# Fluids Homework Basic AP Problem

1. The inside diameters of the larger portions of the horizontal pipe are 2.90 cm. Water flows to the right at a rate of 2.00 10-4 m3/s. Determine the inside diameter of the constriction.

2. P=P0 + pgh
a1v1=a2v2
P + pgh + 1/2pv^2 = constant

3. Since the volumetric flow rate is constant within the big pipe, I used a1v1 = a2v2 in order to solve for the second area of the narrowing pipe. so i have A2= a1v1/v2. Now I don't know exactly where to go from there... I know i could use the formula P+pgh + 1/2pv^2 in order to find the differences in height in the two protruding tubes and solve for the area with that, but I can't seem to link it together, or figure out if I am making any sense...

#### Attachments

• p9_47.gif
2.4 KB · Views: 642

rl.bhat
Homework Helper
P + pgh + 1/2pv^2 = constant

Using the above formula you can write

ρg(h1 - h2) = 1/2*ρ*(v1^2 - v2^2)

Using a1*v1 = a2*v2, substitute v2 = v1*a1/a2 in the above equation and find the expression for v1. Then write v1 = Q/a1 and solve for a2.

Okay that makes alot more sense!

so now i have:

pg(h1+h2)= 1/2p(v1^2 (a1v1/a2)^2)

solved for v1:

V1= (2(h1+h2))/(1-(a1^2/a2^2))^(1/2)

and I plug in v1 into V1A1= Q and get

(2(h1+h2))/(1-(a1^2/a2^2))^(1/2)*(a1)=Q

I don't know how to solve for a2 now.... haha

But the one thing that is tricking me is V1=Q/a1 .... what is Q? Is that the volumetric flow rate that they gave me?

Thank you for your help !