Consider an engine cover that is made with two layers of metal plates. The inner layer is stainless steel \(\left(k_{1}=14 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) with a thickness of \(10 \mathrm{~mm}\), and the outer layer is aluminum \(\left(k_{2}=\right.\) \(237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with a thickness of \(5 \mathrm{~mm}\). Both metal plates have a surface roughness of about \(23 \mu \mathrm{m}\). The aluminum plate is attached on the stainless steel plate by screws that exert an average pressure of \(20 \mathrm{MPa}\) at the interface. The inside stainless steel surface of the cover is exposed to heat from the engine with a convection heat transfer coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at an ambient temperature of \(150^{\circ} \mathrm{C}\). The outside aluminum surface is exposed to a convection heat transfer coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at an ambient temperature of \(40^{\circ} \mathrm{C}\). Determine the heat flux through the engine cover.

Short Answer

Expert verified
The heat flux through the engine cover is approximately 745.63 W/m².

Step by step solution

01

Determine the thermal resistances of conduction

For each metal plate (stainless steel and aluminum), the thermal resistance of conduction can be found using the following formula: \(R_{cond} = \frac{L}{kA}\) Where \(R_{cond}\) is the thermal resistance of conduction, \(L\) is the thickness of the plate, \(k\) is the thermal conductivity, and \(A\) is the surface area. As the area is constant for both, we can denote the thermal resistance of conduction in terms of the thickness and conductivity, avoiding the explicit mention of the surface area. For the stainless steel (inner layer), we have: \(R_{cond1} = \frac{0.01 \mathrm{m}}{14 \mathrm{W/m \cdot K}}\) Similarly, for the aluminum (outer layer), we have: \(R_{cond2} = \frac{0.005 \mathrm{m}}{237 \mathrm{W/m \cdot K}}\)
02

Determine the thermal resistance of convection

The thermal resistance of convection for both surfaces (inner and outer) can be determined using the following formula: \(R_{conv} = \frac{1}{hA}\) Where \(R_{conv}\) represents the thermal resistance of convection, \(h\) is the convection heat transfer coefficient, and \(A\) is the surface area. For the inner stainless steel surface, we have: \(R_{conv1} = \frac{1}{10 \mathrm{W/m^2 \cdot K}}\) For the outer aluminum surface, we have: \(R_{conv2} = \frac{1}{25 \mathrm{W/m^2 \cdot K}}\)
03

Calculate the total thermal resistance for the system

Since the engine cover is made up of two layers of metal plates in contact and has convection at both surfaces, the total thermal resistance can be found by adding the individual resistances: \(R_{total} = R_{conv1} + R_{cond1} + R_{cond2} + R_{conv2}\) Now, we will substitute the values found in Steps 1 and 2: \(R_{total} = \frac{1}{10} + \frac{0.01}{14} + \frac{0.005}{237} + \frac{1}{25}\) Calculate the total thermal resistance: \(R_{total} \approx 0.14754 \mathrm{~W/m^2 \cdot K}\)
04

Determine the temperature difference across the total thermal resistance

The temperature difference across the total thermal resistance can be found by subtracting the ambient temperature of the outer surface from the ambient temperature of the inner surface: \(\Delta T = T_{inside} - T_{outside} = 150^{\circ} \mathrm{C} - 40^{\circ} \mathrm{C} = 110^{\circ} \mathrm{C}\)
05

Apply Fourier's Law to find the heat flux passing through the engine cover

Now that we have the total thermal resistance and the temperature difference across it, we can apply Fourier's Law to determine the heat flux: \(q = \frac{\Delta T}{R_{total}}\) Substitute the values we found in Steps 3 and 4: \(q = \frac{110^{\circ} \mathrm{C}}{0.14754 \mathrm{~W/m^2 \cdot K}} \approx 745.63 \mathrm{~W/m^2}\) Thus, the heat flux through the engine cover is approximately \(745.63 \mathrm{~W/m^2}\).

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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 concept in heat transfer that measures the opposition to the flow of heat across a material. Much like electrical resistance, thermal resistance quantifies how much a material resists the transfer of heat. The higher the thermal resistance, the lower the heat flux for a given temperature difference across the material. Calculating thermal resistance is crucial when designing systems like insulation or heat exchangers, as it helps to predict how effectively heat will be transferred.

For conductive heat transfer, the formula is simple:
\[\begin{equation}R_{cond} = \frac{L}{kA}\end{equation}\]Here, L is the material's thickness, k is its thermal conductivity, and A is the surface area through which heat is being transferred. It's important to make sure that the units are consistent when applying this formula to get an accurate measurement of thermal resistance. In exercises involving heat transfer, you often need to calculate separately for the conduction and convection components and then sum them to find the total thermal resistance, which impedes the heat flow.
Fourier's Law
Fourier's Law is the fundamental principle that governs the conduction of heat. It states that the heat flux, which is the rate of heat transfer per unit area, is proportional to the negative of the temperature gradient. In other words, heat will flow from regions of higher temperature to regions of lower temperature, and the amount of heat transferred is related to how steeply temperature changes in space.

The mathematical expression for Fourier’s Law in one dimension is:
\[\begin{equation}q = -k \frac{dT}{dx}\end{equation}\]In this equation, q is the heat flux, k is the thermal conductivity of the material, \frac{dT}{dx} is the temperature gradient. This principle is often used to analyze heat transfer problems by connecting the temperature difference with the heat flux after determining the total thermal resistance of the system. Fourier's Law serves as the foundation for solving many practical engineering problems involving heat transfer.
Convection Heat Transfer Coefficient
The convection heat transfer coefficient, represented by h, is a measure of the convective heat transfer between a surface and a fluid moving past it. This coefficient is a critical factor in calculating how much heat is transferred by convection, which is the mode of heat transfer involving fluid movement. The higher the value of h, the more efficient the convection process is at transferring heat.

To calculate the thermal resistance due to convection, the following formula is used: \[\begin{equation}R_{conv} = \frac{1}{hA}\end{equation}\]Here, h is the convection heat transfer coefficient, and A is the surface area. A high convection heat transfer coefficient indicates strong convection, which typically corresponds to conditions of turbulent flow, high fluid velocity, or a large temperature difference between the surface and the fluid.
Thermal Conductivity
Thermal conductivity, denoted by k, is a physical property of materials that indicates their ability to conduct heat. It's expressed in units of watts per meter-kelvin \(\frac{W}{m\cdot K}\). Materials with high thermal conductivity, like metals, are excellent conductors of heat, while those with low thermal conductivity, such as insulating materials, resist heat flow. Thermal conductivity is a crucial parameter in many heat transfer calculations because it helps determine how quickly heat moves through a material.

Thermal conductivity is inherently linked to the atomic or molecular structure of a substance. Metals, with their free electrons, tend to have high thermal conductivities, while non-metals and gases, which lack this electron mobility, typically have lower thermal conductivities. Knowing the thermal conductivity of materials helps engineers and designers select the appropriate materials for heat transfer applications, such as the metals used in the creation of an engine cover in the exercise.

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Most popular questions from this chapter

A 4-mm-diameter and 10-cm-long aluminum fin \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is attached to a surface. If the heat transfer coefficient is \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the percent error in the rate of heat transfer from the fin when the infinitely long fin assumption is used instead of the adiabatic fin tip assumption.

A 1-cm-diameter, 30-cm-long fin made of aluminum \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is attached to a surface at \(80^{\circ} \mathrm{C}\). The surface is exposed to ambient air at \(22^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(11 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the fin can be assumed to bery long, its efficiency is (a) \(0.60\) (b) \(0.67\) (c) \(0.72\) (d) \(0.77\) (e) \(0.88\)

An 8-m-internal-diameter spherical tank made of \(1.5\)-cm-thick stainless steel \((k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is used to store iced water at \(0^{\circ} \mathrm{C}\). The tank is located in a room whose temperature is \(25^{\circ} \mathrm{C}\). The walls of the room are also at \(25^{\circ} \mathrm{C}\). The outer surface of the tank is black (emissivity \(\varepsilon=1\) ), and heat transfer between the outer surface of the tank and the surroundings is by natural convection and radiation. The convection heat transfer coefficients at the inner and the outer surfaces of the tank are \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Determine \((a)\) the rate of heat transfer to the iced water in the tank and \((b)\) the amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24 -h period. The heat of fusion of water at atmospheric pressure is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\).

A 10-m-long 5-cm-outer-radius cylindrical steam pipe is covered with \(3-\mathrm{cm}\) thick cylindrical insulation with a thermal conductivity of \(0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). If the rate of heat loss from the pipe is \(1000 \mathrm{~W}\), the temperature drop across the insulation is (a) \(163^{\circ} \mathrm{C}\) (b) \(600^{\circ} \mathrm{C}\) (c) \(48^{\circ} \mathrm{C}\) (d) \(79^{\circ} \mathrm{C}\) (e) \(251^{\circ} \mathrm{C}\)

Explain how the fins enhance heat transfer from a surface. Also, explain how the addition of fins may actually decrease heat transfer from a surface.

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