Viscous work and dissipation of energy

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Discussion Overview

The discussion revolves around the concept of work and energy dissipation in a viscous fluid system, particularly focusing on the interactions between a moving upper plate and a stationary lower plate. Participants explore the implications of work done by the fluid and the surrounding environment, as well as the resulting changes in internal energy.

Discussion Character

  • Exploratory
  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant questions whether the upper plate is inserting work on the fluid system, the system is exerting work on the plate, or both, and why these works do not cancel each other out.
  • Another participant suggests that the works cancel, prompting further inquiry about the analogy with a block dragged over a surface.
  • A participant proposes that if the fluid molecules would move without the upper plate's movement, the fluid would be receiving energy from an external source, while if they do not move independently, the upper plate would be receiving energy transferred into the fluid.
  • It is stated that the fluid moves due to the interaction with the upper plate, supporting the idea that the upper plate is the source of energy input into the fluid.
  • One participant explains that the surroundings are doing work on the fluid, leading to an increase in internal energy, while the fluid does negative work on the surroundings, which they argue does not negate the energy increase.
  • Another participant reiterates the same point about the work done by the surroundings and the fluid, emphasizing that the signs in the work-energy equations account for the dual work issue.
  • One participant expresses confusion about how internal energy can increase if the system is both receiving and delivering work, suggesting that this would imply no net work inflow and thus a constant internal energy.
  • A later reply clarifies that when considering energy balance, only one of the work contributions should be included, similar to how forces are treated in a force balance.

Areas of Agreement / Disagreement

Participants express differing views on the nature of work and energy transfer in the system, with some asserting that the works do not cancel due to the conventions of energy accounting, while others remain uncertain about the implications of simultaneous work being done on and by the system. The discussion does not reach a consensus.

Contextual Notes

Participants highlight the complexity of energy transfer in systems involving viscous fluids and the importance of sign conventions in work-energy relationships. There are unresolved aspects regarding the assumptions made about the fluid's movement and energy sources.

Est120
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TL;DR
viscous dissipation of mechanical energy into internal energy
imagine two flat plates, the upper one moving with speed v in the +z direction, the lower one is stationary
is the upper plate inserting work (per unit time) on the fluid system or the system exerting work on the plate?
or both? why they don't cancel each other e.g if the fluid is receiving work from the upper plate then its internal energy is increasing, but if the fluid exerts work too then the works cancel (?)
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Est120 said:
then the works cancel
In what sense ? Do you have the same problem with a block that is dragged over a surface ?
 
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Would the molecules of the viscous fluid move without the movement of the upper plate?
If yes, the fluid would be receiving energy from an external source (pump or gravity), which would be transferred onto the upper plate.
If not, the upper plate would be receiving energy from an external source (force along z-axis) which would be transferred into the fluid.
 
the fluid is moving because of not split condition with the upper plate so its the second case
 
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in this case, the surroundings are doing work on the fluid. The force of the surrounding and its displacement are both positive. The fluid is doing negative work on the surroundings (the force of the fluid and its displacement are in opposite directions). So, if you write ##\Delta U=Q-W##, where W is the work the fluid does on the surroundings, or ##\Delta U=Q+W##, where W is the work done by the surroundings of the fluid, the result is exactly the same; the internal energy increases.
 
Chestermiller said:
in this case, the surroundings are doing work on the fluid. The force of the surrounding and its displacement are both positive. The fluid is doing negative work on the surroundings (the force of the fluid and its displacement are in opposite directions). So, if you write ##\Delta U=Q-W##, where W is the work the fluid does on the surroundings, or ##\Delta U=Q+W##, where W is the work done by the surroundings of the fluid, the result is exactly the same; the internal energy increases.
so that's why the work doesn't cancel? the signs convention already takes into account the dual work issue (work environment→ system and work system→environment)
 
Est120 said:
so that's why the work doesn't cancel? the signs convention already takes into account the dual work issue (work environment→ system and work system→environment)
It doesn't cancel for the same reason that, if A is exerting a force on B, and B is exerting an equal and opposite reaction force on A, if you do a force balance on A, you only include the force that B exerts on A, not the force that A exerts on B. Didn't they teach you this in freshman Physics?
 
i know that... but it seems no body understands my doubt
if the environment is exerting work on the system, then the system internal energy ↑ increases
but if the system is exerting work on the environment then, energy must come out of the system (in order to deliever work) and that energy decreases internal energy ↓ so internal energy decreases
so if the system is receiving work and also delivering work no net work inflow comes into the system so internal energy stays constant

i know that internal energy must increase but i don't know what i am thinking wrong about work
 
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Est120 said:
i know that... but it seems no body understands my doubt
if the environment is exerting work on the system, then the system internal energy ↑ increases
but if the system is exerting work on the environment then, energy must come out of the system (in order to deliever work) and that energy decreases internal energy ↓ so internal energy decreases
so if the system is receiving work and also delivering work no net work inflow comes into the system so internal energy stays constant

i know that internal energy must increase but i don't know what i am thinking wrong about work
Apparently, you don't understand what I said. If the system does positive work on the surroundings, then the surroundings do a negative amount of work on the system, so the two are equivalent. But, if you are doing an energy balance on the system, then you only include one of these, just as, when you do a force balance, you only include one force of the reaction pair.
 
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i got it now, thank's for all the advice
 
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