Viscous work and dissipation of energy

In summary, the conversation discusses the concept of work being done on a fluid system by an external source, such as an upper plate moving with speed v in the +z direction. It is determined that in this scenario, both the fluid and surroundings are doing work on each other. The signs convention is taken into account to show that the internal energy of the system increases when work is done on it by the surroundings, and decreases when the system does work on the surroundings. It is also noted that when both the system and surroundings are doing work on each other, there is no net work inflow into the system and the internal energy remains constant.
  • #1
Est120
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TL;DR Summary
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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  • #2
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.
 
  • #4
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.
 
  • #6
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 enviroment→ system and work system→enviroment)
 
  • #7
Est120 said:
so that's why the work doesn't cancel? the signs convention already takes into account the dual work issue (work enviroment→ system and work system→enviroment)
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?
 
  • #8
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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  • #9
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.
 
  • #10
i got it now, thank's for all the advice
 
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1. What is the difference between viscous work and dissipation of energy?

Viscous work refers to the energy required to overcome the resistance of a fluid to flow, while dissipation of energy refers to the conversion of mechanical energy into heat due to the internal friction of a fluid.

2. How does viscosity affect the amount of work and energy dissipation in a fluid?

The higher the viscosity of a fluid, the more energy is required to overcome its resistance to flow, resulting in more viscous work. Additionally, higher viscosity also leads to more internal friction and therefore more dissipation of energy into heat.

3. Can you give an example of a real-life application of viscous work and energy dissipation?

One example is the use of hydraulic fluids in machines and engines. The viscous work of the fluid helps to transfer energy and power, while the dissipation of energy through friction helps to dissipate excess heat and prevent overheating of the system.

4. How is the concept of viscous work and energy dissipation related to fluid dynamics?

Viscous work and energy dissipation are important principles in the study of fluid dynamics. The resistance to flow due to viscosity plays a crucial role in determining the behavior of fluids in motion, and the dissipation of energy through internal friction affects the overall energy balance of a fluid system.

5. Is it possible to reduce the amount of viscous work and energy dissipation in a fluid?

While it is not possible to completely eliminate viscous work and energy dissipation in a fluid, it is possible to minimize them through various methods such as using more efficient fluid flow designs, reducing the viscosity of the fluid, and using lubricants to reduce friction. These techniques can help improve the overall efficiency of fluid systems.

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