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

Anna Litical drops a ball from rest from the top of 78.4 m high cliff.

Question 34

Anna Litical drops a ball from rest from the top of 78.4 m high cliff. How much time will it take for the ball to reach the ground and at what height will the ball be after each second of the motion?

Reference: Physics Classroom.com

Solution:

Initial velocity of ball = 0 m s^-1           (at rest)

Total distance travelled (fallen) by ball = 78.4 m

Take the acceleration due to gravity (downward), a = 9.8 m s^-2

Let the time at which the ball reaches the ground be t.

Consider the equation of uniformly accelerated motion:

s = ut + ½ at²

78.4 = 0 + ½×9.8×t²

Time taken to reach ground, t = √[78.4 / (0.5×9.8)] = 4.0 s

We also need to find the height reached after each second (i.e. t = 1, 2, 3, 4 s)

Initial height of the ball = 78.4 m

The distance fallen by the ball can be obtained from s = ut + ½ at² for each second.

At any time t, height of the ball = 78.4 – Distance fallen by ball at time t

Let the distance fallen by the ball at time t be s.

s = ut + ½ at² = 0 + ½×9.8×t²

s = ½ ×9.8 × t²

Using the formula above,

at t = 1 s, distance fallen = 4.9 m, height of ball = 78.4 – 4.9 = 73.5 m

at t = 2 s, distance fallen = 19.6 m, height of ball = 78.4 – 19.6 = 58.8 m

at t = 3 s, distance fallen = 44.1 m, height of ball =78.4 – 44.1 = 34.3 m

at t = 4 s, distance fallen = 78.4 m, height of ball = 78.4 – 78.4 = 0             (reached ground)