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Chapter 3: Problem 28

Two stainless steel plates \(10 \mathrm{~mm}\) thick are subjected to a contact pressure of 1 bar under vacuum conditions for which there is an overall temperature drop of \(100^{\circ} \mathrm{C}\) across the plates. What is the heat flux through the plates? What is the temperature drop across the contact plane?

Short Answer

Expert verified
The heat flux through the stainless steel plates is \(q = -80000A\), where A is the contact area between the plates. The temperature drop across the contact plane between the two plates is 50 K.

Step by step solution

01

Begin by converting the given thickness and temperature drop to the appropriate units for use in the formula. Given thickness = 10 mm = 0.01 m Given temperature drop = \(100^{\circ} \)C = 100 K #Step 2: Substitute values into the formula#

Next, we can substitute the given values into the formula provided earlier, but note that we are given the total temperature drop across both plates. Since both plates are identical, we can assume that the temperature drop across one plate is half the overall temperature drop. Therefore, the temperature drop across one plate is: \(\frac{\Delta T_{total}}{2} = \frac{100}{2} = 50\) K Now, we can calculate the heat flux through one plate: \(q = -kA \frac{\Delta T}{\Delta x} = -16 \cdot A \cdot \frac{50}{0.01}\) #Step 3: Calculate heat flux#
02

Calculate the heat flux through the plates: \(q = -16 \cdot A \cdot 5000\) \(q = -80000A\) The heat flux depends on the contact area between the plates, which is not given in the problem. Thus, our final answer for the heat flux will be in terms of the contact area: \(q = -80000A\) #Step 4: Calculate the temperature drop across the contact plane#

To calculate the temperature drop across the contact plane, we need to find the heat transfer through the contact plane. Since the heat transfer through each plate equals the heat transfer through the contact plane, we can use the same formula: Temperature drop across contact plane = \(\Delta T_{contact}\) \(q = -kA \frac{\Delta T_{contact}}{\Delta x}\) Substituting the values: \(-80000A = -16 A \cdot \frac{\Delta T_{contact}}{0.01}\) Now, solve for \(\Delta T_{contact}\): \(\Delta T_{contact} = -\frac{-80000A \cdot 0.01}{16A}\) \(\Delta T_{contact} = 50\) K The temperature drop across the contact plane between the two plates is 50 K.

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Key Concepts

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

Conduction Heat Transfer
Understanding heat transfer through conduction is fundamental when dealing with thermal processes. Conduction is one of the primary modes of heat transfer and occurs within and between solid materials that are in direct contact. It is governed by Fourier's Law, which tells us that the rate of heat transfer through a material (also known as the heat flux, denoted as q) is proportional to the temperature gradient (螖罢/螖虫) within the material and the area (A) through which heat is flowing.

In the case of the stainless steel plates in our exercise, the conduction heat transfer is taking place from the hot side to the cold side of each plate. The rate at which heat is transferred depends on how well the material conducts heat, which brings us to the concept of thermal conductivity.
Temperature Drop
A temperature drop refers to the change in temperature, or temperature difference, across a material or space. In thermal physics problems, temperature drop is often the driving force for heat transfer. For instance, in the provided exercise, there is an overall temperature drop of 100 degrees Celsius (which is equivalent to 100 Kelvin, since temperature drops are the same in both units) across two stainless steel plates.

Considering Symmetry

When dealing with two identical materials sandwiched together, as in our case, we can infer that the temperature drop is divided evenly unless otherwise noted. This means that each plate experiences half of the total temperature drop, which is 50 K across each plate. This estimation is crucial for calculating the correct heat flux using the conduction heat transfer equation.
Thermal Conductivity
Thermal conductivity, represented by the symbol k, is a material property that measures a material's ability to conduct heat. It appears as a constant in Fourier's law and directly influences the heat transfer rate: the higher the thermal conductivity, the more efficiently a material can transfer heat.

In the textbook problem, we use the thermal conductivity of stainless steel (which for this exercise we can assume as 16 W/m路K) to calculate the heat flux. Combining this with the area of contact (A) and the temperature gradient (螖罢/螖虫), as per Fourier's Law, we determine how much heat is transferred per unit time through a given area. This intrinsic property allows us to make calculations and predictions about thermal processes, such as the amount of heat transferred across engineering materials like the stainless steel plates under scrutiny.

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Most popular questions from this chapter

The temperature of a flowing gas is to be measured with a thermocouple junction and wire stretched between two legs of a sting, a wind tunnel test fixture. The junction is formed by butt-welding two wires of different material, as shown in the schematic. For wires of diameter \(D=125 \mu \mathrm{m}\) and a convection coefficient of \(h=700 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the minimum separation distance between the two legs of the sting, \(L=L_{1}+L_{2}\), to ensure that the sting temperature does not influence the junction temperature and, in turn, invalidate the gas temperature measurement. Consider two different types of thermocouple junctions consisting of (i) copper and constantan wires and (ii) chromel and alumel wires. Evaluate the thermal conductivity of copper and constantan at \(T=300 \mathrm{~K}\). Use \(k_{\mathrm{Ch}}=19 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{\mathrm{Al}}=29 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) for the thermal conductivities of the chromel and alumel wires, respectively.

Annular aluminum fins of rectangular profile are attached to a circular tube having an outside diameter of \(50 \mathrm{~mm}\) and an outer surface temperature of \(200^{\circ} \mathrm{C}\). The fins are \(4 \mathrm{~mm}\) thick and \(15 \mathrm{~mm}\) long. The system is in ambient air at a temperature of \(20^{\circ} \mathrm{C}\), and the surface convection coefficient is \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) What are the fin efficiency and effectiveness? (b) If there are 125 such fins per meter of tube length, what is the rate of heat transfer per unit length of tube?

Consider the composite wall of Example 3.7. In the Comments section, temperature distributions in the wall were determined assuming negligible contact resistance between materials A and B. Compute and plot the temperature distributions if the thermal contact resistance is \(R_{t, c}^{\prime \prime}=10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\).

Consider a tube wall of inner and outer radii \(r_{i}\) and \(r_{o}\), whose temperatures are maintained at \(T_{i}\) and \(T_{o}\), respectively. The thermal conductivity of the cylinder is temperature dependent and may be represented by an expression of the form \(k=k_{o}(1+a T)\), where \(k_{o}\) and \(a\) are constants. Obtain an expression for the heat transfer per unit length of the tube. What is the thermal resistance of the tube wall?

Work Problem \(3.15\) assuming surfaces parallel to the \(x\)-direction are adiabatic.
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