Solving Bernoulli Principle Homework: Water Flows in Pipe

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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?