In a hydroelectric power plant, water enters the turbine nozzles at 880 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 m/s.

39.5 m/s

Explanation:

To solve this question, we would use Bernoulli's equation

P₁/ρ + V₁²/2 + gz₁ = P₂/ρ + V₂²/2 + gz₂,

where V₁ = 0 and z₁ = z₂, so that we have

P₁/ρ = P₂/ρ + V₂²/2

making V₂² subject of formula, we have that

V₂² = 2[P₁/ρ - P₂/ρ]

Then we finally go ahead and substitute the values such that

V₂² = 2 [(880000/1000) - (100000/1000)]

V₂² = 2 (880 - 100)

V₂² = 2(780)

V₂² = 1560

V₂ = √1560

V₂ = 39.5 m/s

answer:

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explanation: