Bernoulli Principle: Homework Solution

[srollin6]

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.

 

[haruspex]

... find the velocity of the water is at both ends ...

Something missing in the problem statement?

P₁+ 1/2 ρv₁²+gy₁=P₂+1/2ρv₂²+gy₂

Should be ρgy₁, ρgy₂, yes?

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)

It doesn't mention any change in area, so presumably A₁=A₂.

Substituting V1 in the Bernoulli Equation:
P₁+1/2ρ(V₂(A₂/A₁))²+ρgy₁=P₂+1/2ρv₂²+gy₂

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

How do you get that last equation from the preceding one? What happened to P₁ and P₂?