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The variation of temperature in a plane wall is determined to be \(T(x)=110-60 x\) where \(x\) is in \(\mathrm{m}\) and \(T\) is in \({ }^{\circ} \mathrm{C}\). If the thickness of the wall is \(0.75 \mathrm{~m}\), the temperature difference between the inner and outer surfaces of the wall is (a) \(30^{\circ} \mathrm{C}\) (b) \(45^{\circ} \mathrm{C}\) (c) \(60^{\circ} \mathrm{C}\) (d) \(75^{\circ} \mathrm{C}\) (e) \(84^{\circ} \mathrm{C}\)

Short Answer

Expert verified
Answer: The temperature difference between the inner and outer surfaces of the wall is 45°C.

Step by step solution

01

Identify the values for the inner and outer surfaces.

The inner and outer surfaces of the wall are located at \(x = 0\) and \(x = 0.75\), respectively.
02

Calculate the temperature at the inner surface.

Substitute \(x = 0\) into the temperature function: \(T(0) = 110 - 60(0) = 110^\circ \mathrm{C}\).
03

Calculate the temperature at the outer surface.

Substitute \(x = 0.75\) into the temperature function: \(T(0.75) = 110 - 60(0.75) = 110 - 45 = 65^\circ \mathrm{C}\).
04

Calculate the temperature difference between the inner and outer surfaces.

Subtract the temperature at the outer surface from the temperature at the inner surface: Temperature difference = \(110^\circ \mathrm{C} - 65^\circ \mathrm{C} = 45^\circ\mathrm{C}\). The temperature difference between the inner and outer surfaces of the wall is (b) \(45^{\circ} \mathrm{C}\).

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Key Concepts

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

Temperature Gradient
The concept of a temperature gradient is essential in understanding how heat transfers through materials. It represents how the temperature changes over a specific distance within a substance. In our plane wall example, the temperature gradient is fundamental to find out how the heat travels from one side to the other.
The formula given, \(T(x) = 110 - 60x\), provides us the temperature for any point \(x\) along the wall's thickness. The number \(-60\) in this equation denotes the rate of change of temperature per meter of the wall's thickness.
A steeper gradient means a higher rate of heat transfer, while a gentler gradient indicates slower heat movement. Therefore, understanding the temperature gradient helps engineers determine how effective a material is in insulating or conducting heat.
Wall Conduction
Wall conduction involves the transfer of heat through a wall from one surface to another. This occurs because of a difference in temperature between the two surfaces, causing heat to flow from the hotter side to the cooler side.
In our example, the inner surface of the wall is hotter than the outer surface, as shown by the temperatures \(110^\circ \mathrm{C}\) and \(65^\circ \mathrm{C}\) respectively. This temperature difference is what drives heat to be conducted from the inside to the outside.
Wall conduction is affected by several factors:
  • The material of the wall, as different materials conduct heat at different rates.
  • The thickness of the wall, since thicker walls generally slow down heat transfer.
  • The temperature difference across the wall; the greater this difference, the more heat flow you get.
Understanding wall conduction is crucial for designing structures with efficient thermal properties.
Thermal Analysis
Thermal analysis refers to evaluating the temperature distribution and heat flow within structures or materials. This allows for the prediction and management of temperatures to prevent overheating or ensure proper insulation.
In our exercise, thermal analysis begins with identifying the temperature function, \(T(x) = 110 - 60x\), and using it to calculate specific temperatures at different points on the wall. This function equips engineers to understand how temperature will change inside the wall material.
Such analysis can help determine:
  • Appropriate materials to use for construction.
  • Insulation performance and needs.
  • Energy efficiency of buildings and systems.
Through thermal analysis, engineers optimize designs to meet both safety and energy requirements.
Engineering Education
Engineering education often includes the study of heat transfer concepts like temperature gradient, wall conduction, and thermal analysis. These fundamental topics equip students with the knowledge needed to tackle real-world problems involving heat flow.
Through practical exercises, like the temperature difference calculation in a wall, students learn to apply mathematical models to predict and control thermal behavior. This builds a foundation for more complex topics in heat transfer such as:
  • Advanced conduction in non-linear systems.
  • Convection and radiation heat transfer.
  • Thermal management solutions in electronics and other industries.
By grasping these foundational principles early in their education, future engineers can design safer, more efficient, and cost-effective thermal systems.

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Most popular questions from this chapter

Consider a steam pipe of length \(L\), inner radius \(r_{1}\), outer radius \(r_{2}\), and constant thermal conductivity \(k\). Steam flows inside the pipe at an average temperature of \(T_{i}\) with a convection heat transfer coefficient of \(h_{i}\). The outer surface of the pipe is exposed to convection to the surrounding air at a temperature of \(T_{0}\) with a heat transfer coefficient of \(h_{o^{*}}\) Assuming steady one-dimensional heat conduction through the pipe, \((a)\) express the differential equation and the boundary conditions for heat conduction through the pipe material, \((b)\) obtain a relation for the variation of temperature in the pipe material by solving the differential equation, and (c) obtain a relation for the temperature of the outer surface of the pipe.

How do you distinguish a linear differential equation from a nonlinear one?

The variation of temperature in a plane wall is determined to be \(T(x)=52 x+25\) where \(x\) is in \(\mathrm{m}\) and \(T\) is in \({ }^{\circ} \mathrm{C}\). If the temperature at one surface is \(38^{\circ} \mathrm{C}\), the thickness of the wall is (a) \(0.10 \mathrm{~m}\) (b) \(0.20 \mathrm{~m}\) (c) \(0.25 \mathrm{~m}\) (d) \(0.40 \mathrm{~m}\) (e) \(0.50 \mathrm{~m}\)

Water flows through a pipe at an average temperature of \(T_{\infty}=90^{\circ} \mathrm{C}\). The inner and outer radii of the pipe are \(r_{1}=\) \(6 \mathrm{~cm}\) and \(r_{2}=6.5 \mathrm{~cm}\), respectively. The outer surface of the pipe is wrapped with a thin electric heater that consumes \(400 \mathrm{~W}\) per \(\mathrm{m}\) length of the pipe. The exposed surface of the heater is heavily insulated so that the entire heat generated in the heater is transferred to the pipe. Heat is transferred from the inner surface of the pipe to the water by convection with a heat transfer coefficient of \(h=85 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming constant thermal conductivity and one-dimensional heat transfer, express the mathematical formulation (the differential equation and the boundary conditions) of the heat conduction in the pipe during steady operation. Do not solve.

A spherical communication satellite with a diameter of \(2.5 \mathrm{~m}\) is orbiting around the earth. The outer surface of the satellite in space has an emissivity of \(0.75\) and a solar absorptivity of \(0.10\), while solar radiation is incident on the spacecraft at a rate of \(1000 \mathrm{~W} / \mathrm{m}^{2}\). If the satellite is made of material with an average thermal conductivity of \(5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and the midpoint temperature is \(0^{\circ} \mathrm{C}\), determine the heat generation rate and the surface temperature of the satellite.

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