Question

\(3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and both natural convection and radiation heat transfer occur at the outer shield surface. To keep the ammonia inside the tube in its liquid state, determine the surrounding temperature that would maintain temperatures of the outer tube and the shield at the specified values. 13-68 PtD A hot liquid is being transported inside a long tube with a diameter of \(25 \mathrm{~mm}\). The hot liquid causes the tube surface temperature to be \(150^{\circ} \mathrm{C}\). To prevent thermal burn hazards, the tube is enclosed with a concentric outer cylindrical cover of \(5 \mathrm{~cm}\) in diameter allowing a vacuumed gap in between the two surfaces. The concentric outer cover has an emissivity of \(0.6\) and the outer surface is exposed to natural convection with a heat transfer coefficient of \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and radiation heat transfer with the surrounding at a temperature of \(20^{\circ} \mathrm{C}\). Determine the necessary emissivity of the inside tube so that the outer cover temperature is below \(45^{\circ} \mathrm{C}\) to prevent thermal burns.

Short answer

Answer: The necessary emissivity of the inside tube needs to be calculated using the given information and derived equations, particularly the last equation: \[\varepsilon_{i} = \frac{Q_{conv} + Q_{rad}}{\sigma \cdot A_{i} \cdot (423.15^{4}\mathrm{~K}^4 - 318.15^{4}\mathrm{~K}^4)}\] By solving this equation, we will obtain the value for the required emissivity of the inside tube.

Step-by-step solution

Assign the given values to variables

Assign all the given values to their respective variables: - Inside tube diameter: \(D_{i} = 25\mathrm{~mm}\) - Inside tube temperature: \(T_{i} = 150^{\circ}\mathrm{C}\) - Outer cover diameter: \(D_{o} = 50\mathrm{~mm}\) - Desired outer cover temperature: \(T_{o} = 45^{\circ}\mathrm{C}\) - Heat transfer coefficient: \(h = 8\mathrm{~W/m^2\cdot K}\) - Emissivity of the outer cover: \(\varepsilon_{o} = 0.6\)

Define heat transfer for convection and radiation

Q_{conv} = h \cdot A \cdot (T_{o} - T_{\infty})\\ Q_{rad} = \sigma \cdot A \cdot \varepsilon_{o} \cdot (T_{o}^{4} - T_{\infty}^{4})\]

Convert temperatures to Kelvin

Convert given temperatures to Kelvin for calculations: - Inside tube temperature: \(T_{i} = 150^{\circ}\mathrm{C} + 273.15 = 423.15\mathrm{~K}\) - Desired outer cover temperature: \(T_{o} = 45^{\circ}\mathrm{C} + 273.15 = 318.15\mathrm{~K}\) - Surrounding temperature: \(T_{\infty} = 20^{\circ}\mathrm{C} + 273.15 = 293.15\mathrm{~K}\)

Calculate areas

Calculate the areas of inside tube and outer cover: - Inside tube area: \(A_{i} = \pi \cdot D_{i} \cdot L\) - Outer cover area: \(A_{o} = \pi \cdot D_{o} \cdot L\)

Calculate heat transfer for outer cover

Calculate the heat transfer for the outer cover using the heat transfer equations from step 2: \[Q_{conv} = h \cdot A_{o} \cdot (318.15\mathrm{~K} - 293.15\mathrm{~K})\] \[Q_{rad} = \sigma \cdot A_{o} \cdot 0.6 \cdot (318.15^4\mathrm{~K}^4 - 293.15^4\mathrm{~K}^4)\]

Calculate necessary emissivity of inside tube

To keep the outer cover temperature below \(45^{\circ} \mathrm{C}\), the heat transfer between the inside tube and the outer cover must be equal. Therefore, we can write the equation: \[Q_{i} = Q_{conv} + Q_{rad}\] Rearrange the equation to find the necessary emissivity (\(\varepsilon_{i}\)) of the inside tube: \[\varepsilon_{i} = \frac{Q_{conv} + Q_{rad}}{\sigma \cdot A_{i} \cdot (423.15^{4}\mathrm{~K}^4 - 318.15^{4}\mathrm{~K}^4)}\] Solve for \(\varepsilon_{i}\). The result will be the necessary emissivity of the inside tube to keep the outer cover temperature below 45°C, preventing thermal burns.

Key concepts

Convection Heat Transfer

Convection heat transfer is a process where heat is carried away by the movement of fluids, such as air or water. When discussing convection in the context of the exercise, we're referring to natural convection, where the fluid motion is caused by buoyancy forces that result from density variations due to temperature differences within the fluid. Specifically, the outer surface of the cylindrical cover in the exercise is subjected to natural convection, where the air surrounding it absorbs heat from the surface.

Applying the concept to the exercise, the heat transfer by natural convection from the outer cover to the surrounding air is calculated using the formula:
\[Q_{conv} = h \cdot A \cdot (T_{o} - T_{\infty})\]
where \(h\) is the heat transfer coefficient, \(A\) is the area of the outer cover, \(T_{o}\) is the outer cover temperature, and \(T_{\infty}\) is the surrounding air temperature. This formula helps predict how effective the cover will be at protecting someone from burns by dissipating heat.

Radiation Heat Transfer

Radiation heat transfer involves the emission of electromagnetic waves that carry energy away from an emitting surface. Unlike convection, this form of heat transfer does not require a medium; hence it can occur even in a vacuum, as is the case in the gap between the inner tube and the outer cover in our exercise.

In this scenario, the heat transfer by radiation is determined using the Stefan-Boltzmann law, expressed as:
\[Q_{rad} = \sigma \cdot A \cdot \varepsilon_{o} \cdot (T_{o}^{4} - T_{\infty}^{4})\]
Here, \(\sigma\) represents the Stefan-Boltzmann constant, \(\varepsilon_{o}\) is the emissivity of the outer cover which indicates its ability to emit thermal radiation, and \(A\) again is the surface area. The radiation heat transfer equation shows the relationship between the temperature of the surface and its surroundings and how it affects the heat loss by radiation.

Emissivity

Emissivity is a measure of a material's ability to emit energy as thermal radiation. It is represented by a value between 0 and 1, with 0 meaning the material does not emit any thermal radiation (perfect reflector), and 1 meaning the material is a perfect emitter. In our exercise, the outer cover has an emissivity of \(0.6\), indicating that it emits 60% of the energy it could if it were a black body (a perfect emitter of radiation at a given temperature). Typically, materials with lower emissivity appear shinier and reflect more heat radiation, while those with higher emissivity appear duller and are better at emitting heat radiation.

Adjusting the emissivity of the inner tube surface can alter the amount of heat radiation emitted, consequently changing the temperature of the outer cover. This is critical in designing systems where maintaining a certain temperature range is essential, such as in the case of the exercise, to ensuring that the outer cover's temperature remains below the threshold to prevent thermal burns.

Thermal Burn Prevention

Preventing thermal burns is paramount when designing heat transfer systems, especially in industries that require handling hot equipment. The exercise demonstrates a practical application where engineering analysis helps prevent burns. The goal in the exercise is to ensure that the temperature of the outer cover remains below \(45^\circ\mathrm{C}\), which is considered a safe touch temperature limit to prevent the likelihood of burns upon contact.

By computing the necessary emissivity of the inner tube, based on the heat transfer equations for convection and radiation, the design can be adjusted to ensure the outer cover remains at a temperature that does not cause harm. This involves a balance between material properties, such as emissivity, and environmental conditions, so that the system can effectively dissipate heat and protect users from burn injuries, demonstrating how thermodynamics and heat transfer principles are directly applied to safety in engineering design.