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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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Determine the angular acceleration of the uniform disk if (a) the rotational inertia of the disk is ignored and (b) the inertia
lukranit [14]

Answer:

α = 7.848 rad/s^2  ... Without disk inertia

α = 6.278 rad/s^2  .... With disk inertia

Explanation:

Given:-

- The mass of the disk, M = 5 kg

- The right hanging mass, mb = 4 kg

- The left hanging mass, ma = 6 kg

- The radius of the disk, r = 0.25 m

Find:-

Determine the angular acceleration of the uniform disk without and with considering the inertia of disk

Solution:-

- Assuming the inertia of the disk is negligible. The two masses ( A & B )  are hung over the disk in a pulley system. The disk is supported by a fixed support with hinge at the center of the disk.

- We will make a Free body diagram for each end of the rope/string ties to the masses A and B.

- The tension in the left and right string is considered to be ( T ).

- Apply newton's second law of motion for mass A and mass B.

                      ma*g - T = ma*a

                      T - mb*g = mb*a

Where,

* The tangential linear acceleration ( a ) with which the system of two masses assumed to be particles move with combined constant acceleration.

- g: The gravitational acceleration constant = 9.81 m/s^2

- Sum the two equations for both masses A and B:

                      g* ( ma - mb ) = ( ma + mb )*a

                      a =  g* ( ma - mb ) / ( ma + mb )

                      a = 9.81* ( 6 - 4 ) / ( 6 + 4 ) = 9.81 * ( 2 / 10 )

                      a = 1.962 m/s^2  

- The rope/string moves with linear acceleration of ( a ) which rotates the disk counter-clockwise in the direction of massive object A.

- The linear acceleration always acts tangent to the disk at a distance radius ( r ).

- For no slip conditions, the linear acceleration can be equated to tangential acceleration ( at ). The correlation between linear-rotational kinematics is given below :

                     a = at = 1.962 m/s^2

                     at = r*α      

Where,

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                    α = at / r

                    α = 1.962 / 0.25

                    α = 7.848 rad/s^2                                

- Take moments about the pivot O of the disk. Apply rotational dynamics conditions:

             

                Sum of moments ∑M = Iα

                 ( Ta - Tb )*r = Iα

- The moment about the pivots are due to masses A and B.

 

               Ta: The force in string due to mass A

               Tb: The force in string due to mass B

                I: The moment of inertia of disk = 0.5*M*r^2

                   ( ma*a - mb*a )*r = 0.5*M*r^2*α

                   α = ( ma*a - mb*a ) / ( 0.5*M*r )

                   α = ( 6*1.962 - 4*1.962 ) / ( 0.5*5*0.25 )

                   α = ( 3.924 ) / ( 0.625 )

                   α = 6.278 rad/s^2

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