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A water jet that leaves a nozzle at 78 m/s at a flow rate of 120 kg/s is to be used to generate power by striking the buckets located on the perimeter of a wheel. Determine the power generation potential of this water jet. Assume that the water jet flows steadily at the specified speed and flow rate. (Round the final answer to two decimal places.) The power generation potential is kW.

User Tenzing
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1 Answer

5 votes

Answer:

The value is
E = 365040 J/s

Step-by-step explanation:

From the question we are told that

The speed of the water jet is
v = 78 \ m/s

The flow rate of the water jet is
\r m = 120 \ kg /s

Generally the mechanically energy possessed by a water flow is mathematically represented as


E = \r m * e

Here here e is the specific mechanical energy generated which is mathematically represented as


e[KJ/kg] = FE + KE + PE

Here

FE - flow energy

KE - kinetic energy

PE - potential energy

Generally give that we are only considering the power generation due to the kinetic energy of the water we have that


E = \r m * KE

Here


KE = (v^2)/(2)


E  =  \r m  * (v^2)/(2)

=>
E  = 120  * (78^2)/(2)

=>
E  = 365040 J/s

User Jose De Gouveia
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