# Homework Help: Venturi tube - Pressure

1. Jul 28, 2011

### Workingitout

1. The problem statement, all variables and given/known data
I have a question involving a venturi tube - Air flows through the channel of a venturi of constant width 0.06m. Compressibility and viscous effects are negligible.

a) Determine the flowrate when water is drawn up 0.10m in a small tube attached to the static pressure tap at the throat where the channel height is 0.02m

b) Determine the channel height, h2, at section (2) where, for the same flowrate as in part (a), the water is drawn up 0.05m

c) Determine the pressure needed at section (1) to produce this flow

(Other information shown in attched picture)

2. Relevant equations
v = (sqrt) (2(P1 - P2)/$\rho$)

The pressure change (P1 - P2) is usually calculated using
P1 - P2 = h x Pw x g
Where h is the height difference between the two points (No picture attached, but if you can imagine the two pressure points being connected using a 'U' shaped pipe with liquid at the bottom), Pw is the density of water and g is the gravity constant

3. The attempt at a solution
I am having a considerable amount of diffuculty due to the fact that I haven't encountered this type of problem before. I have seen similar ones where the 2 pressure points are connected, but not in this way - with an independant "straws" setup.

I intend on calculating 'v' to use it in Q=vA - Am I doing this right? As in, do I do this problem using these formulas? If so, do I use 'h' differently in this example?

Any help would be appreciated, and a link to a site that can guide me through would also be welcome.

#### Attached Files:

• ###### Venturi_Tube.jpg
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2. Jul 29, 2011

### astro_enthu

Isn't this case simpler than having U-shaped tube?

(Here starting from left of the venturi I have considered junction above the first straw as point 2 and one above straw right to previous as point 3.)

Q = v2*a2 = v3*a3
p2-p3 = density* (v2^2 - v3^2)

Solve these equations which gives solution to Q and then v2 and v3 and then everything falls in the place.

I hope this helps.