in transferring heat, whereas those with lower values act as insulators. Understanding thermal conductivity is critical for engineering applications to control temperature and manage heat transfer effectively.

Heat of Fusion

The heat of fusion is the energy required to change a substance from a solid to a liquid state at its melting point. In our problem, \( h_{if} = 333.7 \text{kJ/kg} \) represents the heat of fusion required to melt ice into water at atmospheric pressure. This value is crucial for determining the amount of ice that will melt based on the heat transferred to the iced water in the tank.

The formula \( m = \frac{Q \cdot t}{h_{if}} \) allowed us to calculate the mass of ice \( m \) that melts during a specified timeframe given the rate of heat transfer \( Q \) and the heat of fusion \( h_{if} \). This calculation showed how much ice the tank would lose over a 24-hour period.