Thermodynamics Steam Turbine Specific Entropy

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SUMMARY

The discussion focuses on the thermodynamic analysis of a steam turbine operating at a mass flow rate of 100 kg/s, with steam entering at 12 MPa and 400°C and exiting as a mixture of vapor and liquid at 10 kPa, with 93% of the mass in vapor form. The turbine loses heat at a rate of 3 MW. Key calculations include the power generated by the turbine and the change in specific entropy, with specific entropy values calculated as s1 = 6.0747 kJ/kgK and s2 = 7.625 kJ/kgK. The first law of thermodynamics is applied to derive the energy balance, confirming that kinetic and potential energy effects can be neglected.

PREREQUISITES
  • Understanding of steam tables for water and superheated steam
  • Familiarity with the first law of thermodynamics
  • Knowledge of specific entropy calculations
  • Ability to interpret T-s diagrams
NEXT STEPS
  • Study the application of the first law of thermodynamics in open systems
  • Learn how to use steam tables for various fluids, including water and propane
  • Explore the concept of vapor quality and its impact on thermodynamic properties
  • Investigate the design and efficiency optimization of steam turbines
USEFUL FOR

Mechanical engineers, thermodynamics students, and professionals involved in energy systems and steam turbine design will benefit from this discussion.

ConnorM
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Homework Statement


The mass flow rate through a steam turbine operating under steady conditions is 100 kg/s. Steam enters the turbine at 12 MPa and 400oC. A mixture of vapour and liquid water exits the turbine at 10 kPa. At the exit state 93% of the mass of the water is in vapour form. The turbine loses heat to the surroundings at the rate
of 3 MW. Kinetic and potential energy effects can be ignored.

a) Draw a schematic of the system and clearly indicate the system boundary
using a dashed line. Indicate and label the mass flow inlets and outlets. Using
arrows, illustrate the energy transfers between the system and surroundings
by heat transfer and by work.

b) Write a 1st law energy balance for the system that agrees with your schematic. Cancel all unnecessary terms, providing justification for why these terms can be cancelled.

c) Draw the process on a T-s diagram, clearly indicating the state points and the isobars corresponding to the state points.

d) Calculate the power generated by the turbine in MW.

e) Calculate the change in specific entropy from inlet to exit in kJ/KgK.

Homework Equations



Question 2

s1=sf + x[ sg - sf ]
Also since there is one inlet and one exit,
m=m1=m2

I am pretty sure this equation will be useful as well,
0= Q - W + m[ (h1 - h2) + ((v12 - v22)/2) + g(z1 + z2) ]

but since kinetic and potential energy is negligible I think I can cancel out ((v12 - v22)/2) + g(z1 + z2)]
That would leave me with,

0= Q - W + m[ (h1 - h2) ]

The Attempt at a Solution



For the second question I'm not really sure what to do, I think that what I would have to do is find sg from using 12 MPa, 400oC and looking at the steam table for water.
I'm not sure what I could do next because I would have 2 unknowns for the equation,
s1=sf + x[ sg - sf ]
 
Last edited:
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PF Rules generally ask that posters provide only one problem per thread. This policy is for the benefit of the users and the members since you are not mixing up answering different questions within a single thread.
 
ConnorM said:

Homework Statement



2) The mass flow rate through a steam turbine operating under steady conditions is 100 kg/s. Steam enters the turbine at 12 MPa and 400oC. A mixture of vapour and liquid water exits the turbine at 10 kPa. At the exit state 93% of the mass of the water is in vapour form. The turbine loses heat to the surroundings at the rate
of 3 MW. Kinetic and potential energy effects can be ignored.

a) Draw a schematic of the system and clearly indicate the system boundary
using a dashed line. Indicate and label the mass flow inlets and outlets. Using
arrows, illustrate the energy transfers between the system and surroundings
by heat transfer and by work.

b) Write a 1st law energy balance for the system that agrees with your schematic. Cancel all unnecessary terms, providing justification for why these terms can be cancelled.

c) Draw the process on a T-s diagram, clearly indicating the state points and the isobars corresponding to the state points.

d) Calculate the power generated by the turbine in MW.

e) Calculate the change in specific entropy from inlet to exit in kJ/KgK.

Homework Equations


[/B]
Question 2
For the second question I think this equation may be useful since we are given the quality,
s1=sf + x[ sg - sf ]
Also since there is one inlet and one exit,
m=m1=m2

I am pretty sure this equation will be useful as well,
0= Q - W + m[ (h1 - h2) + ((v12 - v22)/2) + g(z1 + z2) ]

but since kinetic and potential energy is negligible I think I can cancel out ((v12 - v22)/2) + g(z1 + z2)]
That would leave me with,

0= Q - W + m[ (h1 - h2) ]

The Attempt at a Solution



For the second question I'm not really sure what to do, I think that what I would have to do is find sg from using 12 MPa, 400oC and looking at the steam table for water.

Maybe not water. At the inlet, you generally use superheated steam for turbines.

I'm not sure what I could do next because I would have 2 unknowns for the equation,
s1=sf + x[ sg - sf ]

You are given the exit pressure (10 kPa) and the vapor quality of the mixture (93%). What you have to do is supply the saturated properties of the liquid and vapor phases from the steam tables at this exit pressure to determine the specific entropy of the mixture (and the enthalpy).
 
Ok sorry about that! I changed this so it is now only one of the questions. So I should use the pressure table for propane to find sg and sf? Then I would use those values to find the specific entropy once it exits?
Using,

s =sf + x[ sg - sf ]
 
Hey so I used 12 MPa and 400 C to get s1 on the superheated table for water, which was 6.0747 kJ/kgK. Then I used the pressure table for water and the values of entropy at 0.10 bar to get s2. I used the equation s=sf + x[ sg - sf ] and found that s2=7.625 kJ/kgK.

How is that so far?

edit: Just remembered to get the enthalpy's. For h1 I used the superheated table and got 3051.3 kJ/kg. Next I used my pressure table at 0.10 bar and found my hf and hg, from here I used,

h2 = xhg + (1-x)hf

and found h2 = 2417.1991 kJ/kg
 
Last edited:
Hey I got it right! Thanks for the tip =)
 
ConnorM said:
Hey so I used 12 MPa and 400 C to get s1 on the superheated table for water, which was 6.0747 kJ/kgK. Then I used the pressure table for water and the values of entropy at 0.10 bar to get s2. I used the equation s=sf + x[ sg - sf ] and found that s2=7.625 kJ/kgK.

How is that so far?

edit: Just remembered to get the enthalpy's. For h1 I used the superheated table and got 3051.3 kJ/kg. Next I used my pressure table at 0.10 bar and found my hf and hg, from here I used,

h2 = xhg + (1-x)hf

and found h2 = 2417.1991 kJ/kg

The thermo properties look OK. Now you can finish the steam turbine problem.
 

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