Q#37 (Past Exam Paper – November 2013 Paper 11 & 12 Q19)
A turbine at a hydroelectric power station is situated 30 m below the level of the surface of a large lake. The water passes through the turbine at a rate of 340 m3 per minute.
The overall efficiency of the turbine and generator system is 90%.
What is the output power of the power station? (The density of water is 1000 kg m-3.)
A 0.15 MW B 1.5 MW C 1.7 MW D 90 MW
Solution:
Answer: B.
At the top of the lake, the water has gravitational potential energy. As it passes through the turbine, the energy is converted to electrical energy. However, the system is only 90% efficient – not all of the gravitational potential energy becomes electrical energy.
GPE of water = mgh
(Input) Power = energy / time = GPE / time
(Input) Power = mgh / t
The water passes through the turbine at a rate of 340 m3 per minute. This quantity has unit m3 per unit – that is, the unit of volume / unit of time (ΔV/t). It represents the volume flow rate.
Volume flow rate = 340 m3 per minute
ΔV/t = 340 m3 per minute
We need to convert this into SI unit.
In 1 min (60 s), a volume of 340 m3 passes through the turbine.
1 min - - > 340 m3
60 s - - > 340 m3
1 s - - > 340/60 = 5.67 m3
In 1 s, a volume of 5.67 m3 passes through the turbine
(Volume flow rate) ΔV/t = 5.67 m3 s-1
The formula for (input) power above contains the mass, not the volume. We need to convert the volume flow rate into mass flow water (that is, Δm/t).
Density = mass / volume
Mass = Density × Volume
Δm = ρ × ΔV
Divide by time on both sides,
Δm/t = ρ × ΔV/t
Δm/t = 1000 × 5.67 = 5670 kg s-1
This is the equivalent mass flow rate.
(Input) Power = mgh /t = (Δm/t) × gh
(Input) Power = 5670 × 9.8 × 30 = 1.67×106 W = 1.67 MW
The overall efficiency of the system is 90% (= 0.90).
Output power = 0.90 × 1.67 = 1.5 MW
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