Homework Help: Liquid flow problem

1. Mar 3, 2005

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?

2. Mar 3, 2005

xanthym

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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Last edited: Mar 3, 2005
3. Mar 3, 2005

fizziksplaya

thanks alot

4. Oct 30, 2011

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

5. Oct 30, 2011

sandy.bridge

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

6. Oct 30, 2011

KendrickLamar

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

Last edited: Oct 30, 2011
7. Oct 30, 2011

sandy.bridge

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

8. Oct 30, 2011

KendrickLamar

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 doesnt end up with 14.46

9. Oct 30, 2011

sandy.bridge

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