1. Thermodynamic systems and properties

Thermodynamics studies energy, matter, and property changes at the macroscopic scale.

A system is the matter or region selected for analysis. Everything outside the system is the surroundings. The boundary may be real or imaginary, fixed or moving.

Types of systems

System type	Mass crossing boundary?	Energy crossing boundary?	Example
Closed system / control mass	No	Yes	Gas in a piston-cylinder
Open system / control volume	Yes	Yes	Turbine, compressor, nozzle
Isolated system	No	No	Ideal insulated sealed tank

Before solving any problem, identify the system type. This determines the correct balance equation.

Properties

A property describes the state of a system.

Examples:

Pressure, $P$
Temperature, $T$
Volume, $V$
Specific volume, $v$
Density, $\rho$
Internal energy, $U$
Enthalpy, $H$
Entropy, $S$

Extensive and intensive properties

An extensive property depends on system size.

Examples:

m,\quad V,\quad U,\quad H,\quad S

An intensive property does not depend on system size.

Examples:

P,\quad T,\quad v,\quad u,\quad h,\quad s

Specific properties are extensive properties per unit mass:

v = \frac{V}{m}

u = \frac{U}{m}

h = \frac{H}{m}

s = \frac{S}{m}

2. State, equilibrium, and processes

State

The state of a simple compressible system is fixed when enough independent intensive properties are known.

For many simple systems, two independent intensive properties determine the state.

Examples:

P,\ T

P,\ v

T,\ x

where $x$ is quality in a saturated mixture.

Equilibrium

A system is in thermodynamic equilibrium when it has:

Thermal equilibrium: no temperature gradients
Mechanical equilibrium: no unbalanced pressure forces
Phase equilibrium: no net phase change
Chemical equilibrium: no net chemical reaction

Process

A process is a change from one equilibrium state to another.

Common processes:

Process	Meaning
Isothermal	Constant temperature
Isobaric	Constant pressure
Isochoric / isometric	Constant volume
Adiabatic	No heat transfer
Isentropic	Constant entropy
Polytropic	$PV^n = \text{constant}$

Cycle

A cycle is a series of processes that returns the system to its initial state.

For any property over a cycle:

\Delta U = 0

\Delta H = 0

\Delta S = 0

Path functions such as heat and work do not have to be zero over a cycle.

3. Units, dimensions, and sign conventions

Common SI units

Quantity	Symbol	SI unit
Mass	$m$	kg
Temperature	$T$	K
Pressure	$P$	Pa
Energy	$E$	J
Power	$\dot{W}$	W
Specific volume	$v$	m³/kg
Specific energy	$u, h$	J/kg
Entropy	$S$	J/K
Specific entropy	$s$	J/(kg·K)

Pressure

Absolute pressure is measured relative to vacuum.

Gauge pressure is measured relative to the atmosphere.

P_{abs} = P_{gage} + P_{atm}

Vacuum pressure is below atmospheric pressure:

P_{abs} = P_{atm} - P_{vac}

Temperature

Thermodynamic temperature must use an absolute scale.

T(K) = T(^\circ C) + 273.15

Sign convention used in these notes

These notes use the common engineering convention:

Heat transfer into the system is positive.
Work done by the system is positive.

Thus, for a closed system:

\Delta U = Q - W

More generally:

\Delta E = Q - W

where

$$ E = U + KE + PE $$

4. Pure substances and property data

A pure substance has a fixed chemical composition throughout.

Examples:

Water
Nitrogen
Refrigerant-134a
Carbon dioxide

A pure substance can exist in multiple phases.

Phase regions

Region	Description
Compressed liquid / subcooled liquid	Liquid not about to vaporize
Saturated liquid	Liquid about to vaporize
Saturated mixture	Liquid and vapor coexist
Saturated vapor	Vapor about to condense
Superheated vapor	Vapor not about to condense

Saturation temperature and pressure

At a given saturation pressure, a pure substance changes phase at a fixed saturation temperature.

T = T_{sat}(P)

At a given saturation temperature, the corresponding phase-change pressure is:

P = P_{sat}(T)

Quality

Quality is the mass fraction of vapor in a saturated liquid-vapor mixture.

x = \frac{m_g}{m_f + m_g}

where:

$m_f$ = mass of saturated liquid
$m_g$ = mass of saturated vapor

Quality only applies in the saturated mixture region.

Saturated mixture property relation

For a saturated mixture:

y = y_f + x y_{fg}

where $y$ may represent $v$, $u$, $h$, or $s$.

Also:

y_{fg} = y_g - y_f

Examples:

v = v_f + x v_{fg}

u = u_f + x u_{fg}

h = h_f + x h_{fg}

s = s_f + x s_{fg}

Enthalpy

Enthalpy is defined as:

$$ H = U + PV $$

Specific enthalpy is:

$$ h = u + Pv $$

Enthalpy is especially useful for control-volume problems because flow work is naturally included in $h$.

How to choose property data

Use this sequence:

Identify the substance.
Identify known independent properties.
Determine the phase or region.
Use the correct table or equation of state.
Interpolate if needed.

Compressed liquid approximation

For many liquids, if compressed liquid data are unavailable:

v \approx v_f(T)

u \approx u_f(T)

h \approx h_f(T) + v_f(T)\left[P - P_{sat}(T)\right]

Often, for modest pressure changes:

h \approx h_f(T)

5. Ideal gases and equations of state

An equation of state relates pressure, temperature, and volume.

Ideal gas equation

For an ideal gas:

$$ PV = mRT $$

Specific form:

$$ Pv = RT $$

Molar form:

PV = n\bar{R}T

where:

$R$ = gas constant for the specific gas
$\bar{R}$ = universal gas constant
$n$ = number of moles

Relationship between gas constants:

R = \frac{\bar{R}}{M}

where $M$ is molar mass.

Density form

Since

v = \frac{1}{\rho}

the ideal gas law can be written as:

P = \rho RT

Compressibility factor

Real-gas behavior can be estimated using:

$$ Pv = ZRT $$

where $Z$ is the compressibility factor.

For ideal gases:

$$ Z = 1 $$

If $Z$ is close to 1, the ideal gas model is reasonable.

Constant specific heats

For ideal gases with constant specific heats:

\Delta u = c_v(T_2 - T_1)

\Delta h = c_p(T_2 - T_1)

and

$$ c_p - c_v = R $$

Specific heat ratio:

k = \frac{c_p}{c_v}

Ideal gas entropy changes

For constant specific heats:

s_2 - s_1 = c_p \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{P_2}{P_1}\right)

Equivalent form:

s_2 - s_1 = c_v \ln\left(\frac{T_2}{T_1}\right) + R \ln\left(\frac{v_2}{v_1}\right)

Isentropic ideal gas relations

For an ideal gas with constant specific heats undergoing an isentropic process:

$$ PV^k = constant $$

Tv^{k-1} = constant

\frac{T_2}{T_1} = \left(\frac{v_1}{v_2}\right)^{k-1}

\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{(k-1)/k}

\frac{P_2}{P_1} = \left(\frac{v_1}{v_2}\right)^k

6. Heat, work, and energy

Energy

Total energy is:

$$ E = U + KE + PE $$

where

KE = \frac{1}{2}mV^2

$$ PE = mgz $$

Specific total energy:

e = u + \frac{V^2}{2} + gz

Heat

Heat is energy transfer due to temperature difference.

Heat transfer rate:

\dot{Q} = \frac{\delta Q}{dt}

Adiabatic process:

$$ Q = 0 $$

Work

Work is energy transfer due to a generalized force acting through a generalized displacement.

Power:

\dot{W} = \frac{\delta W}{dt}

Boundary work

For a quasi-equilibrium closed-system process:

W_b = \int_{1}^{2} P\,dV

If pressure is constant:

$$ W_b = P(V_2 - V_1) $$

Specific boundary work:

w_b = \int_{1}^{2} P\,dv

Polytropic work

For a polytropic process:

$$ PV^n = C $$

If $n \ne 1$:

W_b = \frac{P_2V_2 - P_1V_1}{1-n}

For an ideal gas:

W_b = \frac{mR(T_2 - T_1)}{1-n}

If $n = 1$:

W_b = P_1V_1 \ln\left(\frac{V_2}{V_1}\right)

For an ideal gas when $n = 1$, the process is isothermal:

W_b = mRT \ln\left(\frac{V_2}{V_1}\right)

Spring work

For a linear spring:

$$ F = kx $$

Spring work:

W = \int_{x_1}^{x_2} kx\,dx

W = \frac{1}{2}k(x_2^2 - x_1^2)

Electrical work

Electrical work:

W_e = \int V I\,dt

For constant voltage and current:

$$ W_e = VIt $$

Electrical power:

\dot{W}_e = VI

7. First Law for closed systems

The First Law is conservation of energy.

For a closed system:

\Delta E = Q - W

Expanded:

\Delta U + \Delta KE + \Delta PE = Q - W

If kinetic and potential energy changes are negligible:

\Delta U = Q - W

On a mass basis:

\Delta e = q - w

If kinetic and potential energy changes are negligible:

\Delta u = q - w

Cyclic closed system

For a cycle:

\Delta E_{cycle} = 0

Therefore:

Q_{net} = W_{net}

Constant-volume closed system

For constant volume boundary work:

W_b = \int P\,dV = 0

If no other work modes occur:

Q = \Delta U

Constant-pressure closed system

For constant pressure:

$$ W_b = P(V_2 - V_1) $$

If only boundary work occurs and kinetic/potential energy changes are negligible:

Q = \Delta U + P\Delta V

Since

$$ H = U + PV $$

for constant pressure:

Q = \Delta H

Rigid tank

For a rigid tank:

$$ V = constant $$

$$ W_b = 0 $$

If the tank is also adiabatic and has no other work modes:

\Delta U = 0

8. First Law for control volumes

A control volume allows mass to cross the boundary.

Conservation of mass

General mass balance:

\frac{dm_{CV}}{dt} = \sum \dot{m}_{in} - \sum \dot{m}_{out}

At steady state:

\frac{dm_{CV}}{dt} = 0

So:

\sum \dot{m}_{in} = \sum \dot{m}_{out}

For one inlet and one outlet:

\dot{m}_1 = \dot{m}_2 = \dot{m}

Mass flow rate

Mass flow rate:

\dot{m} = \rho A V

Using specific volume:

\dot{m} = \frac{AV}{v}

where $V$ is flow speed, not volume.

Energy balance for control volumes

General rate form:

\frac{dE_{CV}}{dt} = \dot{Q} - \dot{W} + \sum \dot{m}_{in}\left(h + \frac{V^2}{2} + gz\right) - \sum \dot{m}_{out}\left(h + \frac{V^2}{2} + gz\right)

At steady state:

0 = \dot{Q} - \dot{W} + \sum \dot{m}_{in}\left(h + \frac{V^2}{2} + gz\right) - \sum \dot{m}_{out}\left(h + \frac{V^2}{2} + gz\right)

For one inlet and one outlet:

\dot{Q} - \dot{W} = \dot{m}\left[(h_2-h_1) + \frac{V_2^2 - V_1^2}{2} + g(z_2-z_1)\right]

Specific form:

q - w = (h_2-h_1) + \frac{V_2^2 - V_1^2}{2} + g(z_2-z_1)

9. Common steady-flow devices

For most steady-flow devices, start with:

\dot{Q} - \dot{W} = \dot{m}\left[(h_2-h_1) + \frac{V_2^2 - V_1^2}{2} + g(z_2-z_1)\right]

Then remove negligible terms.

Turbine

A turbine produces work output.

Typical assumptions:

Steady state
Adiabatic
Negligible kinetic and potential energy changes

Energy balance:

\dot{W}_{out} = \dot{m}(h_1 - h_2)

Specific work output:

w_{out} = h_1 - h_2

Compressor

A compressor requires work input and increases gas pressure.

Typical assumptions:

Steady state
Adiabatic
Negligible kinetic and potential energy changes

Energy balance:

\dot{W}_{in} = \dot{m}(h_2 - h_1)

Specific work input:

w_{in} = h_2 - h_1

Pump

A pump increases liquid pressure.

For an incompressible liquid:

w_{in} \approx v(P_2 - P_1)

Pump power input:

\dot{W}_{in} \approx \dot{m}v(P_2 - P_1)

Nozzle

A nozzle increases flow speed.

Typical assumptions:

Steady state
Adiabatic
No shaft work
Negligible potential energy change

Energy balance:

h_1 + \frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2}

Velocity relation:

V_2 = \sqrt{V_1^2 + 2(h_1-h_2)}

Use SI units carefully: if $h$ is in kJ/kg, multiply by 1000.

Diffuser

A diffuser decreases flow speed and increases pressure.

Typical assumptions are similar to a nozzle:

h_1 + \frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2}

Throttling valve

A throttling valve causes a pressure drop with no work output.

Typical assumptions:

Steady state
Adiabatic
No shaft work
Negligible kinetic and potential energy changes

Energy balance:

$$ h_1 = h_2 $$

Throttling is highly irreversible.

Heat exchanger

A heat exchanger transfers heat between streams.

If the whole heat exchanger is the control volume and heat loss to surroundings is negligible:

\dot{Q}_{CV} \approx 0

For two streams with no work and negligible kinetic/potential energy changes:

\dot{m}_h(h_{h,in}-h_{h,out}) = \dot{m}_c(h_{c,out}-h_{c,in})

Mixing chamber

For an adiabatic mixing chamber with no work and negligible kinetic/potential energy changes:

\sum \dot{m}_{in}h_{in} = \sum \dot{m}_{out}h_{out}

Mass balance:

\sum \dot{m}_{in} = \sum \dot{m}_{out}

10. Second Law of Thermodynamics

The First Law gives energy conservation. The Second Law gives direction and limits.

Thermal reservoir

A thermal reservoir can supply or absorb heat without changing temperature.

Examples:

Atmosphere
Large lake
Furnace
Ocean

Heat engine

A heat engine receives heat from a high-temperature reservoir, produces net work, and rejects heat to a low-temperature reservoir.

Energy balance over a cycle:

W_{net,out} = Q_H - Q_L

Thermal efficiency:

\eta_{th} = \frac{W_{net,out}}{Q_H}

Equivalent form:

\eta_{th} = 1 - \frac{Q_L}{Q_H}

Refrigerator

A refrigerator uses work input to move heat from a low-temperature region to a high-temperature region.

Coefficient of performance:

COP_R = \frac{Q_L}{W_{net,in}}

Since

W_{net,in} = Q_H - Q_L

then:

COP_R = \frac{Q_L}{Q_H - Q_L}

Heat pump

A heat pump uses work input to deliver heat to a high-temperature region.

Coefficient of performance:

COP_{HP} = \frac{Q_H}{W_{net,in}}

Thus:

COP_{HP} = \frac{Q_H}{Q_H - Q_L}

Relationship:

COP_{HP} = COP_R + 1

Kelvin-Planck statement

No heat engine can convert all the heat it receives from a single reservoir into net work while operating in a cycle.

This means no real heat engine has:

\eta_{th} = 1

Clausius statement

No refrigerator or heat pump can move heat from a colder body to a hotter body without work input.

Reversible and irreversible processes

A reversible process can be reversed without leaving any net change in the system or surroundings.

Real processes are irreversible due to effects such as:

Friction
Unrestrained expansion
Mixing
Heat transfer through a finite temperature difference
Electrical resistance
Inelastic deformation

11. Entropy

Entropy is a thermodynamic property that measures energy dispersal and irreversibility.

For an internally reversible process:

dS = \frac{\delta Q_{rev}}{T}

Specific form:

ds = \frac{\delta q_{rev}}{T}

Entropy change

Entropy change between two states is independent of path:

\Delta S = S_2 - S_1

For a closed system:

\Delta S = \int_1^2 \frac{\delta Q_{rev}}{T}

Entropy principle

For an isolated system:

\Delta S_{isolated} \ge 0

For the universe:

\Delta S_{universe} \ge 0

Equality applies to reversible processes:

\Delta S_{universe} = 0

Irreversible processes generate entropy:

\Delta S_{universe} > 0

Entropy balance for a closed system

General entropy balance:

\Delta S_{system} = \sum \frac{Q_k}{T_k} + S_{gen}

where:

$Q_k$ is heat transfer at boundary temperature $T_k$
$S_{gen}$ is entropy generation

Entropy generation cannot be negative:

S_{gen} \ge 0

Entropy balance for a control volume

Rate form:

\frac{dS_{CV}}{dt} = \sum \frac{\dot{Q}_k}{T_k} + \sum \dot{m}_{in}s_{in} - \sum \dot{m}_{out}s_{out} + \dot{S}_{gen}

At steady state:

0 = \sum \frac{\dot{Q}_k}{T_k} + \sum \dot{m}_{in}s_{in} - \sum \dot{m}_{out}s_{out} + \dot{S}_{gen}

Isentropic process

An isentropic process has constant entropy:

$$ s_2 = s_1 $$

For a process to be isentropic, it must be both:

Adiabatic
Reversible

Adiabatic alone does not guarantee isentropic behavior.

T-ds relations

First T-ds relation:

$$ Tds = du + Pdv $$

Second T-ds relation:

$$ Tds = dh - vdP $$

These apply to simple compressible substances.

12. Power and refrigeration cycles

A cycle returns the working fluid to its initial state, so property changes over the cycle are zero.

Carnot cycle

The Carnot cycle is a completely reversible cycle operating between two reservoirs.

Carnot heat engine efficiency:

\eta_{Carnot} = 1 - \frac{T_L}{T_H}

Temperatures must be absolute.

Carnot refrigerator:

COP_{R,Carnot} = \frac{T_L}{T_H - T_L}

Carnot heat pump:

COP_{HP,Carnot} = \frac{T_H}{T_H - T_L}

No engine operating between the same two reservoirs can be more efficient than a Carnot engine.

Otto cycle

The ideal Otto cycle models spark-ignition engines.

Processes:

Isentropic compression
Constant-volume heat addition
Isentropic expansion
Constant-volume heat rejection

Compression ratio:

r = \frac{V_1}{V_2}

Air-standard Otto efficiency:

\eta_{Otto} = 1 - \frac{1}{r^{k-1}}

Diesel cycle

The ideal Diesel cycle models compression-ignition engines.

Processes:

Isentropic compression
Constant-pressure heat addition
Isentropic expansion
Constant-volume heat rejection

Cutoff ratio:

r_c = \frac{V_3}{V_2}

Air-standard Diesel efficiency:

\eta_{Diesel} = 1 - \frac{1}{r^{k-1}} \left[ \frac{r_c^k - 1}{k(r_c - 1)} \right]

Brayton cycle

The ideal Brayton cycle models gas turbine engines.

Processes:

Isentropic compression
Constant-pressure heat addition
Isentropic expansion
Constant-pressure heat rejection

Pressure ratio:

r_p = \frac{P_2}{P_1}

Air-standard Brayton efficiency:

\eta_{Brayton} = 1 - \frac{1}{r_p^{(k-1)/k}}

Rankine cycle

The ideal Rankine cycle models vapor power plants.

Main components:

Pump
Boiler
Turbine
Condenser

Pump work input:

w_{pump,in} = h_2 - h_1

Boiler heat input:

q_{in} = h_3 - h_2

Turbine work output:

w_{turb,out} = h_3 - h_4

Condenser heat rejection:

q_{out} = h_4 - h_1

Net work output:

w_{net,out} = w_{turb,out} - w_{pump,in}

Thermal efficiency:

\eta_{Rankine} = \frac{w_{net,out}}{q_{in}}

Vapor-compression refrigeration cycle

Main components:

Compressor
Condenser
Expansion valve
Evaporator

Compressor work input:

w_{comp,in} = h_2 - h_1

Condenser heat rejection:

$$ q_H = h_2 - h_3 $$

Throttling valve:

$$ h_3 = h_4 $$

Evaporator heat absorption:

$$ q_L = h_1 - h_4 $$

Refrigerator COP:

COP_R = \frac{q_L}{w_{comp,in}}

Interactive visual

T-s diagram explorer

Switch between a process path and a closed cycle to see how temperature and entropy move together on a T-s plane.

Process path process

Path type isobaric Start temperature T1 420 K Entropy span delta s 0.9 Temperature rise delta T 120 K

Hot temperature Th 650 K Cold temperature Tl 320 K Entropy width delta s 0.9

State detail Start T1 = 420 K

State detail Path type = isobaric

Heat proxy End state: T2 = 540 K, s2 = 1.55

Net result q_rev proxy = 108.0 kJ/kg

The filled region gives a simple heat-transfer proxy on the T-s plane.

13. Problem-solving workflow

Use this checklist for most Thermodynamics I problems.

Step 1: Define the system

Decide whether the system is:

Closed
Open
Isolated

Draw a boundary.

Step 2: List knowns and unknowns

Write given values with units.

Convert temperatures to absolute scale when needed.

Step 3: Identify the process model

Examples:

Steady state
Adiabatic
Isothermal
Isentropic
Constant pressure
Constant volume
Rigid tank
Throttling
Polytropic

Step 4: Determine the substance model

Use one of:

Property tables
Ideal gas model
Incompressible substance model
Saturated mixture relation
Real-gas compressibility factor

Step 5: Choose the correct balance

Closed-system energy balance:

\Delta E = Q - W

Control-volume mass balance:

\frac{dm_{CV}}{dt} = \sum \dot{m}_{in} - \sum \dot{m}_{out}

Control-volume energy balance:

\frac{dE_{CV}}{dt} = \dot{Q} - \dot{W} + \sum \dot{m}_{in}\left(h + \frac{V^2}{2} + gz\right) - \sum \dot{m}_{out}\left(h + \frac{V^2}{2} + gz\right)

Entropy balance:

\Delta S = \sum \frac{Q_k}{T_k} + S_{gen}

Step 6: Apply assumptions

Common assumptions:

Negligible kinetic energy change
Negligible potential energy change
Steady state
Adiabatic
No shaft work
Ideal gas
Constant specific heats

Only cancel terms after stating why.

Step 7: Solve symbolically first

Keep equations symbolic as long as possible. Substitute numbers after the equation is arranged.

Step 8: Check physical meaning

Ask:

Does the sign make sense?
Are units consistent?
Is the magnitude reasonable?
Is entropy generation nonnegative?
Is efficiency less than the Carnot limit?

14. Formula sheet

Basic property relations

v = \frac{V}{m}

\rho = \frac{m}{V} = \frac{1}{v}

$$ h = u + Pv $$

$$ E = U + KE + PE $$

e = u + \frac{V^2}{2} + gz

Ideal gas

$$ PV = mRT $$

$$ Pv = RT $$

P = \rho RT

$$ c_p - c_v = R $$

k = \frac{c_p}{c_v}

Saturated mixture

x = \frac{m_g}{m_f + m_g}

y = y_f + xy_{fg}

y_{fg} = y_g - y_f

Closed-system First Law

\Delta E = Q - W

\Delta U + \Delta KE + \Delta PE = Q - W

If kinetic and potential energy changes are negligible:

\Delta U = Q - W

Boundary work

W_b = \int_1^2 P\,dV

Constant pressure:

$$ W_b = P(V_2 - V_1) $$

Polytropic process, $n \ne 1$:

W_b = \frac{P_2V_2 - P_1V_1}{1-n}

Polytropic process, $n = 1$:

W_b = P_1V_1\ln\left(\frac{V_2}{V_1}\right)

Control-volume balances

Mass balance:

\frac{dm_{CV}}{dt} = \sum \dot{m}_{in} - \sum \dot{m}_{out}

Steady-flow energy balance:

\dot{Q} - \dot{W} = \dot{m}\left[(h_2-h_1) + \frac{V_2^2 - V_1^2}{2} + g(z_2-z_1)\right]

Mass flow rate:

\dot{m} = \rho AV

\dot{m} = \frac{AV}{v}

Common devices

Turbine:

\dot{W}_{out} = \dot{m}(h_1-h_2)

Compressor:

\dot{W}_{in} = \dot{m}(h_2-h_1)

Pump:

w_{in} \approx v(P_2-P_1)

Nozzle or diffuser:

h_1 + \frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2}

Throttling valve:

$$ h_1 = h_2 $$

Second Law and entropy

Heat engine efficiency:

\eta_{th} = \frac{W_{net,out}}{Q_H}

\eta_{th} = 1 - \frac{Q_L}{Q_H}

Refrigerator COP:

COP_R = \frac{Q_L}{W_{net,in}}

Heat pump COP:

COP_{HP} = \frac{Q_H}{W_{net,in}}

Carnot efficiency:

\eta_{Carnot} = 1 - \frac{T_L}{T_H}

Entropy inequality:

\Delta S_{universe} \ge 0

Entropy generation:

S_{gen} \ge 0

Closed-system entropy balance:

\Delta S_{system} = \sum \frac{Q_k}{T_k} + S_{gen}

Steady control-volume entropy balance:

0 = \sum \frac{\dot{Q}_k}{T_k} + \sum \dot{m}_{in}s_{in} - \sum \dot{m}_{out}s_{out} + \dot{S}_{gen}

Ideal gas entropy changes

s_2 - s_1 = c_p \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{P_2}{P_1}\right)

s_2 - s_1 = c_v \ln\left(\frac{T_2}{T_1}\right) + R \ln\left(\frac{v_2}{v_1}\right)

Isentropic ideal gas relations

$$ PV^k = constant $$

Tv^{k-1} = constant

\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{(k-1)/k}

\frac{P_2}{P_1} = \left(\frac{v_1}{v_2}\right)^k

Common mistakes to avoid

Using Celsius instead of Kelvin in ideal gas or Carnot equations.
Mixing gauge pressure and absolute pressure.
Applying quality outside the saturated mixture region.
Calling an adiabatic process isentropic without checking reversibility.
Forgetting that $h$ includes flow work.
Using $\Delta U = Q - W$ for open systems without accounting for mass flow.
Forgetting to convert kJ/kg to J/kg when using velocity terms.
Assuming work is always positive. The sign depends on whether work is done by or on the system.
Ignoring entropy generation in irreversible processes.
Comparing efficiencies without checking reservoir temperatures or cycle assumptions.

Sources

OpenStax University Physics
Physics LibreTexts
Halliday, Resnick, and Walker, Fundamentals of Physics
Serway and Jewett, Physics for Scientists and Engineers
Griffiths, Introduction to Electrodynamics
Griffiths, Introduction to Quantum Mechanics
Taylor, Classical Mechanics
Parell GitHub repository