Thermal efficiency of the Rankine cycle

In summary, the task is to calculate the thermal efficiency of an R-C cycle with saturated steam, starting with 4.5MPa entering the HP part of the turbine and finishing at 0.3MPa. After the HP turbine, the steam is separated and overheated before entering the condenser at 4kPa. The efficiency of turbines is 85% and pumps 100%, with no heat loss in the pipeline. The pressure in the deaerator is unknown, but it is known that the separator separates wet steam into a saturated liquid, which goes into the deaerator through a pressure reducing valve. The pressure in the deaerator is at most 0.3MPa, and the saturated steam is
  • #1
Hatyk
6
0

Homework Statement


Calculate thermal efficiency of R-C cycle with saturated steam as shown in the picture. Steam has 4,5MPa when entering into HP part of the turbine (VT in the picture) and expansion on HP is finished at 0,3MPa. Moisture is then removed in the separator and steam is overheated to a temperature 8K less then before HP part of the turbine. Pressure in the condenser is 4kPa. Thermal efficiency of turbines is 85%, pumps 100%. You can neglect heat loss in the pipeline.
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Homework Equations


Hello, I've encountered a problem with this cycle. I seem to miss value of the pressure in the deaerator (NN in the picture). Is there a way to calculate the pressure or calculate enthalpy without this pressure?[/B]

The Attempt at a Solution


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I was calculating the enthalpy values for the thermal efficiency, but the absence of pressure in the deaerator stopped me in my tracks. I was able to calculate enthalpy after both parts of the turbine, but got stuck at a point 5, right after condensate pump. To calculate point 5, I need pressure in the deaerator. If I knew the pressure I would be able to calculate enthalpy after the deaerator and with this enthalpy, I would be able to calculate enthalpy after the condensate pump using the binding equation from the deaerator. [/B]
 

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  • #2
Please describe in more detail your understanding of how the separator is operated, including control of pressure.
 
  • #3
Wet steam goes into a separator at the pressure of 0.3MPAa. The separator separates this wet steam into a saturated liquid (shown as Beta in the picture) which goes into a deaerator through pressure reducing valve. This means, that the pressure in the deaerator is at most 0.3MPa. The saturated steam goes into a superheater at pressure of 0.3MPa and is heated to a temperature 8K less than before HP turbine. Not sure what do you mean by control of pressure. To my understanding, separator is device which does not change the pressure, only separates moisture from the wet steam.
 
  • #4
Hatyk said:
Wet steam goes into a separator at the pressure of 0.3MPAa. The separator separates this wet steam into a saturated liquid (shown as Beta in the picture) which goes into a deaerator through pressure reducing valve. This means, that the pressure in the deaerator is at most 0.3MPa. The saturated steam goes into a superheater at pressure of 0.3MPa and is heated to a temperature 8K less than before HP turbine. Not sure what do you mean by control of pressure. To my understanding, separator is device which does not change the pressure, only separates moisture from the wet steam.
Since the process is continuous, there is no change in enthalpy of the saturated liquid going into the deaerator. Incidentally, when you say deaerator, you are not really implying that there is air in the system, right?
 
  • #5
Oh, no air in the system, I thought it's just called that way even if there is no air in the system. Better translation would be feedwater tank I guess. I have no problem in determining the enthalpy of saturated liquid after the separator. Since I know the pressure in the separator I can just take enthalpy from lower boundary curve for that pressure. The problem is I have no way to calculate how much is the enthalpy changed by the pumps, since I can't interpolate entropy values for the pressure in the feedwater tank. The only solution I came up with so far is to calculate the optimal pressure for the highest thermal efficiency.
 
  • #6
Hatyk said:
Oh, no air in the system, I thought it's just called that way even if there is no air in the system. Better translation would be feedwater tank I guess. I have no problem in determining the enthalpy of saturated liquid after the separator. Since I know the pressure in the separator I can just take enthalpy from lower boundary curve for that pressure. The problem is I have no way to calculate how much is the enthalpy changed by the pumps, since I can't interpolate entropy values for the pressure in the feedwater tank. The only solution I came up with so far is to calculate the optimal pressure for the highest thermal efficiency.
You have a rough estimate of the pressure in the tank, and that should be good enough. Shouldn't the enthalpy change of the pumps just be roughly ##v\Delta P##, where v is the specific volume.
 
  • #7
I'm not sure if I have a rough estimate. I know that the pressure in the tank has to be lower than 0.3MPa and higher than 4kPa. Pressure of 4kPa is not much realistic, so I think I can safely assume that the pressure is between 0.05MPa to 0.3MPa. But I have no further means to guess the right number. Should I just pick some number in the middle, let's say 0.1MPa and run with it?
 
  • #8
Isn't the outlet pressure going to be much higher?
 
  • #9
I don't think so, you can have maximum 0.3MPa in the tank, because the saturated liquid from the separator has that pressure. Continuous mixing has to happen at constant pressure and you can't increase the pressure of saturated liquid from the separator, since it's going only through reducing valve before going into a tank. First pump hast increase pressure from the 4kPa to somewhere below or equal to 0.3Mpa and the second pump has to increase pressure to the 4.5MPa.
 
  • #10
Hatyk said:
I don't think so, you can have maximum 0.3MPa in the tank, because the saturated liquid from the separator has that pressure. Continuous mixing has to happen at constant pressure and you can't increase the pressure of saturated liquid from the separator, since it's going only through reducing valve before going into a tank. First pump hast increase pressure from the 4kPa to somewhere below or equal to 0.3Mpa and the second pump has to increase pressure to the 4.5MPa.
I guess I'm not too clear on the process setup. I thought there was only one pump. Still, the outlet pressures from the pumps seem much higher than the inlet pressures. So maybe inaccuracies in the inlet pressures are negligible.
 

Related to Thermal efficiency of the Rankine cycle

1. What is the Rankine cycle?

The Rankine cycle is a thermodynamic cycle that is commonly used in power plants to convert heat energy into mechanical work. It is named after William Rankine, a Scottish engineer and physicist who first described the cycle in the 19th century.

2. How does the Rankine cycle work?

The Rankine cycle works by using a working fluid, typically water, which is heated in a boiler to produce high-pressure steam. The steam then expands through a turbine, producing mechanical work, and is then condensed back into liquid form in a condenser. The condensed liquid is then pumped back into the boiler to repeat the cycle.

3. What is thermal efficiency?

Thermal efficiency is a measure of how well a system converts heat energy into mechanical work. In the context of the Rankine cycle, it is the ratio of the net work output to the heat input. This is an important factor in determining the overall efficiency of a power plant.

4. How is thermal efficiency calculated for the Rankine cycle?

The thermal efficiency of the Rankine cycle can be calculated using the formula: Efficiency = (Work output/Heat input) x 100%. The work output is the difference between the enthalpy of the steam at the turbine inlet and outlet, and the heat input is the enthalpy of the steam at the boiler inlet.

5. What factors affect the thermal efficiency of the Rankine cycle?

There are several factors that can affect the thermal efficiency of the Rankine cycle, including the pressure and temperature of the steam, the type of working fluid used, and the efficiency of the turbine and condenser. Additionally, any losses due to friction, heat transfer, or incomplete expansion of the steam can also impact the overall efficiency of the cycle.

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