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The change in head across a small turbine is 10 m, the flow rate of water is 1 m3/s, and the efficiency is 80%. The power developed by the turbine is approximately:
  • a)
    100 kW
  • b)
    78 kW
  • c)
    1 MW
  • d)
    50 kW
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The change in head across a small turbine is 10 m, the flow rate of w...
The overall efficiency η0 of turbine = volumetric efficiency (ηv)× hydraulic efficiency (ηn)× mechanical efficiency (ηm)
ηo=ηv×ηh×ηm
Overall efficiency.
Water Power =ρ×Q×g×h
Calculation:
Given, η0=0.8, head h=10m/s and Q=1m3/s
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Most Upvoted Answer
The change in head across a small turbine is 10 m, the flow rate of w...
Given:
- Change in head (Δh) = 10 m
- Flow rate of water (Q) = 1 m^3/s
- Efficiency (η) = 80%

To find:
Power developed by the turbine

Formula:
Power (P) = ρ * g * Q * Δh * η
Where,
ρ = density of water (assumed to be 1000 kg/m^3)
g = acceleration due to gravity (assumed to be 9.8 m/s^2)

Calculation:
Substituting the given values into the formula:
P = 1000 * 9.8 * 1 * 10 * 0.8
P = 78400 W

Converting the power from watts to kilowatts:
P = 78400 / 1000
P ≈ 78 kW

Therefore, the power developed by the turbine is approximately 78 kW, which corresponds to option B.
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The change in head across a small turbine is 10 m, the flow rate of water is 1 m3/s, and the efficiency is 80%. The power developed by the turbine is approximately:a)100 kWb)78 kWc)1 MWd)50 kWCorrect answer is option 'B'. Can you explain this answer?
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