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A hydraulic turbine has 50m of head available at a flow rate of 1.30 m^3/s, and its overall turbine. generator efficiency is 78 percent. Determine the electric power output of this turbine. Given that ρ=1000 kg/m^3

Respuesta :

Answer:

P = 4.96 *10^5 watts

Explanation:

we know that, power is given as

[tex]power =\frac{\dot m gh}{t}[/tex]

[tex]power = \frac{\dot m}{dt} gh[/tex]

 we know that

[tex]1L = 10^{-3} m^3[/tex]

also 1L = 1kg

[tex]\frac{\dot m}{dt} = 1.30 m^3/s[/tex]

therefore

[tex]P = 50 * (\frac{1.30}{10^{-3}}* 9.8[/tex]

P = 6.37*10^5 W

78% of power = 6.37 *0.78

P = 4.96 *10^5 watts

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