Question

A hydroelectric dam holds back a lake of surface area 3.0×106m2 that has vertical sides below the water level. The water level in the lake is 150 m above the base of the dam. When the water passes through turbines at the base of the dam, its mechanical energy is converted to electrical energy with 90% efficiency.
(a) If gravitational potential energy is taken to be zero at the base of the dam, how much energy is stored in the top meter of the water in the lake? The density of water is 1000kg/m3
(b) What volume of water must pass through the dam to produce 1000 kilowatt-hours of electrical energy? What distance does the level of water in the lake fall when this much water passes through the dam?

Answer 1

Answer:

4.41 × 10¹² J, 2.72 × 10³ m³, 0.907 × 10 ⁻³ m

Explanation:

Gravitational potential energy = mgh

where m is mass in kg, g is acceleration due to gravity in m/s², and h is the distance from the base of the dam.

mass of the surface water = density of water × volume of water × 1 m = 1000 kg / m³ × 3.0 × 10⁶ m² × 1 m = 3 × 10⁹ kg

Gravitational potential energy = 3 × 10⁹ kg × 9.81 m/s² × 150 m = 4.41 × 10¹² J

b)what volume of water must pass through the dam to produce 1000 kw-hrs

1 000 kw-hr = 3.6 × 10 ⁹ J

the dam has mechanical energy conversion of 90% to electrical energy

Gravitational potential energy needed = 3.6 × 10 ⁹ J / 0.9 = 4 × 10⁹ J

mass of water needed = Energy  required / g h =  4 × 10⁹ J / (9.81 m/s² × 150 m) = 2.718 × 10 ⁶ kg

density = mass / volume

volume = mass / density =  2.718 × 10 ⁶ kg / (1000 kg/ m³) = 2.72 × 10³ m³

the distance the level of  the water in the lake fell = volume / area =  2.72 × 10³ m³ / (3.0×10⁶ m²) = 0.907 × 10 ⁻³ m

Answer 2

(a) The gravitation potential energy stored in the top meter is 4.41 × 10¹² J

(b) The volume of water required to produce 1000 kW-hr of energy is 2.72 × 10³ m³ and the level of the lake falls by 0.907 × 10 ⁻³ m

Gravitational potential energy

The gravitational potential energy is given by:

PE = mgh

where m is mass,

g is acceleration due to gravity ,

and h is the distance from the base of the dam.

Now, the mass of water for 1m depth

m = density of water × volume(for 1 meter)

m = 1000 × 3.0 × 10⁶× 1 = 3 × 10⁹ kg

Now the Gravitational potential energy stored in the top one meter of the lake is:

PE = mgh = 3 × 10⁹ × 9.81 × 150 = 4.41 × 10¹² J

(b)volume of water that must pass through the dam to produce 1000 kw-hrs

1000the kw-hr = 3.6 × 10 ⁹ J

Given that the dam has mechanical energy conversion rate of 90% to electrical energy

Gravitational potential energy required is 3.6 × 10 ⁹ J which is 90% of the total mechanical energy.

Hence, the total mechanical energy or potential energy is:

PE = 3.6 × 10 ⁹ × 100/90 = 4 × 10⁹J

PE = mgh

therefore, m = PE/gh

mass of water needed =  4 × 10⁹ / (9.81 × 150 ) kg

m = 2.718 × 10 ⁶ kg

volume = mass / density

V =  2.718 × 10 ⁶ / (1000)

V = 2.72 × 10³ m³

Since the area of the lake is 3.0×10⁶ m², as the water passes through the dam, the volume of the lake will decrease, and the level of water will fall by:

2.72 × 10³ m³ / (3.0×10⁶ m²) = 0.907 × 10 ⁻³ m

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