tag:blogger.com,1999:blog-21008936046539514492014-10-05T02:49:29.063-04:00Kiwis and LimesNotes from LifeKaiting Chenhttp://www.blogger.com/profile/06712558527665170123noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-2100893604653951449.post-30559275637483597292011-05-03T00:25:00.004-04:002011-05-03T00:31:32.653-04:00Momentum Transport?<div class="hyphenate" style="text-align: justify; visibility: hidden;">
<p style="font-style: italic;">This post assumes that the reader has knowledge of basic vector calculus and physics.</p>
<p>In the study of fluid dynamics the term ‘momentum transport’ is thrown around quite often and often in conjunction with the idea of ‘momentum accumulation’; it is a rather difficult concept to understand because it is quite impossible to <em>see</em> momentum. Hopefully this post will help to clarify the concept.</p>
<p>We will start with an easy to visualize concept: ‘mass transport’. Let us assume that there is a laminar and steady fluid flow of density <span class="math">\(\rho\)</span> and velocity <span class="math">\(\mathbf{v}\)</span>. We would like to know the rate <span class="math">\(K_T\)</span> at which mass is transported in this flow. The answer—</p>
<span class="math">\[ K_T = \rho |\mathbf{v}| \]</span>
<p>Makes sense until you consider the units of <span class="math">\(K_T\)</span> and realize that they are <span class="math">\(\frac{\text{g}}{\text{cm}^2 \cdot \text{s}}\)</span> which definitely doesn't make much sense. In fact it is difficult to determine at this point what the correct units of the rate of mass transport should even be. To solve this problem we need to consider mass transport from the other side of the equation: ‘mass accumulation’.</p>
<a name='more'></a>
<p>If the total mass transport <span class="math">\(\sum K_T\)</span> to a location is not equal to zero we should see accumulation or dispersion at that location. We can therefore characterize the rate at which mass is accumulating in terms of the mass transport rate such that—</p>
<span class="math">\[ \sum K_T = \sum \frac{dm}{dt} \text{ so } K_T = \frac{dm}{dt} \]</span>
<span class="math">\[ \frac{dm}{dt} = \frac{dm}{dV} \frac{dV}{dt} = \rho\,Q\]</span>
<p>Where \(m\) is mass, \(V\) is volume, and <span class="math">\(Q\)</span> is the <em>volumetric flow rate</em> and is defined as—</p>
<span class="math">\[ Q = \int_A \mathbf{v} \cdot \mathbf{n}\,dS \]</span>
<p>Where <span class="math">\(A\)</span> is the cross sectional surface through which the fluid flow occurs, <span class="math">\(\mathbf{n}\)</span> is the unit outward vector on this surface, and <span class="math">\(dS\)</span> is the surface element. If we then define the average velocity <span class="math">\(\left<v\right>\)</span> as—</p>
<span class="math">\[ \left<v\right> = \frac{1}{A} \int_A \mathbf{v} \cdot \mathbf{n}\,dS \]</span>
<p>And assume that the cross sectional surface is constant we can the mass transport rate as—</p>
<span class="math">\[ \sum K_T = A \sum \rho \left<v\right> \]</span>
<p>Now we have the framework with which to address the issue of momentum transport. Let us assume the same fluid flow as above and describe the rate of momentum transport accumulation—</p>
<span class="math">\[ \sum K_T = \sum \frac{d(m \mathbf{v})}{dt} \text{ so } K_T = \frac{d(m \mathbf{v})}{dt} \]</span>
<span class="math">\[ \frac{d(m \mathbf{v})}{dt} = \frac{d(m \mathbf{v})}{dV} \frac{dV}{dt} \]</span>
<span class="math">\begin{aligned} K_T &= \left(m \frac{d \mathbf{v}}{dV} + \mathbf{v} \frac{dm}{dV}\right) \frac{dV}{dt} \\ &= m \frac{d \mathbf{v}}{dt} + \mathbf{v} \frac{dm}{dV} \frac{dV}{dt} \end{aligned}</span>
<p>We now make the same substitution that we made before and take into account that force \(\mathbf{F}\) and acceleration are \(\mathbf{a}\) then—</p>
<span class="math">\[ \mathbf{F} = m \mathbf{a} \text{ where } \mathbf{a} = \frac{d \mathbf{v}}{dt} \]</span>
<span class="math">\[ \sum K_T = \sum \mathbf{F} + A \sum \rho \left<v\right> \mathbf{v} \]</span>
<p>This result is typically interpreted as that the rate of accumulation of momentum is equal to the sum of the momentum transport rate and the net external force on the system.</p>
</div>Kaiting Chenhttp://www.blogger.com/profile/06712558527665170123noreply@blogger.com0tag:blogger.com,1999:blog-2100893604653951449.post-19018992396659300832010-09-07T20:57:00.066-04:002011-05-02T23:07:36.925-04:00Complex Impedance<div class="hyphenate" style="text-align: justify; visibility: hidden;">
<p style="font-style: italic;">This post assumes a working knowledge of elementary circuit theory as well as Fourier Analysis.</p>
<p>It seems to me that often in an introductory circuits course complex impedance is a major concept but its mathematical basis is never taught. Thus I'd like to take a moment to discuss some of the mathematics surrounding this concept.</p>
<p>The resistor is a purely resistive elementary circuit element possessing a time independent current voltage characteristic described by Ohm's Law.</p>
\[ V(t) = I(t) R \]
<p>The capacitor is a purely reactive elementary circuit element that stores energy in its electric field.</p>
\[ I(t) = C \frac{dV(t)}{dt} \]
<p>The inductor is a purely reactive elementary circuit element that stores energy in its magnetic field.</p>
\[ V(t) = L \frac{dI(t)}{dt} \]
<p>What we aim to do is to derive a linear current voltage characteristic for the two reactive circuit elements by reducing the differential equations to algebraic equations. Using the Fourier Transform we can compute a complex quantity in the frequency domain that is analogous to resistance in the time domain.</p><a name='more'></a>
<p>The resistance is both time independent and frequency independent so its corresponding complex impedance is rather boring.</p>
<span class="math">\begin{aligned}
\newcommand{\FT}[1]{\mathcal F \left\{#1\right\}}
\newcommand{\e}[1]{e^{#1}}
\FT{V(t)} &= \FT{I(t) R} \\
V(j\omega) &= I(j\omega) R \\
\frac{V(j\omega)}{I(j\omega)} &= Z_R = R
\end{aligned}</span>
<p>The capacitance is a reactive quantity so its corresponding complex impedance is a phase shift.</p>
<span class="math">\begin{aligned}
\FT{i(t)} &= \FT{C \frac{dV(t)}{dt}} \\
I(j \omega) &= C j \omega V(j \omega) \\
\frac{V(j \omega)}{I(j \omega)} &= Z_C = \frac{1}{j \omega C}
\end{aligned}</span>
<p>The inductance is a reactive quantity so its corresponding complex impedance is a phase shift.</p>
<span class="math">\begin{aligned}
\FT{V(t)} &= \FT{L \frac{di(t)}{dt}} \\
V(j \omega) &= L j \omega I(j \omega) \\
\frac{V(j \omega)} {I(j \omega)} &= Z_L = j \omega L
\end{aligned}</span>
<p>For this concept of complex impedance to be useful it is necessary to find appropriate expressions for the current and voltage in the frequency domain. It is common for both the input and output to a system be sinusoidal which results in their Fourier Transforms being less than ideal expressions to work it. In this case we determine the coefficients of the Fourier Series and use those as our input and output.</p>
<p>Let us assume that we have a simple series LC circuit driven by a voltage source and we are asked to find the current passing through the resistor. In this case our input voltage is a sinusoid.</p>
\[ V(t) = A \cos(j \omega t + \phi) \]
<p>We can express this input as a complex exponential using Euler's Formula. This representation is the phasor or phase vector of the input.</p>
\[ V(t) = \Re\{A \e{j \omega t} \e{\phi}\} \]
<p>Because we wish to work with the input in the frequency domain by using the coefficients of the Fourier Series we transform the input by taking the coefficients of the time dependent portion of the signal.</p>
\[ \tilde{V} = \Re\{A \e{\phi}\} \]
<p>We can calculate the total impedance of the circuit by combining the impedance of each individual element in the same manner that we use to calculate the total resistance.</p>
\[ \tilde{Z} = R + j \omega L \]
<p>We can express this impedance as a complex exponential using Euler's Formula. This representation is the phasor or phase vector of the impedance.</p>
<span class="math">\[ \tilde{Z} = \left|\tilde{Z}\right| \e{\varphi \tilde{Z}} = \sqrt{R^2 + \omega^2 L^2} \e{\arctan\frac{\omega L}{R}} \]</span>
<p>To calculate the current passing through the resistor we then apply Ohm's Law in the frequency domain.</p>
<span class="math">
\begin{aligned} \tilde{I} &= \frac{\tilde{V}}{\tilde{Z}} =
\Re\left\{ \frac{A \e{\phi}}{
\sqrt{R^2 + \omega^2 L^2} \e{\arctan
\frac{\omega L}{R}}} \right\} \\
&= \frac{A}{\sqrt{R^2 + \omega^2 L^2}}
\Re\left\{ \e{\phi -
\arctan \frac{\omega L}{R}} \right\} \end{aligned}
</span>
<p>At this point we have the frequency domain representation of the output. Usually we wish to convert this back to the time domain representation. In this case the time domain representation is a sinusoid.</p>
<span class="math"> \[ i(t) = \frac{A}{\sqrt{R^2 + \omega^2 L^2}} \cos \left(j \omega t + \phi - \arctan \frac{\omega L}{R} \right) \] </span>
<p>This is the general procedure involved in solving a reactive circuit in the frequency domain. Note that while the example shows a cosinusoidal input the input can be any function. A periodic function can be decomposed into a Fourier Series and solved by taking the coefficients of the terms as the input phasors while a nonperiodic function can be solved by using an input phasor equal to its Fourier Transform. Of course a periodic function can also be solved by using an input phasor equal to its Fourier Transform but this is usually more work because the Fourier Transform of a periodic function usually involves Dirac Delta functions which are not easily manipulated.</p></div>Kaiting Chenhttp://www.blogger.com/profile/06712558527665170123noreply@blogger.com0