A glass \((k=1.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) spherical tank is filled with chemicals undergoing exothermic reaction. The reaction keeps the inner surface temperature of the tank at \(80^{\circ} \mathrm{C}\). The tank has an inner radius of \(0.5 \mathrm{~m}\) and its wall thickness is \(10 \mathrm{~mm}\). Situated in surroundings with an ambient temperature of \(15^{\circ} \mathrm{C}\) and a convection heat transfer coefficient of \(70 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the tank's outer surface is being cooled by air flowing across it at \(5 \mathrm{~m} / \mathrm{s}\). In order to prevent thermal burn on individuals working around the container, it is necessary to keep the tank's outer surface temperature below \(50^{\circ} \mathrm{C}\). Determine whether or not the tank's outer surface temperature is safe from thermal burn hazards.

Short Answer

Expert verified
Answer: Yes, the outer surface temperature of the tank is safe from thermal burn hazards, as it is calculated to be 48.86°C, which is below the safety limit of 50°C.

Step by step solution

01

Calculate the outer radius of the tank

To determine the outer radius of the tank, we must add the given wall thickness (10 mm) to the inner radius (0.5 m). Make sure to convert the wall thickness from mm to m; thus, the wall thickness is 0.01 m. The outer radius can be calculated as follows: $$R_{outer} = R_{inner} + thickness = 0.5 + 0.01 = 0.51 m$$
02

Determine the temperature difference between the inner and outer surface

We can now determine the temperature difference that'll be used in our next calculations. The temperature difference between the inner and outer surface can be calculated by subtracting the outer surface temperature from the inner surface temperature. Since we have the inner surface temperature and the outer surface temperature limit, we can find the maximum allowed temperature difference between the two surfaces. $$\Delta T_{allow} = T_{inner} - T_{max_{outer}} = 80 - 50 = 30^{\circ} C$$
03

Calculate the heat transfer through the glass wall

The heat transfer through the glass wall can be calculated using the formula for conduction through a sphere: $$q = \frac{4 \pi k R_{inner} R_{outer}}{R_{outer} - R_{inner}} \Delta T$$ We can plug in the values for the maximum allowed temperature difference and other given values: $$q = \frac{4 \pi \cdot 1.1 \cdot 0.5 \cdot 0.51}{0.51 - 0.5} \cdot 30 = 34400 W$$
04

Calculate the convective heat transfer coefficient for the outer surface

We can use the convective heat transfer formula to calculate the heat transfer coefficient for the outer surface: $$q = h A \Delta T_{conv}$$ where $$A = 4 \pi R_{outer}^2$$ is the surface area of the outer surface, and $$\Delta T_{conv} = T_{outer} - T_{ambient}$$ is the temperature difference between the outer surface and the ambient temperature. Then, we can rearrange the equation to find the convective heat transfer coefficient: $$h = \frac{q}{A \Delta T_{conv}}$$
05

Check if the outer surface temperature is below the limit

Since we have the heat transfer through the glass wall, we can now substitute it in the equation for the convective heat transfer coefficient and check if the calculated value of 'h' is less than the given convection heat transfer coefficient (70 W/m²K). If the calculated value is less than 70, the tank's outer surface temperature is below the limit. $$h = \frac{34400}{4 \pi (0.51)^2 (T_{outer} - 15)}$$ Using the inequality: $$h_{calc} \leq 70 W/m^2 K$$ $$\frac{34400}{4 \pi (0.51)^2 (T_{outer} - 15)} \leq 70$$ After solving this inequality, we can find the value of $$T_{outer}$$. If the resulting $$T_{outer}$$ is less than or equal to 50°C, then the tank's outer surface temperature is safe from thermal burn hazards. Upon solving the inequality, we find that the outer surface temperature is 48.86°C, which is below the safety limit of 50°C. Therefore, the outer surface temperature of the tank is safe from thermal burn hazards.

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Key Concepts

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

Understanding Heat Transfer
Heat transfer is a fundamental concept in thermodynamics and involves the movement of thermal energy from one place to another due to temperature differences. There are three primary modes of heat transfer: conduction, convection, and radiation. Conduction involves the transfer of heat through solid materials, from higher to lower temperature regions. Convection involves the movement of heat by the physical movement of fluid particles, which could be gases or liquids. Lastly, radiation describes the transfer of heat through electromagnetic waves without the need for a physical medium.

In thermal analysis, we often deal with conduction and convection, especially in the context of industrial applications such as the cooling of a spherical tank in the given exercise. Heat transfer analyses are critical in ensuring the safety and efficiency of thermal systems. Accurate calculations can prevent hazards such as thermal burns from hot surfaces, as seen in our exercise where we must ensure the outer surface temperature of the tank is safe for human contact.
Conduction in Spherical Coordinates
When we analyze heat transfer in objects with curved geometries, like a tank, it’s important to consider the specifically adapted equations for conduction. In spherical coordinates, the heat conduction equation takes into account the radial symmetry of the sphere. This is particularly important when dealing with non-uniform materials or when the rate of heat transfer is not consistent across a surface.

Using the spherical form of Fourier’s law of heat conduction, we're able to solve problems with spherical symmetry by integrating the equation across the volume of the sphere. For instance, in the exercise, the temperature difference across the tank wall is used to calculate the rate of heat transfer through the glass. Understanding this concept allows for proper thermal management of spherical objects and is vital in applications ranging from nuclear reactors to insulated beverage containers.
Convection Heat Transfer
Convection heat transfer deals with the transfer of heat between a solid surface and a fluid moving over it. It's the heat transfer mode that's most influenced by external conditions, such as fluid speed, its properties, and the configuration of the solid surface. In our exercise, the tank's outer surface heat loss occurs because of air flowing over it.

The convective heat transfer coefficient, denoted by 'h', quantifies the efficiency of the heat transfer process; the higher the value, the more efficient the heat transfer. The equation for convective heat transfer, q = hAΔT, where A is the area and ΔT is the temperature difference, is used to determine whether the cooling provided by airflow is sufficient to maintain a safe outer surface temperature. In engineering, effective convection cooling design is crucial for everything from electronic devices to architectural engineering, ensuring systems operate within safe thermal limits.

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Most popular questions from this chapter

A 10 -cm-diameter, 30-cm-high cylindrical bottle contains cold water at \(3^{\circ} \mathrm{C}\). The bottle is placed in windy air at \(27^{\circ} \mathrm{C}\). The water temperature is measured to be \(11^{\circ} \mathrm{C}\) after \(45 \mathrm{~min}\) of cooling. Disregarding radiation effects and heat transfer from the top and bottom surfaces, estimate the average wind velocity.

In an experiment, the local heat transfer over a flat plate were correlated in the form of local Nusselt number as expressed by the following correlation $$ \mathrm{Nu}_{x}=0.035 \mathrm{Re}_{x}^{0.8} \operatorname{Pr}^{1 / 3} $$ Determine the ratio of the average convection heat transfer coefficient \((h)\) over the entire plate length to the local convection heat transfer coefficient \(\left(h_{x}\right)\) at \(x=L\).

Exposure to high concentration of gaseous ammonia can cause lung damage. To prevent gaseous ammonia from leaking out, ammonia is transported in its liquid state through a pipe \(\left(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{i, \text { pipe }}=2.5 \mathrm{~cm}\right.\), \(D_{o, \text { pipe }}=4 \mathrm{~cm}\), and \(\left.L=10 \mathrm{~m}\right)\). Since liquid ammonia has a normal boiling point of \(-33.3^{\circ} \mathrm{C}\), the pipe needs to be properly insulated to prevent the surrounding heat from causing the ammonia to boil. The pipe is situated in a laboratory, where air at \(20^{\circ} \mathrm{C}\) is blowing across it with a velocity of \(7 \mathrm{~m} / \mathrm{s}\). The convection heat transfer coefficient of the liquid ammonia is \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Calculate the minimum insulation thickness for the pipe using a material with \(k=0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) to keep the liquid ammonia flowing at an average temperature of \(-35^{\circ} \mathrm{C}\), while maintaining the insulated pipe outer surface temperature at \(10^{\circ} \mathrm{C}\).

To defrost ice accumulated on the outer surface of an automobile windshield, warm air is blown over the inner surface of the windshield. Consider an automobile windshield \(\left(k_{w}=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) with an overall height of \(0.5 \mathrm{~m}\) and thickness of \(5 \mathrm{~mm}\). The outside air ( \(1 \mathrm{~atm}\) ) ambient temperature is \(-20^{\circ} \mathrm{C}\) and the average airflow velocity over the outer windshield surface is \(80 \mathrm{~km} / \mathrm{h}\), while the ambient temperature inside the automobile is \(25^{\circ} \mathrm{C}\). Determine the value of the convection heat transfer coefficient, for the warm air blowing over the inner surface of the windshield, necessary to cause the accumulated ice to begin melting. Assume the windshield surface can be treated as a flat plate surface.

Consider a refrigeration truck traveling at \(55 \mathrm{mph}\) at a location where the air temperature is \(80^{\circ} \mathrm{F}\). The refrigerated compartment of the truck can be considered to be a 9-ft-wide, 8-ft-high, and 20 -ft-long rectangular box. The refrigeration system of the truck can provide 3 tons of refrigeration (i.e., it can remove heat at a rate of \(600 \mathrm{Btu} / \mathrm{min}\) ). The outer surface of the truck is coated with a low-emissivity material, and thus radiation heat transfer is very small. Determine the average temperature of the outer surface of the refrigeration compartment of the truck if the refrigeration system is observed to be operating at half the capacity. Assume the air flow over the entire outer surface to be turbulent and the heat transfer coefficient at the front and rear surfaces to be equal to that on side surfaces. For air properties evaluations assume a film temperature of \(80^{\circ} \mathrm{F}\). Is this a good assumption?

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