Answer to Question #240825 in Mechanical Engineering for Lala

Question #240825

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.

Expert's answer


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, to fit with the given velocity. mh1 = (1300Btu/lb)(3.33lb/sec) = 953437/4329 Btu/sec

KEi = 1/2(3.33lb/sec)(4002)(ft2/s2) = 266400(ft²lb/sec2³)

Also since the exit enthalpy is given in Btu/min (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

Note: 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 mass unit and one is a speed, it doesn't match

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