How Does Gingerale's Pressure Change in Descending Pipes?

  • Thread starter Thread starter ashwinanand
  • Start date Start date
  • Tags Tags
    Dynamics Fluids
Click For Summary
SUMMARY

The discussion centers on calculating the pressure of gingerale in a descending pipe system using Bernoulli's equation. Given the initial pressure of 1.50x10^5 Pa, the density of gingerale at 1100 kg/m³, and the changes in velocity and height, the correct application of Bernoulli's equation reveals that the pressure at the lower level is 1.40x10^5 Pa. A critical point raised is the need to differentiate the heights in the equation, as both sides should not use the same height when calculating pressure changes due to descent.

PREREQUISITES
  • Understanding of Bernoulli's equation
  • Knowledge of fluid dynamics principles
  • Familiarity with pressure, velocity, and height relationships in fluid flow
  • Basic algebra for solving equations
NEXT STEPS
  • Review Bernoulli's equation applications in fluid mechanics
  • Study the effects of cross-sectional area changes on fluid velocity
  • Learn about hydrostatic pressure and its calculations
  • Explore real-world applications of fluid dynamics in engineering
USEFUL FOR

Students studying fluid mechanics, engineers working with fluid systems, and anyone interested in the practical applications of Bernoulli's equation in real-world scenarios.

ashwinanand
Messages
1
Reaction score
0

Homework Statement


Gingerale (ρ=1100 kg/m3) moves through a system of pipes in a beverage factory. Initially, it travels through a pipe with a cross-sectional area of 3.80 cm2 at a speed of 4.77 m/s. Gradually, the gingerale descends by 8.49 m while the pipe's cross-sectional area increases to 8.15 cm2. The speed of the gingerale at the lower level is 2.22m/s. What is the pressure at the lower level if the pressure in the upper level is known to be 1.50x10^5Pa?


Homework Equations


delta P = P2-P1

P1+(1/2)*rho*(v^2)+rho*g*y

upper level pressure=1.50x10^5Pa

The Attempt at a Solution



P1 + 0.5(1100kg/m^3)(4.77m/s^2)+(1100)(9.8m/s^2)(8.49m)=P2 + 0.5(1100)(2.22m/s^2)+(1100)(9.8m/s^2)(8.49m)
P1 + 1.04x10^5=P2 + 9.42x10^4

(deltaP)P2-P1=9.80x10^3Pa

upper level pressure- deltaP
1.50x10^5Pa-9.80x10^3Pa = 1.40x10^5Pa


I am confused of if I had used bernoulli's equation right and if the proper height was used. I can't seem to get the right pressure.
 
Physics news on Phys.org
P1+(1/2)*rho*(v^2)+rho*g*y

this is part of bernoulli's equation and it's correct.

P1 + 0.5(1100kg/m^3)(4.77m/s^2)+(1100)(9.8m/s^2)(8.49m)=P2 + 0.5(1100)(2.22m/s^2)+(1100)(9.8m/s^2)(8.49m)
P1 + 1.04x10^5=P2 + 9.42x10^4

you applied the same height of "8.49m" on both sides of the equation.
If you do this, it means that "the height of the pipe remains the same throughout. ie. its a horizontal pipe parallel and 8.49m above the ground"
however, The question states that the gingerale decended by 8.49m. (height on both side of the equation shd not be the same then, there should be a height difference of 8.49m)

Therefore, we can take a reference point: the lower level as height 0.00m and upper level as height 8.49m.

therefore, rho*g*y at the right side should be equals 0, because we take reference of height. y=0 at the lower level.

What is the pressure at the lower level if the pressure in the upper level is known to be 1.50x10^5Pa?

Its given that level at upper level is 1.50x10^5 pa, this means that P1 in the bernoulli's equation is 1.50X10^5 and they are asking you to find p2.

Plug in all the relevant numbers and recalculate.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 26 ·
Replies
26
Views
3K
Replies
2
Views
5K
Replies
0
Views
2K
Replies
5
Views
7K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
1
Views
11K
Replies
1
Views
8K
  • · Replies 5 ·
Replies
5
Views
6K