A 50% efficient hydroelectric power plant generates electricity from water falling a height of 110 m and at a rate of x kg/s. If the maximum possible power output of the plant was 27.0 MW, what is x?
2007-10-08 14:43:37 · 4 answers · asked by beer drinkers 2 in Science & Mathematics ➔ Engineering
Power (P) is energy (E) per time (t).
You can get the energy of the water by it's potential at the top,
E = m * g * h
The power the plant can get from the water's energy depends on the efficiency (eff).
So,
P = eff * E / t
. = eff * m * g * h / t
And, rearranging,
m / t = P / ( eff * g * h )
. . . . = ( 27 MW ) / [ ( 0.5 ) * ( 9.81 m/s² ) * ( 110 m ) ]
. . . . = 50,059 kg/s
2007-10-08 16:21:09 · answer #1 · answered by Ben 3 · 0⤊ 0⤋
"x" designates the total flow of water in kg/s needed to produce the 27 MW of electrical power. It is a bit more involved than that, however. That 110 meters is going to have a certain "head pressure" which is part of the equation, as well as the rate of flow. If it takes 1,000,000 kg/s of water to generate the desired plant output, then "x" equals 1,000,000 kg/s.
2007-10-08 21:56:51 · answer #2 · answered by Anonymous · 0⤊ 2⤋
27 MW = 36, 207.6HP = 72,415.2BHP@50% Eff.
110 m = 360.8913 feet
BHP = GPM*H/3969
72,415.2 = GPM*360.8913/3960
GPM = 794,600GPM = 13,243.33 GPSec
13,243.33GPSec 110,463 Lb/sec = 50,105 Kilo/sec
2007-10-08 22:21:58 · answer #3 · answered by gatorbait 7 · 0⤊ 0⤋
P.E. of the water = 'x'kg/s x 9.81m/s² x 110m = 27MJ/s
= 27MW
1,079.1x = 27MW
x = 27,000,000W ÷ 1,079.1 = 25,020kg/s = 50% efficient.
100% efficiency = 25,020 x 2 = 50,040kg/s of water
2007-10-09 00:30:58 · answer #4 · answered by Norrie 7 · 0⤊ 0⤋