/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 A 6-mm-thick stainless steel str... [FREE SOLUTION] | 91Ó°ÊÓ

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A 6-mm-thick stainless steel strip \((k=21 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\rho=8000 \mathrm{~kg} / \mathrm{m}^{3}\), and \(\left.c_{p}=570 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) exiting an oven at a temperature of \(500^{\circ} \mathrm{C}\) is allowed to cool within a buffer zone distance of \(5 \mathrm{~m}\). To prevent thermal burn to workers who are handling the strip at the end of the buffer zone, the surface temperature of the strip should be cooled to \(45^{\circ} \mathrm{C}\). If the air temperature in the buffer zone is \(15^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the maximum speed of the stainless steel strip.

Short Answer

Expert verified
The maximum speed of the stainless steel strip should not exceed approximately 0.0028 m/s to ensure that its temperature cools down to 45°C before workers handle it.

Step by step solution

01

Write Down the Convection Heat Transfer Equation

The convection heat transfer equation is given as: \(q = hA\Delta T\) Where: \(q\) is the heat transfer rate \(h\) is the convection heat transfer coefficient \(A\) is the surface area of the strip \(\Delta T\) is the temperature difference between the strip surface and the air
02

Write Down the Equation for Heat Conducted Through the Strip Thickness

The heat conducted through the strip thickness can be given by Fourier's Law of Heat Conduction as: \(q = \frac{kA\Delta T}{\delta}\) Where: \(\delta\) is the thickness of the strip \(q = m c_p \frac{dT}{dt}\), and \(m = \rho A_1 L\), where \(A_1\) is the cross-sectional area of the strip, and \(L\) is the length of the strip.
03

Equate the Heat Transfer Rates

We can equate the heat transfer rates from the convection heat transfer and the heat conducted to the strip. \(hA\Delta T = \frac{kA\Delta T}{\delta}\) Solving for the speed \(v\), we get: \(v = \frac{ht}{m c_p}\) Where: \(t = \frac{L}{v}\) is the time required to cover the distance of \(5 \mathrm{~m}\)
04

Substitute the Given Values and Solve for v

Substitute the given values into the equation derived in Step 3: \(v = \frac{120 \times 5 }{\frac{8000}{6} \times 570 \times(\frac{500-45}{45-15})}\) After calculating, we get: \(v \approx 0.0028 \mathrm{~m/s}\) Therefore, the maximum speed of the stainless steel strip should not exceed approximately 0.0028 m/s to ensure that its temperature cools down to 45°C before workers handle 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 principle that describes how heat energy is transmitted through materials due to temperature differences. It establishes the relationship between the heat transfer rate and the properties of the material as well as the temperature gradient. According to this law, the rate at which heat is transferred \(q\) through a material is directly proportional to the temperature difference \(\Delta T\) across the material and the area \(A\) normal to the heat transfer direction, and inversely proportional to the material's thickness \(\delta\). Mathematically, it's expressed as \(q = \frac{kA\Delta T}{\delta}\).
Thermal conductivity \(k\) is a property of the material that quantifies its ability to conduct heat. In the context of the exercise, Fourier's Law helps us calculate the rate at which the stainless steel strip loses heat as it conducts through its thickness. It is essential to use this law to determine how quickly the heat should be removed to ensure the steel cools sufficiently before reaching the workers.
Thermal Conductivity
Thermal conductivity \(k\) is a physical property that measures a material's ability to conduct heat. It is a crucial factor in Fourier's Law and is represented in units of watts per meter-kelvin \(W/m\cdot K\). The higher the thermal conductivity of a material, the more efficient it is at transferring heat. Materials with high thermal conductivity, such as metals, are often used where quick heat dissipation is needed, while materials with low thermal conductivity, such as insulators, are used to retain heat.
In our particular exercise, the stainless steel strip has a thermal conductivity of 21 W/m·K, indicating that under a temperature gradient, it is quite effective at transferring heat. Understanding the material's conductivity is critical when solving for the cooling rate required to prevent thermal burns, as it directly impacts the heat transfer rate.
Heat Transfer Coefficient
The heat transfer coefficient \(h\) is a measure of the convective heat transfer between a solid surface and a fluid in contact with it. It is expressed in watts per square meter-kelvin \(W/m^2\cdot K\) and illustrates how well the fluid removes heat from the surface. A high heat transfer coefficient implies efficient cooling, necessitating mechanisms such as forced convection or turbulence to increase the coefficient and subsequently the rate of heat removal.
In the exercise provided, the heat transfer coefficient is 120 W/m²·K. This number helps us understand the effectiveness of air cooling the stainless steel strip. To find the cooling speed, we use the convection heat transfer equation alongside this coefficient. Accurate knowledge of \(h\) is vital because it determines the rate at which heat is transferred from the strip to the surrounding air, hence affecting the cooling rate and ensuring the safety of workers handling the material.
Convection Cooling
Convection cooling refers to the process where heat is removed from an object by the flowing motion of fluids (either liquids or gases). The rate of convective cooling is influenced by factors such as the temperature difference between the object and the fluid, the properties of the fluid, the flow characteristics of the fluid, and the surface area of the object. In convection cooling, heat is carried away by the moving fluid, which can be facilitated by natural convection due to density differences or by forced convection using fans or pumps.
The concept of convection cooling is applied to our textbook problem to calculate the maximum speed at which the hot stainless steel strip can move through the buffer zone while sufficiently cooling down. Utilizing the convection heat transfer equation and the properties given, we quantify how convection ensures that the strip’s surface temperature is safe to touch by the time it reaches the workers.

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Most popular questions from this chapter

4-115 A semi-infinite aluminum cylinder \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\left.\alpha=9.71 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\) of diameter \(D=15 \mathrm{~cm}\) is initially at a uniform temperature of \(T_{i}=115^{\circ} \mathrm{C}\). The cylinder is now placed in water at \(10^{\circ} \mathrm{C}\), where heat transfer takes place by convection with a heat transfer coefficient of \(h=140 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the temperature at the center of the cylinder \(5 \mathrm{~cm}\) from the end surface 8 min after the start of cooling. 4-116 A 20-cm-long cylindrical aluminum block \((\rho=\) \(2702 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=0.896 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=236 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=\) \(\left.9.75 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right), 15 \mathrm{~cm}\) in diameter, is initially at a uniform temperature of \(20^{\circ} \mathrm{C}\). The block is to be heated in a furnace at \(1200^{\circ} \mathrm{C}\) until its center temperature rises to \(300^{\circ} \mathrm{C}\). If the heat transfer coefficient on all surfaces of the block is \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine how long the block should be kept in the furnace. Also, determine the amount of heat transfer from the aluminum block if it is allowed to cool in the room until its temperature drops to \(20^{\circ} \mathrm{C}\) throughout.

A steel casting cools to 90 percent of the original temperature difference in \(30 \mathrm{~min}\) in still air. The time it takes to cool this same casting to 90 percent of the original temperature difference in a moving air stream whose convective heat transfer coefficient is 5 times that of still air is (a) \(3 \mathrm{~min}\) (b) \(6 \mathrm{~min}\) (c) \(9 \mathrm{~min}\) (d) \(12 \mathrm{~min}\) (e) \(15 \mathrm{~min}\)

What are the factors that affect the quality of frozen fish?

The chilling room of a meat plant is \(15 \mathrm{~m} \times 18 \mathrm{~m} \times\) \(5.5 \mathrm{~m}\) in size and has a capacity of 350 beef carcasses. The power consumed by the fans and the lights in the chilling room are 22 and \(2 \mathrm{~kW}\), respectively, and the room gains heat through its envelope at a rate of \(14 \mathrm{~kW}\). The average mass of beef carcasses is \(220 \mathrm{~kg}\). The carcasses enter the chilling room at \(35^{\circ} \mathrm{C}\), after they are washed to facilitate evaporative cooling, and are cooled to \(16^{\circ} \mathrm{C}\) in \(12 \mathrm{~h}\). The air enters the chilling room at \(-2.2^{\circ} \mathrm{C}\) and leaves at \(0.5^{\circ} \mathrm{C}\). Determine \((a)\) the refrigeration load of the chilling room and \((b)\) the volume flow rate of air. The average specific heats of beef carcasses and air are \(3.14\) and \(1.0 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), respectively, and the density of air can be taken to be \(1.28 \mathrm{~kg} / \mathrm{m}^{3}\).

During a fire, the trunks of some dry oak trees (es) \(\left(k=0.17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\) and \(\left.\alpha=1.28 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\) that are initially at a uniform temperature of \(30^{\circ} \mathrm{C}\) are exposed to hot gases at \(520^{\circ} \mathrm{C}\) for a period of \(5 \mathrm{~h}\), with a heat transfer coefficient of \(65 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the surface. The ignition temperature of the trees is \(410^{\circ} \mathrm{C}\). Treating the trunks of the trees as long cylindrical rods of diameter \(20 \mathrm{~cm}\), determine if these dry trees will ignite as the fire sweeps through them. Solve this problem using analytical one-term approximation method (not the Heisler charts).

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