See please the attachRaschjot STU.sm (292,81 KiB) scaricato 1653 volta(e).

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 KiB) scaricato 1640 volta(e).

Best regards.
Alvaro.

Wrote

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 (574,02 KiB) scaricato 1640 volta(e).

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

Wrote

Wrote

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 (574,02 KiB) scaricato 1640 volta(e).

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 KiB) scaricato 1014 volta(e).
Raschjot STU Ideal.sm (308,22 KiB) scaricato 1747 volta(e).

Best regards.
Alvaro.

Wrote

Wrote

Wrote

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 (574,02 KiB) scaricato 1640 volta(e).

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 KiB) scaricato 1014 volta(e).
Raschjot STU Ideal.sm (308,22 KiB) scaricato 1747 volta(e).

Best regards.
Alvaro.

We can use Q=1, 0, 77.85% etc