Total Mechanical Energy: Mass or No Mass?

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SUMMARY

The discussion clarifies that total mechanical energy can be expressed as energy per unit mass, which simplifies the calculations by eliminating the mass variable. The equation for total mechanical energy is redefined as gh + 0.5v² instead of mgh + 0.5mv². This approach is particularly useful for analyzing systems like hydro plants, where energy per unit mass allows for easier calculations of energy flow rates. The key takeaway is that the mass is effectively normalized to one, allowing for a more straightforward analysis of mechanical energy.

PREREQUISITES
  • Understanding of mechanical energy concepts
  • Familiarity with energy equations: potential energy and kinetic energy
  • Knowledge of flow rate and mass flow rate calculations
  • Basic principles of hydroelectric systems
NEXT STEPS
  • Research the concept of energy per unit mass in fluid dynamics
  • Learn about the efficiency of hydroelectric power plants
  • Explore the relationship between flow rate and energy generation
  • Study the derivation of potential and kinetic energy equations
USEFUL FOR

Students in physics or engineering, particularly those studying fluid dynamics and energy systems, will benefit from this discussion. It is also valuable for professionals involved in the design and analysis of hydroelectric power systems.

TyErd
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Homework Statement



In the diagram I have attached the initial question is to find the total mechanical energy. however in the solutions which I have placed under the diagram, they don't use the mass (m). So total mechanical energy i thought was = mgh + 0.5mv^{2} but they use gh + 0.5v^{2} . no mass. why

i think its something to do with the question asking to be in per unit mass but i don't see why.

Homework Equations


The Attempt at a Solution

 

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They ask you for the energy per unit mass that is present, and so they divide both sides of the equation by the mass. It's nothing more complicated than that. It's a more useful quantity to consider in this situation than the amount of energy present, which depends upon how big a "parcel" of water you consider.
 
Think of it this way: every kilogram of water that passes by carries with it 0.887 kJ of mechanical energy. You have the volume flow rate (which is the volume of water that flows by in 1 second). You can use that to determine the mass flow rate (the rate at which water flows by in kg/s). Combine that with your energy per unit mass, and suddenly you have the energy per second being generated (EDIT: or at least the energy per second that could be generated if all of that mechanical energy in the water flow could be transferred to some other system with 100% efficiency -- in a real hydro plant the efficiency is sure to be lower).
 
in essence,

per unit mass = mass = 1

so mgh = 1gh = gh

same with KE
 
thankyou guys. helps a lot
 

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