Flow work and the energy equation

A kilogram of steam entering a turbine brings its internal energy with it, and that is not the whole story, because something had to push it through the inlet.

The question the mass balance left open

Drawing a boundary around a piece of hardware and counting the mass through its ports, as the previous lesson did, says nothing about energy. To get an energy balance we need to know what each kilogram carries as it crosses, and the naive answer is wrong in an interesting way.

The naive answer is internal energy, $u$ , the thing the First Law is written in for a closed system. A kilogram of steam at $3$ MPa and $350$ degrees Celsius has $u = 2844$ kJ kg⁻¹, so surely a kilogram entering a turbine delivers $2844$ kJ into the control volume. It delivers more, and the reason is mechanical rather than thermal.

Picture the inlet port with the fluid behind it. That fluid is at $3$ MPa, and it is pressing on the kilogram in front of it. As the kilogram moves through the port and into the control volume, the fluid behind does work on it, and that work enters the control volume just as surely as the internal energy does. Nobody pushed anything across the boundary of a closed system, so this term has no closed-system counterpart. It is genuinely new, and it is called flow work.

The cost of the push

Compute it. Take a plug of fluid of mass $m$ , occupying volume $V = m v$ , being pushed through a port of area $A$ by the pressure $p$ immediately upstream. The plug is a cylinder of length $L = V / A$ . The force on its rear face is $p A$ , constant while the plug goes through, and it acts through distance $L$ , so the work done on it is

W_{f l o w} = p A \cdot L = p A \cdot \frac{V}{A} = p V = m p v

Per kilogram, the flow work is $p v$ , in joules per kilogram when $p$ is in pascals and $v$ in m³ kg⁻¹. Notice what dropped out: the area cancelled. A narrow port needs a long plug and a wide one a short plug, and the product is the same. The push costs $p v$ per kilogram regardless of the pipework.

The numbers are not small. For steam at $3$ MPa with $v = 0.09056$ m³ kg⁻¹, $p v = 3000 \times 0.09056 = 271.7$ kJ kg⁻¹, which is nearly ten per cent of the internal energy. At the turbine exhaust, $10$ kPa with $v = 12.71$ m³ kg⁻¹, $p v = 127.1$ kJ kg⁻¹, still substantial despite the pressure having fallen by a factor of $300$ , because the volume rose by a factor of $140$ .

At an outlet the same argument runs with the sign flipped: the fluid leaving has to shove the fluid downstream out of its way, so it carries $p v$ per kilogram out of the control volume. Both ports therefore contribute $p v$ alongside $u$ , entering at inlets and leaving at outlets.

Example. Steam at $3$ MPa and $350$ degrees Celsius has $v = 0.09056$ m³ kg⁻¹ and $h = 3116.1$ kJ kg⁻¹. Find the flow work per kilogram at that port and hence the internal energy.

Flow work is $p v = 3000 kPa \times 0.09056 {m³ kg}^{- 1} = 271.7$ kJ kg⁻¹, the units working because a kilopascal times a cubic metre is a kilojoule. Since $h = u + p v$ by definition, $u = 3116.1 - 271.7 = 2844.4$ kJ kg⁻¹, which is what the superheated table lists.

Now you. At the turbine exhaust the steam is a wet mixture at $10$ kPa with $v = 12.71$ m³ kg⁻¹ and $h = 2263.5$ kJ kg⁻¹. Find the flow work and the internal energy.

Answer

$p v = 10 \times 12.71 = 127.1$ kJ kg⁻¹, so $u = 2263.5 - 127.1 = 2136.4$ kJ kg⁻¹. The saturation table gives $u_{f} + x u_{f g} = 191.79 + 0.8661 \times 2245.4 = 2136.5$ kJ kg⁻¹, which agrees.

Enthalpy is what flows

So the energy carried per kilogram through a port is $u + p v$ , and that combination already has a name. It is the enthalpy,

h = u + p v

introduced in the previous course as the correct bookkeeping for a constant-pressure closed process. Here it arrives for a completely different reason and lands on the same quantity. That is worth pausing on. In the closed-system setting, enthalpy looked like an algebraic convenience, a grouping that made $Δ H = Q_{p}$ come out cleanly. In the open-system setting it is not a convenience at all: $u + p v$ is literally the energy per kilogram delivered by a flowing stream, internal energy plus the work of getting it through the door. Enthalpy is the natural energy of flow, and internal energy is the natural energy of a closed box. Once that is seen, the fact that every table in this subject is written in $h$ rather than $u$ stops being an editorial choice.

A stream can also carry bulk kinetic energy $V^{2} / 2$ and potential energy $g z$ per kilogram, so the full content of a kilogram crossing a port is

h + \frac{V^{2}}{2} + g z

with $V$ in m s⁻¹ and $z$ a height above some reference level. Watch the units: $h$ is usually tabulated in kJ kg⁻¹ while $V^{2} / 2$ comes out in J kg⁻¹, so one of them has to be converted, and forgetting is a factor-of-a-thousand error that is at least loud enough to notice.

The steady-flow energy equation

Now assemble the balance. For a control volume, energy accumulates at a rate equal to what comes in minus what goes out, counting heat, work and the streams. Under steady flow nothing inside accumulates, so the rate of change of the control volume's energy is zero and everything must balance:

\dot{Q} - \dot{W} = \sum_{o u t} \dot{m} (h + \frac{V^{2}}{2} + g z) - \sum_{i n} \dot{m} (h + \frac{V^{2}}{2} + g z)

This is the steady-flow energy equation, and it is the single most used result in the subject. The sign convention is the one carried over from the previous course: $\dot{Q}$ is positive when heat flows into the control volume, $\dot{W}$ positive when work is done by it. A turbine has positive $\dot{W}$ , a compressor negative, and it is usually clearer to write ${\dot{W}}_{i n}$ for a compressor and keep the number positive rather than juggle minus signs.

One point about the work term deserves care. $\dot{W}$ here means shaft work, electrical work, and anything else crossing the surface, but not flow work, because flow work has already been absorbed into the enthalpies. Counting it twice is a classic mistake, and the symptom is a turbine that appears to deliver an extra $271$ kJ per kilogram.

For the common case of one inlet and one outlet, divide by $\dot{m}$ and write everything per kilogram:

q - w = h_{2} - h_{1} + \frac{V_{2}^{2} - V_{1}^{2}}{2} + g (z_{2} - z_{1})

That is the working form. Almost every device in this course is analysed by writing this line down and then deleting the terms that do not matter, which is the business of the next lesson.

How big are the velocity and height terms

Deleting terms requires knowing their size, and here the arithmetic is decisive. Kinetic energy per kilogram is $V^{2} / 2$ , so at $45$ m s⁻¹, a typical pipe velocity, it is $45^{2} / 2 = 1013$ J kg⁻¹, or $1.01$ kJ kg⁻¹. Set that beside an enthalpy of $3116$ kJ kg⁻¹ and it is three parts in ten thousand.

The quadratic makes the term wake up quickly. At $100$ m s⁻¹ it is $5.0$ kJ kg⁻¹, at $200$ m s⁻¹ it is $20$ kJ kg⁻¹, and at $450$ m s⁻¹ it is $101$ kJ kg⁻¹. The useful rule is that below about $50$ m s⁻¹ the kinetic term can be dropped without thought, between $50$ and $150$ m s⁻¹ it should be checked against the enthalpy change rather than the enthalpy itself, and above that it must be carried. In a nozzle it is not a correction at all: it is the entire point of the device.

Potential energy is nearly always negligible in a machine. A height difference of $30$ m gives $g z = 9.81 \times 30 = 294$ J kg⁻¹, about $0.29$ kJ kg⁻¹. Even the $100$ m from a boiler drum to a turbine floor buys under $1$ kJ kg⁻¹. The exception is hydraulic machinery, where the height difference is the energy source and the enthalpy change is what is negligible: a hydroelectric station working under a $300$ m head has $g z = 2.94$ kJ kg⁻¹ and no combustion anywhere.

Example. A turbine passes $15.61$ kg s⁻¹ of steam. It enters at $h_{1} = 3116.1$ kJ kg⁻¹ and $45$ m s⁻¹ and leaves at $h_{2} = 2263.5$ kJ kg⁻¹ and $200$ m s⁻¹, with negligible heat loss and no height change. What power does it deliver, and how much does the kinetic term matter?

With $q = 0$ the working form gives $w = (h_{1} - h_{2}) + (V_{1}^{2} - V_{2}^{2}) / 2$ . The enthalpy drop is $3116.1 - 2263.5 = 852.6$ kJ kg⁻¹. The kinetic contribution is $(45^{2} - 200^{2}) / 2 = - 18990$ J kg⁻¹, that is $- 19.0$ kJ kg⁻¹, negative because the steam leaves faster than it arrived and takes that energy away with it. So $w = 852.6 - 19.0 = 833.6$ kJ kg⁻¹ and $\dot{W} = 15.61 \times 833.6 = 13010$ kW, or $13.0$ MW. Ignoring the velocities entirely would have given $13.3$ MW, an overestimate of $2.2$ per cent.

Now you. The same turbine is fitted with a smaller exhaust so the steam leaves at $260$ m s⁻¹ instead, everything else unchanged. What power does it now deliver?

Answer

The kinetic term becomes $(45^{2} - 260^{2}) / 2 = - 32790$ J kg⁻¹, so $w = 852.6 - 32.8 = 819.8$ kJ kg⁻¹ and $\dot{W} = 15.61 \times 819.8 = 12800$ kW, or $12.8$ MW. Two hundred kilowatts have been thrown down the exhaust as unrecovered velocity, which is why real machines fit a diffusing exhaust hood to convert some of it back to pressure.

Reading the equation in both directions

The balance is used two ways, and it is worth being explicit about which.

Read forwards, it is a prediction. Given the inlet state, the outlet state and the flows, it says what power or heat duty the device must have. That is the mode of the example above, and of most textbook problems.

Read backwards, it is a measurement. Given the power on a shaft and the flow through a machine, it says what the outlet state must be, which is how plant instrumentation actually works: you cannot put a thermometer inside a turbine, so the exhaust state is inferred from a load reading and a flow reading. The same reading in reverse is how a device's condition is monitored over years. A turbine whose measured enthalpy drop at rated flow falls by two per cent has fouled or eroded blading, and nobody had to open it to find out.

Example. A compressor takes $0.50$ kg s⁻¹ of air, raising it from $295$ K to $587$ K, and its casing loses $15$ kW to the surroundings. Taking air as an ideal gas with $c_{p} = 1.005$ kJ kg⁻¹ K⁻¹ and ignoring velocities, what shaft power does it need?

The enthalpy rise per kilogram is $c_{p} Δ T = 1.005 \times (587 - 295) = 293.5$ kJ kg⁻¹, so the streams carry away $0.50 \times 293.5 = 146.7$ kW more than they brought in. The heat loss removes a further $15$ kW. The shaft must supply both: ${\dot{W}}_{i n} = 146.7 + 15 = 161.7$ kW. Cooling a compressor does not reduce its shaft power at a fixed outlet temperature; it increases it, because the same temperature rise now has to be paid for twice.

Now you. The same compressor is run at $0.80$ kg s⁻¹ with the same inlet and outlet temperatures, and its heat loss rises to $25$ kW. What shaft power does it need?

Answer

${\dot{W}}_{i n} = 0.80 \times 293.5 + 25 = 234.8 + 25 = 259.8$ kW.

What this equation cannot tell you

Two limits should be stated before the equation is turned loose on hardware.

The first is that it is a balance, not a prediction of performance. Nothing in it says how much work a turbine will produce from a given inlet state, because the outlet state is an input, not an output. Given $3$ MPa and $350$ degrees Celsius at inlet and $10$ kPa at exhaust, the energy equation is equally happy with an exhaust enthalpy of $2264$ kJ kg⁻¹, which is a good turbine, or $3116$ kJ kg⁻¹, which is a turbine that produces nothing at all and simply lets the steam through. Both conserve energy exactly. Choosing between them needs the Second Law, and that is what makes the isentropic device and its efficiency, several lessons ahead, necessary rather than decorative.

The second is that steadiness has been assumed throughout, and with it the promise that nothing inside the control volume is changing. Real plant starts, stops, trips and swings load, and a boiler drum whose level is falling is storing energy in a way this equation cannot see. Handling those is the next lesson, and the answer will turn out to explain something as everyday as why a scuba cylinder is hot to the touch after filling.