---
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
$$c_v = \frac{R}{\gamma - 1}$$
$$c_p = \frac{\gamma}{\gamma -1} R$$
#### Derivation
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*}
### 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
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
## 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:
$$c = \frac{\mathrm{d}q}{\mathrm{d}T}$$
### 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:
$$c_v = \frac{\mathrm{d}q}{\mathrm{d}T} = \frac{\mathrm{d}u}{\mathrm{d}T}$$
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 $\mathrm{d}w = 0$.
### 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:
$$c_p = \frac{\mathrm{d}q}{\mathrm{d}T} = \frac{\mathrm{d}h}{\mathrm{d}T}$$
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*}
### Ratio of Specific Heats
$c_p > c_v$ is always true.
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 $\frac{c_p}{c_v} = \gamma$
# 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.
#### Example 1
Derive
$$\frac{p_2}{p_1} = \left(\frac{T_2}{T_1}\right)^{\frac{n}{n-1}}$$
\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*}
How did you do that last step?
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*}
## 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 $\gamma$.
Derivation
\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*}
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).
# 2nd Law of Thermodynamics
The 2nd Law recognises that processes happen in a certain direction.
It was discovered through the study of heat engines (ones that produce mechanical work from heat).
> Heat does not spontaneously flow from a cooler to a hotter body.
~ Clausius' Statement on the 2nd Law of Thermodynamics
> It is impossible to construct a heat engine that will operate in a cycle and take heat from a
> reservoir and produce an equivalent amount of work.
~ Kelvin-Planck Statement of 2nd Law of Thermodynamics
## Heat Engines
A heat engine must have:
- Thermal energy reservoir --- a large body of heat that does not change in temperature
- Heat source --- a reservoir that supplies heat to the engine
- Heat sink --- a reservoir that absorbs heat rejected from a heat engine (this is usually
surrounding environment)
#### Steam Power Plant
## Thermal Efficiency
For heat engines, $Q_{out} > 0$ so $W_{out} < Q_{in}$ as $W_{out} = Q_{in} - Q_{out}$
$$\eta = \frac{W_{out}}{Q_{in}} = 1 - \frac{Q_{out}}{Q_{in}}$$
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