Control volumes and mass flow

A steam turbine has no fixed quantity of steam in it, which is a problem, because every law in the preceding Thermodynamics course was written for a system containing a fixed quantity of matter.

The closed system runs out

The closed system is a boundary drawn around a definite parcel of matter. It may move, it may deform, it may exchange heat and work with whatever is outside it, but no molecule crosses it. That restriction was harmless while the subject was piston cylinders and sealed vessels, and everything from the First Law to the Clausius inequality was built on it.

Now look at a turbine. Steam enters through a pipe at $3$ MPa and leaves through a much larger one at $10$ kPa, and while it is inside it passes over dozens of blade rows and gives up energy to a shaft. Follow a fixed parcel of that steam and you get a system that enters, expands, accelerates, and leaves, and whose boundary is a moving smear of fluid stretching from the inlet flange to the exhaust hood. Writing $Δ U = Q - W$ for that parcel is not wrong, it is useless: nothing about it is steady, nothing about it is measurable, and there is no instant at which the parcel coincides with the machine.

The same objection kills the closed system for a compressor, a pump, a nozzle, a boiler tube, a condenser, a throttling valve, a jet engine, and a chemical reactor. What every one of these has in common is that the interesting thing is the hardware, which sits still, while the matter streams through it. The hardware is what is bolted down, instrumented and paid for. So the sensible move is to draw the boundary around the hardware and let the matter cross it.

Drawing the boundary in space

A control volume is a region of space, chosen by whoever is doing the analysis, with a boundary called the control surface through which mass may pass. The places where it passes are ports, or inlets and outlets. Nothing about this is physics yet. It is a bookkeeping decision, and like any bookkeeping decision it can be made well or badly.

Made well, it puts the ports where the properties are known or measurable. For a turbine, the natural surface passes through the inlet flange, through the exhaust flange, along the casing, and across the shaft. Pressure and temperature are measured at both flanges, so the two states are known, and the shaft crossing the surface is where the work leaves. For a whole power station the surface can be drawn round the entire plant, cutting the fuel line, the air intake, the stack, the cooling water and the electrical connection. Both are legitimate; they answer different questions.

Made badly, the surface passes through a place where nobody knows what the fluid is doing: halfway along a boiler tube, or across a region where flow is separating and recirculating. The analysis then needs information that does not exist. Choosing the surface is the first real skill in this subject, and it is worth being deliberate about, because everything downstream inherits the choice.

One vocabulary point, since textbooks differ: a control volume with mass crossing it is often called an open system, and the two terms mean the same thing. A control volume whose ports happen to be shut is a closed system, so nothing from the previous course is being discarded here, only generalised.

How much mass crosses a port

Start with the smallest possible question. Fluid flows down a pipe of cross-sectional area $A$ at speed $V$ . In a short time $d t$ , the fluid that gets through a given plane is exactly the fluid that was within a distance $V d t$ upstream of it, a cylindrical slug of volume $A V d t$ . Its mass is that volume divided by the specific volume $v$ of the fluid, or equivalently multiplied by its density $ρ = 1 / v$ . Divide by $d t$ and the mass flow rate is

\dot{m} = \frac{A V}{v} = ρ A V

in kilograms per second when $A$ is in m², $V$ in m s⁻¹ and $v$ in m³ kg⁻¹. The related volume flow rate is $\dot{V} = A V = \dot{m} v$ , in m³ s⁻¹. Volume flow is what a pump curve is plotted against and what a flow meter usually reads; mass flow is what conservation laws are written in. Confusing them is the commonest arithmetic error in this subject, and the two differ by a factor of over a thousand for steam.

The velocity in that formula is an average over the cross-section, and the average is not the centreline value. Real pipe flow is slower at the wall and faster in the middle, turbulent flow having a profile that is fairly flat with a thin steep layer at the wall, laminar flow a parabola whose peak is twice its mean. Writing $\dot{m} = A V / v$ means $V$ is defined as whatever makes that equation true, which is exactly the mass-averaged velocity. That is a definition, not an approximation, and it is why the formula survives the fact that no real flow is uniform.

Example. Steam at $3$ MPa and $350$ degrees Celsius, where the tables give $v = 0.09056$ m³ kg⁻¹, flows down a pipe of internal diameter $200$ mm at $45$ m s⁻¹. What is the mass flow rate?

The area is $A = π d^{2} / 4 = π (0.200)^{2} / 4 = 0.031416$ m². The volume flow is $A V = 0.031416 \times 45 = 1.4137$ m³ s⁻¹, and dividing by the specific volume gives $\dot{m} = 1.4137 / 0.09056 = 15.61$ kg s⁻¹.

Now you. The same steam flows down a $150$ mm pipe at $60$ m s⁻¹. What is the mass flow rate?

Answer

$A = π (0.150)^{2} / 4 = 0.017671$ m², so $\dot{V} = 0.017671 \times 60 = 1.0603$ m³ s⁻¹ and $\dot{m} = 1.0603 / 0.09056 = 11.71$ kg s⁻¹.

The mass balance

Mass is conserved, and for a control volume that statement takes a form worth writing carefully. Over a time interval, the mass inside the region can only change by what came in through the ports minus what went out:

\frac{d m_{c v}}{d t} = \sum_{i n} \dot{m} - \sum_{o u t} \dot{m}

That is the whole law, and it has no exceptions in this course. The term on the left is what distinguishes this from the closed-system case. It is not zero in general: a compressed air receiver being charged, a boiler drum whose level is rising, a fuel tank being drained, all have $d m_{c v} / d t \neq 0$ . Those cases have their own lesson later. What follows here, and what dominates the analysis of running plant, is the case where that term vanishes.

Steady flow

A process is steady when nothing at any fixed point inside the control volume changes with time. Not that nothing changes: the fluid's pressure and temperature change enormously between inlet and outlet of a turbine. What is required is that the pressure at a given point is the same now as it was a minute ago. A turbine at constant load, running for hours, satisfies this to a very good approximation, and so does a compressor, a condenser, a nozzle and a pump.

Steadiness is a strong statement and it has three consequences worth separating.

First, the mass inside the control volume is constant, so $d m_{c v} / d t = 0$ and total mass in equals total mass out:

\sum_{i n} \dot{m} = \sum_{o u t} \dot{m}

Second, for a device with one inlet and one outlet, that reduces to a single number carried through the machine, ${\dot{m}}_{1} = {\dot{m}}_{2} = \dot{m}$ , which combined with the definition gives the relation used constantly:

\frac{A_{1} V_{1}}{v_{1}} = \frac{A_{2} V_{2}}{v_{2}}

Third, the properties at each port are constant in time, so a single value of $p$ , $T$ , $h$ and $s$ describes each port for the whole analysis. That is what makes steady-flow problems arithmetic rather than differential equations.

Steadiness also forbids things. A steady device cannot accumulate energy, so its stored energy is constant, which is the point that makes the next lesson's energy equation as simple as it is. And a steady device cannot be starting up, shutting down, or changing load, which is exactly when real machines break.

Areas, velocities and the size of hardware

The relation $A_{1} V_{1} / v_{1} = A_{2} V_{2} / v_{2}$ looks like a triviality and is not. It is what sets the physical size of turbomachinery, and it explains a shape anyone who has seen a steam turbine has noticed.

Take the steam from the example above, $15.61$ kg s⁻¹ entering at $3$ MPa with $v = 0.09056$ m³ kg⁻¹. Expand it through the turbine to $10$ kPa, at which pressure it emerges as a wet mixture of quality $0.8128$ , and the saturation table gives $v_{f} = 0.001010$ and $v_{g} = 14.670$ m³ kg⁻¹. The mixture's specific volume is $v = v_{f} + x (v_{g} - v_{f}) = 0.001010 + 0.8128 \times 14.669 = 11.92$ m³ kg⁻¹. The steam leaves occupying over $130$ times the volume it entered with, for the same mass.

That factor has to go somewhere, and it goes into area. Holding the velocity at $45$ m s⁻¹ would need an exhaust area of $\dot{m} v / V = 15.61 \times 11.92 / 45 = 4.14$ m², a duct of $2.29$ m diameter for a machine whose inlet pipe is $200$ mm. Real designs let the velocity rise instead, to a few hundred metres per second, which buys back some of the area but not most of it. This is why a large steam turbine is built as a small high-pressure cylinder followed by a physically enormous low-pressure one, often split into several parallel exhaust flows, with last-stage blades over a metre long that are among the most highly stressed components in engineering.

Example. For that turbine, $15.61$ kg s⁻¹ leaving at $v = 11.92$ m³ kg⁻¹, what exhaust area and diameter are needed if the velocity is allowed to reach $180$ m s⁻¹?

$A = \dot{m} v / V = 15.61 \times 11.92 / 180 = 1.034$ m², so $d = \sqrt{4 A / π} = \sqrt{4 \times 1.034 / π} = 1.15$ m. Four times the velocity has cut the diameter by half, and it is still nearly six times the inlet pipe.

Now you. A $600$ MW machine passes $462$ kg s⁻¹ of steam and exhausts it at the same $11.92$ m³ kg⁻¹ and $200$ m s⁻¹, split equally between six parallel low-pressure exhausts. What flow area does each need?

Answer

The total is $A = 462 \times 11.92 / 200 = 27.5$ m², so each of the six carries $27.5 / 6 = 4.59$ m². On an annulus of mean diameter $2.5$ m that is a blade height of $4.59 / (π \times 2.5) = 0.58$ m, which is the right order for real hardware.

Two inlets, and what the balance settles on its own

Where a control volume has several ports, the mass balance alone often fixes something useful before any energy accounting begins. Consider a mixing chamber where hot and cold water streams join, or a feedwater heater in a power station where steam bled from the turbine is mixed into the condensate. Whatever the temperatures, the exit flow is the sum of the inlet flows, and the exit pipe has to be sized for it.

The same reasoning covers a splitter. A turbine passing $462$ kg s⁻¹ that bleeds $23$ per cent of it at an intermediate stage sends $0.23 \times 462 = 106$ kg s⁻¹ to the feedwater heater and $356$ kg s⁻¹ onwards to the next stage. That extraction fraction will turn out to be the central unknown in regenerative cycles, and it is always found from a balance like this one.

Example. A mixing chamber receives $2.0$ kg s⁻¹ of water through one port and $5.0$ kg s⁻¹ through another, and discharges through a single pipe in which the specific volume is $0.001043$ m³ kg⁻¹ and the velocity is $5.0$ m s⁻¹. What exit area is needed?

Steady flow gives ${\dot{m}}_{3} = 2.0 + 5.0 = 7.0$ kg s⁻¹. Then $A = \dot{m} v / V = 7.0 \times 0.001043 / 5.0 = 0.00146$ m², or $14.6$ cm², a pipe of about $43$ mm bore.

Now you. The same chamber now takes $3.0$ and $4.0$ kg s⁻¹, and the exit velocity is raised to $8.0$ m s⁻¹ with the same specific volume. What exit area is needed?

Answer

${\dot{m}}_{3} = 7.0$ kg s⁻¹ still, so $A = 7.0 \times 0.001043 / 8.0 = 0.000913$ m², about $9.1$ cm².

What the model quietly assumes

Three assumptions have been smuggled in, and naming them is what makes the results trustworthy.

The first is one-dimensional flow at the ports: properties are taken as uniform across each port, so one value of $p$ , $T$ and $v$ describes it. Across a well-developed pipe flow this is close to true for pressure and temperature, and the velocity is handled by the mass-averaging above. Across a turbine exhaust annulus, where the flow is swirling and the outer radius sees different conditions from the hub, it is a real approximation, and it is why measured turbine performance never quite matches a one-dimensional calculation.

The second is that the ports are the only places mass crosses. Leakage past shaft seals is small in modern machines but never zero, and boiler tubes do fail.

The third is that nothing has been said about energy yet. Every result here would hold equally for water, mercury or sand, because conservation of mass does not know what is flowing. What makes the analysis thermodynamics is that the mass crossing a port carries energy with it, and that shoving it across the boundary costs work. That is the next lesson, and it produces the single equation the rest of this course runs on.