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Chapter 3: Problem 57

A composite cylindrical wall is composed of two materials of thermal conductivity \(k_{\mathrm{A}}\) and \(k_{\mathrm{B}}\), which are separated by a very thin, electric resistance heater for which interfacial contact resistances are negligible. Liquid pumped through the tube is at a temperature \(T_{\infty, i}\) and provides a convection coefficient \(h_{i}\) at the inner surface of the composite. The outer surface is exposed to ambient air, which is at \(T_{\infty, o}\) and provides a convection coefficient of \(h_{o^{*}}\) Under steady-state conditions, a uniform heat flux of \(q_{h}^{n}\) is dissipated by the heater. (a) Sketch the equivalent thermal circuit of the system and express all resistances in terms of relevant variables. (b) Obtain an expression that may be used to determine the heater temperature, \(T_{h+}\). (c) Obtain an expression for the ratio of heat flows to the outer and inner fluids, \(q_{o}^{\prime} / q_{i}^{\prime}\). How might the variables of the problem be adjusted to minimize this ratio?

Short Answer

Expert verified
The heater temperature expression, given the equivalent thermal circuit, can be found as: \(T_{h+}= \frac{q_{h}^n (R_{k_A}+R_{k_B}) + T_{\infty_o } R_{c_{o^*}}} {R_{c_{o^*}}} \) The ratio of heat flows to the outer and inner fluids is given by: \(\frac{q_{o'}}{q_{i'}} = \frac{(T_{h+}-T_{\infty_o})/R_{c_{o^*}}}{(T_{\infty_i}-T_{h-})/R_{c_i}}\) To minimize this ratio, increase the outer surface convection coefficient \(h_{o^*}\) and decrease the inner surface convection coefficient \(h_i\). Adjusting the thickness and thermal conductivity of the composite materials A and B can also help minimize the ratio.

Step by step solution

01

Understanding the given information

Before diving into the solution, let's briefly understand the given information, which is crucial for the solution. - A composite cylindrical wall is composed of two materials A and B. - Material properties include thermal conductivity (\(k_A\), \(k_B\)) and convection coefficients (\(h_i\), \(h_{o^*}\)). - The thin electric resistance heater is sandwiched between both materials. - The liquid flowing through the tube causes convection on the inner surface, and ambient air causes convection on the outer surface. - All heat losses are uniform and steady-state.
02

Sketch the equivalent thermal circuit

Now, let's create the equivalent thermal circuit for the given information. The thermal circuit represents the temperature and resistance relations for the entire system. The circuit includes the following resistances: 1. Convection resistance (\(R_{c_i}\)) between the inner fluid and material A. 2. Conduction resistance (\(R_{k_A}\)) in material A. 3. Conduction resistance (\(R_{k_B}\)) in material B. 4. Convection resistance (\(R_{c_{o^*}}\)) between material B and the outer fluid. Sketch the above resistances connected in series along with the temperature nodes for \(T_{\infty_i}\), T_h-, T_h+, \(T_{\infty_o}\), and the respective heat fluxes \(q_{i'}\), \(q_{h}^n\), \(q_{o'}\).
03

Express resistances in terms of relevant variables (a)

Use the equations for the convection and conduction resistances in cylindrical coordinates to express each resistance in terms of the given variables. Recall that the convection resistance is given by \(R_{c} = \frac{1}{Ah}\) and the conduction resistance is given by \(R_{k} = \frac{L}{Ak}\). 1. \(R_{c_i} = \frac{1}{2\pi r_i L h_i}\) 2. \(R_{k_A} = \frac{\ln(\frac{r_{A_o}}{r_i})}{2\pi L k_A}\) 3. \(R_{k_B} = \frac{\ln(\frac{r_{o}}{r_{A_o}})}{2\pi L k_B}\) 4. \(R_{c_{o^*}} = \frac{1}{2\pi r_o L h_{o^*}}\)
04

Find the expression for heater temperature T_h+ (b)

Write the heat transfer equations for the entire thermal circuit, starting from the inner fluid to the outer fluid. Assuming energy conservation and that all heat is transferred through the resistance elements, \(q_{i'} = (T_{\infty_i}-T_{h-})/R_{c_i}\) \(q_{h}^n = (T_{h-}-T_{h+})/(R_{k_A}+R_{k_B})\) \(q_{o'} = (T_{h+}-T_{\infty_o})/R_{c_{o^*}}\) Now, solve these equations to get an expression for the heater temperature \(T_{h+}\). \(T_{h+}= \frac{q_{h}^n (R_{k_A}+R_{k_B}) + T_{\infty_o } R_{c_{o^*}}} {R_{c_{o^*}}} \)
05

Obtain the expression for the ratio of heat flows (c)

We need to find the expression for the ratio of heat flows to the outer and inner fluids, defined as \(q_{o'} / q_{i'}\). Using the heat transfer equations obtained in Step 4, we get: \(\frac{q_{o'}}{q_{i'}} = \frac{(T_{h+}-T_{\infty_o})/R_{c_{o^*}}}{(T_{\infty_i}-T_{h-})/R_{c_i}}\) To minimize this ratio, manipulate the variables in the problem so that the resulting ratio is as small as possible. To minimize the ratio, increase the outer surface convection coefficient \(h_{o^*}\) and decrease the inner surface convection coefficient \(h_i\). Adjusting the thickness and thermal conductivity of the composite materials A and B can also help minimize the ratio.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity, denoted by the symbol \(k\), is a measure of a material's ability to conduct heat. It quantifies the rate at which heat can pass through a material when there is a temperature difference across that material. Materials with high thermal conductivity, such as metals, can transfer heat quickly, whereas insulating materials, such as wood or foam, have low thermal conductivity and therefore transfer heat more slowly.

To illustrate, if we consider a cylindrical wall made from two different materials, each with their own thermal conductivity values \(k_{\mathrm{A}}\) and \(k_{\mathrm{B}}\), the heat will flow from one side to the other at rates determined by these values. In the exercise, the concept of thermal conductivity is crucial for determining the conduction resistance of each material segment within the thermal circuit, and thus impacts the estimation of how much heat is dissipated by the heater.
When solving for heat transfer problems, it's paramount to accurately assess these conductivities so that the thermal behavior of the system can be properly understood and controlled.
Convection Coefficient
The convection coefficient, symbolized by \(h\), is a metric that describes the effectiveness of convective heat transfer between a surface and a fluid flowing over it. A high convection coefficient means that the fluid is very effective at removing heat from the surface, like the way a fan boosts air movement, thereby increasing heat transfer from your skin. Conversely, a low convection coefficient indicates that the fluid isn't as efficient in transferring heat.

For instance, in the given exercise, the inner surface of the composite cylinder has a convection coefficient \(h_i\), while the outer surface has a different convection coefficient \(h_{o^*}\). These coefficients are central to computing the convection resistance at both the inner and outer surfaces of the cylinder. The greater the convection coefficient, the lower the thermal resistance to heat flow, facilitating more efficient heat transfer from the heater to the respective fluids.
Understanding the convection coefficient is essential for optimizing heat transfer processes, and in practical settings, it can be influenced by the type of fluid, fluid velocity, and surface characteristics.
Thermal Resistance
Thermal resistance is analogous to electrical resistance, but instead of impeding electrical current, it quantifies the resistance to heat flow within a material or between different materials. It is directly related to both the thermal conductivity of the material and the geometry of the thermal path. The higher the thermal resistance, the less heat passes through per unit of time for a given temperature difference.

In the context of the exercise involving a composite cylindrical wall, there are multiple thermal resistances in play: the conduction resistances \(R_{k_A}\) and \(R_{k_B}\) for materials A and B, and the convection resistances \(R_{c_i}\) and \(R_{c_{o^*}}\) at the inner and outer surfaces. Each resistance plays a role in the overall thermal circuit, which models how heat is dissipated from the heater. By breaking down the thermal circuit into individual resistances, one can solve for temperatures and heat fluxes at different points.
  • Conduction Resistance: It hinders the heat flow through a material and depends on the material's thickness, thermal conductivity, and area perpendicular to the heat flow.
  • Convection Resistance: It slows down the heat transfer between a solid surface and the fluid around it and is dependent on the convection coefficient and surface area.
Minimizing thermal resistance is a key goal in the design of thermal systems to ensure efficient heat transfer. This principle applies whether improving the cooling of electronic devices or optimizing insulation in buildings.

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Most popular questions from this chapter

One method that is used to grow nanowires (nanotubes with solid cores) is to initially deposit a small droplet of a liquid catalyst onto a flat surface. The surface and catalyst are heated and simultaneously exposed to a higher- temperature, low-pressure gas that contains a mixture of chemical species from which the nanowire is to be formed. The catalytic liquid slowly absorbs the species from the gas through its top surface and converts these to a solid material that is deposited onto the underlying liquid-solid interface, resulting in construction of the nanowire. The liquid catalyst remains suspended at the tip of the nanowire. Consider the growth of a 15 -nm-diameter silicon carbide nanowire onto a silicon carbide surface. The surface is maintained at a temperature of \(T_{s}=2400 \mathrm{~K}\), and the particular liquid catalyst that is used must be maintained in the range \(2400 \mathrm{~K} \leq T_{c} \leq 3000 \mathrm{~K}\) to perform its function. Determine the maximum length of a nanowire that may be grown for conditions characterized by \(h=10^{5} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=8000 \mathrm{~K}\). Assume properties of the nanowire are the same as for bulk silicon carbide.

An annular aluminum fin of rectangular profile is attached to a circular tube having an outside diameter of \(25 \mathrm{~mm}\) and a surface temperature of \(250^{\circ} \mathrm{C}\). The fin is \(1 \mathrm{~mm}\) thick and \(10 \mathrm{~mm}\) long, and the temperature and the convection coefficient associated with the adjoining fluid are \(25^{\circ} \mathrm{C}\) and \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. (a) What is the heat loss per fin? (b) If 200 such fins are spaced at \(5-\mathrm{mm}\) increments along the tube length, what is the heat loss per meter of tube length?

A plane wall of thickness \(2 L\) and thermal conductivity \(k\) experiences a uniform volumetric generation rate \(\dot{q}\). As shown in the sketch for Case 1 , the surface at \(x=-L\) is perfectly insulated, while the other surface is maintained at a uniform, constant temperature \(T_{o}\). For Case 2 , a very thin dielectric strip is inserted at the midpoint of the wall \((x=0)\) in order to electrically isolate the two sections, \(\mathrm{A}\) and \(\mathrm{B}\). The thermal resistance of the strip is \(R_{t}^{\prime \prime}=0.0005 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The parameters associated with the wall are \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L=\) \(20 \mathrm{~mm}, \dot{q}=5 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\), and \(T_{o}=50^{\circ} \mathrm{C}\). (a) Sketch the temperature distribution for Case 1 on \(T-x\) coordinates. Describe the key features of this distribution. Identify the location of the maximum temperature in the wall and calculate this temperature. (b) Sketch the temperature distribution for Case 2 on the same \(T-x\) coordinates. Describe the key features of this distribution. (c) What is the temperature difference between the two walls at \(x=0\) for Case 2 ? (d) What is the location of the maximum temperature in the composite wall of Case 2 ? Calculate this temperature.

A composite wall separates combustion gases at \(2600^{\circ} \mathrm{C}\) from a liquid coolant at \(100^{\circ} \mathrm{C}\), with gas- and liquid-side convection coefficients of 50 and 1000 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The wall is composed of a \(10-\mathrm{mm}\)-thick layer of beryllium oxide on the gas side and a 20 -mm-thick slab of stainless steel (AISI 304) on the liquid side. The contact resistance between the oxide and the steel is \(0.05 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). What is the heat loss per unit surface area of the composite? Sketch the temperature distribution from the gas to the liquid.

Consider a plane composite wall that is composed of two materials of thermal conductivities \(k_{\mathrm{A}}=0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{\mathrm{B}}=0.04 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thicknesses \(L_{\mathrm{A}}=10 \mathrm{~mm}\) and \(L_{\mathrm{B}}=20 \mathrm{~mm}\). The contact resistance at the interface between the two materials is known to be \(0.30 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). Material A adjoins a fluid at \(200^{\circ} \mathrm{C}\) for which \(h=10\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and material \(\mathrm{B}\) adjoins a fluid at \(40^{\circ} \mathrm{C}\) for which \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) What is the rate of heat transfer through a wall that is \(2 \mathrm{~m}\) high by \(2.5 \mathrm{~m}\) wide? (b) Sketch the temperature distribution.
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