attempt to better organise thermodynamics page

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@ -116,211 +116,6 @@ c_p &= \frac{c_p}{\gamma} + R \\
c_p &= \frac{\gamma}{\gamma -1} R c_p &= \frac{\gamma}{\gamma -1} R
\end{align*} \end{align*}
</details>
## 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
independant 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> </details>
## Thermodynamic Processes and Cycles ## Thermodynamic Processes and Cycles
@ -355,7 +150,7 @@ Constant volume process
## Heat and Work ## Heat and Work
Heat and Work are different forms of enery transfer. Heat and Work are different forms of energy transfer.
They are both transient phenomena and systems never possess heat or work. 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. Both represent energy crossing boundaries when a system undergoes a change of state.
@ -385,10 +180,6 @@ In thermally insulated systems and isolated systems, heat transfer cannot take p
In thermally isolated systems, work transfer cannot take place. In thermally isolated systems, work transfer cannot take place.
# Process and State Diagrams
Reversible processes are represented by solid lines, and irreversible processes by dashed lines.
# 1st Law of Thermodynamics # 1st Law of Thermodynamics
The 1st Law of Thermodynamics can be thought of as: The 1st Law of Thermodynamics can be thought of as:
@ -411,7 +202,216 @@ The 1st Law of Thermodynamics can be thought of as:
> The internal energy of a closed system remains unchanged if it > The internal energy of a closed system remains unchanged if it
> [thermally isolated](#thermally-insulated-and-isolated-systems) from its surroundings > [thermally isolated](#thermally-insulated-and-isolated-systems) from its surroundings
# Polytropic Processes # 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
independant 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: A polytropic process is one which obeys the polytropic law:
@ -462,7 +462,7 @@ For any values of $x$ and $y$
</details> </details>
</details> </details>
# Isentropic ## Isentropic Processes
*Isentropic* means constant entropy: *Isentropic* means constant entropy:
@ -506,7 +506,7 @@ An isentropic process is an idealization of an actual process, and serves as the
real life processes. real life processes.
They are often desired and often the processes on which device efficiencies are calculated. They are often desired and often the processes on which device efficiencies are calculated.
## Heat Transfer During Isentropic Processes ### Heat Transfer During Isentropic Processes
Assume that the compression 1-2 follows a polytropic process with a polytropic index $n$. Assume that the compression 1-2 follows a polytropic process with a polytropic index $n$.
The work transfer is: The work transfer is: