Energy niagra falls problem water

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Homework Help Overview

The discussion revolves around a problem related to the energy conversion of falling water at Niagara Falls, specifically focusing on calculating the energy obtained from one kilogram of water and determining the mass flow rate required to produce one megawatt of power.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the gravitational energy formula and its application to the problem, questioning the relationship between energy, mass, and time. There is confusion regarding the units of energy and mass flow rate, with attempts to clarify the calculations for energy produced per kilogram of water.

Discussion Status

The discussion is ongoing with participants exploring different interpretations of the problem. Some have attempted calculations and shared their results, while others express uncertainty about the correctness of their approaches. There is no explicit consensus on the final answer, and participants are encouraged to verify their calculations.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available for solving the problem. There is also a noted confusion regarding the power units and the relationship between energy and mass flow rate.

mshah3
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*problem:

In the Niagara Falls hydroelectric generating plant, the energy of falling water is converted into electricity. The height of the falls is about 45 meters. Assuming that the energy conversion is highly efficient, approximately how much energy is obtained from one kilogram of falling water? Therefore, approximately how many kilograms of water must go through the generators every second to produce a megawatt of power (106 watts)?

____kg/s

*formulas:

Gravitational energy = (mass) × g × (height).

*attempts:

I tried to find energy based on the fact one drop of water is one gram.
I ended up pluging in the values and got (45)(9.8) = 441
I found this to be incorrect.
Not sure how to relate mass to seconds to energy to height...
=/
 
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mgh = 1kg*9.8m/s^2*45m = 441 J for the first part was wrong?
 
well they ask for it in kg/s
is this the same units as joules?
 
mshah3 said:
well they ask for it in kg/s
is this the same units as joules?

that's the second part of the question... the first part is:

" how much energy is obtained from one kilogram of falling water?"
 
huh
well, there's only one answer to submit
i think they were simply restating the question to better explain

so are u saying that there are multiple steps to the problem or is it fine to say 441 J = 441 kg/s ?
 
mshah3 said:
huh
well, there's only one answer to submit
i think they were simply restating the question to better explain

so are u saying that there are multiple steps to the problem or is it fine to say 441 J = 441 kg/s ?

Oh, I think I see now. So 1 kg produces 441J.

You need to produce 106 mega watts = 106*10^6 J/s...

so in 1s you need to produce 106*10^6J ?

If 1kg produces 441J, how many kg produces 106*10^6J
 
so then:

x kg = (106E6) / (441) = 2.4036E5
 
mshah3 said:
so then:

x kg = (106E6) / (441) = 2.4036E5

exactly. 2.4036E5 kg/s is your answer.
 
i just submitted that
was incorrect
 
  • #10
mshah3 said:
i just submitted that
was incorrect

I'm sorry. I got confused with the power... for some reason I thought it was 106 megawatts...

the power is 1 megawatt = 10^6 watts = 10^6 J/s

so 10^6/441 = 2267.57

2267.57 kg/s. check yourself also to make sure this makes sense and there aren't any mistakes...
 

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