Laws of Thermodynamics for Engineers with Solved Numerical Examples

Thermodynamics is the accounting system for energy. Engines, compressors, refrigerators, boilers and turbines all obey the same four laws, and almost every thermal engineering problem starts by writing one of them down. This article explains each law in plain terms, lists the formulas you actually use, and solves four numerical problems that cover closed systems, open steady-flow systems and the second-law limits on efficiency.

Systems and Sign Convention

A closed system has a fixed mass, such as gas trapped in a cylinder behind a piston. Energy can cross its boundary as heat or work, but mass cannot. An open system or control volume has mass flowing in and out, such as a turbine, pump or nozzle. An isolated system exchanges neither mass nor energy.

This article uses the common engineering convention: heat added to the system is positive, and work done by the system is positive. Some textbooks use the opposite sign for work, so always check before using a formula from another source.

The Four Laws in Brief

Law	Statement in simple words	What it gives you
Zeroth	If A and B are each in thermal equilibrium with C, they are in equilibrium with each other	The basis of temperature measurement with a thermometer
First	Energy is conserved; heat added equals change in stored energy plus work done	Energy balances for any device
Second	No cyclic engine can convert all heat into work; heat flows naturally from hot to cold	Maximum possible efficiency, entropy
Third	Entropy of a perfect crystal approaches zero as temperature approaches absolute zero	Reference for absolute entropy values

First law, closed system:      Q - W = delta_U
Ideal gas:                     delta_U = m * cv * delta_T,   delta_H = m * cp * delta_T
Steady flow energy equation:   Q_dot - W_dot = m_dot * [ (h2 - h1) + (V2^2 - V1^2)/2 + g(z2 - z1) ]
Carnot efficiency:             eta = 1 - T_L / T_H     (temperatures in kelvin)
Entropy change of reservoir:   delta_S = Q / T
Clausius inequality:           delta_S_total >= 0 for any real process
For air:  cp = 1.005 kJ/kg K,  cv = 0.718 kJ/kg K,  R = 0.287 kJ/kg K,  gamma = 1.4

First Law Worked Examples

Example 1: Heat and work in a closed cylinder

Gas in a piston-cylinder receives 50 kJ of heat and pushes the piston out, doing 20 kJ of work. The change in internal energy is Q - W = 50 - 20 = 30 kJ. The gas temperature rises because only part of the heat left as work.

Example 2: Heating air at constant volume and constant pressure

Two kilograms of air are heated from 300 K to 400 K. Compare the heat needed in a rigid tank (constant volume) with a free piston (constant pressure).

Constant volume:
  W = 0
  Q = m cv dT = 2 * 0.718 * 100 = 143.6 kJ

Constant pressure:
  Q = m cp dT = 2 * 1.005 * 100 = 201.0 kJ
  dU = m cv dT = 143.6 kJ  (same as before, depends only on temperature)
  W = Q - dU = 57.4 kJ
  Check: W = m R dT = 2 * 0.287 * 100 = 57.4 kJ

The extra 57.4 kJ at constant pressure is exactly the work done pushing the piston out. This is why cp is always greater than cv, and their difference equals the gas constant R.

Example 3: Steam turbine with the steady flow energy equation

Steam enters a turbine at 5 kg/s with enthalpy 3200 kJ/kg and velocity 50 m/s, and leaves with enthalpy 2600 kJ/kg and velocity 150 m/s. Heat loss from the casing is 10 kW and the height change is negligible. Find the power output.

Kinetic energy change = (150^2 - 50^2) / 2 = 10,000 J/kg = 10 kJ/kg

W_dot = Q_dot - m_dot * [ (h2 - h1) + dKE ]
      = -10 - 5 * [ (2600 - 3200) + 10 ]
      = -10 - 5 * (-590)
      = 2940 kW

Kinetic energy changed the answer by 50 kW, about 1.7 percent. For turbines it is often neglected, but for nozzles and diffusers it is the main term.

Second Law Worked Examples

Example 4: Checking an inventor's claim

An inventor claims an engine that takes heat from a source at 527 degrees C, rejects heat at 27 degrees C and has a thermal efficiency of 70 percent. Convert to kelvin: 800 K and 300 K. The Carnot limit is 1 - 300/800 = 0.625, or 62.5 percent. No engine between these temperatures can exceed this, so the claim violates the second law.

Example 5: Entropy generated by heat transfer

1000 kJ of heat flows directly from a reservoir at 800 K to one at 300 K with no work produced.

Hot reservoir:   dS = -1000 / 800 = -1.250 kJ/K
Cold reservoir:  dS = +1000 / 300 = +3.333 kJ/K
Total:           dS = +2.083 kJ/K  (> 0, so the process is possible and irreversible)

The positive total entropy change measures the lost opportunity. A Carnot engine placed between the same reservoirs could have produced 625 kJ of work from that heat.

Common Mistakes

Using degrees Celsius in the Carnot formula or entropy calculations; always use kelvin.
Mixing sign conventions for work between formulas from different books.
Using cp for internal energy change. Internal energy of an ideal gas always uses cv, whatever the process.
Forgetting to convert velocity terms from J/kg to kJ/kg in the steady flow equation.
Treating a real cycle as reversible and then being surprised that measured efficiency is much lower.

Summary

The zeroth law defines temperature, the first law balances energy, the second law limits how much heat can become work, and the third law fixes the zero of entropy. With the first law in closed and steady-flow form plus the Carnot and entropy relations, you can analyse the engine cycles, steam plants and refrigerators covered in the rest of this series.