Power generated by the steam turbine

  • #1
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Homework Statement


Shown in the figure

Homework Equations


W= (p2v2-p1v1) / (1-n)

The Attempt at a Solution



I am not sure where to start and where to use all the data given in the table. Please help.
It would be nice if someone could explain the basic nature of the steam turbine, how it works and the process. Is the turbine connected to the compressor?
 

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  • #2
stockzahn
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Homework Statement


Shown in the figure

Homework Equations


W= (p2v2-p1v1) / (1-n)

The Attempt at a Solution



I am not sure where to start and where to use all the data given in the table. Please help.
It would be nice if someone could explain the basic nature of the steam turbine, how it works and the process. Is the turbine connected to the compressor?
Thanks for moving your question. I copy-paste my answer from the original post:

You need the first law of thermodynamics for open systems, since mass enters/leaves the turbine it is not a closed system. The law reads

$$\delta Q +\delta W + \sum \left[\dot{m}_i \left(e_i+h_i \right) \right] = \Delta U + \Delta E $$,

##Q## ... heat
##W## ... work
##h## ... specific enthalpy
##e## ... specific external energy
##U## ...internal energy
##E## ... external energy

Now you can simplify the formula with respect to your problem (i.e. which terms are equals zero?). Start with that, then you will see which data you need from the table.
 
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  • #3
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Thanks for the reply but as it's been a few days since I started this topic(thermodynamics) I still have no idea.

1. Do I need to calculate values for the turbine and compressor separately? (e.g. calculate internal energy separately)
2. How do you get to power from that equation? How to apply the equation to my problem?

Thanks in advance.
 
  • #4
stockzahn
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Thanks for the reply but as it's been a few days since I started this topic(thermodynamics) I still have no idea.

1. Do I need to calculate values for the turbine and compressor separately? (e.g. calculate internal energy separately)
2. How do you get to power from that equation? How to apply the equation to my problem?

Thanks in advance.
Start with question a) (and forget about the compressor at first). Now you define the system turbine by mentally putting it in a "box" called system. Now you only observe what is entering and exiting this system, which is described by the left-hand side (LHS) of the above equation. So ask yourself:

1) Is heat (##\delta Q##) entering or exiting the system?
2) Is work (##\delta W##) entering or exiting the system?
3) Are mass flow rates (##\dot{m}_i##) entering or exiting the system?

The right-hand side (RHS) of the formula describes the change of the energy in the system. So assuming steady state operation:

4) Does the energy content (##\Delta U + \Delta E##) of the system changes with time?
 
  • #5
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Just considering the turbine, work is exiting the system.
As heat goes from hot to cold, it's going from 1 to 2 in the figure.
Mass flow rate remains the same in the table (inlet and outlet) but I'm not sure if it is entering or exiting.

I know you are trying to explain as well as you can and I very much appreciate it but I am having difficulty understanding it.
Could you, if possible, please solve a bit from the beginning (calculations) for me to understand and move on?
 
  • #6
stockzahn
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1) Is heat (##\delta Q##) entering or exiting the system? ##\rightarrow## The text says the turbine can be treated to be adiabatic... so ##\delta Q = ?##
2) Is work (##\delta W##) entering or exiting the system? ##\rightarrow## Correct, work is exiting the system, therefore (##\delta W \neq 0##)
3) Are mass flow rates (##\dot{m}_i##) entering or exiting the system? ##\rightarrow## Mass flow rates are constant, but still they enter and exit the turbine, don't they?
4) Does the energy content (##\Delta U + \Delta E##) of the system changes with time? ##\rightarrow## Steady state means, no change in time, so ##\Delta U + \Delta E = ?##
 
  • #7
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No heat crosses the boundary in the adiabatic system so heat transfer is 0.
Work is exiting the system.
Mass flow rates enter and exit the turbine.
I understand these.

How do you know that it is steady state though? Am I missing something?
ΔU+ΔE = 2456 - 435 = 2021 kJ/kg?
 
  • #8
stockzahn
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How do you know that it is steady state though? Am I missing something?
That's just an assumption to be able to solve the problem with the given data. ##\Delta U + \Delta E = 0## means that at any time the same amount of internal and external energy is contained in the system, which is the case in steady state. If for example you start the turbine, in the beginning the temperature of the contained fluid and of the blades etc. is low. With time they get warmer storing more internal energy and the rotor increases in speed storing external energy. That is described by the RHS of the formula. Since there is no change in the operating status: ##\Delta U + \Delta E = 0##.

No heat crosses the boundary in the adiabatic system so heat transfer is 0.
Work is exiting the system.
Mass flow rates enter and exit the turbine.
I understand these.
Well, then let's simplify the formula:

$$0 +\delta W + \sum \left[\dot{m}_i \left(e_i+h_i \right) \right] = 0 $$

$$\delta W = -\sum \left[\dot{m}_i \left(e_i+h_i \right) \right] $$

The work done by the turbine equals the change of enthalpy and external energy of the entering and exiting mass flows. Since accorindg to the text the external energies are negligible:

$$\delta W = - \sum \left(\dot{m}_i h_i \right) $$

Let's define entering energy positive and exiting energy negative:

$$ \delta W = \dot{m}_{out} h_{out}-\dot{m}_{in} h_{in} $$

Since ##\dot{m}_{out}=\dot{m}_{in}=\dot{m}##:

$$ \delta W = \dot{m} \left(h_{out}-h_{in}\right) $$

Now final question: Where to get the enthalpy from?
 
  • #9
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Thanks, I get the steps.
I guess you need to use the equation h = u + pv?
 
  • #10
stockzahn
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Thanks, I get the steps.
I guess you need to use the equation h = u + pv?
Yes.

I just saw, that the text says only to neglect the potential energy (not the kinetic energy), therefore, to be exact: ##e_i=\frac{v_i^2}{2}\neq 0##. But normally the kinetic energy is negligible compared to the enthalpy in a steam turbine ...
 
  • #11
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Below are my calculations, please check if they are correct.

h(out) = u +pv = 435 kJ/kg + 1 bar x 0.65 m^3/kg = 500 x 10^3 m^2 s^-2
h(in) = u + pv = 2456 kJ/kg + 150 bar x 0.01 m^3/kg = 585 x 10^3 m^2 s^-2

Thus, dW = m( h(out) - h(in)) = 12 kg/s x (-85 x 10^3 m^2 s^-2)
 
  • #12
stockzahn
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h(out) = u +pv = 435 kJ/kg + 1 bar x 0.65 m^3/kg = 500 x 10^3 m^2 s^-2
h(in) = u + pv = 2456 kJ/kg + 150 bar x 0.01 m^3/kg = 585 x 10^3 m^2 s^-2

Thus, dW = m( h(out) - h(in)) = 12 kg/s x (-85 x 10^3 m^2 s^-2)
For ##h_{out}## I obtain the same result, even if at the left-hand side your units are not consistent:

$$h_{out}=u + pv = 435 \cdot 10^3\;J/kg + 1\cdot 10^5 \; N/m^2 \; 0.65\;m^3/kg = 500 \cdot 10^3\;J/kg = 500 \;kJ/kg$$

For ##h_{in}## I obtain a different value, but unfortunately I can't reproduce your result even with the formula you've written down. I suggest that if you write down equations like that to use the correct units in each term, otherwise one makes mistakes.

However, as I mentioned in my last post, if the kinetic energy should be considered, then

$$h_{out}=u + pv = 435 \cdot 10^3\;J/kg + 1\cdot 10^5 \; N/m^2 \; 0.65\;m^3/kg + \frac{50^2}{2}\; m^2/s^2= 501250 \;J/kg = 501.25 \;kJ/kg$$
 
  • #13
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My calculation was wrong. I understand how you calculated the kinetic energy part, using the velocity.
h(in) = u + pv = 2456 x 10^3+ 150 x 10^5 x 0.01 m^3/kg = 2606 x 10^3 = 2606 kJ/kg

Then, h(out) - h(in) = 501.25 - 2606 = -2104.75 which is a negative value.
Substituting the value into the equation,
dW = m(h(out)-h(in)) = 12 kg/s x (-2104.75 kJ/kg) = -25257 kJ/s which is the power generated by the turbine.
Am I correct?
Thanks.
 
  • #14
stockzahn
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My calculation was wrong. I understand how you calculated the kinetic energy part, using the velocity.
h(in) = u + pv = 2456 x 10^3+ 150 x 10^5 x 0.01 m^3/kg = 2606 x 10^3 = 2606 kJ/kg

Then, h(out) - h(in) = 501.25 - 2606 = -2104.75 which is a negative value.
Substituting the value into the equation,
dW = m(h(out)-h(in)) = 12 kg/s x (-2104.75 kJ/kg) = -25257 kJ/s which is the power generated by the turbine.
Am I correct?
Thanks.
The method is correct, but you calculated the enthalpy ##h_{in}## without the kinetic energy - it should be slightly higher, when taking into account the kinetic energy with a fluid velocity of 130 m/s. It's up to you whether you neglect the kinetic energy or not, but it should be the same for the inlet and the outlet.
 
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  • #15
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I get what you mean, thanks. So h(in) would be 2456 x 10^3+ 150 x 10^5 x 0.01 m^3/kg + 130^2 / 2 m^2 s^-2 = 2614450 kJ/kg.
h(out) - h(in) = 501.25 - 2614.45 = -2113.2 and dW (P) = 12 kg/s x (-2113.2 kJ/kg) = -25358.4 kJ/s.
For question (b), I first calculated 20% of the power generated by the turbine.
Then, -25358.4 x 0.2 =-5071.68 kJ/s .
As dW (P) = m(h(out) - h(in)) and the enthalpy drop across the cooling loop is 100 kJ/kg,
-5071.68 kJ/s = m(875+ 0.5 x 12^2 - 350 - 0.5x20^2 -100) to get the mass flow rate of cooling water.
Am I on the right track?
 
  • #16
stockzahn
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So h(in) would be 2456 x 10^3+ 150 x 10^5 x 0.01 m^3/kg + 130^2 / 2 m^2 s^-2 = 2614450 kJ/kg.
h(out) - h(in) = 501.25 - 2614.45 = -2113.2 and dW (P) = 12 kg/s x (-2113.2 kJ/kg) = -25358.4 kJ/s.
I obtained the same numbers, I think that should be the correct answer for a).

Then, -25358.4 x 0.2 =-5071.68 kJ/s .
As far as to this point I agree.

As dW (P) = m(h(out) - h(in)) and the enthalpy drop across the cooling loop is 100 kJ/kg,
-5071.68 kJ/s = m(875+ 0.5 x 12^2 - 350 - 0.5x20^2 -100) to get the mass flow rate of cooling water.
Here you mixed up the enthalpies of the compressed gas and of the cooling water, additionally your units are not consistent (you take the enthalpies into account in kJ/kg and the kinetic energy in J/kg).
 
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  • #17
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How should I consider the cooling loop? Should I also take the work coming inwards (0.2 W) into account?
 
  • #18
stockzahn
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How should I consider the cooling loop? Should I also take the work coming inwards (0.2 W) into account?
Let's start from the beginning:

1) adiabatic ##\rightarrow \delta Q = 0##
2) work ##\rightarrow \delta W = P_{Comp} = 5.072 \; MW## (which has a positive sign now, since it enters the compressor)
3) steady-state ##\Delta U + \Delta E = 0##
4) mass flow rates:

a) gas to be compressed (Subscript ##g##)
b) cooling water (Subscript ##w##)

The 1st law reads:

$$0 + P_{Comp} + \dot{m}_g \left[ \left(h_{g\;in} + e_{g\;in} \right) -\left( h_{g\;out} + e_{g\;out} \right)\right] +\dot{m}_w \left( h_{w\;in} - h_{w\;out} \right) =0$$

Since the density change of water is negligible, the velocity is constant, so no change in the external energy (##e_{w\;in} - e_{w\;out}=0##).
Except the mass flow rate of the cooling water, everything is known.
 
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  • #19
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Thanks for the explanation.
0 + 5.072x 10^6 + 0.1[(350x10^3 + 0.5x20^2)-(875 x10^3 + 0.5x12^2)] + m(water) (-100 x 10^3) = 0
m(water) = 50 kg/s
I double checked, please let me know if my answer is correct.
 
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  • #20
stockzahn
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Thanks for the explanation.
0 + 5.072x 10^6 + 0.1[(350x10^3 + 0.5x20^2)-(875 x10^3 + 0.5x12^2)] + m(water) (-100 x 10^3) = 0
m(water) = 50 kg/s
I double checked, please let me know if my answer is correct.
At least I have the same result ##\dot{m}_w \approx 50\;kg/s## - seems to be a very bad compressor or I've overlooked something. But in general that's how you could solve this kind of problems. Have a nice Sunday!
 
  • #21
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Thanks for your help, I really appreciate it.
I think I can now solve questions similar to this!
Hope you have a nice day :wink:
 

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