Liquid Flow Problem: Bernoulli's Principle & Volume Flow Rate

[fizziksplaya]

"A liquid is flowing through a horizontal pipe whose radius is 0.0200 m. The pipe bends straight upward through a height of 10.0m and joins another horizontal pipe whose radius is 0.0400 m. What volume flow rate will keep the pressures in the two horizontal pipes the same?"

Alright, I was thinking some combination of bernoulli's principle and the volume flow rate law because you've got varying height and varying area. How would I go about doing this? Am I on the right track?

 

[xanthym]

"A liquid is flowing through a horizontal pipe whose radius is 0.0200 m. The pipe bends straight upward through a height of 10.0m and joins another horizontal pipe whose radius is 0.0400 m. What volume flow rate will keep the pressures in the two horizontal pipes the same?"

Alright, I was thinking some combination of bernoulli's principle and the volume flow rate law because you've got varying height and varying area. How would I go about doing this? Am I on the right track?

From Bernoulli's Law for 2 points within a closed steady-state ideal fluid flow system:

:(1): \ \ \ \ P_{1} \ + \ (1/2)\rho v_{1}^{2} + \ \rho gh_{1} = P_{2} \ + \ (1/2)\rho v_{2}^{2} \ + \ \rho gh_{2}

Problem statement indicates {P₁ = P₂}, {h₁ = 0}, and {h₂ = 10 m}, so Eq #1 simplifies to:

:(2): \ \ \ \ (1/2)v_{1}^{2} = (1/2)v_{2}^{2} + (10)g

:(3): \ \ \ \ v_{1}^{2} \ - \ v_{2}^{2} \ = \ 196

From Mass Conservation, {A = πr²}, {r₁ = 0.02 m}, and {r₂ = 0.04 m}:

:(4): \ \ \ \ \rho A_{1} v_{1} = \rho A_{2} v_{2} = \rho (VolumeFlowRate) \ \ \Rightarrow \ \ v_{2} = v_{1} (\frac {r_{1}} {r_{2}})^{2} = v_{1}/4

Placing Eq #4 into Eq #3, solving for v₁, and determining (Volume Flow Rate):

:(5): \ \ \ \ v_{1}^{2} \ - \ (v_{1}/4)^{2} \ = \ 196

:(6): \ \ \ \ v_{1} = (14.46 \ \ m/sec) \ \ \Rightarrow \ \ \color{red} (VolumeFlowRate) = \pi r_{1}^{2} v_{1} = (0.0182 \ \ m^{3}/sec)



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[fizziksplaya]

thanks a lot

 

[KendrickLamar]

^ wait is that right? how did u get 14.46 lol?

shouldn't it be v1 - (1/4 * v1) = 14 and then 14/.75? or am i doing it wrong

 

[sandy.bridge]

@ KendrickLamar
It appears that you tried taking the square root of both sides.
Note that
\sqrt{v_2 ^2-v_1 ^2}\neq v_2 -v_1

 

[KendrickLamar]

@ KendrickLamar
It appears that you tried taking the square root of both sides.
Note that
\sqrt{v_2 ^2-v_1 ^2}\neq v_2 -v_1

wait I am brain dead lol u got to factor it or what, what's the easiest way to get from step 5 to step 6? why can't i figure this out lol!? i tried 1v1^2 - .25v1^2 = .75v1^2 and set that = 196 but that doesn't work either

 

[sandy.bridge]

If you have
v_1 ^2-\frac{1}{4}v_1 ^2
You can take out the common factor and subtract
v_1 ^2(1-\frac{1}{4})=\frac{3}{4}v_1 ^2

 

[KendrickLamar]

If you have
v_1 ^2-\frac{1}{4}v_1 ^2
You can take out the common factor and subtract
v_1 ^2(1-\frac{1}{4})=\frac{3}{4}v_1 ^2

thats what i was doing tho...

3/4 = .75 so if u do 196/.75 then square root both sides u still get like 16 something or even if u square root both sides first then multiply that fraction across its still doesn't end up with 14.46

 

[sandy.bridge]

Yeah, I just calculated it to be 16.17m/s.