  Rank: Advanced Member Groups: Registered, Advanced Member Joined: 06/04/2023(UTC) Posts: 340   Was thanked: 26 time(s) in 25 post(s)
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Rank: Advanced Member Groups: Registered
Joined: 28/08/2014(UTC) Posts: 1,356  Was thanked: 815 time(s) in 516 post(s)
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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 (575kb) downloaded 15 time(s).Best regards. Alvaro. Edited by user 01 June 2023 02:09:13(UTC)
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Rank: Advanced Member Groups: Registered, Advanced Member Joined: 06/04/2023(UTC) Posts: 340 Was thanked: 26 time(s) in 25 post(s)
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Originally Posted by: Razonar  Hi. 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 (575kb) downloaded 15 time(s).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  Edited by user 01 June 2023 09:34:10(UTC)
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Rank: Advanced Member Groups: Registered
Joined: 28/08/2014(UTC) Posts: 1,356 Was thanked: 815 time(s) in 516 post(s)
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Originally Posted by: OchkovVF Originally Posted by: Razonar Hi. 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 (575kb) downloaded 15 time(s).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 (399kb) downloaded 19 time(s). Raschjot STU Ideal.sm (309kb) downloaded 25 time(s).Best regards. Alvaro.
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Rank: Advanced Member Groups: Registered, Advanced Member Joined: 06/04/2023(UTC) Posts: 340 Was thanked: 26 time(s) in 25 post(s)
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Originally Posted by: Razonar Originally Posted by: OchkovVF Originally Posted by: Razonar Hi. 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 (575kb) downloaded 15 time(s).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 (399kb) downloaded 19 time(s). Raschjot STU Ideal.sm (309kb) downloaded 25 time(s).Best regards. Alvaro. We can use Q=1, 0, 77.85% etc
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