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Chapter 3: Problem 48

A thin electrical heater is wrapped around the outer surface of a long cylindrical tube whose inner surface is maintained at a temperature of \(5^{\circ} \mathrm{C}\). The tube wall has inner and outer radii of 25 and \(75 \mathrm{~mm}\), respectively, and a thermal conductivity of \(10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The thermal contact resistance between the heater and the outer surface of the tube (per unit length of the tube) is \(R_{t, c}^{\prime}=\) \(0.01 \mathrm{~m} \cdot \mathrm{K} / \mathrm{W}\). The outer surface of the heater is exposed to a fluid with \(T_{\infty}=-10^{\circ} \mathrm{C}\) and a convection coefficient of \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the heater power per unit length of tube required to maintain the heater at \(T_{o}=25^{\circ} \mathrm{C} .\)

Short Answer

Expert verified
The heater power per unit length required to maintain the heater at \(T_o = 25^\circ\mathrm{C}\) is approximately \(892.86\,\mathrm{W}/\mathrm{m}\).

Step by step solution

01

Identify the given parameters

In this problem, we already have the following given: Inner radius of the tube, \(r_i = 25\,\mathrm{mm} = 0.025\,\mathrm{m}\) Outer radius of the tube, \(r_o = 75 \,\mathrm{mm}= 0.075\,\mathrm{m}\) Thermal conductivity of the tube, \(k = 10\, \mathrm{W}/\mathrm{m}\cdot\mathrm{K}\) Inner surface temperature, \(T_i = 5^\circ\mathrm{C}\) Thermal contact resistance per unit length, \(R_{t,c}' = 0.01 \, \mathrm{m}\cdot\mathrm{K}/\mathrm{W}\) Fluid temperature outside the heater, \(T_\infty = -10^\circ\mathrm{C}\) Convection coefficient, \(h = 100 \, \mathrm{W}/\mathrm{m}^{2}\cdot\mathrm{K}\) Target temperature of the heater, \(T_o = 25^\circ\mathrm{C}\) Now, let's find the heater power per unit length of the tube.
02

Calculate the conduction resistance through the tube

The first step is to compute the conduction resistance through the tube wall. The general equation for conduction resistance in a cylindrical system is: \[R_\text{cond} = \frac{\ln(r_o/r_i)}{2\pi k L}\] Where \(L\) is the length of the tube. In this case, we need to find the resistance per unit length, so we can divide by \(L\): \[R_\text{cond}' = \frac{R_\text{cond}}{L} = \frac{\ln(r_o/r_i)}{2\pi k}\] Now, we can plug in the given values: \[R_\text{cond}' = \frac{\ln(0.075/0.025)}{2\pi(10)} = 0.008\,\mathrm{m}\cdot\mathrm{K}/\mathrm{W}\]
03

Calculate the convection resistance between the heater and the fluid

Next, we need to determine the convection resistance between the heater and the fluid. The general equation for convection resistance in a cylindrical system is: \[R_\text{conv} = \frac{1}{h(2\pi r_o L)}\] Again, we need to find the resistance per unit length, so we can divide by \(L\): \[R_\text{conv}' = \frac{R_\text{conv}}{L} = \frac{1}{h(2\pi r_o)}\] Now, we can plug in the given values: \[R_\text{conv}' = \frac{1}{100(2\pi(0.075))} = 0.0212\,\mathrm{m}\cdot\mathrm{K}/\mathrm{W}\]
04

Calculate the total thermal resistance

To find the total thermal resistance, we add the conduction resistance, convection resistance, and contact resistance per unit length: \[R_\text{total}' = R_\text{cond}' + R_{t,c}' + R_\text{conv}'\] \[R_\text{total}' = 0.008 + 0.01 + 0.0212 = 0.0392\,\mathrm{m}\cdot\mathrm{K}/\mathrm{W}\]
05

Calculate the heater power per unit length required

Now that we have the total thermal resistance, we can use it to calculate the heater power per unit length required to maintain the heater at \(T_o = 25^\circ\mathrm{C}\). We use the formula: \[\dot{Q}' = \frac{T_o - T_\infty}{R_\text{total}'}\] Now, we can plug in the given values: \[\dot{Q}' = \frac{25 - (-10)}{0.0392} = 892.86\,\mathrm{W}/\mathrm{m}\] Therefore, the heater power per unit length required to maintain the heater at \(T_o = 25^\circ\mathrm{C}\) is approximately \(892.86\,\mathrm{W}/\mathrm{m}\).

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Key Concepts

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

Thermal Resistance
Thermal resistance is a crucial concept in understanding heat transfer between various components. It's akin to electrical resistance but applies to the flow of heat rather than electricity. The thermal resistance of a material indicates how effectively it resists the flow of heat. The less resistance, the more heat can pass through it. In our exercise, we calculate thermal resistance to understand how heat moves through the cylindrical tube.
A cylindrical tube's thermal resistance depends on several factors including the material's thermal conductivity, its geometry, and thickness. For cylindrical systems, the conduction thermal resistance can be expressed as:
  • \[ R_{cond} = \frac{\ln(r_o/r_i)}{2\pi k L} \]
Where:
  • \(r_o\) and \(r_i\) are the outer and inner radii respectively.
  • \(k\) is the thermal conductivity.
  • \(L\) represents the length.

This formula helps us evaluate how well the tube will conduct heat from the hotter inside to the cooler outside.
Convection
Convection is a heat transfer mechanism where heat moves through a fluid such as air or liquid. This process occurs when a fluid moves across the surface of a solid that's either heated or cooled. In our exercise, the cylinder鈥檚 outer surface conducts heat to the surrounding fluid through convection.
We measure how effectively convection transfers this heat using the convection coefficient, \(h\). This coefficient varies based on the fluid properties and flow conditions. The convection resistance in cylindrical systems is given by:
  • \[ R_{conv} = \frac{1}{h(2\pi r_o L)} \]
Where:
  • \(h\) is the convection coefficient.
  • \(r_o\) is the outer radius of the cylinder.
  • \(L\) is the length.

The convection process demonstrates how heat energy continues to move from the heater's outer surface into the fluid, significantly affecting the overall energy balance.
Thermal Conductivity
Thermal conductivity, symbolized by \(k\), is a material property that describes its ability to conduct heat. A high thermal conductivity means the material allows heat to pass through it quickly, whereas a low value indicates resistance to heat flow. In our problem, the tube's material has a thermal conductivity of 10 W/m路K.
This property plays a critical role in determining the conduction thermal resistance, which depends directly on \(k\). The better a material conducts heat, the lower its conduction resistance.
  • Materials with high thermal conductivity include metals like copper and aluminum.
  • Materials like rubber or insulation tend to have low thermal conductivity.

Understanding thermal conductivity helps in selecting the best materials for insulating or conducting heat in various applications. In engineering problems, it allows for precise calculations of how heat will distribute across different sections of a system.

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Most popular questions from this chapter

The wind chill, which is experienced on a cold, windy day, is related to increased heat transfer from exposed human skin to the surrounding atmosphere. Consider a layer of fatty tissue that is \(3 \mathrm{~mm}\) thick and whose interior surface is maintained at a temperature of \(36^{\circ} \mathrm{C}\). On a calm day the convection heat transfer coefficient at the outer surface is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), but with \(30 \mathrm{~km} / \mathrm{h}\) winds it reaches \(65 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). In both cases the ambient air temperature is \(-15^{\circ} \mathrm{C}\). (a) What is the ratio of the heat loss per unit area from the skin for the calm day to that for the windy day? (b) What will be the skin outer surface temperature for the calm day? For the windy day? (c) What temperature would the air have to assume on the calm day to produce the same heat loss occurring with the air temperature at \(-15^{\circ} \mathrm{C}\) on the windy day?

A wire of diameter \(D=2 \mathrm{~mm}\) and uniform temperature \(T\) has an electrical resistance of \(0.01 \Omega / \mathrm{m}\) and a current flow of \(20 \mathrm{~A}\). (a) What is the rate at which heat is dissipated per unit length of wire? What is the heat dissipation per unit volume within the wire? (b) If the wire is not insulated and is in ambient air and large surroundings for which \(T_{\infty}=T_{\text {sur }}=20^{\circ} \mathrm{C}\), what is the temperature \(T\) of the wire? The wire has an emissivity of \(0.3\), and the coefficient associated with heat transfer by natural convection may be approximated by an expression of the form, \(h=C\left[\left(T-T_{\infty}\right) / D\right]^{1 / 4}, \quad\) where \(C=1.25\) \(\mathrm{W} / \mathrm{m}^{7 / 4} \cdot \mathrm{K}^{5 / 4}\). (c) If the wire is coated with plastic insulation of 2-mm thickness and a thermal conductivity of \(0.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), what are the inner and outer surface temperatures of the insulation? The insulation has an emissivity of \(0.9\), and the convection coefficient is given by the expression of part (b). Explore the effect of the insulation thickness on the surface temperatures.

A technique for measuring convection heat transfer coefficients involves bonding one surface of a thin metallic foil to an insulating material and exposing the other surface to the fluid flow conditions of interest. By passing an electric current through the foil, heat is dissipated uniformly within the foil and the corresponding flux, \(P_{\text {elec }}^{\prime \prime}\), may be inferred from related voltage and current measurements. If the insulation thickness \(L\) and thermal conductivity \(k\) are known and the fluid, foil, and insulation temperatures \(\left(T_{\infty}, T_{s}, T_{b}\right)\) are measured, the convection coefficient may be determined. Consider conditions for which \(T_{\infty}=T_{b}=25^{\circ} \mathrm{C}, P_{\text {elec }}^{\prime \prime}=2000\) \(\mathrm{W} / \mathrm{m}^{2}, L=10 \mathrm{~mm}\), and \(k=0.040 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) With water flow over the surface, the foil temperature measurement yields \(T_{s}=27^{\circ} \mathrm{C}\). Determine the convection coefficient. What error would be incurred by assuming all of the dissipated power to be transferred to the water by convection? (b) If, instead, air flows over the surface and the temperature measurement yields \(T_{s}=125^{\circ} \mathrm{C}\), what is the convection coefficient? The foil has an emissivity of \(0.15\) and is exposed to large surroundings at \(25^{\circ} \mathrm{C}\). What error would be incurred by assuming all of the dissipated power to be transferred to the air by convection? (c) Typically, heat flux gages are operated at a fixed temperature \(\left(T_{s}\right)\), in which case the power dissipation provides a direct measure of the convection coefficient. For \(T_{s}=27^{\circ} \mathrm{C}\), plot \(P_{\text {elec }}^{\prime \prime}\) as a function of \(h_{o}\) for \(10 \leq h_{o} \leq 1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What effect does \(h_{o}\) have on the error associated with neglecting conduction through the insulation?

Finned passages are frequently formed between parallel plates to enhance convection heat transfer in compact heat exchanger cores. An important application is in electronic equipment cooling, where one or more air-cooled stacks are placed between heat-dissipating electrical components. Consider a single stack of rectangular fins of length \(L\) and thickness \(t\), with convection conditions corresponding to \(h\) and \(T_{\infty}\). (a) Obtain expressions for the fin heat transfer rates, \(q_{f, o}\) and \(q_{f, L}\), in terms of the base temperatures, \(T_{o}\) and \(T_{L}\). (b) In a specific application, a stack that is \(200 \mathrm{~mm}\) wide and \(100 \mathrm{~mm}\) deep contains 50 fins, each of length \(L=12 \mathrm{~mm}\). The entire stack is made from aluminum, which is everywhere \(1.0 \mathrm{~mm}\) thick. If temperature limitations associated with electrical components joined to opposite plates dictate maximum allowable plate temperatures of \(T_{o}=400 \mathrm{~K}\) and \(T_{L}=350 \mathrm{~K}\), what are the corresponding maximum power dissipations if \(h=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=300 \mathrm{~K} ?\)

Rows of the thermoelectric modules of Example \(3.13\) are attached to the flat absorber plate of Problem 3.108. The rows of modules are separated by \(L_{\text {sep }}=0.5 \mathrm{~m}\) and the backs of the modules are cooled by water at a temperature of \(T_{w}=40^{\circ} \mathrm{C}\), with \(h=45 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the electric power produced by one row of thermoelectric modules connected in series electrically with a load resistance of \(60 \Omega\). Calculate the heat transfer rate to the flowing water. Assume rows of 20 immediately adjacent modules, with the lengths of both the module rows and water tubing to be \(L_{\text {row }}=20 W\) where \(W=54 \mathrm{~mm}\) is the module dimension taken from Example 3.13. Neglect thermal contact resistances and the temperature drop across the tube wall, and assume that the high thermal conductivity tube wall creates a uniform temperature around the tube perimeter. Because of the thermal resistance provided by the thermoelectric modules, it is no longer appropriate to assume that the temperature of the absorber plate directly above a tube is equal to that of the water.
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