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In Rankine cycle there's the boiler where water gets boiled into overheated steam. On input there's a pump that delivers more water, and on output there's a turbine that picks up the energy of the compressed steam.

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The steam pressure is roughly identical against the turbine and the pump; the pressure of the boiler section.

What makes the steam power the turbine instead of backing up and forcing the pump to turn backwards? - well, this one is simple, power delivered to the pump. But then how comes the turbine produces more power than the pump takes? The pump, after all, must overcome the same pressure that propels the turbine and deliver the same amount of water that is being ejected as steam. I'm missing some significant element of the device. What is it?

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Pressure is force/area, if the area offered by the pump impeller is smaller than the area against wich the steam must force its way out of the boiler, and both the turbine and the pump are linked, then, the same pressure will result in a smaller force on the pump than on the turbine.

Example :

Lets suppose that the pump is a piston pump, and on the steam side there is a piston engine (for simplicity). In a certain part of the cycle, both the pump piston and the engine piston valves are open towards the boiler (the pump is feeding water to the boiler and the engine is taking steam from the boiler).

The "face" of the pump piston has, lets say, a surface area of 10cm², while the steam engine piston has a surface area of 100cm². Lets suppose the pressure on the boiler is 200kPascal. This means that the pump piston will need to force its way against 200000N/M² * 0.001M² = 200N of force. While this pressure on the steam engine produces 200000N/M² * 0.1M² = 20kN. It's clear that in a direct linkage between the pump and the engine, the steam engine will produce much more force than the pump needs to feed water inside the boiler against the pressure gradient.

For comparision, lets suppose that instead of pumping water into the boiler, the pump takes steam and pumps it into the boiler. If the pump piston is smaller (meaning a smaller stroke or bore), the amount of water mass that gets out of the boiler via the engine would be bigger than the amount of mass that gets back into the boiler via the pump. If both where equal the engine would not generate any power. While if the pump were able to put more steam into the boiler than it were to get out, you would have an above unit power source - i.e. defying the laws of thermodynamics. Soon the boiler would run out of mass to heat.

But, as what gets pumped inside the boiler is liquid water, and water as a liquid has a much higher density than steam (meaning the same volume holds more mass as liquid than as gas - besides the fact that liquids have a fixed volume while the gas can expand), when that small volume is pumped into the boiler as water, the thermal energy added to the system will expand this mass into a steam that has a lot of pressure and much less density, so to get the same mass of water (as steam) out of the boiler via the engine, you will need a much bigger swept volume than the one you used to put that water inside, resulting in a difference in forces exerted by this gas over the piston of the pump (via the incoming water column) versus the engine piston, deciding the direction towards wich the system will move.

Got the idea ?

In other words, the ability of the water to change phase and occupy a bigger volume than the fixed volume liquid results in a net power gain into this system. From where comes the energy to cause the phase change, which then causes thisincrease in pressure? It's from the heat source. Where goes the energy rejected? It goes out of the condenser, where steam becomes liquid, losing volume, to be injected again into the boiler as a smaller and fixed volume liquid, and so on...

The key idea here is pressure equals force divided by area.

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The main reason is that it takes much less energy to compress a liquid than a gas by the same pressure difference. The pump takes a little bit of energy to compress the water, but a huge amount of energy is released when the steam expands in the turbine. This is why a phase change is used in the Rankine and related cycles.

Another thing to consider is that the pump doesn't need to over-match the power of the turbine, as you said - it only needs to over-match the pressure.

You can see the difference in energy between gas and liquid compression when looking at an enthalpy-vs-pressure table.

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  • $\begingroup$ But the pressure on the boiler, travels via water into the pump, thats what he does not understand, how can the same pressure not force against the pump in a way that would cause the pump to stop feeding water ? $\endgroup$ – Jorge Aldo Oct 18 '15 at 1:51
  • $\begingroup$ Ah, the pump is sized so that it will be able to work against the design pressure of the boiler. If the boiler operates at 1000 kPa, a pump should be used that can discharge water at 1100 kPa, for example. That way there's always a favorable pressure gradient in the boiler, moving the steam in the right direction. $\endgroup$ – Carlton Oct 18 '15 at 1:58
  • $\begingroup$ I dont believe so, if both systems are conected, pressure will be equal on all connected vessels. Lets suppose a piston pump, the valve is open towards the boiler that is at 100kpa. How could the pump "discharge" at 110kpa ? The systems are connected, pressure will be communicated via water to the pump piston, simple as that. $\endgroup$ – Jorge Aldo Oct 18 '15 at 2:05
  • $\begingroup$ This is the same principle as the brayton engine, both brayton and rankine need a way to make the fluid/gas go via the power turbine and not back to the compressor/pump. this is solved via size differences in the size of the turbine/compressor (the turbine if rotated by the gas exerts a force that is greated than the drag induced into the compressor...) $\endgroup$ – Jorge Aldo Oct 18 '15 at 2:07
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    $\begingroup$ yes but pressure on the pump discharge is equal as pressure on the engine input. theres no difference in pressure here, the keyword is FORCE, the same pressure on different pistons produce differente forces, and thats what allows the pump to pump water into the system. $\endgroup$ – Jorge Aldo Oct 18 '15 at 2:24
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There is a pressure differential across the turbine which prevents the pressure from backing up.

http://www.mpoweruk.com/images/rankine_pv.gifSomebody's Ideal Rankine Cycle P-vs-vol On the above chart between point 2 and point three the fluid passes through the turbine which expands while it pushes the turbine and reduces pressure. If you ignored the shaft and rotation of the turbine that point could almost be replaced by a nozzle with the same effect on the cycle.

From your description it sounds like you're describing a pumped storage system or you are ignoring the heating/combustion stage (point 4 to point 1 on my chart). This stage is what increases the pressure of the system beyond the pressure that the pump creates.

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  • $\begingroup$ "If you ignore..." - can I ignore it just like that? If there was a nozzle, there would be a pure loss of energy, decompression. If there was a piston steam engine with large valves, the pressure would be backing up after equalizing between the piston chamber and the boiler. Anything in between is a trade-off between harvested power (and holding the pressure back) and releasing the steam through a nozzle, creating a loss. HOW is the pressure not backing up through the turbine, and still leaving enough energy to break even with the pump? $\endgroup$ – SF. Oct 16 '15 at 20:54
  • $\begingroup$ I mean "reductio ad absurdum": put two identical, symmetrical devices that can act both as pumps and as turbines, a heater right in the middle between them on one side, a radiator right in the middle on the other side. Connect their shafts. You may even give it initial push, so that one pumps the water in while the other is propelled by the steam out, but the torques will cancel out. Where does the asymmetry come from? $\endgroup$ – SF. Oct 16 '15 at 21:04
  • $\begingroup$ wait... I think I'm seeing where. I'm not sure if I'm right so correct me if I'm fantasising. Pump/turbine torque is proportional to pressure and volume moved, not mass. The pressure on both ends will be the same (equalized), but there will be much more steam than water, so the water pump can make like one rotation per 100 rotations of the steam turbine and still the amount of water (+steam) on the hot side will remain constant. More volume at the same pressure = more usable work. My machine would need a gearbox: the steam side would produce less torque than the other needs, but more RPM. $\endgroup$ – SF. Oct 16 '15 at 21:19
  • $\begingroup$ "By condensing the working steam vapor to a liquid the pressure at the turbine outlet is lowered and the energy required by the feed pump consumes only 1% to 3% of the turbine output power and these factors contribute to a higher efficiency for the cycle." <Wikipedia> The pump is not whats creating the power that the turbine is being fed by, the boiler/heating stage is. Energy of a fluid isn't stored only by the pressure created by the pump. The energy in this system is built by the boiler/heating stage so there is not really a strong connection between the input of the pump and the turbine. $\endgroup$ – Dopeybob435 Oct 16 '15 at 21:34
  • $\begingroup$ The energy balance is between the heating source and the turbine. upload.wikimedia.org/wikipedia/commons/thumb/b/be/… In the image look at the size of the W-in of the pump vs the Qin of the heat source. $\endgroup$ – Dopeybob435 Oct 16 '15 at 21:36

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