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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

Hot water at an average temperature of \(85^{\circ} \mathrm{C}\) passes through a row of eight parallel pipes that are \(4 \mathrm{~m}\) long and have an outer diameter of \(3 \mathrm{~cm}\), located vertically in the middle of a concrete wall \((k=0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that is \(4 \mathrm{~m}\) high, \(8 \mathrm{~m}\) long, and \(15 \mathrm{~cm}\) thick. If the surfaces of the concrete walls are exposed to a medium at \(32^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat loss from the hot water and the surface temperature of the wall.

Ice slurry is being transported in a pipe \((k=\) \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{i}=2.5 \mathrm{~cm}, D_{o}=3 \mathrm{~cm}\), and \(L=\) \(5 \mathrm{~m}\) ) with an inner surface temperature of \(0^{\circ} \mathrm{C}\). The ambient condition surrounding the pipe has a temperature of \(20^{\circ} \mathrm{C}\), a convection heat transfer coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and a dew point of \(10^{\circ} \mathrm{C}\). If the outer surface temperature of the pipe drops below the dew point, condensation can occur on the surface. Since this pipe is located in a vicinity of high voltage devices, water droplets from the condensation can cause electrical hazard. To prevent such incident, the pipe surface needs to be insulated. Determine the insulation thickness for the pipe using a material with \(k=0.95 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) to prevent the outer surface temperature from dropping below the dew point.

Heat is lost at a rate of \(275 \mathrm{~W}\) per \(\mathrm{m}^{2}\) area of a \(15-\mathrm{cm}-\) thick wall with a thermal conductivity of \(k=1.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The temperature drop across the wall is (a) \(37.5^{\circ} \mathrm{C}\) (b) \(27.5^{\circ} \mathrm{C}\) (c) \(16.0^{\circ} \mathrm{C}\) (d) \(8.0^{\circ} \mathrm{C}\) (e) \(4.0^{\circ} \mathrm{C}\)

Consider a \(1.5\)-m-high and 2 -m-wide triple pane window. The thickness of each glass layer \((k=0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is \(0.5 \mathrm{~cm}\), and the thickness of each air space \((k=0.025 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is \(1 \mathrm{~cm}\). If the inner and outer surface temperatures of the window are \(10^{\circ} \mathrm{C}\) and \(0^{\circ} \mathrm{C}\), respectively, the rate of heat loss through the window is (a) \(75 \mathrm{~W}\) (b) \(12 \mathrm{~W}\) (c) \(46 \mathrm{~W}\) (d) \(25 \mathrm{~W}\) (e) \(37 \mathrm{~W}\)

Consider a house whose attic space is ventilated effectively so that the air temperature in the attic is the same as the ambient air temperature at all times. Will the roof still have any effect on heat transfer through the ceiling? Explain.
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