A spherical tank, with an inner diameter of \(D_{1}=\) \(3 \mathrm{~m}\), is filled with a solution undergoing an exothermic reaction that heats the surface to a uniform temperature of \(120^{\circ} \mathrm{C}\). To prevent thermal burn hazards, the tank is enclosed with a concentric outer cover that provides an evacuated gap of \(5 \mathrm{~cm}\) in the enclosure. Both spherical surfaces have the same emissivity of \(0.5\), and the outer surface is exposed to natural convection with a heat transfer coefficient of \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and radiation heat transfer with the surrounding at a temperature of \(30^{\circ} \mathrm{C}\). Determine whether or not the vacuumed gap is sufficient to keep the outer surface temperature below \(45^{\circ} \mathrm{C}\) to prevent thermal burns. If not, propose a solution to keep the outer surface temperature below \(45^{\circ} \mathrm{C}\).

Short Answer

Expert verified
Short answer: Yes, the vacuumed gap between the inner sphere and outer shell is sufficient to keep the outer surface temperature below the thermal burn threshold of \(45^{\circ} \mathrm{C}\), as the calculated outer surface temperature is approximately \(37.17^{\circ} \mathrm{C}\).

Step by step solution

01

List given variables and find their values in SI units

Let's note down the given variables and their corresponding SI units: 1. Inner diameter of the spherical tank, \(D_{1} = 3\, \mathrm{m}\) 2. Temperature of inner surface, \(T_{1} = 120^{\circ} \mathrm{C} = 393.15\, \mathrm{K}\) 3. Evacuated gap between the tank and outer cover, \(\Delta r = 5\, \mathrm{cm} = 0.05\, \mathrm{m}\) 4. Emissivity of both surfaces, \(\epsilon = 0.5\) 5. Heat transfer coefficient for natural convection, \(h = 5\, \mathrm{W / m^2 \cdot K}\) 6. Surrounding temperature, \(T_{\infty} = 30^{\circ} \mathrm{C} = 303.15\, \mathrm{K}\)
02

Calculate surface areas of the inner and outer spheres

To calculate the heat transfer rates, we need to find the surface areas of the inner and outer spheres. Area of a sphere, \(A = 4 \pi r^2\) Since the evacuated gap is \(\Delta r\), we can find radius of the inner sphere, \(r_1 = D_1 / 2 = 1.5\, \mathrm{m}\), and the radius of the outer sphere, \(r_2 = r_1 + \Delta r = 1.55\, \mathrm{m}\) Now, calculate the surface areas: \(A_1 = 4 \pi r_1^2 = 4 \pi (1.5)^2 = 28.27\, \mathrm{m^2}\) \(A_2 = 4 \pi r_2^2 = 4 \pi (1.55)^2 = 30.19\, \mathrm{m^2}\)
03

Calculate the combined heat transfer from convection and radiation

Here, we need to calculate the heat transfer from the inner sphere to the outer shell considering both radiation and convection. For radiation heat transfer, we'll use the formula: \(q_{radiation} = A_1 \epsilon \sigma (T_1^4 - T_2^4) = A_1 \epsilon \sigma (T_1^4 - T_{\infty}^4)\) For convection heat transfer, we'll use the formula: \(q_{convection} = A_2 h (T_2 - T_{\infty})\) For thermal equilibrium, \(q_{radiation} = q_{convection}\)
04

Solve for \(T_2\) (the outer surface temperature) and check if it's below \(45^{\circ} \mathrm{C}\)

Now, we'll solve for the outer surface temperature (\(T_2\)) by equating the two heat transfer equations above: \(A_1 \epsilon \sigma (T_1^4 - T_{\infty}^4) = A_2 h (T_2 - T_{\infty})\) Solve for \(T_2\): \(T_2 = \frac{A_1 \epsilon \sigma (T_1^4 - T_{\infty}^4)}{A_2 h} + T_{\infty}\), where \(\sigma\) is the Stefan-Boltzmann constant, \(5.67 \times 10^{-8}\, \mathrm{W/m^2 \cdot K^4}\) Plug in the values: \(T_2 = \frac{28.27 \cdot 0.5 \cdot (5.67 \times 10^{-8})(393.15^4 - 303.15^4)}{30.19 \cdot 5} + 303.15\) Solving the equation, we get \(T_2 \approx 310.32\, \mathrm{K} = 37.17^{\circ} \mathrm{C}\). Since the outer surface temperature \(37.17^{\circ} \mathrm{C}\) is below the thermal burn threshold of \(45^{\circ} \mathrm{C}\), the vacuumed gap is sufficient to prevent thermal burns.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radiation Heat Transfer
When it comes to understanding how thermal energy travels through space, radiation heat transfer is a key concept. Unlike convection, it does not require a medium – it moves through a vacuum or any transparent medium in the form of electromagnetic waves. In the case of spherical tanks,
the radiative heat transfer can significantly contribute to the total heat lost or gained by the tank.

The Stefan-Boltzmann law is a fundamental equation in this context, represented as \( q_{radiation} = A \epsilon \sigma (T_1^4 - T_2^4) \), where \( q_{radiation} \) is the heat transfer due to radiation, \( A \) is the surface area, \( \epsilon \) is the emissivity of the material, \( \sigma \) is the Stefan-Boltzmann constant, and \( T_1 \) and \( T_2 \) are the absolute temperatures of the two surfaces involved. In our example, the inner surface of the spherical tank and the much cooler surroundings exchange heat via radiation. The solution's ability to control the heat loss through radiation is imperative for maintaining safety and operational efficiency.
Convection Heat Transfer
Convection heat transfer plays an integral role when dealing with fluids that are in motion around a surface. This movement can occur naturally due to density differences caused by temperature variation, known as natural convection, or it can be forced by an external source like a pump or a fan.

In the context of our spherical tank, the outer surface is subjected to natural convection with the surrounding air. The rate at which heat is transferred by convection can be calculated using the formula \( q_{convection} = A h (T_2 - T_{\infty}) \), where \( h \) is the heat transfer coefficient, and \( T_{\infty} \) is the temperature of the fluid far away from the surface (usually the ambient air temperature). The high importance of this form of heat transfer is due to its role in dissipating excess heat, especially from the outer surface of the tank, to the environment.
Thermal Equilibrium
Thermal equilibrium is a state where there is no net heat transfer between objects in contact or within a closed system because they have reached the same temperature. For a spherical tank, reaching thermal equilibrium is about balancing the heat transfer into and out of the system.

In the example provided, thermal equilibrium implies that the rate of heat transfer from the inner sphere via radiation must equal the rate of heat loss from the outer surface via convection. Mathematically, it can be expressed as \( q_{radiation} = q_{convection} \). Establishing this balance ensures that the temperature of the spherical tank's outer surface remains below the critical temperature. By carefully designing or adjusting the vacuumed gap, materials, or any other relevant parameters, engineers can maintain the tank at an equilibrium state that is both safe and effective.

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Most popular questions from this chapter

A 2-m-internal-diameter double-walled spherical tank is used to store iced water at \(0^{\circ} \mathrm{C}\). Each wall is \(0.5 \mathrm{~cm}\) thick, and the \(1.5\)-cm-thick air space between the two walls of the tank is evacuated in order to minimize heat transfer. The surfaces surrounding the evacuated space are polished so that each surface has an emissivity of \(0.15\). The temperature of the outer wall of the tank is measured to be \(20^{\circ} \mathrm{C}\). Assuming the inner wall of the steel tank to be at \(0^{\circ} \mathrm{C}\), 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.

A thermocouple shielded by aluminum foil of emissivity \(0.15\) is used to measure the temperature of hot gases flowing in a duct whose walls are maintained at \(T_{w}=380 \mathrm{~K}\). The thermometer shows a temperature reading of \(T_{\text {th }}=530 \mathrm{~K}\). Assuming the emissivity of the thermocouple junction to be \(\varepsilon=0.7\) and the convection heat transfer coefficient to be \(h=120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the actual temperature of the gas. What would the thermometer reading be if no radiation shield was used?

Consider a surface at \(0^{\circ} \mathrm{C}\) that may be assumed to be a blackbody in an environment at \(25^{\circ} \mathrm{C}\). If \(300 \mathrm{~W} / \mathrm{m}^{2}\) of radiation is incident on the surface, the radiosity of this black surface is (a) \(0 \mathrm{~W} / \mathrm{m}^{2}\) (b) \(15 \mathrm{~W} / \mathrm{m}^{2}\) (c) \(132 \mathrm{~W} / \mathrm{m}^{2}\) (d) \(300 \mathrm{~W} / \mathrm{m}^{2}\) (e) \(315 \mathrm{~W} / \mathrm{m}^{2}\)

Consider an enclosure consisting of eight surfaces. How many view factors does this geometry involve? How many of these view factors can be determined by the application of the reciprocity and the summation rules?

Two parallel disks of diameter \(D=3 \mathrm{ft}\) separated by \(L=2 \mathrm{ft}\) are located directly on top of each other. The disks are separated by a radiation shield whose emissivity is \(0.15\). Both disks are black and are maintained at temperatures of \(1200 \mathrm{R}\) and \(700 \mathrm{R}\), respectively. The environment that the disks are in can be considered to be a blackbody at \(540 \mathrm{R}\). Determine the net rate of radiation heat transfer through the shield under steady conditions.

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