diff --git a/uni/mmme/1xxx/1026_maths_for_engineering/eigenvalues.md b/uni/mmme/1xxx/1026_maths_for_engineering/eigenvalues.md new file mode 100755 index 0000000..9d3bcc2 --- /dev/null +++ b/uni/mmme/1xxx/1026_maths_for_engineering/eigenvalues.md @@ -0,0 +1,64 @@ +--- +author: Akbar Rahman +date: \today +title: Eigenvalues +tags: [ mmme1026, maths, eigenvalues, uni ] +uuid: f2220395-bc97-432e-a1d2-74085f16991d +--- + +An eigenvalue problem takes the form: + +Find all the values of $\lambda$ for which the equation + +$$A\pmb{x} = \lambda \pmb{x}$$ + +has a nonzero solution $\pmb x$, where $A$ is an $n\times n$ matrix and +$\pmb x$ is a column vector. + +The equation may be written as + +\begin{align*} +A\pmb x &= \lambda I \pmb x \\ +\Leftrightarrow A \pmb x - \lambda I \pmb x & = 0 \\ +\Leftrightarrow (A-\lambda I)\pmb x &= 0 +\end{align*} + +($\Leftrightarrow$ means "if and only if") + +Non-zero solutions will exist if + +$\det(A-\lambda I) = 0$ + +There are infinitely many eigenvectors for a given eigenvalue. +This is because if $\pmb x$ is an eigenvector of $A$ corresponding to the +eigenvalue $\lambda$ and $c$ is a non-zero scalar, then $c\pmb x$ is also +an eigenvector of $A$: + +$$A(c\pmb x) = cA\pmb x = c\lambda \pmb x = \lambda(c\pmb x)$$ + +In general, if $A$ is an $n\times n$ matrix, then $|A-\lambda I|$ is a +polynomial of degree $n$ in $\lambda$, called the characteristic polynomial. +The characteristic equation is: + +$$\lambda^n + c_{n-1}\lambda^{n-1} + c_{n-2}\lambda^{n-2} + \cdots + c_0 = 0$$ + +
+ + +#### Example 1 ($2\times2$ example) + + + +If $A$ is the matrix + +$$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ + +then + +$$|A - \lambda I| = \lambda^2 - (a+d)\lambda + (ad-bc)$$ + +And the standard method for solving a quadratic can be used to find $\lambda$. + +
+ + diff --git a/uni/mmme/1xxx/1028_statics_and_dynamics/dynamics.md b/uni/mmme/1xxx/1028_statics_and_dynamics/dynamics.md index 86aa0b6..40a3cb0 100755 --- a/uni/mmme/1xxx/1028_statics_and_dynamics/dynamics.md +++ b/uni/mmme/1xxx/1028_statics_and_dynamics/dynamics.md @@ -11,3 +11,62 @@ uuid: e6d3a307-b2e6-40e3-83bb-ef73512d69ad $$a_c = r\omega^2$$ $$a = r\alpha \hat{e}_\theta - r\omega^2\hat{e}_r$$ + +## Moment of Inertia + +$$J = mr^2 = \frac{M}{\ddot\theta}$$ + +The unit of $J$ is kgm$^2$. + +Consider a particle of mass $m$ attached to one end of a rigid rod of length $r$. +The rod is pivoting at its other end about point $O$, and experiences a torque $M$. +This torque will cause the mass and the rod to rotate about $O$ with angular velocity +$\dot{\theta}$ an angular acceleration $\ddot{\theta}$. + +![](./images/vimscrot-2022-03-10T14:40:59,716300890+00:00.png) + +What is the expression for $M$? + +Well if break down the moment $M$ into a force, $F$, acting on the mass, we know that the +moment $M = Fr$. +We know $F = ma$, and this case $a = r\ddot{\theta}$ so $M = mr^2\ddot\theta$. + +The moment of inertia is $J = mr^2$ so $M = J\ddot\theta$. + +If multiple torques are applied to a body the *rotational equation* of the motion is + +$$\overrightarrow{M} = \sum_i \overrightarrow{M}_i = J \overrightarrow{\ddot\theta} = J \overrightarrow{\alpha}$$ + +The moment of inertia of any object is found by considering the object to be made up of lots of +small particles and adding the moments of inertia for each small particle. +The moments of inertia for a body depends on the mass and its distribution about the axis in +consideration. + +$$J = \sum_i m_ir^2_i \rightarrow \int\! r^2 \mathrm{d}m$$ + +### Perpendicular Axis Rule + +The perpendicular axis rule states that, for lamina object: + +$$J_z = J_x + J_y$$ + +where $J_x$, $J_y$, and $J_z$ are the moments of inertia along their respective axes. + +### Parallel Axes Rule (Huygens-Steiner Theorem) + +The parallel axes rule states that: + +$$J_A = J_G = md^2$$ + +where $d$ is the perpendicular distance between the two axes. + +![](./images/vimscrot-2022-03-10T15:06:48,355133323+00:00.png) + +### Moment of a Compound Object + +The moment of inertia for any compound object can be calculated by adding and subtracting the +moments of inertia for its 'standard' components. + +### Moment of Inertia of Standard Objects + +// TODO diff --git a/uni/mmme/1xxx/1028_statics_and_dynamics/images/vimscrot-2022-03-10T14:40:32,541011622+00:00.png b/uni/mmme/1xxx/1028_statics_and_dynamics/images/vimscrot-2022-03-10T14:40:32,541011622+00:00.png new file mode 100644 index 0000000..bfb0fc6 Binary files /dev/null and b/uni/mmme/1xxx/1028_statics_and_dynamics/images/vimscrot-2022-03-10T14:40:32,541011622+00:00.png differ diff --git a/uni/mmme/1xxx/1028_statics_and_dynamics/images/vimscrot-2022-03-10T14:40:59,716300890+00:00.png b/uni/mmme/1xxx/1028_statics_and_dynamics/images/vimscrot-2022-03-10T14:40:59,716300890+00:00.png new file mode 100644 index 0000000..2c5f0a9 Binary files /dev/null and b/uni/mmme/1xxx/1028_statics_and_dynamics/images/vimscrot-2022-03-10T14:40:59,716300890+00:00.png differ diff --git a/uni/mmme/1xxx/1028_statics_and_dynamics/images/vimscrot-2022-03-10T15:06:48,355133323+00:00.png b/uni/mmme/1xxx/1028_statics_and_dynamics/images/vimscrot-2022-03-10T15:06:48,355133323+00:00.png new file mode 100644 index 0000000..085f1d1 Binary files /dev/null and b/uni/mmme/1xxx/1028_statics_and_dynamics/images/vimscrot-2022-03-10T15:06:48,355133323+00:00.png differ diff --git a/uni/mmme/1xxx/1048_thermodynamics_and_fluid_mechanics/Example Sheet 10 - 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Q_{out}$ + +$$\eta = \frac{W_{out}}{Q_{in}} = 1 - \frac{Q_{out}}{Q_{in}}$$ + +Early steam engines had efficiency around 10% but large diesel engines nowadays have efficiencies +up to around 50%, with petrol engines around 30%. +The most efficient heat engines we have are large gas-steam power plants, at around 60%. + +## Carnot Efficiency + +The maximum efficiency for a heat engine that operates reversibly between the heat source and heat +sink is known as the *Carnot Efficiency*: + +$$\eta_{carnot} = 1 - \frac{T_2}{T_1}$$ + +where $T$ is in Kelvin (or any unit of absolute temperature, I suppose) + +Therefore to maximise potential efficiency, you want to maximise input heat temperature, and +minimise output heat temperature. + +The efficiency of any heat engine will be less than $\eta_{carnot}$ if it operates between more than +two reservoirs. + +## Reversible and Irreversible Processes + +### Reversible Processes + +A reversible process operate at thermal and physical equilibrium. +There is no degradation in the quality of energy. + +There must be no mechanical friction, fluid friction, or electrical resistance. + +Heat transfers must be across a very small temperature difference. + +All expansions must be controlled. + +### Irreversible Processes + +In irreversible processes, the quality of the energy degrades. +For example, mechanical energy degrades into heat by friction and heat energy degrades into lower +quality heat (a lower temperature), including by mixing of fluids. + +Thermal resistance at both hot sources and cold sinks are an irreversibility and reduce efficiency. + +There may also be uncontrolled expansions or sudden changes in pressure. + +# Energy Quality + +## Quantifying Disorder (Entropy) + +$$S = k\log_eW$$ + +where $S$ is entropy, $k = 1.38\times10^{-23}$ J/K is Boltzmann's constant, and $W$ is the number of +ways of reorganising energy.s