# Steam Turbine: Solve for Velocity

## 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.

Last edited by a moderator:

Related Engineering and Comp Sci Homework Help News on Phys.org
This problem is solvable but it seems that the given problem answer is incorrect

verty
Homework Helper
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.

• Paul Lasdivan
Chestermiller
Mentor
Also, the steam flow should be 200 lb/ min .

• Paul Lasdivan
Got it by changing exit enthalpy to Btu/lb and of course steam flow to 200 lb/min :D Thank you for helping.

• verty