Control Volumes

The question

The steam turbine in our power plant swallows 300 kilograms of steam every second and never holds more than a few hundred of them at a time. Following each kilogram as a closed system would be hopeless. How do you apply energy conservation to things that flow?

Draw the box around the machine

Instead of following the matter, fix the boundary around the device and let matter stream across it. That box is a . Two bookkeeping rules apply to it.

Mass. What comes in either goes out or piles up inside:

$\frac{dm_\text{cv}}{dt} = \sum_\text{in} \dot m - \sum_\text{out} \dot m, \qquad \dot m = \rho A V.$

Energy. The same, plus heat and work — with one subtlety that changes everything.

Flow work: the price of admission

To push a kilogram of fluid into the control volume, the fluid behind it must shove it in against the pressure inside. That takes work: pressure times the volume swept, $pv$ per kilogram. This flow work is done every time mass crosses a boundary, so each kilogram entering brings not just its internal energy $u$ but $u + pv$ — its , $h$ . Here is where enthalpy earns its keep.

The energy rate balance for a control volume is then

$\frac{dE_\text{cv}}{dt} = \cheat{\dot Q}_\text{cv} - \cwork{\dot W}_\text{cv} + \sum_\text{in} \dot m\left(h + \frac{V^2}{2} + gz\right) - \sum_\text{out} \dot m\left(h + \frac{V^2}{2} + gz\right),$

where $\cwork{\dot W}_\text{cv}$ now means only the useful work — shaft power, electric power — since the flow work has been folded into $h$ .

Steady state: the workhorse equation

Most engineering devices run in : nothing inside changes with time. Then $dm/dt = 0$ and $dE/dt = 0$ , and for one inlet and one exit the balance per unit mass becomes the steady-flow energy equation:

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

Every device in the power plant is a special case of this one line. The skill is in knowing which terms to drop:

ExploreFour devices, one equation

Expansion ratio p₁/p₂10.0

Isentropic efficiency88%

Mass flow100 kg/s

ResultsuperheatedAdiabatic, negligible kinetic-energy change: the work is the enthalpy drop.

Check: q − w − Δh − Δke0.00 kJ/kgalways zero

Challenges0/3

Make steam leave a nozzle faster than 1200 m/s.
Expand steam through the turbine until it is less than 88% vapour.
Raise the pressure tenfold with both the pump and the compressor, and compare the work.

Each device is a control volume in steady state: mass in equals mass out, and the steady-flow energy equation q − w = Δh + ΔV²/2 balances the books per kilogram flowing through (potential-energy changes neglected). Steam properties come from the course’s steam tables; air from ideal-gas tables.

Nozzles and diffusers do no work and are too quick for much heat transfer: the enthalpy drop becomes speed. Steam leaving a turbine nozzle can exceed 1 km/s.
Turbines are adiabatic, and the change in kinetic energy across them is small: the shaft work is just the enthalpy drop, $\cwork{w} = h_1 - h_2$ .
Compressors and pumps run the other way: work in, enthalpy up. And here is a lesson worth holding on to: raising the pressure of a liquid costs almost nothing compared with a gas. The pump work is $w \approx v\,\Delta p$ , and liquid water's $v$ is about a thousandth of steam's.

Estimate first

A large steam turbine expands 300 kg/s of steam with an enthalpy drop of about 1300 kJ/kg. What is its power output, in MW?

1101001,00010,000100,000

Drag along the line to place your guess — each tick is ten times the last.

Throttling: same enthalpy, new state

Force fluid through a partly closed valve, a capillary tube or a porous plug, and its pressure drops sharply. No work, no time for heat transfer, negligible change in kinetic energy: the steady-flow equation leaves only

$h_2 = h_1.$

A keeps the enthalpy fixed — but almost nothing else. Wet steam can come out superheated, which lets an engineer measure the wetness of steam in a line with nothing more than a valve and a thermometer. Warm liquid refrigerant flashes into a frigid boiling mixture, which is how every refrigerator, air conditioner and heat pump makes its cold.

ExploreThrough the valve

Line pressure20.0 bar

Hidden quality of the line steam???

move it, then read the outlet thermometer

Outlet pressure1.00 bar

Temperature change-80.2 K

Enthalpyunchangedh₂ = h₁

Challenges0/2

Throttle wet steam until it comes out superheated — then work out its quality from the outlet temperature.
Make the refrigerant leave the valve colder than −20 °C.

Fluid squeezes through a partly open valve or a porous plug. No work is done, there is no time for heat transfer, and the kinetic energy barely changes, so the enthalpy is the same on both sides: h₂ = h₁. That one fact lets you measure the wetness of steam — and it is how every refrigerator makes its cold.

Check your intuition

Air (an ideal gas) is throttled from 10 bar to 1 bar. What happens to its temperature?

Transients: filling a tank

Not everything is steady. Consider an evacuated, insulated, rigid tank opened to a supply line of air at 300 K. With no work, no heat and nothing leaving, the energy balance integrates to

$m_2 u_2 = m_2 h_\text{line} \quad\Rightarrow\quad u_2 = h_\text{line}.$

For an ideal gas with constant specific heats, $c_v T_2 = c_p T_\text{line}$ , so

$T_2 = \gamma\, T_\text{line}.$

Air enters at room temperature and ends up at 420 K — nearly 150 °C — with no heater anywhere. The extra energy is the flow work: the gas in the line did work pushing each parcel in, and that work ended up as internal energy inside the tank.

ExperimentWhy a filled tank gets hot

Initial tank pressureevacuated

Valve opening100%

Tank temperature300 K27 °C

Mass in the tank0.000 kg

Challenges0/3

Fill an evacuated, insulated tank with room-temperature air and get it above 400 K.
Do the same with helium. How hot does it get?
Fill a tank to the line pressure while keeping it below 320 K.

A rigid 100-litre tank is connected to a supply line of gas at 10 bar and 300 K. Open the valve: gas rushes in until the tank pressure reaches the line pressure. The yellow curve is the tank temperature, purple its pressure (scaled to the line pressure). The gas arriving brings its enthalpy, h = u + pv — its internal energy plus the work of the gas behind pushing it in.

Going deeper
Going deeper: putting devices together

Real plants are systems of control volumes, and the art is in choosing the boundaries. Draw one box around the turbine alone and you find its power from the steam states. Draw a box around the whole power plant and the steam vanishes from the books — it goes round and round — leaving only fuel energy in, electricity out and heat to the cooling water. Draw a box around the condenser, and the steam's enthalpy drop equals the cooling water's enthalpy rise, which fixes how much cooling water you need: about 50 kg of it for every kilogram of steam, for a 10 K temperature rise.

Moran and Shapiro call this system integration, and it is how the cycle analyses of Lessons 8–10 are built: one steady-flow energy equation per component, linked by the states between them.

Recap

What to carry forward

A control volume is a fixed region that mass flows through. Mass and energy balances apply to it as rate equations.
Every kilogram crossing the boundary carries flow work $pv$ along with $u$ ; together they make enthalpy, $h$ .
In steady state: $q - w = \Delta h + \Delta V^2/2 + g\Delta z$ . Nozzles trade $h$ for speed; turbines for work; compressors and pumps the reverse.
Pumping a liquid takes about a thousandth of the work of compressing a gas over the same pressure ratio.
Throttling keeps $h$ constant; for refrigerants it produces cold, for ideal gases nothing at all.
Filling an empty insulated tank heats the gas to $\gamma T_\text{line}$ — flow work turned into internal energy.

Next: the first law says energy can't be created. But it doesn't say what we can't do with it. Why must every power plant throw heat away?