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A thin-wall, spherical stainless steel vessel has an outside diameter of \(40 \mathrm{~cm}\) and contains biochemical reactants maintained at \(160^{\circ} \mathrm{C}\). The tank is located in a laboratory where the air is maintained at \(20^{\circ} \mathrm{C}\). A layer of insulation is to be added to prevent workers from being burned by accidental contact with the vessel. The insulation chosen has a thermal conductivity of \(0.1 \mathrm{~W} / \mathrm{m} \mathrm{K}\). How thick should the layer be if the threshold for a skin burn on a nonmetallic surface can be taken as \(55^{\circ} \mathrm{C}\) ? Also calculate the heat loss. The combined convective and radiative heat transfer coefficients on the outside of the insulation is estimated to be \(9 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\).

Step by step solution

01

Identify the given parameters and desired outcome

You are given the outside diameter of the sphere, the temperatures of the vessel and the environment, and the thermal conductivities. You need to find the thickness of the insulation required and the heat loss.

02

Convert given diameter to radius and calculate sphere dimensions

The outside diameter of the vessel is 0.4 m, thus the radius is \( r_1 = \frac{0.4}{2} = 0.2 \) m. If insulation of thickness \( t \) is added, the new outer radius \( r_2 = 0.2 + t \).

03

Set up the heat transfer equation

For steady-state heat transfer through a sphere, the heat transfer equation is \( Q = \frac{T_i - T_o}{\frac{1}{hA_2} + \frac{(r_2 - r_1)}{kA_m}} \), where \( A_2 = 4\pi r_2^2 \) and \( k \) is the thermal conductivity of the insulation.

04

Determine the acceptable surface temperature

The outer surface temperature for safety is specified as 55°C. Thus, \( T_o = 55°C \) is needed for the calculation.

05

Calculate the heat transfer from the inner surface

The area for heat transfer is \( A_{m} = 4\pi r_{m}^2 \), where \( r_m = \frac{r_2 + r_1}{2} \). The thermal resistance through the insulation layer \( R_k = \frac{(r_2 - r_1)}{kA_m} \).

06

Use heat transfer equation to solve for thickness

Rearrange the heat transfer equation to express \( r_2 \), substitute the known temperatures, conductivities, and solve for \( t = r_2 - r_1 \), with \( h = 9 \) W/m²K and\( A_2 = 4\pi r_2^2 \).

07

Calculate the heat loss through the insulation

Once the thickness \( t \) is determined, use the heat transfer formula \( Q = \frac{T_i - T_o}{R_k + \frac{1}{hA_2}} \) to calculate the heat loss through the insulation.

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Key Concepts

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

Thermal Conductivity

Thermal conductivity is a material property that measures a material's ability to conduct heat. This is an important factor when considering insulation because it affects how well the insulation can limit heat transfer.

The thermal conductivity, represented by the symbol \( k \), is measured in watts per meter Kelvin (W/mK). In our exercise, the insulation used around the spherical vessel has a thermal conductivity of 0.1 W/mK. This indicates that the material conducts a small amount of heat, making it a suitable choice for insulation.

• Materials with low thermal conductivity are good insulators, as they resist the flow of heat.
• Conversely, materials with high thermal conductivity allow heat to pass through easily and are therefore better conductors rather than insulators.

Effective insulation must possess a low thermal conductivity to minimize heat transfer, keeping the surface temperature safe and preventing heat loss from the contents of the vessel.

Insulation Thickness

Insulation thickness refers to the depth or width of the insulating material that encases an object. The thicker the insulation, generally, the more effective it is at reducing heat transfer.

In the exercise, we are tasked with determining how much insulating material is required to ensure that the vessel's surface does not exceed 55°C, to avoid burns upon contact. The correct insulation thickness, \( t \), can be found using the heat transfer equation to ensure the outer temperature remains safe and heat loss is minimized.

• The thickness affects the thermal resistance, which is the measure of the ability to resist heat flow.
• With increased thickness, the resistance to heat flow increases, reducing the rate of heat transfer through the insulation.
• An optimal thickness should not only keep the surface temperature within safe limits but also be cost-effective and physically feasible.

Understanding and calculating the right thickness protects users and conserves energy.

Spherical Geometry

Spherical geometry pertains to the shape of the object being insulated, which in this case is a sphere. Spheres have specific properties that affect heat transfer calculations.

The geometry of a spherical vessel implies that both the inner and outer surfaces are curved, which influences how surface area and volume are calculated. The given problem involves a vessel with a known diameter, and hence radius, making it critical to calculate surface areas and volumes appropriately.

• The surface area of a sphere is given by the formula \( A = 4\pi r^2 \), where \( r \) is the radius.
• The heat transfer equations involve these surface area calculations to determine the rates of heat conduction or transfer.
• This understanding of geometry helps to establish the correct thermal dynamics of the system.

Calculating with the correct understanding of spherical geometry ensures accurate results in insulation discussions.

Convective and Radiative Heat Transfer Coefficients

The convective and radiative heat transfer coefficients are factors that represent how effectively heat is transferred from a surface to the surrounding environment via convection and radiation.

These coefficients, often denoted as \( h \), measure how well heat is carried away or absorbed. In our exercise, the combined convection and radiation effect is estimated at 9 W/m²K.

• Convection refers to the heat transfer due to fluid motion, typically air or liquid, moving past the object and removing or adding heat.
• Radiation is the transfer of heat through electromagnetic waves, which can occur in a vacuum.
• The total heat transfer coefficient accounts for both these mechanisms combined, providing a holistic view of heat transfer from the vessel surface to the ambient environment.

Understanding these coefficients allows us to estimate how heat is lost or gained from the system, essential in designing effective insulation.