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Chapter 3: Problem 72

A plane wall of thickness \(0.1 \mathrm{~m}\) and thermal conductiv ity \(25 \mathrm{~W} / \mathrm{m}\), K having uniform volumetric heat generation of \(0.3 \mathrm{MW} / \mathrm{m}^{3}\) is insulated on one side, while the other side is exposed to a fluid at \(92^{\circ} \mathrm{C}\). The convection heat transfer coefficient between the wall and the fluid is \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the maximum temperature in the wall.

Short Answer

Expert verified
The maximum temperature in the wall is 182掳C.

Step by step solution

01

Identify Known Values

We start by identifying all known variables from the problem statement:- Thickness of the wall, \( L = 0.1 \, \text{m} \)- Thermal conductivity, \( k = 25 \, \text{W/m} \cdot \text{K} \)- Volumetric heat generation, \( q'' = 0.3 \, \text{MW/m}^3 = 300,000 \, \text{W/m}^3 \)- Fluid temperature, \( T_{\infty} = 92^{\circ} \, \text{C} \)- Convection heat transfer coefficient, \( h = 500 \, \text{W/m}^2 \cdot \text{K} \)
02

Use Heat Equation for Maximum Temperature

Apply the equation for maximum temperature in an insulated wall with uniform generation:\[T_{\text{max}} = T_s + \frac{q''L^2}{2k}\text{where, } T_s\text{ is the surface temperature of the wall exposed to fluid.}\]
03

Calculate Surface Temperature with Heat Balance

The balance between conduction and convection at the surface gives:\[q' = h(T_s - T_{\infty}) = q'L/k\]Solving for \(T_s\):\[T_s = T_{\infty} + \frac{q'L}{h}\]We first need to find the heat flux \(q'\):\[q' = q'' \times \frac{L}{2} = 300,000 \, \text{W/m}^3 \times 0.05 \, \text{m} = 15,000 \, \text{W/m}^2\]Calculate \(T_s:\)\[T_s = 92^{\circ} \, \text{C} + \frac{15,000 \, \text{W/m}^2}{500 \, \text{W/m}^2 \cdot \text{K}}\]\[T_s = 122^{\circ} \, \text{C}\]
04

Determine Maximum Temperature in Wall

Plug \(T_s\) back into the equation for \(T_{\text{max}}\):\[T_{\text{max}} = 122^{\circ} \, \text{C} + \frac{300,000 \, \text{W/m}^3 \times (0.1)^2}{2 \times 25 \, \text{W/m} \cdot \text{K}}\]\[T_{\text{max}} = 122^{\circ} \, \text{C} + 60^{\circ} \, \text{C} = 182^{\circ} \, \text{C}\]
05

Review and Summarize

The maximum temperature occurs within the wall due to internal heat generation, higher than the surface temperature exposed to the fluid.

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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-specific property that plays a crucial role in the transfer of heat through materials. It describes a material's ability to conduct heat. A high thermal conductivity means that the material conducts heat efficiently, while a low thermal conductivity means that the material is a good insulator.

In the context of our exercise, the plane wall with a thermal conductivity of 25 W/m路K can efficiently transfer heat generated inside it. The thermal conductivity value is integral to calculating the temperature within the wall. It determines how quickly heat can be conducted away from the point of generation to the surface of the wall.

To apply it mathematically, when calculating the maximum temperature in the wall with internal heat generation, we use the equation \[ T_{\text{max}} = T_s + \frac{q''L^2}{2k} \]where \( q'' \) is the volumetric heat generation, \( L \) is the thickness of the material, and \( k \) is the thermal conductivity. This relationship shows how thermal conductivity moderates the increase of temperature inside the material due to heat generation.
Volumetric Heat Generation
Volumetric heat generation represents the energy produced per unit volume within a material. It is crucial in situations where heat is generated internally, such as in electronic equipment or exothermic chemical processes.

In the example provided, the wall experiences a uniform volumetric heat generation of 0.3 MW/m鲁. This heat generation produces internal heat which must be managed to prevent excessive temperatures that could damage materials or components.

The formula involving volumetric heat generation is used to calculate the internal temperature rise in the wall:\[ T_{\text{max}} = T_s + \frac{q''L^2}{2k} \]where \( q'' \) indicates the volumetric heat generation, showing how this constant input of heat affects the maximum temperature when other factors like thermal conductivity and wall thickness are considered.
Convection Heat Transfer Coefficient
The convection heat transfer coefficient, often denoted by \( h \), measures the convective heat transfer between a surface and a fluid such as air or water. It is dependent on both the properties of the fluid and the condition of the surface, including factors like fluid velocity and surface roughness.

In this problem, the wall is insulated on one side, while the other is exposed to a fluid at 92掳C with a convection heat transfer coefficient of 500 W/m虏路K. This means that a moderate flow of heat can move from the wall's surface into the fluid. The convection heat transfer coefficient is critical in determining the wall's surface temperature in equilibrium conditions, which subsequently affects the maximum temperature within the wall.

To find the surface temperature, use the formula:\[ h(T_s - T_{\infty}) = q'L/k \]where \( T_{\infty} \) is the fluid temperature, \( q' \) is the heat flux, and \( L \) is wall thickness. This equation highlights how the heat transfer coefficient works in conjunction with surface temperature and heat flux to ensure adequate heat dissipation into the surrounding fluid.

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Most popular questions from this chapter

A high-temperature, gas-cocled nuclear reactor consists of a composite cylindrical wall for which a thorium fuel elerment \((k=57 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is encused in graphite \((k=3\) \(W / m \cdot K)\) and gaseous helium flows through an annular coolant channel. Consider conditions for which the helium temperature is \(T_{w}=600 \mathrm{~K}\) and the convection coefficien at the outer surface of the graphite is \(h=2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If thermal energy is uniformly gencrated in the fuel element at a rate \(q=10^{\text {n }} \mathrm{W} / \mathrm{m}^{3}\), what are the temperatures \(T_{1}\) and \(T_{2}\) at the inner and outer surfaces, respectively, of the fuel element? (b) Compute and plot the temperature distribution in the composite wall for selected values of \(\dot{q}\). What is the maximum allowable value of \(q\) ?

Unique characteristics of biologically active materials such as fruits, vegetables, and cther products require special care in handling. Following harvest and separation from producing plants, glucose is catabolized to produce carbon diexide, water vapor, and heat, with attendant internal energy generation. Consider a carton of apples, each of 80-mm diameter, which is ventilated with air at \(5^{\circ} \mathrm{C}\) and a velocity of \(0.5 \mathrm{~m} / \mathrm{s}\). The corresponding value od the heat transfer coefficient is \(7.5 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\). Within each apple thermal energy is uniformly generated at a total rate of \(4000 \mathrm{~J} / \mathrm{kg}\) - diry. The density and thermal conductivity of the apple are \(840 \mathrm{~kg} / \mathrm{m}^{3}\) and \(0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), respectively.(a) Determine the apple center and surface temperatures. (b) For the stacked arrangement of apples within the crate, the convection coefficient depends on the velocity as \(h=C_{1} v^{0.02 s}\), where \(C_{1}=10.1 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K} \cdot(\mathrm{m} / \mathrm{s})^{0.05}\). Compute and plot the center and surface temperatures as a function of the air velocity for \(0.1 \leq V \leq 1 \mathrm{~m} / \mathrm{s}\).

Superheated steam at \(575^{\circ} \mathrm{C}\) is routed from a boiler to the turbine of an electric power plant through steel tubes \((k=35 \mathrm{~W} / \mathrm{m}+\mathrm{K})\) of \(300 \mathrm{~mm}\) inner diameter and \(30 \mathrm{~mm}\) wall thickness. To reduce heat loss to the surroundings and to maintain a safe-to-touch outer surface temperature, a layer of calcium silicate insulation \((k=0.10 \mathrm{~W} / \mathrm{m}-\mathrm{K})\) is applied to the tubes, while degradation of the insulation is reduced by wrapping it in a thin sheet of aluminum having an emissivity of \(\varepsilon=0.20\). The air and wall temperatures of the power plant are \(27^{\circ} \mathrm{C}\). (a) Assuming that the inner surface temperature of a steel tube corresponds to that of the steam and the convection coefficient cutside the aluminum sheet is \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the minimum insulation thickness needed to insure that the teriperature of the aluminum does not exceed \(50^{\circ} \mathrm{C}\) ? What is the corresponding heat loss per meter of tube length? (b) Explore the effect of the insulation thickness on the temperature of the aluminum and the heat loss per unit tube length.

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 \(10 \mathrm{~W} / \mathrm{m}+\mathrm{K}\). The thermal contact resistance between the heater and the outer surface of the tube (per unit length of the tube) is \(R_{\text {Le }}^{\prime}=0.01 \mathrm{~m} \cdot \mathrm{K} / \mathrm{W}\). The outer surfice of the heater is exposed to a fluid with \(T_{n}=-10^{\circ} \mathrm{C}\) and a convecticn cocfficient of \(h=\) \(100 \mathrm{~W} / \mathrm{m}^{2}\) - K. Determine the heater power per unit length of tube required to maintain the heater in \(T_{e}=25^{\circ} \mathrm{C}\).

The radiation heat gage shown in the diagram is made from constantan metal foil, which is coated black and is in the form of a circular disk of radius \(R\) and thickness \(t\). The page is located in an evacuated enclosure. The incident radiation flux absorbed by the foil, \(q_{n}^{*}\) diffuses toward the cuter circumference and into the larger copper ring, which acts as a heat sink at the constant temperature \(T(R)\). Tuo copper lead wires are attached to the center of the foil and to the ring to complete a thermocouple circuit that allows for measurement of the temperature difference between the foil center and the foil edge, \(\Delta T=T(0)-T(R)\). Obtain the differential equation that determines \(T(r)\), the temperature distribution in the foil, under steady- state conditions. Solve this equation to obtuin an expression relating \(\Delta T\) to \(q^{\prime \prime}\). You may neglect radiation exchange between the foil and its surroundings.
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