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OlgaM077 [116]
3 years ago
14

An inventor claims that he wants to build a dam to produce hydroelectric power. He correctly realizes that civilization uses a l

ot more electricity during the day than at night, so he thinks he has stumbled upon a great untapped energy supply. His plan is to install pumps at the bottom of the dam so that he can pump some of the water that flows out from the generators back up into the reservoir using the excess electricity generated at night. He reasons that if he did that, the water would just flow right back down through the generators the next day producing power for free. What is wrong with his plan?
Engineering
1 answer:
Mekhanik [1.2K]3 years ago
6 0

Answer:

The problem is that the pumps would consume more energy than the generators would produce.

Explanation:

Water has a potential energy associated with the height it is at. The higher it is, the higher the potential energy. When water flows down into the turbines that energy is converted to kinetic energy and then into electricity.

A pump uses electricity to add energy to the water to send it to a higher potential energy state.

Ideally no net energy woul be hgenerate or lost, because the generators would release the potential energy and pumps would store it again in the water. However the systems are not ideal, everything has an efficiency and losses. The losses would accumulate and the generator would be generating less energy than the pumps consume, so that system wastes energy.

What should be done is closing the floodgates to keep the water up in the dam at night producing only the power that is needed and releasing more water during the day.

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Two dogbone specimens of identical geometry but made of two different materials: steel and aluminum are tested under tension at
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Answer:

\dot L_{steel} = 3.448\times 10^{-4}\,\frac{in}{min}

Explanation:

The Young's module is:

E = \frac{\sigma}{\frac{\Delta L}{L_{o}} }

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Let assume that both specimens have the same geometry and load rate. Then:

E_{aluminium} \cdot \dot L_{aluminium} = E_{steel} \cdot \dot L_{steel}

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\dot L_{steel} = \frac{E_{aluminium}}{E_{steel}}\cdot \dot L_{aluminium}

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A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. If the gas
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Answer:

gauge pressure is 133 kPa

Explanation:

given data

initial temperature T1 = 27°C = 300 K

gauge pressure = 300 kPa = 300 × 10³ Pa

atmospheric pressure = 1 atm

final temperature T2 = 77°C = 350 K

to find out

final pressure

solution

we know that gauge pressure is = absolute pressure - atmospheric pressure so

P (gauge ) = 300 × 10³ Pa - 1 × 10^{5} Pa

P (gauge ) = 2 × 10^{5} Pa

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{P2} = \frac{2*10^5*350}{300}

P2 = 2.33 × 10^{5} Pa

so gauge pressure = absolute pressure - atmospheric pressure

gauge pressure = 2.33 × 10^{5}  - 1.0 × 10^{5}

gauge pressure = 1.33 × 10^{5} Pa

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