プロセス工学数学A
Takuya Kawanishi
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Solutions for Quizzes and Exercises

Macroscopic Balance

Solution, Exercise 1

\[\frac{dC(t)}{dt}+ \frac FV C(t) = \frac FV c(1 + \sin t)\]

which is a first order linear (ordinary) differential equation.

Integrating factor

\[\mu(t) = e^{\int^t \frac{F}{V} \, dt} = e^{tF/V}\]
\[\frac{1}{\mu(t)} \left(\int^t e^{tF/V} \frac FV c (1 + \sin t) \, dt + A \right), \quad A\text{: arbitrary constant}\]

  • Equality

    \[\int e^{at} \sin {bt} \, dt = \frac{e^{at}}{a^2 + b^2}\left( a \sin bt - b \cos bt \right)\]
\[\int e^{tF/V} \sin t dt = \frac{e^{tF/V}}{(F/V)^2 + 1} \left(\frac{F}{V}\sin t - \cos t \right)\]
\[\int e^{tF/V} \,dt = \frac{V}{F} e^{tF/V}\]
\[\int^t e^{tF/V} \frac FV c (1 + \sin t) \, dt = \frac{cF}{V} \left\{ \frac{e^{tF/V}}{(F/V)^2 + 1} \left(\frac{F}{V}\sin t - \cos t \right) + \frac{V}{F} e^{tF/V} \right\}\]

The general solution \(y\) is

\[y = \frac{1}{e^{tF/V}} \left\{ \frac {cF}{V} \frac{e^{tF/V}}{(F/V)^2 + 1} \left(\frac{F}{V} \sin t - \cos t \right) + ce^{tF/V} + A \right\}\]
\[y = \frac{cF/V}{(F/V)^2 + 1}\left(\frac{F}{V} \sin t - \cos t \right) + c + A e^{-tF/V}\]
\[A = - c + c \frac{F/V}{(F/V)^2 + 1}\]
\[C(t) = \frac{cF/V}{(F/V)^2 + 1}\left(\frac{F}{V}\sin t - \cos t + e^{-tF/V}\right) + c\left\{ 1 - \exp\left( - \frac{F}{V} t \right) \right\}\]
\[\frac{dC(t)}{dt}+ \frac FV C(t) = \frac FV c(1 + \sin t)\]

which is a first order linear (ordinary) differential equation.

Integrating factor

\[\mu(t) = e^{\int^t \frac{F}{V} \, dt} = e^{tF/V}\]
\[\frac{1}{\mu(t)} \left(\int^t e^{tF/V} \frac FV c (1 + \sin t) \, dt + C \right)\]

  • Equality

    \[\int e^{at} \sin {bt} \, dt = \frac{e^{at}}{a^2 + b^2}\left( a \sin bt - b \cos bt \right)\]
\[\int e^{tF/V} \sin t dt = \frac{e^{tF/V}}{(F/V)^2 + 1} (\sin t - \cos t)\]
\[\int e^{tF/V} \,dt = \frac{V}{F} e^{tF/V}\]
\[\int \cdot \, dt = \frac{cF}{V} \left\{ \frac{e^{tF/V}}{(F/V)^2 + 1} (\sin t - \cos t) + \frac{V}{F} e^{tF/V} \right\}\]

The general solution \(y\) is

\[y = \frac{1}{e^{tF/V}} \left\{ \frac {cF}{V} \frac{e^{tF/V}}{(F/V)^2 + 1} (\sin t - \cos t) + ce^{tF/V} + A \right\}\]
\[y = \frac{cF/V}{(F/V)^2 + 1}(\sin t - \cos t) + c + A e^{-tF/V}\]
\[A = - c + c \frac{F/V}{(F/V)^2 + 1}\]
\[C(t) = \frac{cF/V}{(F/V)^2 + 1}(\sin t - \cos t + e^{-tF/V}) + c\left\{ 1 - \exp\left( - \frac{F}{V} t \right) \right\}\]
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Linear Ordinary Differential Equations

Solution, Quiz 1

(a.1) Linear

(a.2) Linear

(a.3) Nonlinear, term \(y^{(2)} y\) is nonlinear.

(a.4) Linear

(a.5) Linear

(a.6) Linear, equivalent to \(y^{(2)} + y^\prime + (1 + x^2) y = 0\)

(a.7) Nonlinear, term \(y^2\) is nonlinear.

(a.8) Nonlinear, term \(y^\prime y\) is nonlinear.

Linear Partial Differential Equations

Solution, Quiz 1

  • First, dividing the both sides of (xp. 1) by \(T_h - T_l\) does not affect the coefficient.

  • That is,

    \begin{equation} \frac{\partial u}{\partial t} = \frac{k}{\rho C_p} \frac{\partial^2 u}{\partial x^2} \end{equation}

  • Next,

\begin{equation*} \frac{\partial^2 u}{\partial s^2} = \frac{\partial^2 u}{\partial (pL)^2} = \frac{\partial^2 u}{\partial x^2} \frac{1}{L^2} \end{equation*}

  • Note that constants can be freely moved between inside and outside of the integral.

  • Then we have

    \begin{equation} \frac{\partial u}{\partial t} = \frac{k}{\rho C_p} \frac{1}{L^2} \frac{\partial^2 u}{\partial x^2} \end{equation}

  • Thus,

    \[\alpha^2 = \frac{k}{\rho C_p L^2}\]

  • Dimension of \(\alpha^2\)

    \[\left[ \frac{k L^2}{\rho C_p} \right] = \mathrm{\frac{\left[\frac{J}{s~m~K}\right]}{\left[\frac{kg}{m^3}\right]\left[\frac{J}{kg~K}\right]\left[m^2\right]} = \left[\frac{1}{s} \right]}\]

Solution, Quiz 2

  • Substitute \(t = s t_0\) for \(t\) in (xp. 1),

    \[\frac{\partial u}{\partial (s t_0)} = \alpha^2 \frac{\partial^2 u}{\partial p^2}\]
    \[\frac{\partial u(p, s)}{\partial s} = t_0 \alpha^2 \frac{\partial^2 u(p, t)}{\partial p^2}\]

  • That is, \(t_0 \alpha^2 = 1\), hence

    \[t_0 = \frac{1}{\alpha^2} = \frac{\rho C_p L^2}{k}\]

  • The dimension of \(t_0 = [ \mathrm s ]\)


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