mmme2046 lecture 1

uni/mmme/1xxx/1048_thermodynamics_and_fluid_mechanics/fluid_dynamics.md

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+---
+author: Alvie Rahman
+date: \today
+title: MMME1048 // Fluid Dynamics
+tags:
+- uni
+- nottingham
+- mechanical
+- engineering
+- fluid_mechanics
+- mmme1048
+- fluid_dynamics
+uuid: b88f78f8-a358-460b-9dbb-812e7b1ace92
+---
+
+\newcommand\Rey{\text{Re}}
+\newcommand\textRey{$\Rey$}
+
+# Introductory Concepts
+
+These are ideas you need to know about to know what's going on, I guess?
+
+## Control Volumes
+
+A control volume is a volume with an imaginary boundary to make it easier to analyse the flow of a
+fluid.
+The boundary is drawn where the properties and conditions of the fluid is known, or where an
+approximation can be made.
+Properties which may be know include:
+
+- Velocity
+- Pressure
+- Temperature
+- Viscosity
+
+The region in the control volume is analysed in terms of energy and mass flows entering and leaving
+the control volumes.
+You don't have to understand what's going on inside the control volume.
+
+<details>
+<summary>
+
+### Example 1
+
+The thrust of a jet engine on an aircraft at rest can be analysed in terms of the changes in
+momentum or the air passing through the engine.
+
+</summary>
+
+![](./images/vimscrot-2021-11-03T21:51:51,497459693+00:00.png)
+
+The control volume is drawn far enough in front of the engine that the air velocity entering can
+be assumed to be at atmospheric pressure and its velocity negligible.
+
+At the exit of the engine the boundary is drawn close where the velocity is known and the air
+pressure atmospheric.
+
+The control volume cuts the material attaching the engine to the aircraft and there will be a force
+transmitted across the control volume there to oppose the forces on the engine created by thrust
+and gravity.
+
+The details of the flows inside the control volume do not need to be known as the thrust can be 
+determined in terms of forces and flows crossing the boundaries drawn.
+However, to understand the flows inside the engine in more detail, a more detailed analysis would
+be required.
+
+</details>
+
+## Ideal Fluid
+
+The actual flow pattern in a fluid is usually complex and difficult to model but it can be
+simplified by assuming the fluid is ideal.
+The ideal fluid has the following properties:
+
+- Zero viscosity
+- Incompressible
+- Zero surface tension
+- Does not change phases
+
+Gases and vapours are compressible so can only be analysed as ideal fluids when flow velocities are
+low but they can often be treated as ideal (or perfect) gases, in which case the ideal gas equations
+apply.
+
+## Steady Flow
+
+Steady flow is a flow which has *no changes in properties with respect to time*.
+Properties may vary from place to place but in the same place the properties must not change in
+the control volume to be steady flow.
+
+Unsteady flow does change with respect to time.
+
+## Uniform Flow
+
+Uniform flow is when all properties are the same at all points at any given instant but can change
+with respect to time, like the opposite of steady flow.
+
+## One Dimensional Flow
+
+In one dimensional (1D) flow it is assumed that all properties are uniform over any plane
+perpendicular to the direction of flow (e.g. all points along the cross section of a pipe have
+identical properties).
+
+This means properties can only flow in one direction---usually the direction of flow.
+
+1D flow is never achieved exactly in practice as when a fluid flows along a pipe, the velocity at
+the wall is 0, and maximum in the centre of the pipe.
+Despite this, assuming flow is 1D simplifies the analysis and often is accurate enough.
+
+## Flow Patterns
+
+There are multiple ways to visualize flow patterns.
+
+### Streamlines
+
+A streamline is a line along which all the particle have, at a given instant, velocity vectors
+which are tangential to the line.
+
+Therefore there is no component of velocity of a streamline.
+
+A particle can never cross a streamline and *streamlines never cross*.
+
+They can be constructed mathematically and are often shown as output from CFD analysis.
+
+For a steady flow there are no changes with respect to time so the streamline pattern does not.
+The pattern does change when in unsteady flow.
+
+Streamlines in uniform flow must be straight and parallel.
+They must be parallel as if they are not, then different points will have different directions and
+therefore different velocities.
+Same reasoning with if they are not parallel.
+
+### Pathlines
+
+A pathline shows the route taken by a single particle during a given time interval.
+It is equivalent to a high exposure photograph which traces the movement of the particle marked.
+You could track pathlines with a drop of injected dye or inserting a buoyant solid particle which
+has the same density as the solid.
+
+Pathlines may cross.
+
+### Streaklines
+
+A streakline joins, at any given time, all particles that have passed through a given point.
+Examples of this are line dye or a smoke stream which is produced from a continuous supply.
+
+## Viscous (Real) Fluids
+
+### Viscosity
+
+A fluid offers resistance to motion due to its viscosity or internal friction.
+The greater the resistance to flow, the greater the viscosity.
+
+Higher viscosity also reduces the rate of shear deformation between layers for a given shear stress.
+
+Viscosity comes from two effects:
+
+- In liquids, the inter-molecular forces act as drag between layers of fluid moving at different
+  velocities
+- In gases, the mixing of faster and slower moving fluid causes friction due to momentum transfer.
+  The slower layers tend to slow down the faster ones
+
+### Newton's Law of Viscosity
+
+Viscosity can be defined in terms of rate of shear or velocity gradient.
+
+![](./images/vimscrot-2021-11-17T14:14:05,079195275+00:00.png)
+
+Consider the flow in the pipe above.
+Fluid in contact with the surface has a velocity of 0 because the surface irregularities trap the
+fluid particles.
+A short distance away from the surface the velocity is low but in the middle of the pipe the
+velocity is $v_F$.
+
+Let the velocity at a distance $y$ be $v$ and at a distance $y + \delta y$ be $v + \delta v$.
+
+The ratio $\frac{\delta v}{\delta y}$ is the average velocity gradient over the distance
+$\delta y$.
+
+But as $\delta y$ tends to zero, $\frac{\delta v}{\delta y} \rightarrow$ the value of the
+differential $\frac{\mathrm{d}v}{\mathrm{d}y}$ at a point such as point A.
+
+For most fluids in engineering it is found that the shear stress, $\tau$, is directly proportional
+to the velocity gradient when straight and parallel flow is involved:
+
+$$\tau = \mu\frac{\mathrm{d}v}{\mathrm{d}y}$$
+
+Where $\mu$ is the constant of proportionality and known as the dynamic viscosity, or simply the
+viscosity of the fluid.
+
+This is Newton's Law of Viscosity and fluids that obey it are known as Newtonian fluids.
+
+### Viscosity and Lubrication
+
+Where a fluid is a thin film (such as in lubricating flows), the velocity gradient can be
+approximated to be linear and an estimate of shear stress obtained:
+
+$$\tau = \mu \frac{\delta v}{\delta y} \approx \mu \frac{v}{y}$$
+
+From the shear stress we can calculate the force exerted by a film by the relationship:
+
+$$\tau = \frac F A$$
+
+# Fluid Flow
+
+## Types of flow
+
+There are essentially two types of flow:
+
+- Smooth (laminar) flow
+
+  At low flow rates, particles of fluid are moving in straight lines and can be considered to be
+  moving in layers or laminae.
+
+- Rough (turbulent) flow
+
+  At higher flow rates, the paths of the individual fluid particles are not straight but disorderly
+  resulting in mixing taking place
+
+Between fully laminar and fully turbulent flows is a transition region.
+
+## The Reynolds Number
+
+### Development of the Reynolds Number
+
+In laminar flow the most influential factor is the magnitude of the viscous forces:
+
+$$viscous\, forces \propto \mu\frac v l l^2 = \mu vl$$
+
+where $v$ is a characteristic velocity and $l$ is a characteristic length.
+
+In turbulent flow viscous effects are not significant but inertia effects (mixing, momentum
+exchange, acceleration of fluid mass) are.
+Inertial forces can be represented by $F = ma$
+
+\begin{align*}
+m &\propto \rho l^3 \\
+a &= \frac{dv}{dt} \\
+&\therefore a \propto \frac v t \text{ and } t = \frac l v \\
+&\therefore a \propto \frac {v^2} l \\
+&\therefore \text{Interial forces} \propto \rho l^2\frac{v^2} l = \rho l^2v^2
+\end{align*}
+
+The ratio of internal forces to viscous forces is called the Reynolds number and is abbreviated to
+Re:
+
+$$\Rey = \frac{\text{interial forces}}{\text{viscous forces}} = \frac {\rho l^2v^2}{\mu vl} = \frac {\rho vl} \mu$$
+
+where $\rho$ and $\mu$ are fluid properties and $v$ and $l$ are characteristic velocity and length.
+
+- During laminar flow, $\Rey$ is small as viscous forces dominate.
+- During turbulent flow, $\Rey$ is large as inertial forces dominate.
+
+\textRey is a non dimensional group.
+It has no units because the units cancel out.
+
+Non dimensional groups are very important in fluid mechanics and need to be considered when scaling
+experiments.
+
+If \textRey is the same in two different pipes, the flow will be the same regardless of actual
+diameters, densities, or other properties.
+
+#### \textRey for a Circular Section Pipe
+
+The characteristic length for pipe flow is the diameter $d$ and the characteristic velocity is
+mean flow in the pipe, $v$, so \textRey of a circular pipe section is given by:
+
+$$\Rey = \frac{\rho vd} \mu$$
+
+For flow in a smooth circular pipe under normal engineering conditions the following can be assumed:
+
+- $\Rey < 2000$ --- laminar flow
+- $2000 < \Rey < 4000$ --- transition
+- $\Rey > 4000$ --- fully turbulent flow
+
+These figures can be significantly affected by surface roughness so flow may be turbulent below
+$\Rey = 4000$.
+
+# Euler's Equation
+
+In a static fluid, pressure only depends on density and elevation.
+In a moving fluid the pressure is also related to acceleration, viscosity, and shaft work done on or
+by the fluid.
+
+$$\frac 1 \rho \frac{\delta p}{\delta s} + g\frac{\delta z}{\delta s} + v\frac{\delta v}{\delta s} = 0$$
+
+## Assumptions / Conditions
+
+The Euler equation applies where the following can be assumed:
+
+- Steady flow
+- The fluid is inviscid
+- No shaft work
+- Flow along a streamline
+
+# Bernoulli's Equation
+
+Euler's equation comes in differential form, which is difficult to apply.
+We can integrate it to make it easier
+
+\begin{align*}
+\frac 1 \rho \frac{\delta p}{\delta s} + g\frac{\delta z}{\delta s} + v\frac{\delta v}{\delta s} &= 0
+ & \text{(Euler's equation)} \\
+\int\left\{\frac{\mathrm{d}p} \rho + g\mathrm{d}z + v\mathrm{d}v \right\} &= \int 0 \,\mathrm{d}s \\
+\therefore \int \frac 1 \rho \,\mathrm{d}p + g\int \mathrm{d}z + \int v \,\mathrm{d}v &= \int 0 \,\mathrm{d}s \\
+\therefore \int \frac 1 \rho \,\mathrm{d}p + gz + \frac{v^2}{2} &= \text{constant}_1
+\end{align*}
+
+The first term of the equation can only be integrated if $\rho$ is constant as then:
+
+$$\int \frac 1 \rho \,\mathrm{d}p = \frac 1 \rho \int \mathrm{d}p = \frac p \rho$$
+
+So, if density is constant:
+
+$$\frac p \rho + gz + \frac{v^2}{2} = \text{constant}_2$$
+
+## Assumptions / Conditions
+
+All the assumptions from Euler's equation apply:
+
+- Steady flow
+- The fluid is inviscid
+- No shaft work
+- Flow along a streamline
+
+But also one more:
+
+- Incompressible flow
+
+## Forms of Bernoulli's Equation
+
+### Energy Form
+
+This form of Bernoulli's Equation is known as the energy form as each component has the units
+energy/unit mass:
+
+$$\frac p \rho + gz + \frac{v^2}{2} = \text{constant}_2$$
+
+It is split into 3 parts:
+
+- Pressure energy ($\frac p \rho$) --- energy needed to move the flow against the pressure
+  (flow work)
+- Potential energy ($gz$) --- elevation
+- Kinetic energy ($\frac{v^2}{2}$) --- kinetic energy
+
+### Elevation / Head Form
+
+Divide the energy form by $g$:
+
+$$\frac p {\rho g} + z + \frac{v^2}{2g} = H_T$$
+
+where $H_T$ is constant and:
+
+- $\frac{p}{\rho g}$ --- static/pressure head
+- $z$ --- elevation head
+- $\frac{v_2}{2g}$ --- dynamic/velocity head
+- $H_T$ --- total head
+
+- Each term now has units of elevations
+- In fluids the elevation is sometimes called head
+- This form of the equation is also useful in some applications
+
+### Pressure Form
+
+Multiply the energy form by $\rho$ to give the pressure form:
+
+$$p + \rho gz + \frac 1 2 \rho v^2 = \text{constant}$$
+
+where:
+
+- $p$ --- static pressure (often written as $p_s$)
+- $\rho gz$ --- elevation pressure
+- $\frac 1 2 \rho v^2$ --- dynamic pressure
+
+- Density is constant
+- Each term now has the units of pressure
+- This form is useful is we are interested in pressures
+
+### Comparing two forms of the Bernoulli Equation (Piezometric)
+
+$$\text{piezometric} = \text{static} + \text{elevation}$$
+
+Pressure form:
+
+\begin{align*}
+p_s + \rho gz + \frac 1 2 \rho v^2 &= \text{total pressure} \\
+p_s + \rho gz  &= \text{piezometric pressure}
+\end{align*}
+
+Head form:
+
+\begin{align*}
+\frac{p_s}{\rho g} + z + \frac{v^2}{2g} &= \text{total head} \\
+\frac{p_s}{\rho g} + z &= \text{piezometric head}
+\end{align*}
+
+# Steady Flow Energy Equation (SFEE) and the Extended Bernoulli Equation (EBE)
+
+SFEE is a more general equation that can be applied to **any fluid** and also is also takes
+**heat energy** into account.
+This is useful in applications such as a fan heater, jet engines, ICEs, and steam turbines.
+
+The equation deals with 3 types of energy transfer:
+
+1. Thermal energy transfer (e.g. heat transfer from central heating to a room)
+2. Work energy transfer (e.g. shaft from car engine that turns wheels)
+3. Energy transfer in fluid flows (e.g. heat energy in a flow, potential energy in a flow, kinetic
+   energy in a flow)
+
+## Derivation of Steady Flow Energy Equation
+
+#### Consider a control volume with steady flows in and out and steady transfers of work and heat.
+
+The properties don't change with time at any any location and are considered uniform over inlet and
+outlet areas $A_1$ and $A_2$.
+
+For steady flow, the mass, $m$, of the fluid **within the control volume** and the total energy, $E$,
+must be constant.
+
+$E$ includes **all forms for energy** but we only consider internal, kinetic, and potential energy.
+
+#### Consider a small time interval $\delta t$.
+
+During $\delta t$, mass $\delta m_1$ enters the control volume and $\delta m_2$ leaves:
+
+![](./images/vimscrot-2022-03-01T22:47:31,932087932+00:00.png)
+
+The specific energy $e_1$ of fluid $\delta m_1$ is the sum of the specific internal energy, specific
+kinetic energy, and specific potential energy:
+
+$$e_1 = u_1 + \frac{v_1^2}{2} gz_1$$
+$$e_2 = u_2 + \frac{v_2^2}{2} gz_2$$
+
+Since the mass is constant in the control volume, $\delta m_1 = \delta m_2$.
+
+#### Applying the First Law of Thermodynamics
+
+The control volume is a system for which $\delta E_1$ is added and $\delta E_2$ is removed::
+
+$$\delta E = \delta E_2 - \delta E_1$$
+
+$E$ is constant so applying the
+[first law of thermodynamics](thermodynamics.html#st-law-of-thermodynamics)
+we know that:
+
+$$\delta Q + \delta W = \delta E$$
+
+We can also say that:
+
+$$\delta E = \delta E_2 - \delta E_1 = \delta m(e_2 - e_1)$$
+
+#### The Work Term
+
+The work term, $\delta W$, is made up of shaft work **and the work necessary to deform the system**
+(by adding $\delta m_1$ at the inlet and removing $\delta m_2$ at the outlet):
+
+$$\delta W = \delta W_s + \text{net flow work}$$
+
+Work is done **on** the system by the mass entering and **by** the system on the mass leaving.
+
+For example, at the inlet:
+
+![](./images/vimscrot-2022-03-01T22:59:14,129582752+00:00.png)
+
+$$\text{work done on system} = \text{force} \times \text{distance} = p_1A_1\delta x = p_1\delta V_1$$
+
+Knowing this, we can write:
+
+$$\delta W = \delta W_s + (p_1\delta V_1 - p_2\delta V_2)$$
+
+#### Back to the First Law
+
+Substituting these equations:
+
+$$\delta E = \delta E_2 - \delta E_1 = \delta m(e_2 - e_1)$$
+$$\delta W = \delta W_s + (p_1\delta V_1 - p_2\delta V_2)$$
+
+into:
+
+$$\delta Q + \delta W = \delta E$$
+
+gives us:
+
+$$\delta Q + \left[ \delta W_s + (p_1\delta V_1 - p_2\delta V_2)\right] = \delta m (e_2-e_1)$$
+
+Dividing everything by $\delta m$ and with a bit of rearranging we get:
+
+$$q + w_s = e_2-e_1 + \frac{p_2}{\rho_2} - \frac{p_1}{\rho_1}$$
+
+#### Substitute Back for $e$
+
+$$e = u + \frac{v^2}{2} + gz$$
+
+This gives us:
+
+$$q + w_s + \left[ u_2 + \frac{p_2}{\rho_2} + gz_2 + \frac{v_2^2}{2} \right] - \left[ u_1 + \frac{p_1}{\rho_1} + gz_1 + \frac{v_1^2}{2} \right]$$
+
+#### Rearrange and Substitute for Enthalpy
+
+By definition, enthalpy $h = u + pv = u + \frac p \rho$.
+This gives us the equation:
+
+$$q + w_s = (h_2 - h_1) + g(z_2-z_1) + \frac{v_2^2-v_1^2}{2}$$
+
+This equation is in specific energy form.
+
+Multiplying by mass flow rate will give you the power form.
+
+## Application of the Steady Flow Energy Equation
+
+#### Heat Transfer Devices
+
+Like heat exchangers, boilers, condensers, and furnaces.
+
+In this case, $\dot W = 0$, $\delta z ~ 0$, and $\delta v^2 ~ 0$ so the equation can be simplified
+to just
+
+$$\dot Q = \dot m(h_2-h_1) = \dot m c_p(T_2-T_1)$$
+
+#### Throttle Valve
+
+No heat and work transfer.
+Often you can neglect potential and kinetic energy terms, giving you:
+
+$$0 = h_2-h_1)$$
+
+#### Work Transfer Devices
+
+e.g. Turbines, Pumps, Fans, and Compressors
+
+For these there is often no heat transfer ($\dot Q = 0$) and we can neglect potential
+($\delta z ~ 0$) and kinetic ($\delta v^2 ~ 0$) energy terms, giving us the equation
+
+$$\dot W = \dot m (h_2-h_1) = \dot m c_p(T_2-T_1)$$
+
+#### Mixing Devices
+
+e.g. Hot and cold water in a shower
+
+In these processes, work and heat transfers are not important and you can often
+neglect potential and kinetic energy terms, giving us the same equation as for the throttle valve
+earlier:
+
+$$0 = h_2-h_1$$
+
+which you may want to write more usefully as:
+
+$$\sum \dot m h_{out} = \sum \dot m h_{in}$$
+
+## SFEE for Incompressible Fluids and Extended Bernoulli Equation
+
+$$\frac{w_s}{g} = H_{T2} - H_{T1} + \left[ \frac{(u_2-u_1)-1}{g}\ \right]$$
+
+or
+
+$$w_s = g(H_{T2}-H_{T1}+H_f$$
+
+If we assume shaft work, $w_s$, is 0, then we can get this equation:
+
+$$H_{T1}-H_{T2} = H_f$$
+
+This is very similar to the Bernoulli equation.
+The difference is that it considers friction so it can be applied to real fluids, not just ideal
+ones.
+It is called the *Extended Bernoulli Equation*.
+
+The assumptions remain:
+
+- Steady flow
+- No shaft work
+- Incompressible
+
+### $H_f$ for Straight Pipes
+
+$$H_f = \frac{4fL}{D} \frac{v^2}{2g}$$
+
+$$\Delta p = \rho g H_f \text{ (pressure form)}$$
+
+This equation applies to long, round and straight pipes.
+It applies to both laminar and turbulent flow.
+
+However be aware that in North America the equation is:
+
+$$H_f = f \frac{L}{D} \frac{v^2}{2g}$$
+
+Their $f$ (the Darcy Friction Factor) is four times our $f$ (Fanning Friction Factor).
+In mainland Europe, they use $\lambda = 4f_{Fanning}$, which is probably the least confusing version
+to use.
+
+### Finding $f$
+
+#### $f$ for Laminar Flow
+
+$$f = \frac{16}{\Rey}$$
+
+#### $f$ for Turbulent Flow
+
+For turbulent flow, the value defends on relative pipe roughness ($k' = \frac k d$) and Reynolds
+number.
+
+Note when calculating $k'$ that **both $k$ and $d$ are measured in mm** for some reason.
+
+A *Moody Chart* is used to find $f$:
+
+![A Moody Chart](./images/vimscrot-2022-03-08T09:28:38,519555620+00:00.png)
+
+### Hydraulic Diameter
+
+$$D_h = \frac{4 \times \text{duct area}}{\text{perimeter}}$$
+
+### Loss Factor $K$
+
+There are many parts of the pipe where losses can occur.
+
+It is convenient to represent these losses in terms of loss factor, $K$, times the velocity head:
+
+$$H_f = K \frac{v^2}{g}$$
+
+Most manufacturers include loss factors in their data sheets.
+
+#### Loss Factor of Entry
+
+![](./images/vimscrot-2022-03-08T10:01:31,557158164+00:00.png)
+
+#### Loss Factor of Expansion
+
+$$K = \left( \frac{A_2}{A_1} - 1\right)^2$$
+
+This also tells us the loss factor on exit is basically 1.
+
+For conical expansions, $K ~ 0.08$ (15 degrees cone angle), 
+$K ~ 0.25$ (30 degrees).
+For cones you use the inlet velocity.
+
+#### Loss Factor of Contraction
+
+$\frac{d_2}{d_1}$ | K
+----------------- | ----
+0                 | 0.5
+0.2               | 0.45
+0.4               | 0.38
+0.6               | 0.28
+0.8               | 0.14
+1.0               | 0
+
+#### Loss Factor of Pipe Bends
+
+On a sharp bend, $K ~ 0.9$.
+
+On a bend with a radius, $K ~ 0.16-0.35$.
+
+#### Loss Factor of Nozzle
+
+$$K ~ 0.05$$
+
+But you use the outlet velocity, increasing losses.

uni/mmme/1xxx/1048_thermodynamics_and_fluid_mechanics/fluid_mechanics.md

@@ -0,0 +1,512 @@
+---
+author: Alvie Rahman
+date: \today
+title: MMME1048 // Fluid Mechanics Intro and Statics
+tags:
+- uni
+- nottingham
+- mechanical
+- engineering
+- fluid_mechanics
+- mmme1048
+- fluid_statics
+uuid: 43e8eefa-567f-438b-b93d-63ae08e61d8f
+---
+
+# Properties of Fluids
+
+## What is a Fluid?
+
+- A fluid may be liquid, vapour, or gas
+- No permanent shape
+- Consists of atoms in random motion and continual collision
+- Easy to deform
+- Liquids have fixed volume, gasses fill up container
+- **A fluid is a substance for which a shear stress tends to produce unlimited, continuous
+  deformation**
+
+## Shear Forces
+
+- For a solid, application of shear stress causes a deformation which, if not too great (elastic),
+  is not permanent and solid regains original position
+- For a fluid, continuous deformation takes place as the molecules slide over each other until the
+  force is removed
+
+## Density
+
+- Density: $$ \rho = \frac m V $$
+- Specific Density: $$ v = \frac 1 \rho $$
+
+### Obtaining Density
+
+- Find mass of a given volume or volume of a given mass
+- This gives average density and assumes density is the same throughout
+
+  - This is not always the case (like in chocolate chip ice cream)
+  - Bulk density is often used to refer to average density
+
+### Engineering Density
+
+- Matter is not continuous on molecular scale
+- For fluids in constant motion, we take a time average
+- For most practical purposes, matter is considered to be homogeneous and time averaged
+
+## Pressure
+
+- Pressure is a scalar quantity
+- Gases cannot sustain tensile stress, liquids a negligible amount
+
+- There is a certain amount of energy associated with the random continuous motion of the molecules
+- Higher pressure fluids tend to have more energy in their molecules
+
+### How Does Molecular Motion Create Force?
+
+- When molecules interact with each other, there is no net force
+- When they interact with walls, there is a resultant force perpendicular to the surface
+- Pressure caused my molecule: $$ p = \frac {\delta{}F}{\delta{}A} $$
+- If we want total force, we have to add them all up
+- $$ F = \int \mathrm{d}F = \int p\, \mathrm{d}A $$
+
+  - If pressure is constant, then this integrates to $$ F = pA $$
+  - These equations can be used if pressure is constant of average value is appropriate
+  - For many cases in fluids pressure is not constant
+
+### Pressure Variation in a Static Fluid
+
+- A fluid at rest has constant pressure horizontally
+- That's why liquid surfaces are flat
+- But fluids at rest do have a vertical gradient, where lower parts have higher pressure 
+
+### How Does Pressure Vary with Depth?
+
+![From UoN MMME1048 Fluid Mechanics Notes](./images/vimscrot-2021-10-06T10:51:51,499044519+01:00.png)
+
+Let fluid pressure be p at height $z$, and $p + \delta p$ at $z + \delta z$.
+
+Force $F_z$ acts upwards to support the fluid, countering pressure $p$.
+
+Force $F_z + \delta F_z$acts downwards to counter pressure $p + \delta p$ and comes from the weight
+of the liquid above.
+
+Now:
+
+\begin{align*}
+F_z &= p\delta x\delta y \\
+F_z + \delta F_z &= (p + \delta p) \delta x \delta y \\
+\therefore \delta F_z &= \delta p(\delta x\delta y)
+\end{align*}
+
+Resolving forces in z direction:
+
+\begin{align*}
+F_z - (F_z + \delta F_z) - g\delta m &= 0 \\
+\text{but } \delta m &= \rho\delta x\delta y\delta z \\
+\therefore -\delta p(\delta x\delta y) &= g\rho(\delta x\delta y\delta z) \\
+\text{or } \frac{\delta p}{\delta z} &= -\rho g \\
+\text{as } \delta z \rightarrow 0,\, \frac{\delta p}{\delta z} &\rightarrow \frac{dp}{dz}\\
+\therefore \frac{dp}{dz} &= -\rho g\\
+\Delta p &= \rho g\Delta z
+\end{align*}
+
+The equation applies for any fluid.
+The -ve sign indicates that as $z$, height, increases, $p$, pressure, decreases.
+
+### Absolute and Gauge Pressure
+
+- Absolute Pressure is measured relative to zero (a vacuum)
+- Gauge pressure = absolute pressure - atmospheric pressure
+
+  - Often used in industry
+
+- If absolute pressure = 3 bar and atmospheric pressure is 1 bar, then gauge pressure = 2 bar
+- Atmospheric pressure changes with altitude
+
+## Compressibility
+
+- All fluids are compressible, especially gasses
+- Most liquids can be considered **incompressible** most of the time (and will be in MMME1048, but
+  may not be in future modules)
+
+## Surface Tension
+
+- In a liquid, molecules are held together by molecular attraction
+- At a boundary between two fluids this creates "surface tension"
+- Surface tension usually has the symbol $$\gamma$$
+
+## Ideal Gas
+
+- No real gas is perfect, although many are similar
+- We define a specific gas constant to allow us to analyse the behaviour of a specific gas:
+
+  $$ R = \frac {\tilde R}{\tilde m} $$
+
+  (Universal Gas Constant / molar mass of gas)
+
+- Perfect gas law
+
+  $$pV=mRT$$
+
+  or
+
+  $$ p = \rho RT$$
+
+  - Pressure always in Pa
+  - Temperature always in K
+
+## Units and Dimensional Analysis
+
+- It is usually better to use SI units
+- If in doubt, DA can be useful to check that your answer makes sense
+
+# Fluid Statics
+
+## Manometers
+
+![](./images/vimscrot-2021-10-13T09:09:32,037006075+01:00.png)
+
+$$p_{1,gauge} = \rho g(z_2-z_1)$$
+
+- Manometers work on the principle that pressure along any horizontal plane through a continuous
+  fluid is constant
+- Manometers can be used to measure the pressure of a gas, vapour, or liquid
+- Manometers can measure higher pressures than a piezometer
+- Manometer fluid and working should be immiscible (don't mix)
+
+![](./images/vimscrot-2021-10-13T09:14:59,628661490+01:00.png)
+
+\begin{align*}
+p_A &= p_{A'} \\
+p_{bottom} &= p_{top} + \rho gh \\
+\rho_1 &= density\,of\,fluid\,1 \\
+\rho_2 &= density\,of\,fluid\,2
+\end{align*}
+
+Left hand side:
+
+$$p_A =  p_1 + \rho_1g\Delta z_1$$
+
+Right hand side:
+
+$$p_{A'} = p_{at} + \rho_2g\Delta z_2$$
+
+Equate and rearrange:
+
+\begin{align*}
+p_1 + \rho_1g\Delta z_1 &= p_{at} + \rho_2g\Delta z_2 \\
+p_1-p_{at} &= g(\rho_2\Delta z_2 - \rho_1\Delta z_1) \\
+p_{1,gauge} &= g(\rho_2\Delta z_2 - \rho_1\Delta z_1)
+\end{align*}
+
+If $\rho_a << \rho_2$:
+
+$$\rho_{1,gauge} \approx \rho_2g\Delta z_2$$
+
+### Differential U-Tube Manometer
+
+![](./images/vimscrot-2021-10-13T09:37:02,070474894+01:00.png)
+
+- Used to find the difference between two unknown pressures
+- Can be used for any fluid that doesn't react with manometer fluid
+- Same principle used in analysis
+
+\begin{align*}
+p_A &= p_{A'} \\
+p_{bottom} &= p_{top} + \rho gh \\
+\rho_1 &= density\,of\,fluid\,1 \\
+\rho_2 &= density\,of\,fluid\,2
+\end{align*}
+
+Left hand side:
+
+$$p_A = p_1 + \rho_wg(z_C-z_A)$$
+
+Right hand side:
+
+$$p_B = p_2 + \rho_wg(z_C-z_B)$$
+
+Right hand manometer fluid:
+
+$$p_{A'} = p_B + \rho_mg(z_B - z_a)$$
+
+\begin{align*}
+p_{A'} &= p_2 + \rho_mg(z_C - z_B) + \rho_mg(z_B - zA)\\
+       &= p_2 + \rho_mg(z_C - z_B) + \rho_mg\Delta z \\
+\\
+p_A &= p_{A'} \\
+p_1 + \rho_wg(z_C-z_A) &= p_2 + \rho_mg(z_C - z_B) + \rho_mg\Delta z \\
+p_1 - p_2 &= \rho_wg(z_C-z_B-z_C+z_A) + \rho_mg\Delta z \\
+&= \rho_wg(z_A-z_B) + \rho_mg\Delta z \\
+&= -\rho_wg\Delta z + \rho_mg\Delta z 
+\end{align*}
+
+### Angled Differential Manometer
+
+![](./images/vimscrot-2021-10-13T09:56:15,656796805+01:00.png)
+
+- If the pipe is sloped then
+
+  $$p_1-p_2 = (\rho_m-\rho_w)g\Delta z + \rho_wg(z_{C2} - z_{C1})$$
+
+- $p_1 > p_2$ as $p_1$ is lower
+- If there is no flow along the tube, then
+
+  $$p_1 = p_2 + \rho_wg(z_{C2} - z_{C1})$$
+
+<details>
+<summary>
+
+### Exercise Sheet 1
+
+</summary>
+
+1. If 4 m$^3$ of oil weighs 35 kN calculate its density and relative density.
+   Relative density is a term used to define the density of a fluid relative 
+
+   > $$ \frac{35000}{9.81\times4} = 890 \text{ kgm}^{-3} $$
+   >
+   > $$1000 - 891.9... = 108 \text{ kgm}^{-3}$$
+
+2. Find the pressure relative to atmospheric experienced by a diver 
+   working on the sea bed at a depth of 35 m.
+   Take the density of sea water to be 1030 kgm$^{-3}$.
+
+   > $$
+   > \rho gh = 1030\times 9.81 \times 35 = 3.5\times10^5
+   > $$
+
+3. An open glass is sitting on a table, it has a diameter of 10 cm.
+   If water up to a height of 20 cm is now added calculate the force exerted onto the table by
+   the addition of the water.
+
+   > $$V_{cylinder} = \pi r^2h$$
+   > $$m_{cylinder} = \rho\pi r^2h$$
+   > $$W_{cylinder} = \rho g\pi r^2h = 1000\times9.8\pi\times0.05^2\times0.2 = 15.4 \text{ N} $$
+
+4. A closed rectangular tank with internal dimensions 6m x 3m x 1.5m 
+   high has a vertical riser pipe of cross-sectional area 0.001 m2 in  
+   the upper surface (figure 1.4).  The tank and riser are filled with 
+   water such that the water level in the riser pipe is 3.5 m above the 
+
+   Calculate:
+
+   i. The gauge pressure at the base of the tank. 
+
+      > $$\rho gh = 1000\times9.81\times(1.5+3.5) = 49 \text{ kPa}$$
+
+   ii. The gauge pressure at the top of the tank.
+
+       > $$\rho gh = 1000\times9.81\times3.5 = 34 \text{ kPa}$$
+
+   iii. The force exerted on the base of the tank due to gauge water pressure.
+
+        > $$F = p\times A = 49\times10^3\times6\times3 = 8.8\times10^5 \text{ N}$$
+
+   iv. The weight of the water in the tank and riser.
+
+       > $$V = 6\times3\times1.5 + 0.001\times3.5 = 27.0035$$
+       > $$W = \rho gV = 1000\times9.8\times27.0035 = 2.6\times10^5\text{ N}$$
+
+   v. Explain the difference between (iii) and (iv). 
+
+      *(It may be helpful to think about the forces on the top of the tank)*
+
+      > The pressure at the top of the tank is higher than atmospheric pressure because of the
+      > riser.
+      > This means there is an upwards force on the top of tank.
+      > The difference between the force acting up and down due to pressure is equal to the
+      > weight of the water.
+
+6. A double U-tube manometer is connected to a pipe as shown below.
+ 
+   Taking the dimensions and fluids as indicated; calculate
+   the absolute pressure at point A (centre of the pipe).
+
+   Take atmospheric pressure as 1.01 bar and the density of mercury as 13600 kgm$^{-3}$.
+
+   ![](./images/vimscrot-2021-10-13T10:42:52,999793176+01:00.png)
+
+   > \begin{align*}
+        P_B &= P_A + 0.4\rho_wg          &\text{(1)}\\
+        P_C = P_{C'} &= P_B - 0.2\rho_mg &\text{(2)}\\
+        P_D = P_{at} &= P_A + 0.4\rho_wg &\text{(3)}\\
+        \\
+        \text{(2) into (3)} &\rightarrow P_B -0.2\rho_mg-0.1\rho_wg = P_{at} &\text{(4)}\\
+        \text{(1) into (4)} &\rightarrow P_A + 0.4\rho_wg - 0.2\rho_mg - 0.1\rho_wg = P_{at} \\
+        \\
+        P_A &= P_{at} + 0.1\rho_wg + 0.2\rho_mg - 0.4\rho_wg \\
+            &= P_{at} + g(0.2\rho_m - 0.3\rho_w) \\
+            &= 1.01\times10^5 + 9.81(0.2\times13600 - 0.3\times1000)\\
+            &= 124.7\text{ kPa}
+   > \end{align*}
+
+</details>
+
+## Submerged Surfaces
+
+### Preparatory Maths
+
+#### Integration as Summation
+
+#### Centroids
+
+- For a 3D body, the centre of gravity is the point at which all the mass can be considered to act
+- For a 2D lamina (thin, flat plate) the centroid is the centre of area, the point about which the
+  lamina would balance
+
+To find the location of the centroid, take moments (of area) about a suitable reference axis:
+
+$$moment\,of\,area = moment\,of\,mass$$
+
+(making the assumption that the surface has a unit mass per unit area)
+
+$$moment\,of\,mass = mass\times distance\,from\,point\,acting\,around$$
+
+Take the following lamina:
+
+![](./images/vimscrot-2021-10-20T10:01:30,080819382+01:00.png)
+
+1. Split the lamina into elements parallel to the chosen axis
+2. Each element has area $\delta A = w\delta y$
+3. The moment of area ($\delta M$) of the element is $\delta Ay$
+4. The sum of moments of all the elements is equal to the moment $M$ obtained by assuming all the
+   area is located at the centroid or:
+
+   $$Ay_c = \int_{area} \! y\,\mathrm{d}A$$
+
+   or:
+
+   $$y_c = \frac 1 A \int_{area} \! y\,\mathrm{d}A$$
+
+   - $\int y\,\mathrm{d}A$ is known as the first moment of area
+
+<details>
+<summary>
+
+##### Example 1
+
+Determine the location of the centroid of a rectangular lamina.
+
+</summary>
+
+###### Determining Location in $y$ direction
+
+![](./images/vimscrot-2021-10-20T10:14:17,688774145+01:00.png)
+
+1. Take moments for area about $OO$
+
+   $$\delta M = y\delta A = y(b\delta y)$$
+
+2. Integrate to find all strips
+
+   $$M = b\int_0^d\! y \,\mathrm{d}y = b\left[\frac{y^2}2\right]_0^d = \frac{bd^2} 2$$
+
+   ($b$ can be taken out the integral as it is constant in this example)
+
+   but also $$M = (area)(y_c) = bdy_c$$
+
+   so $$y_c = \frac 1 {bd} \frac {bd} 2 = \frac d 2$$
+
+###### Determining Location in $x$ direction
+
+![](./images/vimscrot-2021-10-20T10:24:48,372189101+01:00.png)
+
+1. Take moments for area about $O'O'$:
+
+   $$\delta M = x\delta A = x(d\delta x)$$
+
+2. Integrate
+
+   $$M_{O'O'} = d\int_0^b\! x \,\mathrm{d}x = d\left[\frac{x^2} 2\right]_0^b = \frac{db^2} 2$$
+
+   but also $$M_{O'O'} = (area)(x_c) = bdx_c$$
+
+   so $$x_c = \frac{db^2}{2bd} = \frac b 2$$
+
+</details>
+
+### Horizontal Submerged Surfaces
+
+![](./images/vimscrot-2021-10-20T10:33:16,783724117+01:00.png)
+
+Assumptions for horizontal lamina:
+
+- Constant pressure acts over entire surface of lamina
+- Centre of pressure will coincide with centre of area
+- $total\,force = pressure\times area$
+
+![](./images/vimscrot-2021-10-20T10:36:12,520683729+01:00.png)
+
+### Vertical Submerged Surfaces
+
+![](./images/vimscrot-2021-10-20T11:05:33,235642932+01:00.png)
+
+- A vertical submerged plate does experience uniform pressure
+- Centroid of pressure and area are not coincident
+- Centroid of pressure is always below centroid of area for a vertical plate
+- No shear forces, so all hydrostatic forces are perpendicular to lamina
+
+![](./images/vimscrot-2021-10-20T11:07:52,929126609+01:00.png)
+
+Force acting on small element:
+
+\begin{align*}
+\delta F &= p\delta A \\
+&= \rho gh\delta A \\
+&= \rho gh w\delta h
+\end{align*}
+
+Therefore total force is
+
+$$F_p = \int_{area}\! \rho gh \,\mathrm{d}A = \int_{h_1}^{h_2}\! \rho ghw\,\mathrm{d}h$$
+
+#### Finding Line of Action of the Force
+
+![](./images/vimscrot-2021-10-20T11:15:51,200869760+01:00.png)
+
+\begin{align*}
+\delta M_{OO} &= \delta Fh = (\rho gh\delta A)h \\
+&= \rho gh^2\delta A = \rho gh^2w\delta h \\
+\\
+M_{OO} &= F_py_p = \int_{area}\! \rho gh^2 \,\mathrm{d}A \\
+&= \int_{h1}^{h2}\! \rho gh^2w \,\mathrm{d}h \\
+\\
+y_p = \frac{M_{OO}}{F_p}
+\end{align*}
+
+## Buoyancy
+
+### Archimedes Principle
+
+> The resultant upwards force (buoyancy force) on a body wholly or partially immersed in a fluid is
+> equal to the weight of the displaced fluid.
+
+When an object is in equilibrium the forces acting on it balance.
+For a floating object, the upwards force equals the weight:
+
+$$mg = \rho Vg$$
+
+Where $\rho$ is the density of the fluid, and $V$ is the volume of displaced fluid.
+
+### Immersed Bodies
+
+As pressure increases with depth, the fluid exerts a resultant upward force on a body.
+There is no horizontal component of the buoyancy force because the vertical projection of the body
+is the same in both directions.
+
+### Rise, Sink, or Float?
+
+- $F_B = W$ \rightarrow equilibrium (floating)
+- $F_B > W$ \rightarrow body rises
+- $F_B < W$ \rightarrow body sinks
+
+### Centre of Buoyancy
+
+Buoyancy force acts through the centre of gravity of the volume of fluid displaced.
+This is known as the centre of buoyancy.
+The centre of buoyancy does not in general correspond to the centre of gravity of the body.
+
+If the fluid density is constant the centre of gravity of the displaced fluid is at the centroid of
+the immersed volume.
+
+![](./images/vimscrot-2021-12-21T15:08:22,285753421+00:00.png)
+

uni/mmme/1xxx/1048_thermodynamics_and_fluid_mechanics/thermodynamics.md

@@ -0,0 +1,549 @@
+---
+author: Alvie Rahman
+date: \today
+title: MMME1048 // Thermodynamics
+tags:
+- uni
+- nottingham
+- mechanical
+- engineering
+- mmme1048
+- thermodynamics
+uuid: db8abbd9-1ef4-4a0d-a6a8-54882f142643
+---
+
+# What is Thermodynamics?
+
+Thermodynamics deals with the transfer of heat energy and temperature.
+
+# Concepts and Definitions
+
+## System
+
+A region of space, marked off by its boundary.
+It contains some matter and the matter inside is what we are investigating.
+
+There are two types of systems:
+
+- Closed systems
+
+  - Contain a fixed quantity of matter
+  - Work and heat cross boundaries
+  - Impermeable boundaries, some may be moved
+  - Non-flow processes (no transfer of mass)
+
+- Open systems
+
+  - Boundary is imaginary
+  - Mass can flow in an out (flow processes)
+  - Work and heat transfer can occur
+
+## Equilibrium
+
+The system is in equilibrium if all parts of the system are at the same conditions, such as pressure
+and temperature.
+
+The system is not in equilibrium if parts of the system are at different conditions.
+
+#### Adiabatic
+
+A process in which heat does not cross the system boundary
+
+## Perfect (Ideal) Gasses
+
+A perfect gas is defined as one in which:
+
+- all collisions between molecules are perfectly elastic
+- there are no intermolecular forces
+
+Perfect gases do not exist in the real world and they have two requirements in thermodynamics:
+
+### The Requirements of Perfect Gasses
+
+#### Obey the Perfect Gas Equation
+
+$$pV = n \tilde R T$$
+
+where $n$ is the number of moles of a substance and $\tilde R$ is the universal gas constant
+
+or
+
+$$pV =mRT$$
+
+where the gas constant $R = \frac{\tilde R}{\tilde m}$, $\tilde m$ is molecular mass
+
+or
+
+$$pv = RT$$
+
+(using the specific volume)
+
+#### $c_p$ and $c_v$ are constant
+
+This gives us the equations:
+
+$$u_2 - u_1 = c_v(T_2-T_1)$$
+
+$$h_2 - h_1 = c_p(T_2-T_1)$$
+
+### Relationship Between Specific Gas Constant and Specific Heats
+
+$$c_v = \frac{R}{\gamma - 1}$$
+
+$$c_p = \frac{\gamma}{\gamma -1} R$$
+
+<details>
+<summary>
+
+#### Derivation
+
+</summary>
+
+We know the following are true (for perfect gases):
+
+$$\frac{c_p}{c_v} = \gamma$$
+
+$$u_2 - u_1 = c_v(T_2-T_1)$$
+
+$$h_2 - h_1 = c_p(T_2-T_1)$$
+
+So:
+
+\begin{align*}
+h_2 - h_1 &= u_2 - u_1 + (p_2v_2 - p_1v_1) \\
+c_p(T_2-T_1) &= c_v(T_2-T_1) + R(T_2-T_1) \\
+c_p &= c_v + R \\
+\\
+c_p &= c_v \gamma \\
+c_v + R &= c_v\gamma \\
+c_v &= \frac{R}{\gamma - 1} \\
+\\
+\frac{c_p}{\gamma} &= c_v \\
+c_p &= \frac{c_p}{\gamma} + R \\
+c_p &= \frac{\gamma}{\gamma -1} R
+\end{align*}
+
+</details>
+
+### The Specific and Molar Gas Constant
+
+The molar gas constant is represented by $\tilde R = 8.31 \text{JK}^{-1}\text{mol}^{-1}$.
+
+The specific gas constant is $R = \frac{\tilde{R}}{M}$.  
+The SI unit for the specific gas constant is J kg$^{-1}$ mol$^{-1}$.  
+The SI unit for molar mass is kg mol$^{-1}$.
+
+## Thermodynamic Processes and Cycles
+
+When a thermodynamic system changes from one state to another it is said to execute a *process*.
+An example of a process is expansion (volume increasing).
+
+A *cycle* is a process or series of processes in which the end state is identical to the beginning.
+And example of this could be expansion followed by a compression.
+
+### Reversible and Irreversible Processes
+
+During reversible processes, the system undergoes a continuous succession of equilibrium states.
+Changes in the system can be defined and reversed to restore the initial conditions
+
+All real processes are irreversible but some can be assumed to be reversible, such as controlled
+expansion.
+
+### Constant _____ Processes
+
+#### Isothermal
+
+Constant temperature process
+
+#### Isobaric
+
+Constant pressure process
+
+#### Isometric / Isochoric
+
+Constant volume process
+
+## Heat and Work
+
+Heat and Work are different forms of energy transfer.
+
+They are both transient phenomena and systems never possess heat or work.
+Both represent energy crossing boundaries when a system undergoes a change of state.
+
+By convention, the transfer of energy into the system from the surroundings is positive (work is
+being done *on* the system *by* the surroundings).
+
+### Heat
+
+*Heat* is defined as:
+
+> The form of energy that is transferred across the boundary of a system at a given temperature to
+> another system at a lower temperature by virtue of the temperature difference between the two
+
+### Work
+
+*Work* is defined as:
+
+$$W = \int\! F \mathrm{d}x$$
+
+(the work, $W$, done by a force, $F$, when the point of application of the force undergoes a
+displacement, $\mathrm{d}x$)
+
+## Thermally Insulated and Isolated Systems
+
+In thermally insulated systems and isolated systems, heat transfer cannot take place.
+
+In thermally isolated systems, work transfer cannot take place.
+
+# 1st Law of Thermodynamics
+
+The 1st Law of Thermodynamics can be thought of as:
+
+> When a closed system is taken through a cycle, the sum of the *net* work transfer ($W$) and *net*
+> heat transfer ($Q$) equals zero:
+>
+> $$W_{net} + Q_{net} = 0$$
+>
+
+## 1st Corollary
+
+> The change in internal energy of a closed system is equal to the sum of the heat transferred
+> and the work done during any change of state
+>
+> $$W_{12} + Q_{12} = U_2 - U_1$$
+
+## 2nd Corollary
+
+> The internal energy of a closed system remains unchanged if it
+> [thermally isolated](#thermally-insulated-and-isolated-systems) from its surroundings
+
+# Properties of State
+
+*State* is defined as the condition of a system as described by its properties.
+The state may be identified by certain observable macroscopic properties.
+These properties are the *properties of state* and they always have the same values for a given
+state.
+
+A *property* can be defined as any quantity that depends on the *state* of the system and is
+independent of the path by which  the system arrived at the given state.
+Properties determining the state of a thermodynamic system are referred to as *thermodynamic
+properties* of the *state* of the system.
+
+Common properties of state are:
+
+- Temperature
+- Pressure
+- Mass
+- Volume
+
+And these can be determined by simple measurements.
+Other properties can be calculated:
+
+- Specific volume
+- Density
+- Internal energy
+- Enthalpy
+- Entropy
+
+## Intensive vs Extensive Properties
+
+In thermodynamics we distinguish between *intensive*, *extensive*, and *specific* properties:
+
+- Intensive --- properties which do not depend on mass (e.g. temperature)
+- Extensive --- properties which do depend on the mass of substance in a system (e.g. volume)
+- Specific (extensive) --- extensive properties which are reduced to unit mass of substance
+  (essentially an extensive property divided by mass) (e.g. specific volume)
+
+## Units
+
+<div class="tableWrapper">
+
+Property        | Symbol | Units           | Intensive | Extensive
+--------------- | ------ | --------------- | --------- | ---------
+Pressure        | p      | Pa              | Yes       |
+Temperature     | T      | K               | Yes       |
+Volume          | V      | m$^3$           |           | Yes
+Mass            | m      | kg              |           | Yes
+Specific Volume | v      | m$^3$ kg$^{-1}$ |  Yes      |
+Density         | $\rho$ | kg m$^{-3}$     |  Yes      |
+Internal Energy | U      | J               |           | Yes
+Entropy         | S      | J K$^{-1}$      |           | Yes
+Enthalpy        | H      | J               |           | Yes
+
+</div>
+
+## Density
+
+For an ideal gas:
+
+$$\rho = \frac{p}{RT}$$
+
+## Enthalpy and Specific Enthalpy
+
+Enthalpy does not have a general physical interpretation.
+It is used because the combination $u + pv$ appears naturally in the analysis of many
+thermodynamic problems.
+
+The heat transferred to a closed system undergoing a reversible constant pressure process is equal
+to the change in enthalpy of the system.
+
+Enthalpy is defined as:
+
+$$H = U+pV$$
+
+and Specific Enthalpy:
+
+$$h = u + pv$$
+
+## Entropy and Specific Entropy
+
+Entropy is defined as the following, given that the process s reversible:
+
+$$S_2 - S_1 = \int\! \frac{\mathrm{d}Q}{T}$$
+
+### Change of Entropy of a Perfect Gas
+
+Consider the 1st corollary of the 1st law:
+
+$$\mathrm dq + \mathrm dw = \mathrm du$$
+
+and that the process is reversible:
+
+\begin{align*}
+\mathrm ds &= \frac{\mathrm dq} T \bigg|_{rev} \\
+\mathrm dq = \mathrm ds \times T \\
+\mathrm dw &= -p\mathrm dv \\
+\end{align*}
+
+The application of the 1st corollary leads to:
+
+$$T\mathrm ds - p\mathrm dv = \mathrm du$$
+
+Derive the change of entropy
+
+\begin{align*}
+\mathrm ds &= \frac{\mathrm du}{T} + \frac{p \mathrm dv}{T} \\
+\\
+\mathrm du &= c_v \mathrm{d}T \\
+\frac p T &= \frac R v \\
+\\
+\mathrm ds &= \frac{c_v\mathrm{d}T}{T} \frac{R\mathrm dv}{v} \\
+s_2 - s_1 &= c_v\ln\left(\frac{T_2}{T_1}\right) + R\ln\left(\frac{v_2}{v_1}\right)
+\end{align*}
+
+There are two other forms of the equation that can be derived:
+
+$$s_2 - s_1 = c_v\ln\left(\frac{p_2}{p_1}\right) + c_p\ln\left(\frac{v_2}{v_1}\right)$$
+$$s_2 - s_1 = c_p\ln\left(\frac{T_2}{T_1}\right) - R\ln\left(\frac{p_2}{p_1}\right)$$
+
+## Heat Capacity and Specific Heat Capacity
+
+Heat capacity is quantity of heat required to raise the temperature of a system by a unit
+temperature:
+
+$$C = \frac{\mathrm{d}Q}{\mathrm{d}T}$$
+
+Specific heat capacity is the quantity of heat required to raise the temperature of a unit mass
+substance by a unit temperature:
+
+$$c = \frac{\mathrm{d}q}{\mathrm{d}T}$$
+
+<details>
+<summary>
+
+### Heat Capacity in Closed Systems and Internal Energy
+
+The specific heat transfer to a closed system during a reversible constant **volume** process is
+equal to the change in specific **internal energy** of the system:
+
+$$c_v = \frac{\mathrm{d}q}{\mathrm{d}T} = \frac{\mathrm{d}u}{\mathrm{d}T}$$
+
+</summary>
+
+This is because if the change in volume, $\mathrm{d}v = 0$, then the work done, $\mathrm{d}w = 0$
+also.
+So applying the (1st Corollary of the) 1st Law to an isochoric process:
+
+$$\mathrm{d}q + \mathrm{d}w = \mathrm{d}u \rightarrow \mathrm{d}q = \mathrm{d}u$$
+
+since $\mathrm{d}w = 0$.
+
+</details>
+
+<details>
+<summary>
+
+### Heat Capacity in Closed Systems and Enthalpy
+
+The specific heat transfer to a closed system during a reversible constant **pressure** process is
+equal to the change in specific **enthalpy** of the system:
+
+$$c_p = \frac{\mathrm{d}q}{\mathrm{d}T} = \frac{\mathrm{d}h}{\mathrm{d}T}$$
+
+</summary>
+
+This is because given that pressure, $p$, is constant, work, $w$, can be expressed as:
+
+$$w = -\int^2_1\! p \,\mathrm{d}v = -p(v_2 - v_1)$$
+
+Applying the (1st corollary of the) 1st law to the closed system:
+
+\begin{align*}
+  q + w &= u_2 - u_1 \rightarrow q = u_2 - u_1 + p(v_2 - v_1) \\
+  q     &= u_2 + pv_2 - (u_1 + pv_1) \\
+        &= h_2 - h_1 = \mathrm{d}h \\
+  \therefore \mathrm{d}q &= \mathrm{d}h
+\end{align*}
+
+</details>
+
+<details>
+<summary>
+
+### Ratio of Specific Heats
+
+$c_p > c_v$ is always true.
+
+</summary>
+
+Heating a volume of fluid, $V$, at a constant volume requires specific heat $q_v$ where
+
+$$q_v = u_2 - u_1 \therefore c_v = \frac{q_v}{\Delta T}$$
+
+Heating the same volume of fluid but under constant pressure requires a specific heat $q_p$ where
+
+$$q_p =u_2 - u_1 + p(v_2-v_1) \therefore c_p = \frac{q_p}{\Delta T}$$
+
+Since $p(v_2-v_1) > 0$, $\frac{q_p}{q_v} > 1 \therefore q_p > q_v \therefore c_p > c_v$.
+
+The ratio $\frac{c_p}{c_v} = \gamma$
+
+</details>
+
+# Process and State Diagrams
+
+Reversible processes are represented by solid lines, and irreversible processes by dashed lines.
+
+# Isentropic and Polytropic Processes
+
+## Polytropic Processes
+
+A polytropic process is one which obeys the polytropic law:
+
+$$pv^n = k \text{ or } p_1v_1^n = p_2v_2^n$$
+
+where $n$ is a constant called the polytropic index, and $k$ is a constant too.
+
+A typical polytropic index is between 1 and 1.7.
+
+<details>
+<summary>
+
+#### Example 1
+
+Derive
+
+$$\frac{p_2}{p_1} = \left(\frac{T_2}{T_1}\right)^{\frac{n}{n-1}}$$
+
+</summary>
+
+\begin{align*}
+p_1v_1^n &= p_2v_2^n \\
+pv &= RT \rightarrow v = R \frac{T}{p} \\
+\frac{p_2}{p_1} &= \left( \frac{v_1}{v_2} \right)^n \\
+                &= \left(\frac{p_2T_1}{T_2p_1}\right)^n \\
+                &= \left(\frac{p_2}{p_1}\right)^n \left(\frac{T_1}{T_2}\right)^n \\
+\left(\frac{p_2}{p_1}\right)^{1-n} &= \left(\frac{T_1}{T_2}\right)^n \\
+\frac{p_2}{p_1} &= \left(\frac{T_1}{T_2}\right)^{\frac{n}{1-n}} \\
+                &= \left(\frac{T_2}{T_1}\right)^{\frac{n}{n-1}} \\
+\end{align*}
+
+<details>
+<summary>
+
+How did you do that last step?
+
+</summary>
+
+For any values of $x$ and $y$
+
+\begin{align*}
+\frac x y &= \left(\frac y x \right) ^{-1} \\
+\left(\frac x y \right)^n &= \left(\frac y x \right)^{-n} \\
+\left(\frac x y \right)^{\frac{n}{1-n}} &= \left(\frac y x \right)^{\frac{-n}{1-n}} \\
+ &= \left(\frac y x \right)^{\frac{n}{n-1}} \\
+\end{align*}
+
+</details>
+</details>
+
+## Isentropic Processes
+
+*Isentropic* means constant entropy:
+
+$$\Delta S = 0 \text{ or } s_1 = s_2 \text{ for a precess 1-2}$$
+
+A process will be isentropic when:
+
+$$pv^\gamma = \text{constant}$$
+
+This is basically the [polytropic law](#polytropic-processes) where the polytropic index, $n$, is
+always equal to $\gamma$.
+
+<details>
+<summary>
+
+Derivation
+
+</summary>
+
+\begin{align*}
+0 &= s_2 - s_1 = c_v \ln{\left(\frac{p_2}{p_1}\right)} + c_p \ln{\left( \frac{v_2}{v_1} \right)} \\
+0 &= \ln{\left(\frac{p_2}{p_1}\right)} + \frac{c_p}{c_v} \ln{\left( \frac{v_2}{v_1} \right)} \\
+  &= \ln{\left(\frac{p_2}{p_1}\right)} + \gamma \ln{\left( \frac{v_2}{v_1} \right)} \\
+  &= \ln{\left(\frac{p_2}{p_1}\right)} + \ln{\left( \frac{v_2}{v_1} \right)^\gamma} \\
+  &= \ln{\left[\left(\frac{p_2}{p_1}\right)\left( \frac{v_2}{v_1} \right)^\gamma\right]} \\
+ e^0  = 1 &= \left(\frac{p_2}{p_1}\right)\left( \frac{v_2}{v_1} \right)^\gamma \\
+&\therefore p_2v_2^\gamma = p_1v_1^\gamma \\
+\\
+pv^\gamma = \text{constant}
+\end{align*}
+
+</details>
+
+During isentropic processes, it is assumed that no heat is transferred into or out of the cylinder.
+It is also assumed that friction is 0 between the piston and cylinder and that there are no energy
+losses of any kind.
+
+This results in a reversible process in which the entropy of the system remains constant.
+
+An isentropic process is an idealization of an actual process, and serves as the limiting case for
+real life processes.
+They are often desired and often the processes on which device efficiencies are calculated.
+
+### Heat Transfer During Isentropic Processes
+
+Assume that the compression 1-2 follows a polytropic process with a polytropic index $n$.
+The work transfer is:
+
+$$W  = - \frac{1}{1-n} (p_2V_2 - p_1V_1) = \frac{mR}{n-1} (T_2-T_1)$$
+
+Considering the 1st corollary of the 1st law, $Q + W = \Delta U$, and assuming the gas is an ideal
+gas [we know that](#obey-the-perfect-gas-equation) $\Delta U = mc_v(T_2-T_1)$ we can deduce:
+
+\begin{align*}
+Q &= \Delta U - W = mc_v(T_2-T_1) - \frac{mR}{n-1} (T_2-T_1) \\
+  &= m \left(c_v - \frac R {n-1}\right)(T_2-T-1)
+\end{align*}
+
+Now, if the process was *isentropic* and not *polytropic*, we can simply substitute $n$ for
+$\gamma$ so now:
+
+$$Q = m \left(c_v - \frac R {\gamma-1}\right)(T_2-T-1)$$
+
+But since [we know](#relationship-between-specific-gas-constant-and-specific-heats) $c_v = \frac R {\gamma - 1}$:
+
+$$Q = m (c_v-c_v)(T_2-T_1) = 0 $$
+
+This proves that the isentropic version of the process adiabatic (no heat is transferred across the
+boundary).