Troubleshooting Steam Plant Output: Tips for Evening Class Assignment

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SUMMARY

The discussion focuses on calculating steam output for a steam plant assignment, emphasizing the use of the overall efficiency of 77%. The formula Q=M(h2-h1) is confirmed as the correct approach to determine steam output. Participants clarify that the efficiency can be applied directly without considering the economizer and super heater. The final steam output calculated is 6244.09 kg/h, confirming the accuracy of the method used.

PREREQUISITES
  • Understanding of thermodynamic principles related to steam generation
  • Familiarity with the formula Q=M(h2-h1) for calculating heat transfer
  • Knowledge of system efficiency concepts, specifically in steam plants
  • Basic calculations involving energy and mass flow rates
NEXT STEPS
  • Research the role of economizers and super heaters in steam plant efficiency
  • Learn about advanced thermodynamic calculations for steam output
  • Explore methods to improve overall efficiency in steam generation
  • Study the impact of fuel type on steam output calculations
USEFUL FOR

Students in engineering programs, steam plant operators, and anyone involved in thermal system design and optimization will benefit from this discussion.

anthonyk2013
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Im attempting this question for an evening class I am attending. hope I am posting in the right place. I'm not sure where I have gone wrong think its where I rearranged my formula to find steam output.

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The work is a bit unclear. I'll get back to you in a day or so.
 
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Ok Thanks
 
I think I calculated the steam output wrong, I have the over all efficiently of the system 77%, what formula do I use the calculate steam output?
Q=M(h2-h1)
 
Yeah that's the one..!
For calculating the steam output, you can directly use the 77% efficiency without involving the economizer and super heater (if I'm not wrong).
So 77% of the heat obtained from the combustion of coal is used to convert water to steam.
 
siddharth23 said:
Yeah that's the one..!
For calculating the steam output, you can directly use the 77% efficiency without involving the economizer and super heater (if I'm not wrong).
So 77% of the heat obtained from the combustion of coal is used to convert water to steam.

77% is the overall efficiency am I right ? So did I divide .77 by h2-h1 to get steam output?
 
anthonyk2013 said:
77% is the overall efficiency am I right ? So did I divide .77 by h2-h1 to get steam output?
Yes! And equate it with the energy provided by the combustion of coal. Tell me if that gives the right answer.
 
6244.09 kg/h
 
I meant did you get the right answer?
 

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