Solving Bernoulli Principle Homework: Water Flows in Pipe

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SUMMARY

The discussion centers on solving a Bernoulli Principle homework problem involving water flow in a pipe. The water flows upward at a rate of 96 L/min, with a pressure of 80 kPa at the lower end and height differences of h1 = 10 m and h2 = 13 m. The continuity equation and Bernoulli's equation are applied to find the velocities at both ends of the pipe, but the lack of diameter information complicates the calculations. The volume flow rate is calculated as 1.6 x 10^-3 m³/s, and further clarification on the pipe's diameter and orientation is necessary for accurate velocity determination.

PREREQUISITES
  • Understanding of Bernoulli's equation and its components
  • Familiarity with the continuity equation in fluid dynamics
  • Basic knowledge of fluid mechanics concepts such as pressure, velocity, and height
  • Ability to perform unit conversions, particularly for flow rates
NEXT STEPS
  • Learn how to apply Bernoulli's equation in various fluid flow scenarios
  • Study the implications of pipe diameter changes on fluid velocity using the continuity equation
  • Explore the effects of gravitational potential energy on fluid flow in vertical pipes
  • Investigate how to determine flow rates and velocities in tapered versus constant diameter pipes
USEFUL FOR

Students studying fluid dynamics, engineers working with hydraulic systems, and anyone needing to solve practical problems involving the Bernoulli Principle and fluid flow in pipes.

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Homework Statement



Water Flows upward throw the pipe shown in the diagram at 96 L/Min. If the pressure at the lower end is 80kPa, find the velocity of the water is at both ends and the pressure at the upper end. Assume that the density of water remains constant throughout the tube and that h1= 10 m and h2 = 13m

Homework Equations


P1+ 1/2 ρv^2+gy1=P2+1/2ρv2^2+gy2

Continuity Equ:
A1V1=A2V2
(Tried to use this equation to hep me find the velocity but given their is no diameter or radius given to find the are it was a waste of time)

The Attempt at a Solution


volume flow rate up the pipe:
96L/min (1.0X10^3 cm^3/ 1.00L)(1.00m/100cm)^3(1.00min/60sec) = 1.6x10^3 m^3/s


Attempt to tried to use the Continuity Equation as substitution for one of the velocities:


A1V1=A2V2

V2(A2/A1)= V1

Substituting V1 in the Bernoulli Equation:
P1+1/2ρ(V2(A2/A1))^2+ρgy1=P2+1/2ρv2^2+gy2

2g(y^2-y1)=v2[1-(A2/A1)]

sqrt(2gh)/sqrt(1-(A2/A1)^2) =v2



Any help or guidance will be appreciated. Thank you.
 
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I can't see your diagram so if you could attached it maybe I can help.

I refer to the bernoulli's theorem as energy per unit weight:
(P_1/(rho*g))+(V_1^2/2g)+h1 = (P_2/(rho*g))+(V_2^2/2g)+h2
Q = 1.6x10^-3 m^3/s

By inspection of the pipeline, is it tapered or is the diameter constant? If it is constant what would be the relationship between V_1 and V_2?

A change of diameter should however should be stated if there is one... Is it a horizontal, angled or vertical pipe?
 

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