Thermodynamics Steam Turbine Specific Entropy

In summary, the mass flow rate through a steam turbine is 100 kg/s and it operates under steady conditions. Steam enters at 12 MPa and 400oC, and exits as a mixture of vapor and liquid water at 10 kPa. The turbine loses heat to the surroundings at a rate of 3 MW and kinetic and potential energy effects can be ignored. The specific entropy at the inlet is 6.0747 kJ/kgK and at the exit it is 7.625 kJ/kgK. The enthalpy at the inlet is 3051.3 kJ/kg and at the exit it is 2417.1991 kJ/kg.
  • #1
ConnorM
79
1

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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  • #2
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.
 
  • #3
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).
 
  • #4
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 ]
 
  • #5
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:
  • #6
Hey I got it right! Thanks for the tip =)
 
  • #7
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.
 

Related to Thermodynamics Steam Turbine Specific Entropy

1. What is the specific entropy of steam in a turbine?

The specific entropy of steam in a turbine is a measure of the amount of entropy per unit mass of steam. It is defined as the change in entropy per unit mass of steam as it undergoes a process, such as passing through a turbine.

2. How does specific entropy impact the efficiency of a steam turbine?

The specific entropy of steam plays a critical role in the efficiency of a steam turbine. The higher the specific entropy, the lower the efficiency of the turbine, as more energy is lost due to the increase in entropy during the expansion process.

3. What factors affect the specific entropy of steam in a turbine?

The specific entropy of steam in a turbine is affected by several factors, including the pressure and temperature of the steam, the type of turbine, and the amount of moisture present in the steam. Higher pressure and temperature, as well as lower moisture content, result in lower specific entropy and higher efficiency.

4. Can the specific entropy of steam be controlled in a turbine?

Yes, the specific entropy of steam can be controlled in a turbine by adjusting the pressure and temperature of the steam, as well as by removing moisture through the use of separators. This allows for the optimization of turbine efficiency and performance.

5. How is the specific entropy of steam calculated in a turbine?

The specific entropy of steam in a turbine is calculated using the steam tables, which provide values for specific entropy at different pressures and temperatures. It can also be calculated using the first and second law of thermodynamics, taking into account the specific heat and work done by the steam in the turbine.

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