Water moves through a constricted pipe in steady, ideal flow. At the lower point

Question

Water moves through a constricted pipe in steady, ideal flow. At the lower point shown in the figure below, the pressure is P1 = 1.95 104 Pa, and the pipe diameter is 8.0 cm. At another point y = 0.30 m higher, the pressure is P2 = 1.15 104 Pa and the pipe diameter is 4.00 cm.

(a) Find the speed of flow in the lower section.

_______________m/s

(b) Find the speed of flow in the upper section.

______________m/s

(c) Find the volume flow rate through the pipe.

______________m/s

I HAVE BEN TRYING TO FIGURE THIS PROBLEM OUT BUT I DO NOT UNDERSTAND HOW TO GET A1 AND A2

I USED V2 = RADICAL -pg(h1-h2)-(P1-P2)/ 1/2 P [A2^2-A1^2/A1^2]

WHIUCHEVER STYLE YOU DO, PLEASE LET ME KNOW HOW YOU GET A1 & A2  

Explanation / Answer

A1 and A2 are simply calculated as R² and R2² respectively

I would give a overall view of solving the problem....

Use Bernoulli's equation
p/ + gy + (1/2)v² = constant
=>
p/ + gy + (1/2)v² = p/ + gy + (1/2)v²

Conservation of mass of requires that mass flow through every section of the pipe is the same, that means
dm/dt = dV/dt = Av = R²v = constant
<=>
R²v = R²v
So the th velocities are related as:
R²v = R²v
<=>
v = (R/R)²v

Substitute v in the Bernoulli equation and solve for v
p/ + gy + (1/2)v² = p/ + gy + (1/2)[(R/R)²v]²
<=>
(p - p)/ + g(y - y) = (1/2)(R/R) - )v²
=>
v = [ 2((p - p)/ + g(y - y)) / ((R/R) - ) ]

PLS DO MATH
As you wanted Method of solving only.....

(b)
v = (R/R)²v


(c)
dV/dt = Av = R²v

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