A turbine at a hydroelectric power station is situated 30 m below the level of the surface of a large lake.

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