Calculating Kinetic & Potential Energy: Is it Correct?

In summary, In this problem i don't find any way to obtain de kinetic energy in KJ/Kg because when i resolve the kinetic energy formula the result its: 1/2 (1300 kg/s) (9 m/s)^2 = 5850 kg * m/s (i don't obtain m^2/s^2, so KJ/Kg its not possible)
  • #1
Krokodrile
45
3
Homework Statement
The permanent winds of a wind farm travel at 9 m/s, determine the kinetic energy of the winds (KJ/Kg) as well as their potential (w) if the mass flow (m) is 1300 Kg/s.
Relevant Equations
K.E = 1/2 mv^2 ; P.E = m*g*h
In this problem i don't find any way to obtain de kinetic energy in KJ/Kg because when i resolve the kinetic energy formula the result its:
1/2 (1300 kg/s) (9 m/s)^2 = 5850 kg * m/s (i don't obtain m^2/s^2, so KJ/Kg its not possible)
In the potential energy (w) part i obtain this:
m*g ( i don't have height) = (5850 kg * m/s) (9.8 m/s^2) = 57. 3 kg/s * m^2/s^2 = 57. 3 kg/s * KJ/Kg = 57. 3 KJ/s = 57,300 W

Its correct my answer?
 
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  • #2
Krokodrile said:
i don't find any way to obtain de kinetic energy in KJ/Kg because when i resolve the kinetic energy formula the result its:
1/2 (1300 kg/s) (9 m/s)^2
For this part of the question you are not expected to use the 1300 kg/s. Just use the 9m/s.

The second part is not asking about potential energy. It is asking for the potential to generate power. This is an everyday usage of potential, not a scientific one.
Gravity is irrelevant.

I'm unsure whether you are expected to take into account the Betz limit.
See https://energyeducation.ca/encyclopedia/Betz_limit. Has this been mentioned in your course?
 
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  • #3
haruspex said:
For this part of the question you are not expected to use the 1300 kg/s. Just use the 9m/s.

The second part is not ask about potential energy. It is asking for the potential to generate power. This is an everyday usage of potential, not a scientific one.
Gravity is irrelevant.

I'm unsure whether you are expected to take into account the Betz limit.
See https://energyeducation.ca/encyclopedia/Betz_limit. Has this been mentioned in your course?
Now i see it :), so i obtain 40,500 KJ/Kg ( 1 m^2/s^2 = 1000 KJ/Kg, true?. In the second part its possible multiply KJ/Kg * m/s^2 and obtain (w)? i tried to use 40.5 m^2/s^2 to multiply with 9.81 m/s^2 and convert to kj/kg, but i failed...
 
  • #4
Krokodrile said:
1 m^2/s^2 = 1000 KJ/Kg
No. ## 1J = 1 kg m^2/s^2## . ## 1kJ = 1000J = 1000 kg m^2/s^2##.
Divide both sides by kg.
Krokodrile said:
multiply with 9.81 m/s^2
As I wrote, gravity has nothing to do with it. The wind is moving horizontally, not falling.
The first part gives you the kJ per kg of the wind mass that passes through, and now you are told the mass per second. So the kJ per second is...?
 
  • #5
haruspex said:
No. ## 1J = 1 kg m^2/s^2## . ## 1kJ = 1000J = 1000 kg m^2/s^2##.
Divide both sides by kg.

As I wrote, gravity has nothing to do with it. The wind is moving horizontally, not falling.
The first part gives you the kJ per kg of the wind mass that passes through, and now you are told the mass per second. So the kJ per second is...?
So, first if 1KJ/Kg = 1000 m^2/s^2 when i divide both sides by kg, kinetic energy of 9 m/s is 0.04 KJ/Kg? (40.5 m^2/s^2 multiply by 1 KJ/Kg and divide by 1000 m^2/s^2)? And, if the mass per second of the mass flow is a velocity unit, i only converte the 0.04 KJ/Kg in W?
 
  • #6
Krokodrile said:
kinetic energy of 9 m/s is 0.04 KJ/Kg?
Yes.
Krokodrile said:
if the mass per second of the mass flow is a velocity unit, i only converte the 0.04 KJ/Kg in W?
It is not a question of converting kJ/kg to W. They are dimensionally different.
kJ/kg is dimensionally L2T-2, W are ML2T-3.

I gave you a very strong hint before. The question setter has complicated things by introducing kJ. Let's take it back to plain J.
A Watt is a unit of power, i.e. energy per unit time. 1W = 1J/s. You have energy per unit mass of 40.5 J/kg, and a mass per unit time of 1300 kg/s.
Given something that's energy/mass and something else that's mass/time, how can you combine them to get a quantity that's energy/time?
 
  • #7
haruspex said:
Yes.

It is not a question of converting kJ/kg to W. They are dimensionally different.
kJ/kg is dimensionally L2T-2, W are ML2T-3.

I gave you a very strong hint before. The question setter has complicated things by introducing kJ. Let's take it back to plain J.
A Watt is a unit of power, i.e. energy per unit time. 1W = 1J/s. You have energy per unit mass of 40.5 J/kg, and a mass per unit time of 1300 kg/s.
Given something that's energy/mass and something else that's mass/time, how can you combine them to get a quantity that's energy/time?
so, i only multiply and got 52.6 W (eliminating the kg and converting J/S to W)
 
  • #8
Krokodrile said:
so, i only multiply and got 52.6 W (eliminating the kg and converting J/S to W)
Nearly right, but the J v. kJ seems to have thrown you.
You can multiply 40.5 J/kg by 1300 kg/s to get .. what units? Or 0.0405 kJ/kg by 1300 kg/s to get .. what units?

By the way, the k in kJ, kg etc. should be lowercase.
 
  • #9
multiply 40.5 J/kg by 1300 kg/s obtain 52,650 J/s = 52,650 W, and 0.0405 kJ/kg by 1300 kg/s = 52.65 kj/s =52,650 J/s = 52,650 W the same result...;)?
 
  • #10
Krokodrile said:
multiply 40.5 J/kg by 1300 kg/s obtain 52,650 J/s = 52,650 W, and 0.0405 kJ/kg by 1300 kg/s = 52.65 kj/s =52,650 J/s = 52,650 W the same result...;)?
Yes, both are valid, so should produce the same result.
My point in post #8 is that in post #7 you got 52.6W instead of 52.6kW.
 
  • #11
haruspex said:
Yes, both are valid, so should produce the same result.
My point in post #8 is that in post #7 you got 52.6W instead of 52.6kW.
oooh, yes, that my mistake :/. So, my reasoning from the beginning is: "m" is not used in the formula of k.e because the mass of the air is too small and is omitted. And finally, I obtained w by multiplying j / kg and kg / s, by logic (as you mentioned) but this is due to formula / reasoning?
 
  • #12
Krokodrile said:
I obtained w by multiplying j / kg and kg / s, by logic (as you mentioned) but this is due to formula / reasoning?
The units give a good hint, but the proof is in the reasoning.
You found that for each kg of air that comes through, the farm generates 40.5J.
You are told that 1300kg/s of air blows through.
Therefore the wind farm generates 40.5J/kg * 1300 kg each second.
 
  • #13
haruspex said:
The units give a good hint, but the proof is in the reasoning.
You found that for each kg of air that comes through, the farm generates 40.5J.
You are told that 1300kg/s of air blows through.
Therefore the wind farm generates 40.5J/kg * 1300 kg each second.
Now i i understand this problem, thank you thank you so much ;) and thank for your time to read it.
 

Related to Calculating Kinetic & Potential Energy: Is it Correct?

1. What is kinetic energy?

Kinetic energy is the energy an object possesses due to its motion. It is calculated using the formula KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity.

2. How is potential energy calculated?

Potential energy is the energy an object possesses due to its position or state. It is calculated using the formula PE = m * g * h, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object.

3. Is it important to consider both kinetic and potential energy?

Yes, it is important to consider both kinetic and potential energy because they are interrelated and together make up the total energy of an object.

4. Can kinetic energy be converted into potential energy?

Yes, kinetic energy can be converted into potential energy and vice versa. This is known as the principle of conservation of energy.

5. What are some real-life examples of kinetic and potential energy?

Examples of kinetic energy include a moving car, a bouncing ball, and a swinging pendulum. Examples of potential energy include a stretched rubber band, a raised weight, and a stretched spring.

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