/*! 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 14 An aluminum alloy (2024) plate, ... [FREE SOLUTION] | 91Ó°ÊÓ

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An aluminum alloy (2024) plate, heated to a uniform temperature of \(227^{\circ} \mathrm{C}\), is allowed to cool while vertically suspended in a room where the ambient air and surroundings are at \(27^{\circ} \mathrm{C}\). The plate is \(0.3 \mathrm{~m}\) square with a thickness of \(15 \mathrm{~mm}\) and an emissivity of \(0.25\). (a) Develop an expression for the time rate of change of the plate temperature, assuming the temperature to be uniform at any time. (b) Determine the initial rate of cooling (K/s) when the plate temperature is \(227^{\circ} \mathrm{C}\). (c) Justify the uniform plate temperature assumption. (d) Compute and plot the temperature history of the plate from \(t=0\) to the time required to reach a temperature of \(30^{\circ} \mathrm{C}\). Compute and plot the corresponding variations in the convection and radiation heat transfer rates.

Short Answer

Expert verified
We developed an expression for the time rate of change of the plate temperature by considering both convective and radiative heat transfers: \[\rho C_p V \frac{\mathrm{d} T_{plate}}{\mathrm{d} t} = -hA(T_{plate} - T_{ambient}) - e\sigma A(T_{plate}^4 - T_{ambient}^4)\] The initial cooling rate (K/s) can be determined by evaluating the derivative at the initial temperature of \(227^{\circ} \mathrm{C}\) (or \(500 K\)) using given values and assumptions. The uniform plate temperature assumption can be justified because of the aluminum's high thermal conductivity and the relatively small thickness of the plate. Finally, we compute and plot the temperature history of the plate, as well as the corresponding variations in the convection and radiation heat transfer rates, using numerical methods and plotting libraries in Python, MATLAB, or similar software programs.

Step by step solution

01

Newton's Law of Cooling

Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between its temperature and the ambient temperature. For convective heat transfer, we will use the formula: \[q_{conv} = hA(T_{plate} - T_{ambient})\] For radiative heat transfer, we can use the formula (assuming a small body in a large enclosure): \[q_{rad} = e\sigma A(T_{plate}^4 - T_{ambient}^4)]\] where \(q_{conv}\) and \(q_{rad}\) are the rates of convective and radiative heat transfers, respectively, \(h\) is the heat transfer coefficient, \(e\) is the emissivity, \(A\) is the area, \(\sigma\) is the Stefan-Boltzmann constant, and \(T_{plate}\) and \(T_{ambient}\) are the temperatures of the plate and the surroundings, respectively.
02

Energy balance

We have to consider both convective and radiative heat transfers. We can write the energy balance equation as: \[\rho C_p V \frac{\mathrm{d} T_{plate}}{\mathrm{d} t} = -q_{conv} - q_{rad}\] where \(\rho\) is the density of the material, \(C_p\) is the specific heat capacity, \(V\) is the volume, and \(t\) is time. Now, substituting the expressions for \(q_{conv}\) and \(q_{rad}\) from Step 1: \[\rho C_p V \frac{\mathrm{d} T_{plate}}{\mathrm{d} t} = -hA(T_{plate} - T_{ambient}) - e\sigma A(T_{plate}^4 - T_{ambient}^4)\]
03

Initial cooling rate

To find the initial cooling rate, we need to calculate the derivative of the plate temperature with respect to time when the plate temperature is \(227^{\circ} \mathrm{C}\). Using the given values (with temperature converted to Kelvin) and assuming an average heat transfer coefficient of \(h \approx 15~\mathrm{W/m^2K}\) for natural convection: \[ \frac{\mathrm{d} T_{plate}}{\mathrm{d} t} \Bigg|_{T_{plate} = 500~K} = -\frac{hA(500 - 300) + e\sigma A(500^4 - 300^4)}{\rho C_p V} \] Now, plug in the given values and calculate the initial cooling rate.
04

Justifying the uniform plate temperature assumption

To justify the assumption of uniform plate temperature, we can argue that the following factors will make the temperature distribution inside the plate nearly uniform: 1. The Aluminum 2024 alloy has a high thermal conductivity (around \(120~\mathrm{W/mK}\)), which means that heat is quickly distributed throughout the material, ensuring a near-uniform temperature distribution. 2. The thickness of the plate is only \(15~\mathrm{mm}\), which is relatively small compared to the overall dimensions, so the temperature across the thickness is likely to be very close.
05

Computer and plot the results

To find the temperature history and corresponding heat transfer rates, we need to integrate the energy balance equation numerically. This can be achieved using numerical methods like the Euler method, the Runge-Kutta method, or built-in solvers in mathematical software like MATLAB or Python. Once we obtain the temperature history, we can plot it and compute the convection and radiation heat transfer rates that correspond to those temperatures using the expressions derived in Step 1. The plots can be generated using plotting libraries in Python, MATLAB, or similar software programs.

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Key Concepts

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

Newton's Law of Cooling
Understanding Newton's Law of Cooling is crucial when analyzing how quickly a heated object cools down in an environment. According to this principle, the rate at which an object's temperature changes is proportional to the difference between its own temperature and the ambient temperature. This means that if an object is significantly hotter than its surroundings, it will cool faster, and as it approaches the temperature of the environment, the rate of cooling will decrease. This law is applicable in a variety of practical situations, such as cooling of electronics, food storage, and even forensic science to estimate the time of death.
Convection Heat Transfer
Convection heat transfer is a mechanism of energy movement in fluids (liquid or gas) where heat is carried away by the physical movement of the fluid itself. It occurs when a surface at a certain temperature comes into contact with a fluid moving at a different temperature. This transfer is described by the equation
\(q_{conv} = hA(T_{plate} - T_{ambient})\)
where
\(h\) is the heat transfer coefficient which depends on the properties of the fluid and the flow conditions,
\(A\) is the surface area, and
\(T_{plate}\) and \(T_{ambient}\) are the temperatures of the solid surface and the fluid, respectively. It's one of the primary modes of thermal energy transfer, alongside conduction and radiation.
Radiation Heat Transfer
Radiation heat transfer occurs through electromagnetic waves and does not require a medium, unlike convection or conduction. It is the reason you feel warmth without touching a hot object nearby. The key formula for radiation transfer, when considering a small body in a large enclosure, looks like this:
\(q_{rad} = e\sigma A(T_{plate}^4 - T_{ambient}^4)\)
where
\(e\) is the emissivity of the surface, \(\sigma\) is the Stefan-Boltzmann constant, and \(T_{plate}\) and \(T_{ambient}\) are the fourth powers of the absolute temperatures of the body and the surroundings, respectively. The emissivity factor is a measure of how effectively a surface emits thermal radiation.
Energy Balance Equation
The energy balance equation is a fundamental concept in thermodynamics that ensures the energy transferring into a system is equal to the energy transferring out, taking into account the energy stored within the system. In the context of cooling, the equation represents the change in internal energy of the object as a result of heat transfers by convection and radiation. The generalized form of this equation is an expression of the first law of thermodynamics and is vital in predicting how the temperature of a body changes over time. It enables us to calculate the rate of temperature change for the aluminum alloy plate as it cools.
Numerical Integration Methods
Numerical integration methods are mathematical tools used to solve differential equations when an analytical solution is difficult or impossible to obtain. Common methods include the Euler method, the Midpoint method, and the Runge-Kutta methods, which approximate the solution by breaking the problem into small steps and calculating the change over each step. In our scenario, these methods can be used to compute the temperature history of the cooling plate by integrating the energy balance equation. Numerical software, such as MATLAB or Python, often comes with built-in functions that make this process more efficient and accurate.

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Most popular questions from this chapter

Consider an experiment to investigate the transition to turbulent flow in a free convection boundary layer that develops along a vertical plate suspended in a large room. The plate is constructed of a thin heater that is sandwiched between two aluminum plates and may be assumed to be isothermal. The heated plate is \(1 \mathrm{~m}\) high and \(2 \mathrm{~m}\) wide. The quiescent air and the surroundings are both at \(25^{\circ} \mathrm{C}\). (a) The exposed surfaces of the aluminum plate are painted with a very thin coating of high emissivity \((\varepsilon=0.95)\) paint. Determine the electrical power that must be supplied to the heater to sustain the plate at a temperature of \(T_{s}=35^{\circ} \mathrm{C}\). How much of the plate is exposed to turbulent conditions in the free convection boundary layer? (b) The experimentalist speculates that the roughness of the paint is affecting the transition to turbulence in the boundary layer and decides to remove the paint and polish the aluminum surface ( \(\varepsilon=0.05\) ). If the same power is supplied to the plate as in part (a), what is the steady- state plate temperature? How much of the plate is exposed to turbulent conditions in the free convection boundary layer?

Beginning with the free convection correlation of the form given by Equation 9.24, show that for air at atmospheric pressure and a film temperature of \(400 \mathrm{~K}\), the average heat transfer coefficient for a vertical plate can be expressed as $$ \begin{array}{ll} \bar{h}_{L}=1.40\left(\frac{\Delta T}{L}\right)^{1 / 4} & 10^{4}

Consider the conveyor system described in Problem \(7.24\), but under conditions for which the conveyor is not moving and the air is quiescent. Radiation effects and interactions between boundary layers on adjoining surfaces may be neglected. (a) For the prescribed plate dimensions and initial temperature, as well as the prescribed air temperature, what is the initial rate of heat transfer from one of the plates? (b) How long does it take for a plate to cool from \(300^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) ? Comment on the assumption of negligible radiation.

At the end of its manufacturing process, a silicon wafer of diameter \(D=150 \mathrm{~mm}\), thickness \(\delta=1 \mathrm{~mm}\), and emissivity \(\varepsilon=0.65\) is at an initial temperature of \(T_{i}=325^{\circ} \mathrm{C}\) and is allowed to cool in quiescent, ambient air and large surroundings for which \(T_{\infty}=T_{\text {sur }}=25^{\circ} \mathrm{C}\). (a) What is the initial rate of cooling? (b) How long does it take for the wafer to reach a temperature of \(50^{\circ} \mathrm{C}\) ? Comment on how the relative effects of convection and radiation vary with time during the cooling process.

As discussed in Section 5.2, the lumped capacitance approximation may be applied if \(B_{i}<0.1\), and, when implemented in a conservative fashion for a long cylinder, the characteristic length is the cylinder radius. After its extrusion, a long glass rod of diameter \(D=15 \mathrm{~mm}\) is suspended horizontally in a room and cooled from its initial temperature by natural convection and radiation. At what rod temperatures may the lumped capacitance approximation be applied? The temperature of the quiescent air is the same as that of the surroundings, \(T_{\infty}=T_{\text {sur }}=27^{\circ} \mathrm{C}\), and the glass emissivity is \(\varepsilon=0.94\).

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