In a hydroelectric power plant, water flows from an elevation of 400 ft to a turbine, where electric power is generated. For an overall turbine-generator efficiency of 85 percent, determine the flow rate of water to generate 100 kW of electricity if the irreversible head loss of the piping system between the free surfaces of the inlet and the outlet is 36 ft. Take the density of water to be rho= 62.4 lbm/ft ^3.The flow rate of water is 348 ft^3/s.

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cjmejiab cjmejiab · Answer 1 · 2019-11-05T23:50:06+01:00

Answer:

[tex]V=3.475ft^3/s=3.48ft^3/s[/tex]

Explanation:

We have here values from SI and English Units. I will convert the units to English Units.

We hace for the power P,

[tex]P= 100kW \rightarrow 100kW*(\frac{737.56ft.lbf/s}{1kW})[/tex]

[tex]P= 73.7*10^{3}ft.lbf/s[/tex]

we have other values such [tex]h=400ft[/tex] and [tex]\gamma= 62.42lb/ft^3[/tex] (specific weight of the water), and 0.85 for \eta

We need to figure the flow rate of the water (V) out, that is,

[tex]V=\frac{P}{\gamma h \eta_0}[/tex]

Where [tex]\eta_0[/tex] is the turbine efficiency, at which is,

[tex]\eta_0 = \frac{P}{\gamma Vh}[/tex]

Replacing,

[tex]V=\frac{73.7*10^{3}}{62.42*400*0.85}[/tex]

[tex]V=3.475ft^3/s[/tex]

With this value (the target of this question) we can also calculate the mass flow rate of the waters,

through the density and the flow rate,

[tex]m=\rho V\\m= 3.475*1.94 \\m=6.7415 slugs/s[/tex]

converting the slugs to lbm, 1slug = 32.174lbm, we have that the mass flow rate of the water is,

[tex]m= 217lbm/s[/tex]