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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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