It is common knowledge in thermodynamics that Closed FeedWater Heaters(CFWH) do not directly mix the hot stream (bled from turbine) and cold stream (coming from the condenser via 1st pump), do not necessarily need a 2nd pump after heat exchanging for every CFWH addition in a power plant, as opposed to an Open FeedWater Heater(OFWH) which require a 2nd pump to reach the pressure of the exhaust of the turbine bled steam.
However we always see a mixing chamber at the exit of the CFWH, where the two streams (hot and cold) are mixed. So my question is this
1. What kind of mixing takes place at this mixing chamber.
2. Why are the enthalpies considered constant across the mixing chamber (points 4,5,9 in the first figure).
3. Doesn't the mixing chamber necessarily act as an OFWH where mixing of the two streams takes place, and if so
4. Why isn't the mass balance taken across the a.CFWH as well as b.Mixing chamber, since the streams are being mixed in both of them.
Also, what is the significance of a steam trap (second figure) ? Does it still give a hot stream to the condenser? What are the enthalpies across the steam trap. and how does it effect the overall thermal efficiency of the cycle.
Looking forward to your constructive feedbacks
Regards
https://doi.org/10.1016/B978-008044529-8.50014-8
Chapter Exergy Analysis of Steam Power Plants
https://www.sciencedirect.com/search?qs=role%20of%20Mixing%20Chamber%20and%20Steam%20Trap%20in%20a%20Closed%20Feedwater%20Heater%20containing%20Power%20Plant&show=25&sortBy=relevance
Many thanks for the share Md.Washim., but I do not have the access to the book that you shared. If you have, can you please share here. Or atleast the pages that are related to my questions.
Regards
Waqas Tanveer , the role of FWH is removing air from feed water. The air is going into steam/water using looseness in pipelines. The air makes difficult to get nessesary pressure in the condenser and therefore get nessesary power.
To prevent it you need constantly removing air form steam.
The idea lies in the fact that soluble the gases in water on the saturation line is zero. Therearefore , if yoo look your first image you can see that steam from extraction in the turbine goes to FWH. The goal of it is heating feed water to saturation temperature, if feed water achieved saturation temperature all gases exit from water.
The role of mixing chamber is heating feed water before boiler . If you heating feed water you have to need spend less energy in the boiler so efficience of cycle is raise.
Thank you very much for a detailed answer.
In an OFWH the dearation is achieved by the dearating unit(DA) built inside the OFWH chamber. So removal of air takes place there. But in case of CFWH we use a "Shell and Tube Heat Exchanger" and the DA takes place in the "Trap". That is clear now.
However, I am still confused about the mixing chamber (MC). If the purpose of the MC is to reach a single high temperature at constant pressure for the two streams, then still
1. There should be an increase in "Enthalpy" from entrance (9,4) to exit (5) of the mixing chamber.
2. Why isn't the Energy mass balance taken across the mixing chamber?
Waqas Tanveer , in point 7 we take part of stem from turbine for deaerator (FWH). We understand that in p.7 - steam because p.7 over saturation line in T-S diagrams.
After deaerator (FWH) in point 3 steam condenses into water. We understand that because point 3 stay on saturation line in T-S diagram.
The vertical line 3-4 in T-S diagram is work Pump II.
In Mixing chamber we mixed two streams of water. First - water from condenser (from point 9). This water is colder. Second - water from Pump II (from point 4). This water is more hot.
When this streams of water will be mixing their temperature will align. More hot water from point 4 will become more colder , more cold waterfrom point 9 will become more hot. This happens because more hot bodies give heat to more colder (this physical law usually call "The second beginning of thermodynamics").
So, as a result we will have one stream of water with temperature in point 5.
Excellent explanation @Vladimir Dorofeev
Agreed with all points from you
Let me put the question in another way,
Can heat transfer at constant pressure (but infinite temperatures from, (4,5,9) in the T-S dia) happen at constant enthalpy ? There has to be some change in the internal energy of the system while the second law of thermodynamics is at work.
Is it possible, that in the single phase (compressed liquid region- left side of the TS dia) the change in the internal energy of the system (H=U+PV) at constant pressure is negligible? I mean I cannot absorb the fact that temperature exchange of the hot(4) and cold(9) stream to make one stream(5) results in no change in enthalpy of the system. Then how does the internal energy change (from 4,9->5) express itself.
The cycle on your first image is ideal Rankine cycle.We understand that because line 6-7 and line 1-2 are vertical - entropy is not rising. In real life ideal cycle is impossible. All process going with the grown of entropy. This physical law usually call "The first beginning of thermodynamics".
On the attach images you can see ideal Rankine cycle and real Rankine cycle. Images copy fom book :"Shelaev A. V. Steam turbines. Theory of thermal process and turbine design.T 1" pages 27 and 29.
Ok Vladimir Dorofeev , many thanks
The answer to my question of enthalpy change in a compressed liquid region is still waiting your precious views :) Perhaps it might have been explained in your books better.
Waqas Tanveer , I really understand what you mean, but if you call horizontal line 8-1 in bottow of cycle "liquid region" you are dont right. There is line of condensing steam into water. When water change physical state from steam to water the temperature dont change. That all know.
On your first image (T-S diagramm) in point 8 you have wet steam. Line 8-1 is condensing. Only in point 1 you have water.
I never said anything about line 8-1...
my question is about enthalpies of , 4,5 and 9
Are they same? The book says they are same, yet doesn't elaborate why. If there is a temperature increase, the internal energy of the system in a single phase must be described by 2 or more states independently . So
1. There should be an increase in "Enthalpy" from entrance (9,4) to exit (5) of the mixing chamber.
What do you think ?
- Badges
- Science topic
- Similar topics
- Engineering
- Petroleum Engineering
- Gas