An earth dam is in danger of failing. A pump is needed to quickly drain the small lake behind the dam. The total energy head required to move water over the top of the dam is 2.43 m. The only pump available is an old 10-cm diameter propeller pump. The power requirement for the motor is 1,000 w and the pump-motor combination has a low efficiency of 50%. Estimate the drawdown (how many centimeters the lake goes down) in the first 2-hour period if the lake has a surface area of 5,000 m2. (Note: Assume that the energy head the pump must overcome does not change much as the lake is lowered in the first 2 hours.)

the drawdown of the lake in centimeters is 3.02 cm

Explanation:

Given data;

Power of motor = 1000 W

efficiency M = 50% = 0.5

so Effective power = 1000 × 0.5 = 500 W

time t = 2 hrs = 2 × 60 × 60 = 7200 secs

so

Energy = p × t = 500 × 7200 = 3,600,000 J

Now Energy required to pump water to height h = mgh (J)

3,600,000 J = mgh

m = 3,600,000 / gh

given that; h is 2.43 and we know that g is 9.81, we substitute

m = 3,600,000 / (9.81 × 2.43)

m = 3,600,000 / 23.8383

m = 151017.48 kg

Now we know that, Volume = Mass / Density          [density of water]

so, Volume = 151017.48  / 1000

Volume = 151.01748 kg/m³

given that surface area = 5000 m²

Drawdown = 151.01748 / 5000

Drawdown = 0.3020 m

we convert to centimeter

Drawdown = 0.3020 m × 100

Drawdown = 3.02 cm

Therefore the drawdown of the lake in centimeters is 3.02 cm

how is this a question? in order for me to answer it i have to understand the question

explanation: