Control volumes and steady flow

Everything up to here has been written for a closed system: a quantity of matter chosen once, followed wherever it goes, with a boundary that stretches and moves to keep the same molecules inside. That is a good description of gas under a piston and a poor description of a jet engine. The change made in this topic is small to state and large in consequence. Nail the boundary to the hardware, let it be crossed, and the two balances that have been quietly true all along — mass and energy — both acquire terms.

Mass crosses the boundary

A control volume is a region of space, chosen by you, with a surface drawn around it. The surface is usually put where the answer is convenient rather than where nature suggests: around a turbine casing, around one side of a heat exchanger, around a whole power plant. Nothing about the region has to be physical. What matters is that once it is fixed, the accounting is done for that region and for nothing else, and every quantity that crosses the surface has to be written down.

For a closed system, conservation of mass was silent, because the mass was constant by construction. Now it has to be stated, and it is a rate balance rather than a total:

The mass inside grows at exactly the rate at which mass arrives minus the rate at which it leaves. To use this, the flow rate has to be turned into something measurable. Take a small patch of the boundary of area \(A\), with fluid of density \(\rho\) approaching it at speed \(V\). In one second the fluid that gets through is the column of length \(V\) standing on that patch, so

in kilograms per second, with the volume flow rate \(\dot{\mathcal V} = A V\) and \(\dot m = \rho \dot{\mathcal V}\) as the bridge between the two. The second form, dividing by specific volume, is the one to reach for when the substance is being looked up in a table, because \(v\) is what the table gives.

The velocity in that expression is not the speed of the fluid. It is the component of the velocity normal to the area. An area is a door, and fluid sliding parallel to a door does not go through it, so what belongs in the product is \(V\cos\theta\), the part aimed at the surface. Almost every textbook writes \(\dot m = \rho A V\) without a subscript, which is safe only because the inlet plane is nearly always drawn perpendicular to the pipe. Draw it at an angle, as happens the moment a control surface is cut across a bend or a blade passage, and the cosine is not optional.

When the flow is steady the left-hand side is zero, because the mass inside a region whose contents never change cannot be changing. For one inlet and one outlet this collapses to

and if the fluid is a liquid, so that the density is effectively the same at both ends, to \(A_1 V_1 = A_2 V_2\). That is the garden hose: cover half the opening with a thumb and the water leaves twice as fast. The error to avoid is carrying that intuition into a gas. Mass flow rate is conserved through a steady device; volume flow rate is not, and for a compressor the difference is the entire point. A compressor swallows a large volume every second and delivers a small one, and the two are the same kilograms.

Flow work, and why enthalpy

There is a question worth asking before any energy equation is written for a flowing system. In a closed system the energy stored per unit mass is \(u\). So why do flow problems use \(h\) everywhere instead, and why was enthalpy introduced in the property tables with nothing more than the remark that the combination keeps appearing? The answer is that mass does not wander across a boundary of its own accord. It has to be pushed, and pushing is work.

Follow one plug of fluid sitting in the pipe just outside the inlet: cross-sectional area \(A\), length \(L\), mass \(m = \rho A L\). The fluid queueing behind it is at pressure \(P\) and presses on its back face with a force \(F = PA\). To get the whole plug inside, that force must act through the plug’s own length, so the work done is

and per unit mass, dividing by \(m\) and using \(v = \mathcal{V}/m\), simply \(Pv\). This is flow work, and it is the strangest kind of work in the subject because no shaft turns and no one does it deliberately. The fluid upstream does it, and the fluid leaving at the far end does the same to whatever is beyond the outlet. It is nonetheless work, force through distance, exactly as in the accounting of heat and work.

So the energy that arrives with a kilogram of fluid is not its internal energy alone. It is the internal energy it holds, plus the work spent shoving it through the door, plus whatever kinetic and potential energy it happens to have:

The first two terms travel together, always, for every stream at every inlet and outlet. Writing them out in full for the rest of the subject would be intolerable, so they are given a name:

That is the whole of enthalpy. It is not a deeper form of energy and it is not conserved by any special law; it is a bookkeeping device that happens to be a property, because \(u\), \(P\) and \(v\) are all properties and any combination of properties is one. Being a property is what lets it be tabulated, and being tabulated is what makes flow problems tractable.

Two consequences follow, and both are worth holding onto. The first is that \(u\) belongs to closed systems and \(h\) belongs to flowing ones, and using the wrong one is not a small slip. Solve a rigid sealed tank with enthalpies and the answer includes a \(Pv\) that was never transferred. The second is arithmetic: pressure in kilopascals times specific volume in cubic metres per kilogram gives kilojoules per kilogram directly, which is why \(100 \times 1.6941 = 169.4\) is exactly the gap between \(u_g\) and \(h_g\) for saturated steam at one bar. No conversion factor hides in there.

The steady flow energy equation

Steady flow has a precise meaning and a loose one, and the loose one causes trouble. Precisely, it means that no property at any fixed point inside the control volume changes with time. The temperature at a chosen spot on the turbine blade is whatever it is, and it is the same value a minute later. It does not mean the properties are the same everywhere — a boiler is emphatically not uniform inside — and it does not mean nothing is moving. It means the picture is frozen even though the fluid is not.

From that, three things follow at once. The mass inside is constant, so \(dm_{cv}/dt = 0\) and the flow rates in and out must balance. The energy inside is constant, so \(dE_{cv}/dt = 0\) and the first law is a statement about what crosses the surface and nothing else. And the boundary transfers — heat and work — must be steady too, since a fluctuating heat rate would put a fluctuation somewhere inside. Applying the first law to the region gives

reading, in words, that the net rate of energy in as heat and work equals the net rate at which the streams carry energy out. The sign convention is the classical one carried over unchanged: heat into the system is positive, work done by the system is positive. The \(\dot W\) here is shaft and electrical work only. Flow work is not listed separately because it has already been absorbed into every \(h\), and counting it twice is a real and common mistake.

For the single-stream case, which covers most devices, divide through by the mass flow rate to get everything per kilogram:

Now the practical question: when do the last two terms matter? Put numbers on them rather than guessing. A velocity of \(45\) m/s contributes \(45^{2}/2 = 1012\) J/kg, which is about \(1\) kJ/kg. A height change of \(100\) m contributes \(9.81 \times 100 = 981\) J/kg, again about \(1\) kJ/kg. Enthalpy changes across a turbine or a boiler run to several hundred kilojoules per kilogram. Against that, a kilojoule is noise, and dropping both terms is not laziness but proportion.

The exception is any device whose purpose is the velocity. A nozzle exit at \(300\) m/s carries \(45\) kJ/kg of kinetic energy, which is nobody’s rounding error, and in that case the enthalpy change and the kinetic energy change are the same size because one is being turned into the other. Potential energy survives only in hydroelectric work and in long vertical pipes. So the working rule is to keep the kinetic term when velocities are named in the problem or when the device exists to change them, and otherwise to keep \(h\) alone.

One arithmetic warning, because it costs more marks than any concept in this topic. Enthalpies come out of tables in kilojoules per kilogram; \(V^{2}/2\) in metres and seconds comes out in joules per kilogram. They differ by a factor of a thousand, and adding them raw produces an answer wrong by three orders of magnitude that still looks plausible on the page. Divide the kinetic term by \(1000\) before it joins the others, every time.

One inlet, one outlet

What usually gets memorised as a table of device formulas is one equation with different terms crossed out, and the useful skill is knowing which deletion each device earns and why. Every case below starts from the single-stream equation of the previous section and removes only what can be argued away. A nozzle accelerates a fluid and a diffuser slows it down, and they are the same duct pointed the other way. No shaft crosses the boundary, so \(w = 0\). The fluid is inside for a few milliseconds and there is little surface area, so \(q \approx 0\). The device is horizontal and short, so the elevation term goes. What is left is the trade the device exists to make:

Speed is bought with enthalpy and nothing else. It is worth noticing that the shape does not follow from this equation alone: a subsonic nozzle converges, but past the speed of sound the density falls faster than the velocity rises and a nozzle must diverge to keep accelerating the flow. Continuity, not energy, decides the geometry.

A turbine takes work out and a compressor puts work in. A shaft crosses the boundary, so \(w\) is the whole object of the exercise. The casing is usually insulated, or the fluid passes through so quickly that any heat loss is small beside the work, so \(q \approx 0\), and the velocity and height terms are negligible unless the problem names them. That leaves \(w = h_1 - h_2\) for the turbine, positive because work leaves, and \(w_{\text{in}} = h_2 - h_1\) for the compressor. A pump is a compressor for a liquid, and because a liquid barely changes volume its work reduces to \(w \approx v(P_2 - P_1)\) — the reason a Rankine cycle spends so little of its output driving the feed pump.

A throttle is a restriction: a valve, a porous plug, a length of capillary. It does no work, it is too small and too fast to exchange meaningful heat, and the velocities on either side are modest even though the pressure drops sharply. Everything cancels, and

A throttling process is isenthalpic. It is not isothermal, and the difference is what makes refrigeration possible. Since \(h = u + Pv\) is fixed while the fluid expands, a rise in \(Pv\) must be paid for by a fall in \(u\), and internal energy is what a thermometer reports. For an ideal gas the two effects cancel exactly, because \(h\) depends on temperature alone and a constant enthalpy therefore means a constant temperature; throttle an ideal gas and nothing happens to \(T\). For a refrigerant flashing from liquid into a wet mixture, the temperature falls steeply, and that drop is the cold end of every vapour-compression cycle.

A heat exchanger has no moving parts at all, so \(w = 0\) without argument, and the velocity and height terms are negligible. What it does have is a choice of boundary, and the choice changes the answer. Draw the control volume around one stream only and the heat crossing to the other stream is a \(\dot Q\) in the balance, giving \(\dot Q = \dot m (h_{\text{out}} - h_{\text{in}})\). Draw it around the whole insulated unit and there is no heat crossing anything, and the balance becomes a statement that one stream gains what the other loses:

Neither version is more correct; they answer different questions. Saying out loud where the boundary has been drawn, before writing anything down, resolves most of the confusion this device causes. A mixing chamber is the same device with the wall removed so the streams meet, and the balance is the same too: \(\dot m_1 h_1 + \dot m_2 h_2 = \dot m_3 h_3\), alongside \(\dot m_1 + \dot m_2 = \dot m_3\), which must be solved together because the split of the flow is usually part of what is being asked.

Unsteady flow

Charging a scuba cylinder, filling a gas bottle, blowing down a pressure vessel through a relief valve: in each of these the contents of the control volume are changing while mass crosses the boundary, and the steady assumption is gone. Both balances revert to their general form, integrated over the whole event rather than written per second:

The terms in the bracket are the change in what the vessel holds, which is internal energy because the contents are not going anywhere. The stream terms carry enthalpy, for the reason established two sections ago. The difficulty is that \(h_{\text{in}}\) and \(h_{\text{out}}\) may drift while the process runs, in which case they belong inside an integral nobody can evaluate without knowing the whole history. The uniform-flow model assumes they do not: the state at each opening is taken as one fixed value for the duration. Filling from a large supply line is well described that way, because the line is unaffected by the small amount drawn off. Discharging is described less well, because the fluid leaving is the fluid inside, and the inside is changing.

The cleanest case is worth working in full because the result is genuinely surprising. Take a rigid, insulated, evacuated tank connected by a valve to a line carrying gas at \(P_i\) and \(T_i\). Rigid, so \(W = 0\). Insulated, so \(Q = 0\). Evacuated, so \(m_1 = 0\). Nothing leaves, so the outflow term goes. Five terms out of the balance vanish and the survivors are

and since everything that ended up inside came through the valve, \(m_i = m_2\) and the masses cancel:

The gas inside ends with an internal energy equal to the enthalpy of the line. It is hotter than the gas it came from, and nothing heated it. The explanation sits in the term that was added at the start of this topic: every kilogram that entered was pushed in by the fluid behind it, at a cost of \(P_i v_i\) per kilogram, and once inside the tank there is no shaft to carry that work away and no insulation breach to let it out. It has nowhere to go but into internal energy. Flow work went in at the door and came out as temperature.

For an ideal gas with constant specific heats the result is blunt. Writing \(u_2 = c_v T_2\) and \(h_i = c_p T_i\),

with \(k = c_p/c_v\). Fill an evacuated bottle from an air line at \(20\,^\circ\text{C}\) and the air inside reaches \(1.4 \times 293 = 410\) K, about \(137\,^\circ\text{C}\). Note what is absent from that expression: the line pressure. A higher supply pressure puts more mass in the tank but leaves the final temperature alone, which is not what intuition offers and is a good check that the algebra was followed rather than remembered. Anyone who has filled a diving cylinder and found it warm to the touch has met this result, and reversing it explains why a cylinder being emptied quickly grows cold enough to frost. That reversal deserves one caution of its own. As a tank empties, the gas left inside expands and cools, so the enthalpy of the stream leaving falls throughout the process and no single value is right. The usual approximation takes the average of the initial and final enthalpy of the contents, which is defensible when the change is modest and dishonest when it is not. Whether the approximation is good enough is a question about how far the state moved, and answering it properly needs the second law, which is the next thing to arrive.