Steam Turbine: Solve for Velocity

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SUMMARY

The discussion centers on calculating the exit velocity of a steam turbine given specific inlet conditions and enthalpy values. The original problem states that the steam flow should be in lb/min rather than Btu/min, leading to confusion in the calculations. The correct exit velocity is determined to be 50 ft/s, but participants highlight discrepancies in the provided enthalpy values, particularly questioning the use of 800 Btu/min instead of 800 Btu/lb. The energy equation used is mh1 + KEi = mh2 + KEf + W, with heat losses neglected.

PREREQUISITES
  • Understanding of thermodynamics, specifically energy equations in turbines
  • Familiarity with units of measurement in thermodynamics (Btu/lb, lb/min)
  • Knowledge of steam properties and flow calculations
  • Proficiency in converting between units (Btu/min to lb/sec)
NEXT STEPS
  • Review the principles of energy conservation in thermodynamic systems
  • Learn about steam turbine performance metrics and calculations
  • Study the conversion of thermal energy units in engineering contexts
  • Explore common errors in enthalpy and flow rate measurements in thermodynamics
USEFUL FOR

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

Paul Lasdivan
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Homework Statement


Homework Statement [/B]
A steam turbine developed 2372.20 Hp when its inlet condition is 1300 Btu/lb enthalpy and 400 ft/s velocity and steam flow of 200 Btu/min :The exit enthalpy is 800 Btu/min. Find the exit velocity.

That is the original problem statement but i think the steam flow should be in lb/min.
The answer to the problem is 50 feet per second but no solution was provided.

Homework Equations


Energy process in turbine is mh1 + KEi = mh2 + KEf + W + Q , Heat is neglected so Q = 0.

The Attempt at a Solution


I assumed that steam flow isn't in Btu/min but in lb/min so I converted this to lb/sec to fit with the given velocity.
mh1 = (1300Btu/lb)(3.33lb/sec) = 4329 Btu/sec
KEi = 1/2(3.33lb/sec)(4002)(ft2/s2) = 266400(ft2lb/sec3)
Also since the exit enthalpy is given in Btu/min, I assumed this to be mh2
mh2 = 800Btu/min(1min/60sec) = 13.33Btu/sec
W = 2372.20Hp ((0.707Btu/sec)/Hp) = 1677.14 Btu/sec
1lbf = 32.2lbm-ft/sec so KE1 = 8273.29ft-lbf/sec = 10.63Btu/sec
Solving for KEf, KEf = 4329+10.63 - 13.33 -1677.14 = 2649.16btu/sec
I solved Vf by converting btu again to ft-lbf and then to ft-lbm from this but the answer is much greater than 50fps.
 
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This problem is solvable but it seems that the given problem answer is incorrect
 
Since no one has answered you just yet, I will do my best to help you. I'm thinking that the 800 Btu/min is wrong because enthalpy is a total amount of heat, and Btu/lb is heat per pound, it's a mass unit. But 800 Btu/min is a measure of the speed of heat loss. One is a total and one is a speed, it doesn't match up.

I would recalculate it with 800 Btu/lb and see if you get the right answer.
 
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Also, the steam flow should be 200 lb/ min .
 
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Got it by changing exit enthalpy to Btu/lb and of course steam flow to 200 lb/min :D Thank you for helping.
 
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