The laws of thermodynamics, steam tables, power and refrigeration cycles, fluid statics, Bernoulli, pipe losses, heat transfer, R-values and heat exchangers, with Python exercises.

Thermodynamics & Fluids

## Why it matters

Every engine, power plant, refrigerator, boiler and chemical reactor is an energy bookkeeping problem. Thermodynamics gives you the rules for that bookkeeping. Civil engineers meet it in building energy and water heating, mechanical engineers in engines and HVAC, chemists in reaction energies, and materials scientists in phase changes and heat treatment. The laws are short; the skill is applying them carefully, with the right system, the right signs and the right units.

## Systems and sign conventions

Start every problem by drawing a boundary around a **system**:

- A **closed system** (control mass) lets energy cross the boundary but not mass. Example: gas trapped under a piston.
- An **open system** (control volume) lets mass flow in and out. Example: a pump, turbine or heat exchanger.

Energy crosses a boundary in two ways: **heat** Q (driven by a temperature difference) and **work** W (everything else, such as a moving piston or a spinning shaft). This course uses the common engineering convention:

- Q is **positive** when heat flows **into** the system.
- W is **positive** when the system does work **on the surroundings**.

Some chemistry and physics texts use W positive *into* the system, which flips a sign in the first law. Always check which convention your textbook uses.

## The four laws in one paragraph each

- **Zeroth law.** If A is in thermal equilibrium with B, and B with C, then A is with C. This is why thermometers work.
- **First law.** Energy is conserved. For a closed system: `Q − W = ΔU` (plus changes in kinetic and potential energy if they matter).
- **Second law.** Heat flows on its own only from hot to cold, and no engine can turn all of its heat input into work. The best possible (Carnot) efficiency between a hot reservoir at T_H and a cold one at T_C is `η = 1 − T_C / T_H`, with temperatures in **kelvin**.
- **Third law.** The entropy of a perfect crystal approaches zero as temperature approaches absolute zero. You rarely use it in engineering calculations, but it anchors entropy tables.

## Worked example 1: heating gas under a piston

A cylinder holds air under a frictionless piston that keeps the pressure at 200 kPa. You add 50 kJ of heat, and the gas expands from 0.10 m³ to 0.25 m³. What is the change in internal energy?

At constant pressure, the boundary work is pressure times the change in volume. Note that kPa × m³ = kJ:

```text
W  = P · (V2 − V1) = 200 kPa × (0.25 − 0.10) m³ = 30 kJ   (done by the gas, so positive)
ΔU = Q − W = 50 kJ − 30 kJ = 20 kJ
```

Only 20 kJ of the 50 kJ stayed in the gas as internal energy; the other 30 kJ left as work pushing the piston.

## Worked example 2: the Carnot limit

A power plant boiler supplies heat at 600 K and the cooling water sits at 300 K. The second law caps any engine between these temperatures at:

```text
η_Carnot = 1 − T_C / T_H = 1 − 300 / 600 = 0.50   (50 %)
```

Real plants reach much less, because real processes have friction, heat losses and finite temperature differences. If someone claims an efficiency above the Carnot value, there is an error somewhere.

## The first law for open systems

For a steady-flow device with one inlet and one outlet, the first law becomes a rate balance using **enthalpy** h = u + Pv, which bundles the internal energy with the "flow work" needed to push fluid across the boundary:

```text
Q̇ − Ẇ = ṁ · [ (h2 − h1) + (V2² − V1²)/2 + g (z2 − z1) ]
```

For many turbines, pumps and boilers the kinetic and potential terms are small, so `Q̇ − Ẇ ≈ ṁ (h2 − h1)`. You will use this form for cycles in lesson 3. Watch the units: h is usually in kJ/kg, but V²/2 comes out in J/kg (m²/s²), so divide it by 1,000 before adding.

## Common mistakes

- **Using °C in a temperature ratio.** Carnot efficiency, ideal gas law and radiation all need kelvin. 300 °C is 573.15 K.
- **Mixing sign conventions.** Decide once, write it at the top of your work, and stick to it.
- **Forgetting that work is a path quantity.** `P·ΔV` is only right at constant pressure. For other processes, W is the area under the P–V curve.
- **Mixing kJ and J.** Pressure in Pa times volume in m³ gives J; in kPa it gives kJ.

## Recap

- Draw the system boundary first: closed or open.
- First law (closed): `Q − W = ΔU`. Open, steady flow: `Q̇ − Ẇ = ṁ Δh` when KE and PE are small.
- Second law: `η ≤ 1 − T_C/T_H` with kelvin temperatures.
- These are teaching examples. Real equipment is designed to codes and standards, and engineering designs that affect public safety must be reviewed by a licensed professional engineer.

## Try it

On the Do tab, finish a Python function that applies the first law to a closed system, then use it on the piston problem.


Use the four laws of thermodynamics and apply the first law to a closed system, with sign conventions and the Carnot limit.

Laws of thermodynamics

The First & Zeroth Laws of Thermodynamics: Crash Course Engineering #9

The First Law of Thermodynamics: Internal Energy, Heat, and Work

## Why it matters

Water and steam do a huge share of the world's engineering work: power plants, district heating, sterilisers, process heating and cooling towers. Unlike an ideal gas, water's properties near boiling cannot be written as one tidy equation you would use by hand, so engineers look them up in **property tables** (steam tables) or software built on the international IAPWS formulation. Reading these tables, and interpolating between rows, is a core skill tested in every thermodynamics course and on the FE exam.

## State principle and phases

For a simple pure substance, **two independent intensive properties** fix the state. Temperature and pressure work for compressed liquid or superheated vapour, but not inside the dome: while water boils at a given pressure, its temperature stays at the saturation temperature. Inside the dome you need a second property such as the **quality** x, the mass fraction that is vapour (x = 0 saturated liquid, x = 1 saturated vapour).

The main table families are:

- **Saturated water, by temperature** and **by pressure**: properties of saturated liquid (subscript f) and saturated vapour (subscript g), plus the difference fg (for example hfg, the enthalpy of vaporisation).
- **Superheated vapour**: blocks of rows at fixed pressure, for temperatures above saturation.
- **Compressed liquid**: often approximated by saturated liquid at the same temperature.

## Reading a saturated table

Here are a few rows of a saturated-water table (values rounded, from standard textbook tables based on IAPWS):

```text
T (°C)   Psat (kPa)   hf (kJ/kg)   hfg (kJ/kg)   hg (kJ/kg)
100      101.42       419.17       2256.4        2675.6
105      120.90       440.28       2243.1        2683.4
110      143.38       461.42       2229.7        2691.1
115      169.18       482.59       2216.0        2698.6
120      198.67       503.81       2202.1        2706.0
```

Check the consistency: hf + hfg = hg (419.17 + 2256.4 = 2675.6). For a wet mixture at quality x:

```text
h = hf + x · hfg
```

So water at 100 °C with x = 0.5 has h = 419.17 + 0.5 × 2256.4 = 1547.4 kJ/kg.

## Linear interpolation

Tables list values at steps. For a value between two rows, assume a straight line:

```text
y = y1 + (x − x1) · (y2 − y1) / (x2 − x1)
```

**Worked example:** hf at 112 °C.

```text
y = 461.42 + (112 − 110) · (482.59 − 461.42) / (115 − 110)
  = 461.42 + 2 · 21.17 / 5
  = 461.42 + 8.468 = 469.89 kJ/kg
```

The same formula works in either direction: you can find the temperature for a given enthalpy by swapping which column is x and which is y. Linear interpolation is accurate enough for most table steps, but it gets worse near the critical point, where properties change sharply.

## Deciding which table to use

Given temperature T and pressure P:

1. Look up Psat at T (or Tsat at P).
2. If P > Psat, the water is a **compressed liquid**.
3. If P < Psat, it is a **superheated vapour**.
4. If P = Psat, it is a saturated mixture, and you need one more property (x, v, h or s) to fix the state.

For example, water at 120 °C and 500 kPa: Psat at 120 °C is 198.67 kPa, and 500 > 198.67, so it is a compressed liquid, and h ≈ hf at 120 °C ≈ 503.8 kJ/kg.

## Other sources of properties

- **NIST Chemistry WebBook** offers free fluid property calculators for water and many refrigerants: https://webbook.nist.gov/chemistry/fluid/ (Checked on 2026-10-23.)
- **CoolProp** is an open-source property library you can call from Python on your own computer.
- Textbook tables differ slightly in the last digit and in the reference state (the zero point of h and s). Always take every property in a calculation from the same source.

## Common mistakes

- **Using the superheated table inside the dome.** Check Psat first.
- **Interpolating between rows of different pressure blocks** in the superheated table without doing a double interpolation.
- **Mixing reference states** from two different tables or programs.
- **Forgetting units.** Tables give v in m³/kg, h and u in kJ/kg, s in kJ/(kg·K), P in kPa or MPa.

## Recap

- Two independent properties fix a state; inside the dome T and P are not independent.
- h = hf + x·hfg for a wet mixture.
- Linear interpolation: y = y1 + (x − x1)(y2 − y1)/(x2 − x1).
- Compare P with Psat to choose the right table.

## Try it

On the Do tab, write the interpolation formula and use it to look up saturated-water properties between table rows.


Read saturated-water tables, decide which phase a state is in, and interpolate between table rows.

Properties and tables

How to Use Steam Tables

Steam Tables: Interpolation

## Why it matters

Most of the world's electricity comes from **heat engines** that run a fluid around a cycle: steam plants (coal, nuclear, solar thermal, geothermal) use the **Rankine cycle**, and gas turbines use the Brayton cycle. Every fridge, air conditioner and heat pump runs a **vapour-compression refrigeration cycle**, which is a heat engine run backwards. Computing a cycle's efficiency, or its coefficient of performance (COP), tells you how much useful output you get for the energy you pay for.

## The ideal Rankine cycle

Four steady-flow devices, numbered by state:

1. **Pump (1 → 2):** saturated liquid from the condenser is pumped to boiler pressure. Small work input.
2. **Boiler (2 → 3):** heat added at constant pressure turns the water into superheated steam.
3. **Turbine (3 → 4):** steam expands and produces shaft work.
4. **Condenser (4 → 1):** heat is rejected to cooling water at constant pressure, back to saturated liquid.

Apply the open-system first law (`q − w = Δh` per kg, KE and PE neglected) to each device:

```text
w_pump    = h2 − h1          (work in)
q_in      = h3 − h2          (boiler)
w_turbine = h3 − h4          (work out)
q_out     = h4 − h1          (condenser)
w_net     = w_turbine − w_pump
η_th      = w_net / q_in
```

## Worked example: a simple steam plant

A textbook-style ideal Rankine cycle runs between a boiler at 3 MPa, 350 °C and a condenser at 75 kPa. From steam tables (with an isentropic turbine and pump):

```text
h1 = 384.44 kJ/kg   (saturated liquid at 75 kPa)
h2 = 387.47 kJ/kg   (after the pump; w_pump ≈ 3.03 kJ/kg)
h3 = 3116.1 kJ/kg   (superheated steam, 3 MPa, 350 °C)
h4 = 2403.0 kJ/kg   (wet steam at 75 kPa, quality about 0.886)
```

```text
w_turbine = 3116.1 − 2403.0  = 713.1 kJ/kg
w_pump    = 387.47 − 384.44  = 3.03 kJ/kg
w_net     = 713.1 − 3.03     = 710.1 kJ/kg
q_in      = 3116.1 − 387.47  = 2728.6 kJ/kg
η_th      = 710.1 / 2728.6   ≈ 0.260   (26 %)
```

Notice how small the pump work is: pumping a liquid takes far less work than the turbine produces from the vapour. That is the Rankine cycle's big advantage. Compare 26 % with the Carnot limit between 350 °C (623 K) and the condenser's saturation temperature, about 92 °C (365 K): 1 − 365/623 ≈ 41 %. Real plants raise efficiency with higher boiler pressure and temperature, reheat and feedwater heating.

## The vapour-compression refrigeration cycle

A refrigerant runs through:

1. **Compressor (1 → 2):** low-pressure vapour is compressed to a hot, high-pressure vapour (work in).
2. **Condenser (2 → 3):** heat is rejected to the room or outdoors; the refrigerant condenses.
3. **Expansion valve (3 → 4):** throttling drops the pressure. A throttle does no work and, ideally, no heat transfer, so h4 = h3.
4. **Evaporator (4 → 1):** the cold refrigerant absorbs heat from the space being cooled.

The figure of merit is the **coefficient of performance**:

```text
COP_R  = q_L / w_in = (h1 − h4) / (h2 − h1)        (refrigerator)
COP_HP = q_H / w_in = (h2 − h3) / (h2 − h1)        (heat pump)
```

**Worked example** with approximate R-134a values: h1 = 240 kJ/kg, h2 = 275 kJ/kg, h3 = h4 = 95 kJ/kg.

```text
COP_R = (240 − 95) / (275 − 240) = 145 / 35 ≈ 4.1
```

The fridge moves about 4.1 J of heat for every joule of electricity. A COP above 1 does not break any law: a refrigerator moves heat rather than creating it.

## Common mistakes

- **Forgetting pump or compressor work** when computing net work or COP.
- **Using q_out instead of q_in** in the efficiency denominator.
- **Assuming the throttle drops enthalpy.** It drops pressure and temperature, not enthalpy.
- **Treating ideal-cycle numbers as real performance.** Real turbines and compressors have isentropic efficiencies around 70–90 %, so real cycle numbers are lower.

## A note on real systems

Refrigerants are regulated (for example under the EPA's Section 608 rules in the US), and handling them requires certified technicians. Real plant and HVAC designs follow codes and standards and are reviewed by licensed professionals.

## Recap

- Rankine: η = (w_turbine − w_pump) / q_in, with each term a difference of enthalpies.
- Refrigeration: COP_R = q_L / w_in; the throttle keeps h constant.
- Ideal-cycle results are upper bounds for real equipment.

## Try it

On the Do tab, write the Rankine efficiency formula in Python and check it against the worked example.


Compute the thermal efficiency of an ideal Rankine cycle and the COP of a vapour-compression refrigerator from enthalpies.

Power and refrigeration cycles

Thermodynamics RANKINE CYCLE in 10 Minutes!

Refrigeration Cycle Tutorial: Step by Step, Detailed and Concise!

Energy bookkeeping for engineers: the laws of thermodynamics and the first law for closed systems, reading and interpolating steam tables, and the efficiency of power and refrigeration cycles.