The irreversible losses in the penstock of a hydroelectric dam are estimated to be 7 m. The elevation difference between the reservoir surface upstream of the dam and the surface of the water exiting the dam is 140 m. If the flow rate through the turbine is 4000 L/min, determine (a) the power loss due to irreversible head loss, (b) the efficiency of the piping, and (c) the electric power output if the turbine-generator efficiency is 84 percent.

a) the power loss due to irreversible head loss is 4.57 kW

b) the efficiency of the piping is 95%

c) the electric power output is 72.9918 kW

Explanation:

Given the data in the question below;

Irreversible loses = 7m

Total head, H = 140 m

flow rate Q = 4000 L/min = 0.0666 m³/s

Generator efficiency n₀ = 84% = 0.84

we know that density of water is 1000 kg/m³

g = 9.81 m/s²

a) power loss due to irreversible head loss is;    

= p × Q × g ×

we substitute

= 1000 × 0.0666 × 9.81 × 7

= 4573.422 W

= 4573.422 / 1000

= 4.57 kW

Therefore, the power loss due to irreversible head loss is 4.57 kW

b) the efficiency of the piping n is;

n = (Actual head / maximum head) × 100

n = (( H - ) / H) × 100

so we substitute

n = (( 140 - 7 ) / 140) × 100

n = (133/140) × 100

n = 0.95 × 100

n = 95%

Therefore, the efficiency of the piping is 95%

c) the electric power output if the turbine-generator efficiency is 84 percent;

n₀ = /

= n₀ ×

= n₀ × ( pQg( H - ))

so we substitute

= 0.84 × ( 1000 × 0.0666 × 9.81( 140 - 7 ))

= 0.84 × 653.346( 133)

= 0.84 × 86895.018

= 72991.815 W

= 72991.815 / 1000

= 72.9918 kW

Therefore, the electric power output is 72.9918 kW

answer:

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explanation:

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