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Ideal Rankine cycle - with T, s diagram - Сообщения
See please the attachRaschjot STU.sm (292,81 КиБ) скачан 358 раз(а).
Hi. I don't understand how you get Q5 = 77.8518 % given that at T5 = 32.88°C and P5 = 5 kPa it must tu be Q5 = 50 %.
This is a more direct way of solving the problem, in the sense that the same results are obtained. A great advantage of using functions of two variables for thermodynamic variables is that now you don't have to look them up in tables, and therefore you no longer need to separate the vapor and liquid parts to find them in tables.
Raschjot STU.sm (574,02 КиБ) скачан 333 раз(а).
Best regards.
Alvaro.
This is a more direct way of solving the problem, in the sense that the same results are obtained. A great advantage of using functions of two variables for thermodynamic variables is that now you don't have to look them up in tables, and therefore you no longer need to separate the vapor and liquid parts to find them in tables.
Raschjot STU.sm (574,02 КиБ) скачан 333 раз(а).
Best regards.
Alvaro.
WroteHi. I don't understand how you get Q5 = 77.8518 % given that at T5 = 32.88°C and P5 = 5 kPa it must be Q5 = 50 %.
Raschjot STU.sm (574,02 КиБ) скачан 333 раз(а).
Best regards.
Alvaro.
At T5 = 32.88°C and P5 = 5 kPa it Q5 must be from 0 % to 100%.
If Q5=50% - it is not a steam turbine - it is a 50% steam turbine and a 50% water turbine ;-)
Calculation with WaterSteamPro

WroteWroteHi. I don't understand how you get Q5 = 77.8518 % given that at T5 = 32.88°C and P5 = 5 kPa it must be Q5 = 50 %.
Raschjot STU.sm (574,02 КиБ) скачан 333 раз(а).
Best regards.
Alvaro.
At T5 = 32.88°C and P5 = 5 kPa it Q5 must be from 0 % to 100%.
If Q5=50% - it is not a steam turbine - it is a 50% steam turbine and a 50% water turbine ;-) ...
You're right. I put the numbers backwards. The correct question is why do you use Q5 = 50% since the correct value is Q5 = 77.85%?
In the attached file, in addition to showing the most direct way to calculate all the states, the T-s and T-h diagrams are added.
Raschjot STU Ideal.pdf (398,4 КиБ) скачан 275 раз(а).
Raschjot STU Ideal.sm (308,22 КиБ) скачан 342 раз(а).
Best regards.
Alvaro.
WroteWroteWroteHi. I don't understand how you get Q5 = 77.8518 % given that at T5 = 32.88°C and P5 = 5 kPa it must be Q5 = 50 %.
Raschjot STU.sm (574,02 КиБ) скачан 333 раз(а).
Best regards.
Alvaro.
At T5 = 32.88°C and P5 = 5 kPa it Q5 must be from 0 % to 100%.
If Q5=50% - it is not a steam turbine - it is a 50% steam turbine and a 50% water turbine ;-) ...
You're right. I put the numbers backwards. The correct question is why do you use Q5 = 50% since the correct value is Q5 = 77.85%?
In the attached file, in addition to showing the most direct way to calculate all the states, the T-s and T-h diagrams are added.
Raschjot STU Ideal.pdf (398,4 КиБ) скачан 275 раз(а).
Raschjot STU Ideal.sm (308,22 КиБ) скачан 342 раз(а).
Best regards.
Alvaro.
We can use Q=1, 0, 77.85% etc
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