A stainless steel plate is connected to a copper plate by...

A stainless steel plate is connected to a copper plate by long ASTM B98 copper-silicon bolts of $9.5 \mathrm{~mm}$ in diameter. The portion of the bolts exposed to convection heat transfer with air is $5 \mathrm{~cm}$ long. The air temperature for convection is at $20^{\circ} \mathrm{C}$ with a convection heat transfer coefficient of $5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. The thermal conductivity of the bolt is known to be $36 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. The copper plate has a uniform temperature of $70^{\circ} \mathrm{C}$. If each bolt is estimated to have a rate of heat loss to convection of $5 \mathrm{~W}$, determine the temperature $T_{b}$ at the upper surface of the stainless steel plate. According to the ASME Code for Process Piping (ASME B31.3-2014, Table A-2M), the maximum use temperature for the ASTM B 98 copper-silicon bolt is $149^{\circ} \mathrm{C}$. Does the use of the ASTM B 98 bolts in this condition comply with the ASME code?

Key Concepts

Extended Surface (Fin) Analysis

When a component like a bolt is exposed to fluid flow along a significant portion of its surface, its behavior can be analyzed similarly to a fin. Here, conduction along the bolt couples with distributed convective losses from its surface, leading to a temperature profile along its length. This fin analysis is crucial for accurately predicting the temperature at specific points along an extended surface that is subject to convective cooling.

Thermal Resistance Networks

This idea involves modeling different modes of heat transfer—like conduction and convection—as resistances in series or parallel. Each mode contributes a thermal resistance that causes a portion of the overall temperature drop. By summing these resistances, it is possible to perform an energy balance and solve for unknown temperatures in a system where heat flows sequentially through different materials or processes.

Design Code Compliance

In many engineering applications, components must operate within prescribed temperature limits set by design codes or standards. Assessing code compliance involves comparing the maximum temperature reached in the component during operation with the allowable maximum use temperature specified in the code. Ensuring that actual temperatures are well below the code limitations is a crucial step in demonstrating that a material or design is safe for its intended service conditions.

Conduction Heat Transfer

This concept describes the transfer of thermal energy through a solid material as a result of a temperature gradient. Using the material’s thermal conductivity, the heat transfer rate can be determined based on the cross?sectional area and the length over which the temperature drop occurs. In engineering applications, conduction analysis is employed to estimate how much temperature difference occurs along a component that is used as a thermal bridge between two environments at different temperatures.

Convection Heat Transfer

Convection refers to the process by which a fluid (such as air) carries heat away from a surface. It is quantified by the convection heat transfer coefficient and the surface area exposed to the fluid. The temperature difference between the surface and the ambient fluid drives the convective heat loss, making this concept essential when analyzing components that are cooled by their surrounding environment.

Transcript

00:01 Hi for this question.

00:02 We're given two metal plates are sold out together.

00:04 We're given the data for the two and this is this is formal.

00:09 It's what to use so the other q is the change is the temperature is heat transferred rather divided by the time per time and kt is the terminal conductivity a is the area delta t that the big t is the temperature difference which is the trend between the temperatures on the two faces.

00:31 L is the so we can assume that the heat flowing the two through the two plates are the same to delta q delta t will be the same for both plates so to plate one to equating those two so we have k to c 1a1 t and delta t will be the t2 minus t1 but in this case we have t as the temperature in the in the middle so we're going to have t minus temperature at the junction rather we're going to have t1 minus t divided by l so this is for the um first part first plate and for the second plate we have k t2 a2 up for this side we wouldn't have t minus t 2 divided by l is l1 and and the area will be the same for both so we have a the same a on both sides and we also have the same l the same length because l1 is equal to l2 we have a a it's called this l l so this is cancel out a cancels l cancels we have k t1 t1 minus t also k t2 t minus t2 and what we want to find first of all is t which is the temperature i junction so k t one t one minus k t one t is equal to k t two t minus k t two two two two so collecting like damage with t we have k t two t two t plus k t one t you go to k t one t t1 plus k t2 t 2 and so this t will be equal to gay t 1 t 1 plus k t 2 divided by k t2 plus k t 1 for this is equal to substituting we have 48 times 100 100 plus 68 .2 .1 here times zero divided by 468 .2 plus 48 .1...

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