Question 37
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
Reference: Past Exam Paper – November 2013 Paper 11 & 12 Q19
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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