1.

Water enters in a turbine at a speed of 500 m/s and leaves at 400 m/s. If 2 xx 10^(3) kg/s of water flows and efficiency is 75%, then output power is

Answer»

`6.75xx10^(7) W`
1000 KW
100 KW
400 W

Solution :Here, CHANGE in K.E./sec.
`E=1/2mv_1^2-1/2mv_2^2`
=`1/2m(v_1^2-v_2^2)`
=`1/2xx2xx10^3[(500)^2-(400)^2]`
=`10^3xx10^4[25-16]`
=`9xx10^7 J//sec or WALTS`.
`:.` Real output POWER=`9xx10^7xx(75)/(100)=6.75xx10^7W`


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