mmme/1048 add notes on lectures for 1st dec
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@ -42,6 +42,84 @@ The system is not in equilibrium if parts of the system are at different conditi
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A process in which does not cross the system boundary
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## Perfect (Ideal) Gasses
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A perfect gas is defined as one in which:
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- all collisions between molecules are perfectly elastic
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- there are no intermolecular forces
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Perfect gases do not exist in the real world and they have two requirements in thermodynamics:
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### The Requirements of Perfect Gasses
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#### Obey the Perfect Gas Equation
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$$pV = n \tilde R T$$
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where $n$ is the number of moles of a substance and $\tilde R$ is the universal gas constant
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or
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$$pV =mRT$$
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where the gas constant $R = \frac{\tilde R}{\tilde m}$, $\tilde m$ is molecular mass
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or
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$$pv = RT$$
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(using the specific volume)
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#### $c_p$ and $c_v$ are constant
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This gives us the equations:
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$$u_2 - u_1 = c_v(T_2-T_1)$$
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$$h_2 - h_1 = c_p(T_2-T_1)$$
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### Relationship Between Specific Gas Constant and Specific Heats
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$$c_v = \frac{R}{\gamma - 1}$$
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$$c_p = \frac{\gamma}{\gamma -1} R$$
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<details>
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<summary>
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#### Derivation
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</summary>
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We know the following are true (for perfect gases):
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$$\frac{c_p}{c_v} = \gamma$$
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$$u_2 - u_1 = c_v(T_2-T_1)$$
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$$h_2 - h_1 = c_p(T_2-T_1)$$
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So:
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\begin{align*}
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h_2 - h_1 &= u_2 - u_1 + (p_2v_2 - p_1v_1) \\
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c_p(T_2-T_1) &= c_v(T_2-T_1) + R(T_2-T_1) \\
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c_p &= c_v + R \\
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\\
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c_p &= c_v \gamma \\
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c_v + R &= c_v\gamma \\
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c_v &= \frac{R}{\gamma - 1} \\
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\\
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\frac{c_p}{\gamma} &= _v \\
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c_p &= \frac{c_p}{\gamma} + R \\
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c_p &= \frac{\gamma}{\gamma -1} R
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\end{align*}
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</details>
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## Properties of State
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*State* is defined as the condition of a system as described by its properties.
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@ -79,7 +157,7 @@ In thermodynamics we distinguish between *intensive*, *extensive*, and *specific
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- Specific (extensive) --- extensive properties which are reduced to unit mass of substance
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(essentially an extensive property divided by mass) (e.g. specific volume)
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## Units
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### Units
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Property | Symbol | Units | Intensive | Extensive
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--------------- | ------ | --------------- | --------- | ---------
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@ -87,15 +165,128 @@ Pressure | p | Pa | Yes |
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Temperature | T | K | Yes |
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Volume | V | m$^3$ | | Yes
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Mass | m | kg | | Yes
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Specific Volume | $\nu$ | m$^3$ kg$^{-1}$ | Yes |
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Specific Volume | v | m$^3$ kg$^{-1}$ | Yes |
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Density | $\rho$ | kg m$^{-3}$ | Yes |
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Internal Energy | U | J | | Yes
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Entropy | S | J K$^{-1}$ | | Yes
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Enthalpy | H | J | | Yes
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### Density
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For an ideal gas:
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$$\rho = \frac{p}{RT}$$
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### Enthalpy and Specific Enthalpy
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Enthalpy does not have a general physical interpretation.
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It is used because the combination $u + pv$ appears naturally in the analysis of many
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thermodynamic problems.
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The heat transferred to a closed system undergoing a reversible constant pressure process is equal
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to the change in enthalpy of the system.
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Enthalpy is defined as:
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$$H = U+pV$$
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and Specific Enthalpy:
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$$h = u + pv$$
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### Entropy and Specific Entropy
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Entropy is defined as the following, given that the process s reversible:
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$$S_2 - S_1 = \int\! \frac{\mathrm{d}Q}{T}$$
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### Heat Capacity and Specific Heat Capacity
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Heat capacity is quantity of heat required to raise the temperature of a system by a unit
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temperature:
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$$C = \frac{\mathrm{d}Q}{\mathrm{d}T}$$
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Specific heat capacity is the quantity of heat required to raise the temperature of a unit mass
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substance by a unit temperature:
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$$c = \frac{\mathrm{d}q}{\mathrm{d}T}$$
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<details>
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<summary>
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#### Heat Capacity in Closed Systems and Internal Energy
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The specific heat transfer to a closed system during a reversible constant **volume** process is
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equal to the change in specific **internal energy** of the system:
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$$c_v = \frac{\mathrm{d}q}{\mathrm{d}T} = \frac{\mathrm{d}u}{\mathrm{d}T}$$
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</summary>
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This is because if the change in volume, $\mathrm{d}v = 0$, then the work done, $\mathrm{d}w = 0$
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also.
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So applying the (1st Corollary of the) 1st Law to an isochoric process:
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$$\mathrm{d}q + \mathrm{d}w = \mathrm{d}u \rightarrow \mathrm{d}q = \mathrm{d}u$$
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since $\mathrm{d}w = 0$.
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</details>
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<details>
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<summary>
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#### Heat Capacity in Closed Systems and Enthalpy
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The specific heat transfer to a closed system during a reversible constant **pressure** process is
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equal to the change in specific **enthalpy** of the system:
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$$c_p = \frac{\mathrm{d}q}{\mathrm{d}T} = \frac{\mathrm{d}h}{\mathrm{d}T}$$
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</summary>
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This is because given that pressure, $p$, is constant, work, $w$, can be expressed as:
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$$w = -\int^2_1\! p \,\mathrm{d}v = -p(v_2 - v_1)$$
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Applying the (1st corollary of the) 1st law to the closed system:
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\begin{align*}
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q + w &= u_2 - u_1 \rightarrow q = u_2 - u_1 + p(v_2 - v_1) \\
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q &= u_2 + pv_2 - (u_1 + pv_1) \\
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&= h_2 - h_1 = \mathrm{d}h \\
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\therefore \mathrm{d}q &= \mathrm{d}h
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\end{align*}
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</details>
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<details>
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<summary>
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#### Ratio of Specific Heats
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$c_p > c_v$ is always true.
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</summary>
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Heating a volume of fluid, $V$, at a constant volume requires specific heat $q_v$ where
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$$q_v = u_2 - u_1 \therefore c_v = \frac{q_v}{\Delta T}$$
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Heating the same volume of fluid but under constant pressure requires a specific heat $q_p$ where
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$$q_p =u_2 - u_1 + p(v_2-v_1) \therefore c_p = \frac{q_p}{\Delta T}$$
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Since $p(v_2-v_1) > 0$, $\frac{q_p}{q_v} > 1 \therefore q_p > q_v \therefore c_p > c_v$.
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The ratio $\frac{c_p}{c_v} = \gamma$
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</details>
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## Thermodynamic Processes and Cycles
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When a thermodynaic system changes from one state to another it is said to execute a *process*.
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When a thermodynamic system changes from one state to another it is said to execute a *process*.
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An example of a process is expansion (volume increasing).
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A *cycle* is a process or series of processes in which the end state is identical to the beginning.
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@ -130,7 +321,8 @@ Heat and Work are different forms of enery transfer.
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They are both transient phenomena and systems never possess heat or work.
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Both represent energy crossing boundaries when a system undergoes a change of state.
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By convention, the transfer of energy into the system from the surroundings is positive.
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By convention, the transfer of energy into the system from the surroundings is positive (work is
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being done *on* the system *by* the surroundings).
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### Heat
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@ -146,8 +338,36 @@ By convention, the transfer of energy into the system from the surroundings is p
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$$W = \int\! F \mathrm{d}x$$
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(the work, $W$, done by a force, $F$, when the point of application of the force undergoes a
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displacement, $\mathrm dx$)
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displacement, $\mathrm{d}x$)
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## Thermally Insulated and Isolated Systems
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In thermally insulated systems and isolated systems, heat transfer cannot take place.
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In thermally isolated systems, work transfer cannot take place.
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# Process and State Diagrams
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Reversible processes are represented by solid lines, and irreversible processes by dashed lines.
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# 1st Law of Thermodynamics
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The 1st Law of Thermodynamics can be thought of as:
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> When a closed system is taken through a cycle, the sum of the *net* work transfer ($W$) and *net*
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> heat transfer ($Q$) equals zero:
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>
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> $$W_{net} + Q_{net} = 0$$
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>
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## 1st Corollary
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> The change in internal energy of a closed system is equal to the sum of the heat transferred
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> and the work done during any change of state
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>
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> $$W_{12} + Q_{12} = U_2 - U_1$$
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## 2nd Corollary
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> The internal energy of a closed system remains unchanged if it
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> [thermally isolated](#thermally-insulated-and-isolated-systems) from its surroundings
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