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Chapter 3: Problem 14

Consider steady one-dimensional heat transfer through a multilayer medium. If the rate of heat transfer \(Q\) is known, explain how you would determine the temperature drop across each layer.

Short Answer

Expert verified
Answer: To find the temperature drop across each layer, first calculate the thermal resistance (R_th) of each layer using the formula R_th = L/(kA), where L is the thickness, k is the thermal conductivity, and A is the area of heat transfer. Then, use the formula 螖T = QR_th to determine the temperature drop across each layer.

Step by step solution

01

Understand the problem

Consider a multilayer medium in which heat is being transferred at a steady one-dimensional rate. We are given that the rate of heat transfer is Q. The objective is to determine the temperature drop across each layer of the medium.
02

Fourier's Law of Heat Conduction

Fourier's Law of heat conduction states that the rate of heat transfer through a material or layer is proportional to the temperature difference across the layer and the area of heat transfer, and inversely proportional to the layer's thickness. Mathematically, it is given by \[-Q = kA\frac{\Delta T}{L},\] where \(Q\) is the heat transfer rate, \(k\) is the thermal conductivity of the material, \(A\) is the area of heat transfer, \(\Delta T\) is the temperature difference across the layer, and \(L\) is the thickness of the layer.
03

Thermal Resistance

The concept of thermal resistance (\(R_\text{th}\)) relates the heat transfer rate to the temperature drop across the layer. The total thermal resistance for a multilayer medium can be defined as the sum of the thermal resistance of each individual layer. To find the thermal resistance of a layer, we can rearrange Fourier's Law like so: \[R_\text{th} = \frac{\Delta T}{Q} = \frac{L}{kA},\] where \(R_\text{th}\) is the thermal resistance of the layer.
04

Find the Thermal Resistance for Each Layer

Using the formula for thermal resistance, calculate the thermal resistance for each layer of the multilayer medium. You will need to know the thermal conductivity of each layer, as well as the area of heat transfer and the thickness of each layer.
05

Determine the Temperature Drop across Each Layer

Given the thermal resistance of each layer and the known heat transfer rate, we can now determine the temperature drop across each layer. Simply rearrange the formula for thermal resistance to solve for the temperature difference: \[\Delta T = Q R_\text{th},\] Calculate this value for each layer to find the temperature drop across it.

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Key Concepts

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

Fourier's Law of Heat Conduction
Fourier's Law of Heat Conduction is a fundamental concept in understanding how heat moves through materials. The essence of this law is that the rate of heat transfer along a material is directly related to the temperature difference across the material, its area, and inversely to the material's thickness. The equation \[-Q = kA\frac{\Delta T}{L}\]helps us quantify this. Here,
  • \(Q\) represents the heat transfer rate,
  • \(k\) denotes the thermal conductivity of the material鈥攈ow well the material conducts heat,
  • \(A\) is the area through which the heat transfers, and
  • \(L\) is the thickness of the material.
Simply put, a higher thermal conductivity or larger area increases heat transfer, whereas greater thickness reduces it. Remember, this law applies to steady-state conditions, meaning the rate of heat transfer doesn't change over time.
Thermal Resistance
Thermal resistance is like a barrier to heat flow. It helps us understand how hard it is for heat to pass through a material. Imagine it like electrical resistance but for heat. When you know the rate of heat transfer, you can find the total thermal resistance by adding together the resistances of each layer. The formula for a single layer's thermal resistance is:\[R_\text{th} = \frac{L}{kA}\]In this equation:
  • \(R_\text{th}\) is the thermal resistance,
  • \(L\) is the thickness of the material,
  • \(k\) is the thermal conductivity, and
  • \(A\) is the area of heat transfer.
Adding up the thermal resistances of all layers gives you the total thermal resistance, which allows us to calculate how temperature changes as heat flows through the multilayer system.
Multilayer Medium Heat Transfer
In multilayer systems, heat encounters several layers of different materials, each with its own thermal characteristics. Understanding heat transfer through such a system requires evaluating the thermal resistance of each layer. This approach is crucial because each layer adds resistance to the heat flow, similar to how multiple resistors affect current in an electrical circuit. To determine the temperature drop across each layer, we use the relationship from thermal resistance:\[\Delta T = Q R_\text{th}\]For each layer, calculate its thermal resistance using its properties, then apply the formula to find the temperature difference. Heat transfer across a multilayer medium involves:
  • Identifying each layer's material properties,
  • Calculating their individual thermal resistances, and
  • Using the heat transfer rate to find the temperature drop across each layer.
This systematic approach lets us understand how temperature varies from one side of the system to the other.

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Most popular questions from this chapter

When plotting the thermal resistance network associated with a heat transfer problem, explain when two resistances are in series and when they are in parallel.

Two flow passages with different cross-sectional shapes, one circular another square, are each centered in a square solid bar of the same dimension and thermal conductivity. Both configurations have the same length, \(T_{1}\), and \(T_{2}\). Determine which configuration has the higher rate of heat transfer through the square solid bar for \((a) a=1.2 b\) and \((b) a=2 b\).

Consider a 3-m-high, 6-m-wide, and \(0.3\)-m-thick brick wall whose thermal conductivity is \(k=0.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). On a certain day, the temperatures of the inner and the outer surfaces of the wall are measured to be \(14^{\circ} \mathrm{C}\) and \(2^{\circ} \mathrm{C}\), respectively. Determine the rate of heat loss through the wall on that day.

A 5-m-wide, 4-m-high, and 40-m-long kiln used to cure concrete pipes is made of 20 -cm-thick concrete walls and ceiling \((k=0.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The kiln is maintained at \(40^{\circ} \mathrm{C}\) by injecting hot steam into it. The two ends of the kiln, \(4 \mathrm{~m} \times 5 \mathrm{~m}\) in size, are made of a 3 -mm-thick sheet metal covered with 2 -cm-thick Styrofoam \((k=0.033 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The convection heat transfer coefficients on the inner and the outer surfaces of the kiln are \(3000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Disregarding any heat loss through the floor, determine the rate of heat loss from the kiln when the ambient air is at \(-4^{\circ} \mathrm{C}\).

Consider heat conduction through a plane wall. Does the energy content of the wall change during steady heat conduction? How about during transient conduction? Explain.
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