---
author: Alvie Rahman
date: \today
title: MMME1048 // Thermodynamics
tags: [ uni, nottingham, mechanical, engineering, mmme1048, thermodynamics ]
---

# 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 sysems:

- Closed systems

  - Contain a fixed quantity of matter
  - Work and heat cross bounaries
  - 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>

## 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 Proccesses

During reversible processes, the system undergoes a continuous succession of equilibrium states.
Changes in the system can be defined and reversed to restore the intial 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
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:

$$pv^n = k \text{ or } p_1v_1^n = p_2v_2^n$$

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).
