/*! 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 34 The rate of heat flow (conductio... [FREE SOLUTION] | 91Ó°ÊÓ

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The rate of heat flow (conduction) between two points on a cylinder heated at one end is given by \(\frac{d Q}{d t}=\lambda A \frac{d T}{d x}\) where \(\lambda=\mathrm{a}\) constant, \(A=\) the cylinder's cross-sectional area, \(Q=\) heat flow, \(T=\) temperature, \(t=\) time, and \(x=\) distance from the heated end. Because the equation involves two derivatives, we will simplify this equation by letting \(\frac{d T}{d r}=\frac{100(L-x)(20-t)}{100-x t}\) where \(L\) is the length of the rod. Combine the two equations and compute the heat flow for \(t=0\) to 25 s. The initial condition is \(Q(0)=0\) and the parameters are \(\lambda=0.5 \mathrm{cal} \cdot \mathrm{cm} / \mathrm{s}, A=12 \mathrm{cm}^{2}\).

Short Answer

Expert verified
In summary, we can compute the heat flow between two points on a cylinder by combining the given equations for rate of heat flow and temperature derivative, substituting the given constant values, and calculating the definite integral from \(t=0\) to \(t=25\) seconds. The final equation to be solved numerically is: \[Q(t) = \int_0^{25} 0.5 \cdot 12 \frac{100(L-x)(20-t)}{100-xt} dt\] Use a numerical integration method to obtain the heat flow within the specified time interval.

Step by step solution

01

Combining the given equations

We are given the equation for the rate of heat flow: \[\frac{dQ}{dt} = \lambda A \frac{dT}{dx}\] And another equation for the temperature derivative: \[\frac{dT}{dx} = \frac{100(L-x)(20-t)}{100-xt}\] Next, we will substitute the second equation into the first equation to create a single equation.
02

Substitute and Simplify

Substituting the second equation into the first equation, we get: \[\frac{dQ}{dt} = \lambda A \frac{100(L-x)(20-t)}{100-xt}\] Given that \(\lambda = 0.5 cal \cdot cm / s\) and \(A = 12 cm^2\), replace these values in the above equation: \[\frac{dQ}{dt} = 0.5 \cdot 12 \frac{100(L-x)(20-t)}{100-xt}\] Now, we have a single equation representing the heat flow rate.
03

Solve for Heat Flow from t=0 to 25 s

To find the heat flow \(Q(t)\), we need to integrate the rate of heat flow with respect to time from 0 to 25 seconds: \[Q(t) = \int_0^{25} \frac{dQ}{dt} dt\] \[Q(t) = \int_0^{25} 0.5 \cdot 12 \frac{100(L-x)(20-t)}{100-xt} dt\] We will solve this definite integral using an appropriate numerical integration method, as the integral is not easily solvable analytically.
04

Initial Condition and Parameters

Don't forget the initial condition \(Q(0) = 0\). The integral gives us the change in heat flow, so by adding the initial condition, we get the total heat flow: \[Q(t) = Q(0) + \int_0^{25} 0.5 \cdot 12 \frac{100(L-x)(20-t)}{100-xt} dt\] \[Q(t) = 0 + \int_0^{25} 0.5 \cdot 12 \frac{100(L-x)(20-t)}{100-xt} dt\] Now, you can use a numerical method to solve the integral, for example Simpson's rule or a numerical integration function in a software like MATLAB or Python. The result will give you the heat flow within the time interval from \(t=0\) to \(t=25\) seconds.

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Key Concepts

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

Heat Transfer
Heat transfer is a fundamental concept in thermodynamics, involving the movement of thermal energy from one object to another caused by a temperature difference. In the context of the problem, we're discussing conduction, a mode of heat transfer through a material without the movement of the material itself. It typically occurs in solids, where particles are closely packed, allowing energy to transfer efficiently.

The key equation presented, \(\frac{dQ}{dt} = \lambda A \frac{dT}{dx}\), reflects this process. Here, \(\frac{dQ}{dt}\) represents the rate of heat flow, \(\lambda\) is the thermal conductivity of the material (a constant that indicates how well the material conducts heat), \(A\) is the cross-sectional area, and \(\frac{dT}{dx}\) is the temperature gradient along the path of heat transfer.
  • Greater thermal conductivity (\(\lambda\)) means higher heat transfer efficiency.
  • A larger cross-sectional area allows more heat to be conducted at once.
These parameters interact to determine how effectively heat moves through the cylinder, making it vital to understand each when evaluating or solving heat transfer problems like this one.
Differential Equations
Differential equations are mathematical expressions that describe relationships involving rates of change. They are crucial in engineering and physics, helping to model dynamic systems that evolve over time.

In this problem, we have a partial differential equation shown as \(\frac{dQ}{dt} = \lambda A \frac{dT}{dx}\). This equation highlights the interplay between variables like temperature change over distance \(\left(\frac{dT}{dx}\right)\) and time-based changes \(\left(\frac{dQ}{dt}\right)\).

Solving differential equations often involves several steps:
  • First, substituting expressions to unify equations, as seen when substituting \(\frac{dT}{dx}\) in the heat flow expression.
  • Next, simplifying the equation to isolate desired variables or rates.
  • Finally, integrating over the relevant interval, here time from \(t=0\) to \(t=25\) seconds, to find the cumulative effect on the system, represented as \(Q(t)\).
Understanding these steps allows engineers and scientists to predict how systems behave under various conditions, making differential equations an indispensable tool in fields involving modeling and simulations.
Engineering Mathematics
Engineering mathematics involves applying mathematical methods and techniques to solve practical problems in engineering fields. It combines knowledge from calculus, algebra, and numerical methods to provide powerful tools for analysis and design in engineering contexts.

In the provided exercise, engineering mathematics is used through the application of calculus – specifically, integration – to find a precise solution for heat flow over time. The problem involves:
  • Formulating equations from given problem conditions.
  • Using calculus (integrals) to calculate changes and predict outcomes.
  • Employing numerical integration methods like Simpson's rule or software tools for non-analytic integrals.
These techniques allow engineers to solve complex equations that arise in real-world applications where analytical solutions are not feasible. Mastery of engineering mathematics ensures accurate modeling, efficient design processes, and problem-solving capabilities, which are essential in developing innovative and effective engineering solutions.

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Most popular questions from this chapter

A forced damped spring-mass system (Fig. \(\mathrm{P} 28.38\) ) has the following ordinary differential equation of motion: \\[ m \frac{d^{2} x}{d t^{2}}+a\left|\frac{d x}{d t}\right| \frac{d x}{d t}+k x=F_{o} \sin (\omega t) \\] where \(x=\) displacement from the equilibrium position, \(t=\) time, \(m=2 \mathrm{kg}\) mass \(, a=5 \mathrm{N} /(\mathrm{m} / \mathrm{s})^{2}\), and \(k=6 \mathrm{N} /(\mathrm{m} / \mathrm{s})^{2}\). The damping term is nonlinear and represents air damping. The forcing function \(F_{0} \sin (\omega t)\) has values of \(F_{o}=2.5 \mathrm{N}\) and \(\omega=0.5 \mathrm{rad} / \mathrm{sec} .\) The initial conditions are Initial velocity' \\[ \frac{d x}{d t}=0 \mathrm{m} / \mathrm{s} \\] Initial displacement \\[ x=1 \mathrm{m} \\] Solve this equation using a numerical method over the time period \(0 \leq t \leq 15\) s. Plot the displacement and velocity versus time, and plot the forcing function on the same curve. Also, develop a separate plot of velocity versus displacement.

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