The steady-flow devices

Almost every component in a power station, a refrigerator or a jet engine is analysed by writing down the same equation and then deleting most of it.

One equation, six deletions

The previous lesson produced, for a control volume with one inlet and one outlet running steadily,

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

with $q$ and $w$ per kilogram of fluid, heat positive inwards and work positive outwards. It has five terms and no device uses all five. What distinguishes a nozzle from a turbine is not a different physical law, it is which terms survive.

The deletions are not arbitrary, and each one is a claim about the hardware that can be checked. Saying a turbine is adiabatic is a claim that its casing loses less than about one per cent of the power passing through, which is true of a lagged machine and false of an uninsulated one. Saying a throttle has no kinetic term is a claim about the pipe diameters either side. The habit worth building is to write the full equation, then justify each deletion out loud, because the deletions are where the physics is.

Nozzles and diffusers

A nozzle is a duct shaped to raise a fluid's velocity at the expense of its pressure; a diffuser is the same device used in reverse, slowing the flow and raising the pressure. Neither has any moving part, so $w = 0$ . The fluid is inside for a few milliseconds and the surface area is small, so $q \approx 0$ . There is no height change worth counting. Every term dies except two:

h_{1} + \frac{V_{1}^{2}}{2} = h_{2} + \frac{V_{2}^{2}}{2}

Enthalpy converts to velocity, one for one. The quantity on each side, called the stagnation enthalpy, is what a nozzle conserves.

The velocities are large by construction, so this is the one device where the kinetic term is never negligible. It is also the device where the mass balance from the earlier lesson bites hardest: since $\dot{m} = A V / v$ is fixed, the area must follow $A = \dot{m} v / V$ , and whether a duct that accelerates a fluid gets narrower or wider depends on whether $v$ grows faster than $V$ does. For a liquid, or for a gas below the speed of sound, $V$ wins and the duct converges. Above the speed of sound $v$ wins and the duct must diverge, which is why a rocket nozzle has a throat and then flares out. That result belongs to compressible flow rather than to thermodynamics, but the energy equation above is what it is built on.

Example. Air enters a nozzle at $200$ degrees Celsius and $30$ m s⁻¹ and leaves at $100$ degrees Celsius. Taking air as an ideal gas with $c_{p} = 1.005$ kJ kg⁻¹ K⁻¹, what is the exit velocity?

For an ideal gas $h_{2} - h_{1} = c_{p} (T_{2} - T_{1}) = 1.005 \times (- 100) = - 100.5$ kJ kg⁻¹, a fall of $100500$ J kg⁻¹. So $V_{2}^{2} / 2 = V_{1}^{2} / 2 + 100500 = 450 + 100500 = 100950$ J kg⁻¹ and $V_{2} = \sqrt{2 \times 100950} = 449$ m s⁻¹. A hundred kelvin of temperature is worth four hundred and fifty metres per second, which is the trade every jet and rocket lives on.

Now you. Air enters a nozzle at $400$ degrees Celsius and $20$ m s⁻¹ and leaves at $250$ degrees Celsius. What is the exit velocity?

Answer

$c_{p} Δ T = 1.005 \times 150 = 150.75$ kJ kg⁻¹, so $V_{2}^{2} / 2 = 200 + 150750 = 150950$ J kg⁻¹ and $V_{2} = 549$ m s⁻¹.

Run the same equation backwards for a diffuser. An aircraft intake takes air at $250$ m s⁻¹ relative to the aeroplane and slows it to $30$ m s⁻¹ before the compressor. The temperature rise is $(V_{1}^{2} - V_{2}^{2}) / 2 c_{p} = (250^{2} - 30^{2}) / (2 \times 1005) = 30.6$ K, achieved with no machinery at all. At supersonic speeds this ram effect does a large part of the compression, which is why a ramjet needs no compressor and why it cannot work standing still.

Turbines

A turbine lets a fluid expand across rotating blade rows and takes work out through a shaft. The kinetic terms are small compared with the enthalpy drop, at least once the exhaust hood has been included in the control volume, and a lagged casing loses well under one per cent of the throughput as heat. So $q = 0$ , the velocity and height terms go, and

w = h_{1} - h_{2}

The work per kilogram is the enthalpy drop, and nothing else. A turbine passing $15.61$ kg s⁻¹ of steam from $h_{1} = 3116.1$ to $h_{2} = 2263.5$ kJ kg⁻¹ delivers $15.61 \times 852.6 = 13.3$ MW, subject to the velocity correction computed in the previous lesson.

The equation is silent about how the enthalpy drop is achieved, which is exactly right: it holds for a steam turbine, a gas turbine, a hydraulic turbine and a wind turbine, and knows nothing about blades. What it does not tell you is how large the drop will be for a given inlet state and back pressure, and that gap is the reason the Second Law has to be brought in later.

Compressors, fans and pumps

Reverse the shaft and the same analysis gives a machine that raises pressure at the cost of work. The three names are the same device at different duties: a fan moves a lot of gas against a small pressure rise, a compressor raises gas pressure substantially, and a pump does the same to a liquid. Dropping the same terms,

w_{i n} = h_{2} - h_{1}

with the sign taken care of by naming the work as an input. Compressors are sometimes deliberately cooled, in which case $q$ is not zero and must be carried, as in the worked example of the previous lesson.

The interesting comparison is between a pump and a compressor working over the same pressure range, because it explains a fact that dominates the design of power stations. Water at $10$ kPa has a specific volume of $0.001010$ m³ kg⁻¹ and saturated steam at the same pressure has $14.67$ m³ kg⁻¹, a ratio of $14500$ . Raising the water from $10$ kPa to $3$ MPa costs about $3.0$ kJ kg⁻¹, which is $0.35$ per cent of the $852.6$ kJ kg⁻¹ the turbine delivers over the same pressure range. Compressing the vapour instead would cost of the same order as the turbine gives back. Pressurise the liquid, never the vapour: it is the whole reason the Rankine cycle exists in the form it does, and the reason a feed pump on a $600$ MW station absorbs about $1.4$ MW.

Throttling valves

A throttling valve is any restriction that drops pressure without any intention of extracting work: a partly open valve, an orifice plate, a porous plug, a capillary tube. There is no shaft, so $w = 0$ . The device is small and fast, so $q = 0$ . The pipes either side are usually similar, so the velocities roughly cancel, and there is no height change. Everything is gone except

h_{1} = h_{2}

Throttling is isenthalpic. That is a strange-looking result: pressure falls by a large factor and enthalpy does not move at all. What happens internally is that flow work is converted into internal energy by friction and turbulence, and the sum $u + p v$ comes out unchanged even though both parts of it change a lot.

For an ideal gas, whose enthalpy depends on temperature alone, constant enthalpy means constant temperature, and throttling does nothing thermally. For a real fluid it can do a great deal. The temperature change per unit pressure drop at constant enthalpy is the Joule-Thomson coefficient, and its sign flips at a substance's inversion temperature. Nitrogen's is $621$ K, so nitrogen throttled from room temperature cools, which is the basis of the Linde process for liquefying air. Hydrogen's is $202$ K, so hydrogen throttled from room temperature warms, and any attempt to liquefy it by throttling alone must precool it first, a fact discovered expensively.

The dramatic case is a saturated liquid, where throttling causes flash evaporation and a large temperature drop. This is what happens in the expansion valve of every refrigerator.

Example. Saturated liquid R-134a at $800$ kPa, where $h_{f} = 95.47$ kJ kg⁻¹ and the saturation temperature is $31.31$ degrees Celsius, is throttled to $132.8$ kPa, where $h_{f} = 25.49$ and $h_{f g} = 212.91$ kJ kg⁻¹ at a saturation temperature of $- 20$ degrees Celsius. What emerges?

Enthalpy is unchanged, so the exit state has $h = 95.47$ kJ kg⁻¹ at $132.8$ kPa. Since that lies between $h_{f} = 25.49$ and $h_{g} = 238.40$ , it is a wet mixture, of quality

x = \frac{h - h_{f}}{h_{f g}} = \frac{95.47 - 25.49}{212.91} = 0.329

So a third of the liquid flashes to vapour, and the mixture sits at $- 20$ degrees Celsius. A valve costing a few pounds has produced a $51$ K temperature drop with no moving parts, no work and no heat. The evaporating third is what absorbs heat from the cold space, and the other two thirds are along for the ride.

Now you. The same saturated liquid at $800$ kPa is throttled to $200$ kPa instead, where $h_{f} = 38.43$ and $h_{f g} = 206.03$ kJ kg⁻¹ at $- 10.1$ degrees Celsius. What quality results?

Answer

$x = (95.47 - 38.43) / 206.03 = 0.277$ . Less flashing, because less cooling is being asked for, and a smaller share of the flow is doing useful work in the evaporator.

Heat exchangers

A heat exchanger brings two streams into thermal contact without mixing them: a shell full of tubes, a plate stack, a coil in a tank. Draw the control surface around the whole unit and there is no shaft, no significant velocity change and, since the outside is insulated, no heat crossing the outer boundary. What is left is that whatever one stream loses, the other gains:

{\dot{m}}_{A} (h_{A 2} - h_{A 1}) = {\dot{m}}_{B} (h_{B 1} - h_{B 2})

The heat transferred between them does not appear, because it never crossed the surface we drew: it went from one stream to the other inside it. Draw the surface around one stream only and the same heat reappears explicitly as $\dot{Q}$ . Both control volumes are correct, and choosing between them is choosing which question to answer.

Example. The condenser of the turbine above receives $15.61$ kg s⁻¹ of steam at $h = 2263.5$ kJ kg⁻¹ and returns it as saturated liquid at $10$ kPa, where $h_{f} = 191.81$ kJ kg⁻¹. Cooling water enters at $18$ degrees Celsius and may rise by $10$ K, with $c = 4.18$ kJ kg⁻¹ K⁻¹. What is the heat duty and how much cooling water is needed?

Per kilogram of steam, $2263.5 - 191.81 = 2071.7$ kJ is rejected, so the duty is $15.61 \times 2071.7 = 32340$ kW, or $32.3$ MW. The water side must absorb the same, so ${\dot{m}}_{w} = 32340 / (4.18 \times 10) = 774$ kg s⁻¹.

Now you. Environmental limits cut the permitted water temperature rise to $8$ K. What flow is now needed?

Answer

${\dot{m}}_{w} = 32340 / (4.18 \times 8) = 967$ kg s⁻¹, a quarter more water for the same duty.

Scale that up and the number becomes a siting constraint. A $600$ MW steam plant passing $462$ kg s⁻¹ rejects about $957$ MW in its condenser, needing roughly $23$ m³ s⁻¹ of cooling water for a $10$ K rise. That is why large thermal stations sit on rivers, estuaries or coastlines, or else pay an efficiency penalty for cooling towers.

Mixing chambers, and choosing the surface

A mixing chamber is a heat exchanger with the wall removed: the streams join. There is no work and, if it is insulated, no heat, so the balance is enthalpy in equals enthalpy out,

{\dot{m}}_{1} h_{1} + {\dot{m}}_{2} h_{2} = {\dot{m}}_{3} h_{3}

alongside ${\dot{m}}_{1} + {\dot{m}}_{2} = {\dot{m}}_{3}$ from the mass balance. Two equations, and typically two unknowns. Mixing $2.0$ kg s⁻¹ of water at $80$ degrees Celsius with $5.0$ kg s⁻¹ at $20$ degrees Celsius gives $7.0$ kg s⁻¹ at $(2 \times 80 + 5 \times 20) / 7 = 37.1$ degrees Celsius, taking the specific heat as constant. A domestic mixer tap is exactly this calculation. So is the open feedwater heater that will turn out to raise the efficiency of a power station by several points.

That leaves the point this lesson has been circling. The equations above are not six different results; they are one result plus six defensible sets of deletions. What actually varies between problems is where the control surface goes, and drawing it in the right place is worth more than any formula. Put it around a condenser and the heat vanishes into an internal transfer. Put it around one stream and the heat reappears. Put it around a whole power station and every internal flow disappears, leaving fuel in, air in, stack out, cooling water out and electricity out, which is the right surface for asking what the plant does and the wrong one for asking which component is at fault.

Two limits are worth naming before moving on. First, every deletion above assumed steady operation, and none of these balances survives a machine that is filling, emptying or changing load. Second, and more seriously, not one of these equations decides how well a device performs. The turbine relation $w = h_{1} - h_{2}$ is satisfied by a perfect machine and by a broken one; the compressor relation is satisfied by an efficient compressor and by a heater with an impeller in it. Energy accounting alone cannot tell them apart, and the missing ingredient is entropy.