# physics

Water moves through a constricted pipe in steady, ideal ﬂow. At the lower point shown in ﬁgure below, the pressure is P1 = 1.75X104 Pa and the pipe diameter is 6.00 cm. At another point y = 0.250 m higher, the pressure is P2 = 1.20X104 Pa and the pipe diameter is 3.00 cm. Find the speed of ﬂow (A) in the lower section and (B) in the upper section. (C) Find the volume ﬂow rate through the pipe.

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1. p1 + rho g h1 + (1/2) rho v1^2 = p1 + rho g h1 + (1/2) rho v1^2
and v2 A2 = v1 A1 = Q , because Q, vol flow, is constant

density = rho = 10^3 kg/m^3
g = 9.8

p1 = 1.75*10^4 N/m^2
r1 = 0.030 m so A1 = pi (.03)^2 = 2.83*10^-3 m^2
h1 = 0, select as base height
p2 = 1.20*10^4
r2 = 0.015 m so A2 = pi (0.015)^2 = 0.707*10^-3 m^2
h2 = 0.250 m
------------------------------------
v2 = v1 (A1/A2) = v1 (2.83/.707) = 4 v1
so
1.75*10^4 + 0 + 500 v1^2 = 1.2*10^4 + 10^3 * 9.8*0.25 + 500*16v1^2
solve for v1, then go back for the rest
v2 = 4 v1
Q = A1 v1 = A2 v2

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Damon

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