Venturi meter with 2 tubes

Linus Pauling

1.

Find , the speed of the fluid in the left end of the main pipe.
Express your answer in terms of h_1, h_2 , g , A_1, A_2.

2. Bernoulli and continuity equations

3.
p_1 = rho*gh_1
p_2 = rho*gh_2

p_2 = p_1 + 0.5rho*v_1^2 + 0.5rho*(A_1/A_2)^2 * v_1^2

I know those are correct.

Substituting for p_1 and p_2, and solving for v_1 I obtain:

2g(h_2 - h_1) / [1-(A_1/A_2)]

all that under a square root.

No idea why I am getting it wrong. Should h_2 - h_1 just be h??

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Linus Pauling

Yeah, I am totally stuck on this one. Solving for V_1 I get the following, and I don't see where I could have made a mistake:

v_1 = sqrt[2g(h_2 - h_1) / (1-(A_1/A_2))]

Blsd

:D this question had me stumped for a bit but ur equations helped a bit ... sorta >.>

kk First i got my own equations xD lol

Q=v1A1=v2A2

from this we get that v2=(v1A1)/A2

P1-P2=$$\rho$$/2* (v2-v1)

Now sub in v1 for v2

this gives us
P1-P2=$$\rho$$/2* ((v1A1)/A2-v1)

Now do a little math ... >.> well maybe not a little

but you'll get that v1=

sqrt( (g(h1-h2)) / ((A1/A2)^2-1))

o.o' wish i knew how to code that up so it would look pretty

but the only difference between this and your answer is in the demoninator

first off the A1/A2 part is supposed to be squared

and ur signs are backwards :D hope this helps ^^

o.o ... now to go back and make this reads properly

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