dabezt
Answered

At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Ask your questions and receive detailed answers from professionals with extensive experience in various fields. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

Water is stored in a reservoir at an average vertical height of 350 m above the turbines of a hydroelectric power station. During a 7.0 hour period, 1.8 × 10^6 m3 of water flows down from the reservoir to the turbines. (a) The density of water is 1000 kg / m3. For this 7.0 hour period, calculate (i) the mass of water that flows from the reservoir to the turbines, find the mass

the gravitational potential energy transformed as the water flows to the turbines, find the energy

the maximum possible average output power.

find the power ​

Sagot :

Answer:

(a) (i) 1) The mass of the water is 1.8 × 10⁹ kg

2) The gravitational potential energy is 6.174 × 10¹² J

4) The energy given to the turbine by the falling water is 6.174 × 10¹² J

5) The maximum possible average output power is 245 MW

Explanation:

In the question we are required to convert the total initial potential energy, P.E., of the water to power

The given parameters are;

The height of the reservoir above the turbine, h = 350 m

The time duration the volume of water takes to flow, t= 7.0 hours

The volume of water that flows down from the reservoir, V = 1.8 × 10⁶ m³

The density of the water, ρ = 1000 kg/m³

(a) (i) Density = Mass/Volume

∴ Mass = Density × Volume

The mass of the water that flows down from the reservoir = m = V × ρ

By substituting the known values, we have

The mass of the water, m = 1.8 × 10⁶ m³ × 1000 kg/m³ = 1.8 × 10⁹ kg

2) The gravitational potential energy, P.E. = m × g × h

where;

m = The mass of the water = 1.8 × 10⁹ kg

g = The acceleration due to gravity = 9.8 m/s²

h = The height from which the water falls = 350 m

Which gives;

The gravitational potential energy,

P.E. = 1.8 × 10⁹ kg × 9.8 m/s² × 350 m = 6.174 × 10¹² J

The gravitational potential energy = 6.174 × 10¹² J

4) The energy given to the turbine by the falling water, E = The gravitational potential energy = 6.174 × 10¹² J

5) The maximum possible average output power, P = (Energy given to the turbine, E)/(Time taken to produce the energy = The time duration the water takes to flow, t)

The time duration the water takes to flow, t = 7.0 hours = 7 × 60 × 60 seconds =  25,200 seconds

∴ P = 6.174 × 10¹² J/(25,200 s) = 245,000,000 watts = 245 MW

Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.