Closed systems, control volumes and energy balances

Give each crossing a sign

For a closed system, use ΔE=Q−W, with Q positive into the system and W positive for work done by the system. E includes internal, kinetic and potential energy. When changes in bulk motion and elevation are negligible, ΔU=Q−W. Compression work done on the system has negative W; heat leaving has negative Q.

Heat and work describe energy transfer during a process. They are not properties stored inside a body. U is a state property: its change depends on endpoints. The separate Q and W amounts depend on the route. Adiabatic means no heat transfer through the chosen boundary, not constant temperature, no work, or no change of internal energy.

Build pressure work from force and displacement

Take an idealised frictionless piston with face area A. A constant opposing pressure p produces opposing force pA. During a slow, mechanically balanced outward displacement Δx, swept volume is ΔV=AΔx, so work done by the sample is Wby=(pA)Δx=pΔV. This uses constant pressure and compatible near-equilibrium states. A changing-pressure route requires adding its successive pressure–volume contributions; the same endpoint volume change does not determine that work.

Expansion gives positive ΔV and positive work out. Compression gives negative ΔV and negative Wby, indicating work into the sample. The units carry the physical connection: Pa·m³=(N/m²)·m³=N·m=J. Do not insert a pressure in kPa into a calculation labelled joules without its factor of one thousand.

Pressure also pushes moving matter across a control-volume port. Pushing a slice of volume ΔV across a port at pressure p entails pΔV of work. For its mass m, dividing by m gives pv, with v=ΔV/m. Incoming matter carries u and is pushed in; outgoing matter must be pushed out. Grouping these contributions gives h=u+pv at each stream state. This constructs the flow-work term rather than counting it again alongside h.

Let matter carry energy across an open boundary

A control volume tracks stored energy plus energy carried by incoming and outgoing matter. For a steady single-inlet, single-outlet stream, mass flow rate ṁ is the same at both ports. The useful energy relationship is Q̇−Ẇs=ṁ[(h2−h1)+(c2²−c1²)/2+g(z2−z1)]. Dots mean rates, c is speed, z elevation, and Ẇs is shaft work out.

Specific enthalpy h=u+pv combines internal energy per kilogram with the pressure-volume flow-work term. Here v is specific volume. Once h is used, adding the same flow work separately would double-count it. Enthalpy is a property; it is not a second kind of heat. Kinetic terms in m²/s² are J/kg, so convert them before combining with h in kJ/kg.

Remove terms only with reasons

Steady operation means no time accumulation within the control volume, although properties can differ from inlet to outlet. It does not imply uniform temperature or zero heat transfer. A startup, filling tank or changing inventory requires storage terms and cannot silently use a steady-flow equation.

An adiabatic turbine can obtain shaft work from an enthalpy decrease. A compressor can require shaft work while raising enthalpy. A nozzle may instead convert enthalpy into kinetic energy. The same balance serves these cases; the justified simplifications differ. A numerical balance does not establish material limits or actual component efficiency.

Check a fixed sample and a separate flow case

The fixed sample receives Q=120 kJ and does W=50 kJ, with negligible bulk-motion changes. Its internal-energy increase is 120−50=70 kJ. If instead 50 kJ were work supplied to it, W would be−50 kJ and the increase 170 kJ.

For a separate constructed steady adiabatic turbine, ṁ=0.2 kg/s, h1=500 kJ/kg and h2=350 kJ/kg. Neglect speed and elevation changes. Shaft output is 0.2×(500−350)=30 kJ/s=30 kW. The absolute enthalpy reference cancels; the difference matters.

In a separate piston illustration, constant opposing pressure is 100 kPa and expansion is 0.001 m³. The work out is 100000×0.001=100 J. If the sample has a supplied internal-energy increase of 200 J and negligible bulk-motion changes, Q=ΔU+Wby=300 J enters. A second supplied route between the same initial and final sample states could have Wby=200 J and Q=400 J: both give ΔU=200 J. These are illustrative transfer accounts, not pressure histories inferred from the endpoints. They show why internal energy belongs to the state while heat and work describe the route.

Try the reasoning

1. Can adiabatic compression raise temperature?

   Show the answer and reasoning

   Yes. Work can increase internal energy even when Q is zero.

2. Why not add pressure-flow work again when using enthalpy?

   Show the answer and reasoning

   The pv flow-work contribution is already included in h=u+pv.

3. Can the steady-flow balance describe a filling tank without a storage term?

   Show the answer and reasoning

   No. Changing inventory and stored energy require an unsteady account.

4. A sample is compressed by 0.002 m³ against a constant 150 kPa opposing pressure in the piston model. Its supplied ΔU is +200 J. Find signed work by the sample and heat into it.

   Show the answer and reasoning

   ΔV=−0.002 m³, so Wby=150000×(−0.002)=−300 J. Thus Q=ΔU+Wby=200−300=−100 J: 300 J of work enters and 100 J of heat leaves, leaving the 200 J increase.

5. Two supplied routes have the same ΔU=200 J. One has Q=300 J, Wby=100 J; the other is reported as Q=400 J, Wby=100 J. Can both reports describe the specified endpoints with no other energy terms?

   Show the answer and reasoning

   No. The first gives 300−100=200 J, but the second gives 400−100=300 J. If the second heat amount is correct, its work must be 200 J for the same ΔU. Route dependence does not permit violating the energy balance.

What to carry forward

Choose fixed matter or a control volume, declare signs, then justify every omitted storage or transfer term.