In a hydroelectric power plant, water enters the turbine nozzles at 780 kpa absolute with a low velocity. if the nozzle outlets are exposed to atmospheric pressure of 100 kpa, determine the maximum velocity to which water can be accelerated by the nozzles before striking the turbine blades.

the maximum velocity is approximately 36.878 m/s

Explanation:

We can apply a balance of mechanical energy.  

P1 + ρgy1 + 1/2ρv1² = P2 + ρgy1 + 1/2ρv2² + Fr

where P=pressure , ρgy = hydrostatic pressure , 1/2ρv² = dynamic pressure , Fr= friction

we can neglect the variations of height in the nozzle and depreciate the low velocity term, therefore

P1 - P2  - Fr= 1/2ρv2²

the maximum velocity is found in case of no friction , then

P1 - P2 = 1/2 ρ vmax²

therefore

vmax =√2(P1-P2)/ ρ

assuming the density of water as ρ= 1000 kg/m³

vmax =√2(P1-P2)/ ρ = √(2*(780000 Pa - 100000 Pa)/1000 Kg/m³ ) = 36.878 m/s

answer:

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