Thermodynamic Systems and the First Law
Systems, properties, state and processes, work and heat, the first law for closed systems and steady-flow devices (nozzles, turbines, compressors, throttles) with solved examples.
Thermodynamics is about energy and its conversion between heat and work. Its first law is energy conservation; most numerical questions are first-law balances on a system or a flow device.
Basic terms
- System: closed (fixed mass), open (control volume with mass flow) or isolated (no mass or energy crosses).
- Property: a quantity that depends only on the state (p, T, v, u, h, s). Intensive properties do not depend on size (p, T); extensive ones do (V, U).
- Process: change of state. A quasi-static process passes through equilibrium states; a cycle returns to the start.
- Zeroth law: two bodies each in thermal equilibrium with a third are in equilibrium with each other (the basis of thermometers).
Work and heat
Both are energy in transit, and both are path functions (their values depend on the process, not just the end states).
Boundary work W = ∫p dV:
| Process | Work |
|---|---|
| Constant volume | 0 |
| Constant pressure | p(V₂ − V₁) |
| Isothermal (ideal gas) | p₁V₁ ln(V₂/V₁) |
| Polytropic pVⁿ = C | (p₁V₁ − p₂V₂)/(n − 1) |
Sign convention used here: heat into the system and work by the system are positive.
First law for a closed system
Q − W = ΔU (for a cycle: ∮δQ = ∮δW).
For an ideal gas: ΔU = m c_v ΔT, ΔH = m c_p ΔT, c_p − c_v = R, γ = c_p/c_v.
First law for steady flow (SFEE)
q − w = (h₂ − h₁) + (V₂² − V₁²)/2 + g(z₂ − z₁) (per unit mass)
| Device | Simplification |
|---|---|
| Turbine (adiabatic) | w = h₁ − h₂ |
| Compressor / pump (adiabatic) | w_in = h₂ − h₁ |
| Nozzle (adiabatic, no work) | V₂ = √(2(h₁ − h₂) + V₁²) |
| Throttle valve | h₂ = h₁ (isenthalpic) |
| Heat exchanger | Σ ṁh in = Σ ṁh out |
Worked example
Air (c_p = 1.005 kJ/kg·K) enters a nozzle at 400 K with negligible velocity and leaves at 350 K. Find the exit velocity.
- Δh = 1.005 × 50 = 50.25 kJ/kg.
- V₂ = √(2 × 50,250) = 317 m/s.
Practice
2 kg of an ideal gas (c_v = 0.718 kJ/kg·K) is heated at constant volume from 300 K to 400 K. How much heat (kJ) is added?
Show answer and explanation
Answer: 143.6W = 0 at constant volume, so Q = ΔU = mc_vΔT = 2 × 0.718 × 100 = 143.6 kJ.
Steam enters an adiabatic turbine with h = 3200 kJ/kg and leaves with h = 2500 kJ/kg at 10 kg/s. What is the power output in MW? (Neglect KE and PE.)
Show answer and explanation
Answer: 7P = ṁ(h₁ − h₂) = 10 × 700 = 7000 kW = 7 MW.
In a throttling process, which property stays constant?
Show answer and explanation
Answer: CA throttle has no work, negligible heat transfer and negligible change in KE, so h₂ = h₁. Entropy increases.
A gas expands at a constant pressure of 200 kPa from 0.1 m³ to 0.4 m³. What is the work done in kJ?
Show answer and explanation
Answer: 60W = pΔV = 200 × 0.3 = 60 kJ.
Which of these are path functions? (Select all that apply.)
Show answer and explanation
Answer: A, BHeat and work depend on the process; U and H are properties (point functions).