Q#36 (Past Exam Paper – June 2017 Paper 11 Q19)
Water flows from a lake into a turbine that is a vertical distance of 90 m below the lake, as shown.
The mass flow rate of the water is 2400 kg min-1. The turbine has an efficiency of 75%.
What is the output power of the turbine?
A 26 kW B 35 kW C 1.6 MW D 2.1 MW
Solution:
Answer: A.
Power = energy / time
Water in the lake (at top) has gravitational potential energy.
Power = GPE / time
Power = mgh / t = (m/t) gh
The mass flow rate of the water is 2400 kg min-1. This represents the quantity (m/t).
m/t = 2400 kg min-1
To be used in the above formula for the power of water, it should be converted into SI units.
2400 kg min-1 means that 2400 kg of water flows in 1 min.
1 min - - > 2400 kg
60 s - - > 2400 kg
1 s - - > 2400/60 = 40 kg
The mass flow rate is 40 kg s-1.
Power of water = (m/t) gh = 40 × 9.8 × 90 = 35 280 W
This is the power available from the water. But the turbine has an efficiency of only 75%.
Power output of turbine = (75/100) × 35 280 = 26 460 = 26 kW
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