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Chapter 7: Problem 28

In the production of sheet metals or plastics, it is customary to cool the material before it leaves the production process for storage or shipment to the customer. Typically, the process is continuous, with a sheet of thickness \(\delta\) and width \(W\) cooled as it transits the distance \(L\) between two rollers at a velocity \(V\). In this problem, we consider cooling of an aluminum alloy (2024-T6) by an airstream moving at a velocity \(u_{\infty}\) in counter flow over the top surface of the sheet. A turbulence promoter is used to provide turbulent boundary layer development over the entire surface. (a) By applying conservation of energy to a differential control surface of length \(d x\), which either moves with the sheet or is stationary and through which the sheet passes, derive a differential equation that governs the temperature distribution along the sheet. Because of the low emissivity of the aluminum, radiation effects may be neglected. Express your result in terms of the velocity, thickness, and properties of the sheet \(\left(V, \delta, \rho, c_{p}\right)\), the local convection coefficient \(h_{x}\) associated with the counter flow, and the air temperature. For a known temperature of the sheet \(\left(T_{i}\right)\) at the onset of cooling and a negligible effect of the sheet velocity on boundary layer development, solve the equation to obtain an expression for the outlet temperature \(T_{a}\). (b) For \(\delta=2 \mathrm{~mm}, V=0.10 \mathrm{~m} / \mathrm{s}, L=5 \mathrm{~m}, W=1 \mathrm{~m}\), \(u_{\infty}=20 \mathrm{~m} / \mathrm{s}, T_{\infty}=20^{\circ} \mathrm{C}\), and \(T_{i}=300^{\circ} \mathrm{C}\), what is the outlet temperature \(T_{a}\) ?

Short Answer

Expert verified
The outlet temperature \(T_a\) for the aluminum sheet is approximately \(71.61^{\circ} C\).

Step by step solution

01

Derivation of the Differential Equation

Using the conservation of energy principle for the differential control surface of length \(d x\), we can write the energy balance equation: \(V \delta \rho c_{p} \left( T(x)-T_{i} \right) = -h_x (T(x)-T_\infty)d xW\) Now, let's find a differential equation for the temperature distribution: \(\frac{d (T(x)-T_{i})}{d x} = \frac{-h_x (T(x)-T_\infty)}{V \delta \rho c_{p}}\) Since the sheet velocity has negligible effect on boundary layer development, we can assume that \(h_{x}\) is a constant, and rewrite the equation as: \(\frac{d (T(x)-T_{i})}{d x} = -k(T(x)-T_\infty)\), where \(k = \frac{h_x}{V \delta \rho c_{p}}\)
02

Solving the Differential Equation

Now, let's solve this first-order linear ordinary differential equation: \((T(x)-T_{i})'= -k(T(x)-T_\infty)\) We can rewrite the equation in the form: \(\frac{d(T(x)-T_{i})}{(T(x)-T_\infty)} = -kdx\) Now, integrate both sides: \[\int \frac{d(T(x)-T_{i})}{(T(x)-T_\infty)}= -\int kdx\] \[\ln{(T(x) - T_\infty)} = -kx + C\] To eliminate the logarithm, exponentiate both sides of the equation: \(T(x) - T_\infty = e^{-kx} e^{C}\) \(T(x) = T_\infty + e^{-kx} e^{C}\) Now, let's find the constant \(C\). We know that when \(x=0\), the temperature is \(T_i\). So, we have: \(T_i = T_\infty + e^{-k(0)} e^{C} \Rightarrow e^C = T_i - T_\infty\) Thus, the temperature distribution is given by: \(T(x) = T_\infty + (T_i - T_\infty)e^{-kx}\) The outlet temperature when the sheet exits the cooling zone is \(T_a = T(L)\), so we have: \(T_a = T_\infty + (T_i - T_\infty)e^{-kL}\) Now, we will substitute the given values and find \(T_{a}\).
03

Substituting the given parameters

Using the given problem parameters: \(L = 5 \ m\), \(V = 0.10 \ \frac{m}{s}\), \(\delta = 2 \ mm\), \(W = 1 \ m\), \(u_{\infty} = 20 \ \frac{m}{s}\), \(T_{i} = 300^{\circ} C\) and \(T_{\infty} = 20^{\circ} C\), we can find the outlet temperature \(T_{a}\). First, we need to find the convection coefficient \(h_x\) for the aluminum sheet. The properties of the aluminum alloy are: \(\rho = 2700 \ \frac{kg}{m^3}\), \(c_p = 903 \ \frac{J}{kg \cdot K}\), and the thermal conductivity \(k_{al} = 167 \ \frac{W}{m \cdot K}\). Assuming that the convection coefficient is roughly equal to the heat transfer coefficient, based on mixing-length theory, we can approximate it as: \(h_x = \frac{k_{al}}{(Pr \delta / 2)}\), where \(Pr\) is the Prandtl number. For air, we can assume the Prandtl number to be approximately \(Pr = 0.72\). Then, we can find \(h_x\): \(h_x = \frac{167}{(0.72 \cdot 0.002 / 2)} \approx 23194 \ \frac{W}{m^2 \cdot K}\). Now, we can find the value of \(k\): \(k = \frac{23194}{(0.10 \cdot 0.002 \cdot 2700 \cdot 903)} \approx 0.354 \ \frac{1}{m}\). Finally, we can find the outlet temperature \(T_a\): \(T_a = 20 + (300 - 20)e^{-(0.354)(5)} \approx 20 + 280e^{-1.77} \approx 20 + 51.61 \approx 71.61^{\circ} \text{C}\). Hence, the outlet temperature \(T_{a}\) for the aluminum sheet is about \(71.61^{\circ} \text{C}\).

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Key Concepts

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

Conservation of Energy
The principle of Conservation of Energy is a fundamental concept in physics stating that energy cannot be created or destroyed, only transformed from one form to another. In the context of this exercise, conservation of energy is applied to the cooling process of an aluminum sheet as it moves through a control surface. The energy balance considers the rate at which energy enters, leaves, and accumulates within the sheet as it exchanges heat with the surrounding air. This kind of analysis is crucial for determining how much the sheet's temperature changes over its transit between the rollers. The initial mathematical expression for the energy balance is:
  • The term \(V \delta \rho c_{p} (T(x) - T_i)\) represents the heat stored in the sheet per unit length.
  • The term \(-h_x (T(x) - T_\infty)d xW\) accounts for the convective heat loss to the air across an infinitesimal section of the sheet.
By equating these expressions, we preserve the conservation of energy by ensuring that the heat lost through convection equals the heat reduced from the sheet.
Differential Equation
A Differential Equation is an equation involving derivatives of a function, which in this case, describes how the temperature of the aluminum sheet changes as it moves. The derived differential equation for temperature distribution along the sheet is expressed as:\[ \frac{d(T(x) - T_i)}{d x} = -k(T(x) - T_\infty) \]By solving this type of first-order linear ordinary differential equation, we gain insight into how the temperature of the moving sheet declines due to cooling. The solution involves integrating both sides of the equation to find a relationship for the temperature at any position \(x\) on the sheet. The process of solving this differential equation involves finding a specific function that describes the behavior of \(T(x)\) depending on initial boundary conditions provided (like the initial sheet temperature \(T_i\) at \(x=0\)). These conditions help solve for any constants of integration and provide a complete solution for temperature distribution.
Convection Coefficient
The Convection Coefficient \(h_x\) is a measure of the convective heat transfer between the surface of the sheet and the moving air above it. It represents how efficiently heat is transferred away from the sheet. The value of \(h_x\) influences how much heat is lost and consequently affects the final temperature of the sheet. To determine \(h_x\), we estimate it using properties like the Prandtl number \(Pr\) for air and the turbulent flow conditions. The calculation is given by:\[h_x = \frac{k_{al}}{(Pr \delta / 2)}\]Where \(k_{al}\) is the thermal conductivity of the aluminum. With a known \(Pr=0.72\) and sheet thickness \(\delta\), this semi-empirical formula gives an approximation of \(h_x\), which helps in predicting the cooling rate and outlet temperature of the sheet over its length.
Temperature Distribution
Understanding the Temperature Distribution along the length of the sheet is essential for ensuring the final product meets desired temperature specifications. The temperature distribution is expressed through the equation derived from solving the differential equation:\[T(x) = T_\infty + (T_i - T_\infty)e^{-kx}\]This equation illustrates how the sheet's temperature adapts as it moves and cools. Here, \(T(x)\) represents the temperature at position \(x\), \(T_i\) is the initial temperature, and \(T_\infty\) is the ambient air temperature. The exponential term \(e^{-kx}\) indicates how quickly the temperature approaches the air temperature, controlled by the rate coefficient \(k\) which was previously calculated.Finally, the outlet temperature \(T_a\) is simply \(T(L)\), the temperature at the end of the cooling process, which helps inform whether further cooling adjustments are needed for optimal product specifications.

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Most popular questions from this chapter

Packed beds of spherical particles can be sintered at high temperature to form permeable, rigid foams. A foam sheet of thickness \(t=10 \mathrm{~mm}\) is comprised of sintered bronze spheres, each of diameter \(D=0.6 \mathrm{~mm}\). The metal foam has a porosity of \(\varepsilon=0.25\), and the foam sheet fills the cross section of an \(L=40 \mathrm{~mm} \times W=40 \mathrm{~mm}\) wind tunnel. The upper and lower surfaces of the foam are at temperatures \(T_{s}=80^{\circ} \mathrm{C}\), and the two other foam edges (the front edge shown in the schematic and the corresponding back edge) are insulated. Air flows in the wind tunnel at an upstream temperature and velocity of \(T_{i}=20^{\circ} \mathrm{C}\) and \(V=10 \mathrm{~m} / \mathrm{s}\), respectively. (a) Assuming the foam is at a uniform temperature \(T_{s}\), estimate the convection heat transfer rate to the air. Do you expect the actual heat transfer rate to be equal to, less than, or greater than your estimated value? (b) Assuming one-dimensional conduction in the \(x=\) direction, use an extended surface analysis to estimate the heat transfer rate to the air. To do so, show that the effective perimeter associated with Equation \(3.70\) is \(P_{\mathrm{eff}}=A_{p, \mathrm{r}} / L\). Determine the effective thermal conductivity of the foam \(k_{\text {eff }}\) by using Equation 3.25. Do you expect the actual heat transfer rate to be equal to, less than, or greater than your estimated value?

Mass transfer experiments have been conducted on a naphthalene cylinder of \(18.4-\mathrm{mm}\) diameter and \(88.9-\mathrm{mm}\) length subjected to a cross flow of air in a low-speed wind tunnel. After exposure for \(39 \mathrm{~min}\) to the airstream at a temperature of \(26^{\circ} \mathrm{C}\) and a velocity of \(12 \mathrm{~m} / \mathrm{s}\), it was determined that the cylinder mass decreased by \(0.35 \mathrm{~g}\). The barometric pressure was recorded at \(750.6 \mathrm{~mm} \mathrm{Hg}\). The saturation pressure \(p_{\text {sat }}\) of naphthalene vapor in equilibrium with solid naphthalene is given by the relation \(p_{\text {sat }}=p \times 10^{E}\), where \(E=8.67-(3766 / T)\), with \(T(\mathrm{~K})\) and \(p\) (bar) being the temperature and pressure of air. Naphthalene has a molecular weight of \(128.16 \mathrm{~kg} / \mathrm{kmol}\). (a) Determine the convection mass transfer coefficient from the experimental observations. (b) Compare this result with an estimate from an appropriate correlation for the prescribed flow conditions.

Air at \(27^{\circ} \mathrm{C}\) with a free stream velocity of \(10 \mathrm{~m} / \mathrm{s}\) is used to cool electronic devices mounted on a printed circuit board. Each device, \(4 \mathrm{~mm} \times 4 \mathrm{~mm}\), dissipates \(40 \mathrm{~mW}\), which is removed from the top surface. A turbulator is located at the leading edge of the board, causing the boundary layer to be turbulent. (a) Estimate the surface temperature of the fourth device located \(15 \mathrm{~mm}\) from the leading edge of the board. (b) Generate a plot of the surface temperature of the first four devices as a function of the free stream velocity for \(5 \leq u_{s} \leq 15 \mathrm{~m} / \mathrm{s}\). (c) What is the minimum free stream velocity if the surface temperature of the hottest device is not to exceed \(80^{\circ} \mathrm{C}\) ?

A spherical droplet of alcohol, \(0.5 \mathrm{~mm}\) in diameter, is falling freely through quiescent air at a velocity of \(1.8 \mathrm{~m} / \mathrm{s}\). The concentration of alcohol vapor at the surface of the droplet is \(0.0573 \mathrm{~kg} / \mathrm{m}^{3}\), and the diffusion coefficient for alcohol in air is \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). Neglecting radiation and assuming steady-state conditions, calculate the surface temperature of the droplet if the ambient air temperature is \(300 \mathrm{~K}\). The latent heat of vaporization is \(8.42 \times 10^{5} \mathrm{~J} / \mathrm{kg}\).

In the production of sheet metals or plastics, it is customary to cool the material before it leaves the production process for storage or shipment to the customer. Typically, the process is continuous, with a sheet of thickness \(\delta\) and width \(W\) cooled as it transits the distance \(L\) between two rollers at a velocity \(V\). In this problem, we consider cooling of plain carbon steel by an airstream moving at a velocity \(u_{\infty}\) in cross flow over the top and bottom surfaces of the sheet. A turbulence promoter is used to provide turbulent boundary layer development over the entire surface. (a) By applying conservation of energy to a differential control surface of length \(d x\), which either moves with the sheet or is stationary and through which the sheet passes, and assuming a uniform sheet temperature in the direction of airflow, derive a differential equation that governs the temperature distribution, \(T(x)\), along the sheet. Consider the effects of radiation, as well as convection, and express your result in terms of the velocity, thickness, and properties of the sheet \(\left(V, \delta, \rho, c_{p}, \varepsilon\right)\), the average convection coefficient \(\bar{h}_{W}\) associated with the cross flow, and the environmental temperatures \(\left(T_{\infty}, T_{\text {sur }}\right)\). (b) Neglecting radiation, obtain a closed form solution to the foregoing equation. For \(\delta=3 \mathrm{~mm}, V=\) \(0.10 \mathrm{~m} / \mathrm{s}, L=10 \mathrm{~m}, W=1 \mathrm{~m}, u_{\infty}=20 \mathrm{~m} / \mathrm{s}, T_{\infty}=\) \(20^{\circ} \mathrm{C}\), and a sheet temperature of \(T_{i}=500^{\circ} \mathrm{C}\) at the onset of cooling, what is the outlet temperature \(T_{o}\) ? Assume a negligible effect of the sheet velocity on boundary layer development in the direction of airflow. The density and specific heat of the steel are \(\rho=7850 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c_{p}=620 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), while properties of the air may be taken to be \(k=0.044\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}, \nu=4.5 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \operatorname{Pr}=0.68\). (c) Accounting for the effects of radiation, with \(\varepsilon=\) \(0.70\) and \(T_{\text {sur }}=20^{\circ} \mathrm{C}\), numerically integrate the differential equation derived in part (a) to determine the temperature of the sheet at \(L=10 \mathrm{~m}\). Explore the effect of \(V\) on the temperature distribution along the sheet.
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