Thermodynamics: Pass-out Turbine Question

In summary, the conversation is about a person seeking help with a physics question. They are struggling with understanding certain parameters and equations, and are unsure if they have used the correct equations for parts b and c. The other person agrees with their approach of using p=1((h1-h4)+(1-0.32)(h4-h2)) to find the power output, and requests for a rationale and flow diagram for further clarification.
  • #1
PaxFinnica96
14
0
Homework Statement
Steam at 30 bar and 500°C enters a pass-out turbine at a rate of 18000 kg h–1. The bulk of this steam is expanded isentropically to a condenser pressure of 0.03 bar, but 32% is bled from the turbine at a pressure of 3 bar for process heating. If the condensate from the process heater is directly pumped back to the boiler, find:
a) Power output from the turbine
(b) the rate of process heating
(c) the efficiency of the plant.
Relevant Equations
m=Qp1/Qp2
Q=m_p (h2-h5)
η=(Qs-Q)/q
Hello Physics Forums,
I have been struggling with this question for some time now and I'm not sure my method is correct - please see attached.
Any help you can give to check I'm on the right lines would be very much appreciated!
Cheers, C
 

Attachments

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  • #2
I don't understand what you're doing in the section starting with h5=hu-wp, and I don't understand what the parameters represent. Also Vu, and what it's used for.
 
  • #3
Hi Chestermiller, I'm not really sure where I was going with this either and as a result I'm quite confused on where to begin.
I think I should use:
p=1((h1-h4)+(1-0.32)(h4-h2))
to find the power output from the turbine but not sure if I've used the correct equations for (b) and (c).
 
  • #4
PaxFinnica96 said:
Hi Chestermiller, I'm not really sure where I was going with this either and as a result I'm quite confused on where to begin.
I think I should use:
p=1((h1-h4)+(1-0.32)(h4-h2))
to find the power output from the turbine but not sure if I've used the correct equations for (b) and (c).
If h4 is the 3 bars and h2 is the 0.03 bars, then I agree.

Please say in words your rationale for parts b and c. Also, a complete flow diagram would be helpful.
 

FAQ: Thermodynamics: Pass-out Turbine Question

What is a pass-out turbine in thermodynamics?

A pass-out turbine is a type of steam turbine that is used in power plants to generate electricity. It is designed to extract energy from the steam that is produced by the boiler, and then release the remaining steam into a separate chamber for further use.

How does a pass-out turbine work?

A pass-out turbine works by using the energy from high-pressure steam to rotate a series of blades, which in turn rotate a shaft connected to a generator. The steam is then released into a separate chamber, where it can be used for other purposes, such as heating or powering other turbines.

What are the advantages of using a pass-out turbine?

One of the main advantages of using a pass-out turbine is that it allows for more efficient use of steam, as the remaining steam can be used for other purposes instead of being wasted. Additionally, pass-out turbines are more reliable and have a longer lifespan compared to other types of turbines.

What are the limitations of a pass-out turbine?

One limitation of a pass-out turbine is that it requires a larger and more complex system compared to other types of turbines. It also requires regular maintenance and can be expensive to operate. Additionally, pass-out turbines are not suitable for all types of power plants and may not be as efficient in certain conditions.

How is the efficiency of a pass-out turbine measured?

The efficiency of a pass-out turbine is measured by the amount of energy that is converted into electricity. This is known as the thermal efficiency and is calculated by dividing the electrical output by the total energy input. The higher the thermal efficiency, the more efficient the pass-out turbine is in converting steam energy into electricity.

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