anonymous
  • anonymous
what is the particular reason for that we can write an equation for change of mechanical energy....? and a different equation for change of total energy?
Physics
chestercat
  • chestercat
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anonymous
  • anonymous
mechanical energy\[\frac{\partial}{\partial t} (\frac{1}{2}\rho\text{v}^2)=...\]total energy\[\frac{\partial}{\partial t} (\text{U}+\frac{1}{2}\rho\text{v}^2)=...\]
anonymous
  • anonymous
i think my question is not clear...
anonymous
  • anonymous
It is clear enough. The best reason I can think of is scope. If we are designing the aiming system for a tank, the increase in temperature of the projectile flying through the air wouldn't have much of an affect on how the projectile travels. On the other hand, if we are designing the frame of a car and we want to make it safe during an impact, temperature may be an important factor. Consider that when metal is deformed its internal energy increases. We need to make sure that this temperature increase doesn't adversely affect the mechanical properties of the steel.

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anonymous
  • anonymous
it makes sense ... :)
anonymous
  • anonymous
are there some such examples about fluids u can think of?
anonymous
  • anonymous
Also, order of magnitude plays in some times. In Newtonian Mechanics, the above reason holds. Physics 101 just doesn't care about internal energies. In Thermodynamics, we don't care about mechanical energy when dealing with fluids because the order of magnitude of the mechanical terms are less. Let's look at the units on a complete energy equation: \[E = U + KE + PE\]\[kJ = kJ + {kg m^2 \over 2 s^2} + {kg m^2 \over s^2}\] The kinetic and potential energies are in Joules. Consider that 1kg of water at STP has an internal energy of ~100 kJ. We would need to be moving that water at supersonic speeds for the mechanical energy to have much impact.
anonymous
  • anonymous
In response to your question: We consider both internal and mechanical energy of fluids a lot when dealing with centrifugal pumps and turbines. Take a turbo charger, the turbine side gathers energy from both the fluid flow (change in momentum across the turbine fins) and the fluids internal energy (we can expand the gas across the fins and create beneficial pressure gradients).
anonymous
  • anonymous
so helpful.thank u very much indeed :)
anonymous
  • anonymous
One more note: In centrifugal equipment, internal energy is actually enthalpy because we have an open system. Enthalpy (H) is\[H = U + pV\]This is an entirely different discussion.
anonymous
  • anonymous
You're very welcome.

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