Fluid Mechanics-Bernoulli and Energy Equations

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SUMMARY

The discussion focuses on applying Bernoulli's Equation to determine the force required to maintain the motion of a piston in a sealed water-filled cylinder. The piston moves at a constant speed of 4 mm/s while the pressure in the right compartment is held constant at 50 kPa gage. The calculations reveal that the velocity of the fluid through the holes is significantly higher than the velocity of the piston, resulting in a calculated force of 40 N needed to maintain this motion. Key equations utilized include Bernoulli's Equation and the flow rate relationship between the inlet and outlet of the system.

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  • Understanding of Bernoulli's Equation and its applications
  • Knowledge of fluid dynamics principles
  • Familiarity with flow rate calculations and area relationships
  • Basic skills in algebra and unit conversions
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  • Learn about flow rate calculations in various geometries
  • Explore the implications of pressure differences in fluid systems
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Joseph95
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Homework Statement


media-ea1-ea184f27-2552-4f66-9ba1-210cc540694a-phpWLbN6e.png


A well-fitting piston with 4 small holes in a sealed water-filled cylinder,shown in Fig is pushed to the right at a constant speed of 4 mm /s while the pressure in the right compartment remains constant at 50 kPa gage. Disregarding the frictional effects,determine the force F that needs to be applied to the piston to maintain this motion.

Homework Equations


Bernoulli Equation: P1/ρ + V1^2/2 + gz1 = P2/ρ + V2^2/2 + gz2

Flow Rate : Q1 = Q2 V1.A1=V2.A2

The Attempt at a Solution



I set two points.The first point (1) is back of the piston and the other point (2) front of the piston.So applied Bernoulli Equation:

P1/ρ + V1^2/2 +gz1 = P2/ρ + V2^2 / 2 + gz2 No elevation difference z1=z2

P1/ρ + V1^2/2 = P2/ρ + V2^2/2 V1=4 mm/s =0.004 m/s
V2= V1.A1/A2 = 0.004 x 0.0113 / 0.0000031 m/s
V2=3.64 m/s but there are 4 small holes so V2/4= 14.58/ 4 m/s=3.64 m/s

P1= P2 + ρV2^2/2 (V1 can be neglected)

P1= 50 kPa + 1000 (3.64)^2 /2 Pa = 56624.8 Pa

F/πDr^2/4=56624.8 Pa

F=40 N
 
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I'm not following this. (1) is back of piston and (2) is front of piston. So why is V1 so small and v2 so big ?
Joseph95 said:
F/πDr^2/4=56624.8 Pa

F=40 N
What does it say here ? Could you explain the symbols and place appropriate brackets ?
 

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