Water moves through a constricted pipe in steady, ideal flow. At the lower point
ID: 2044040 • Letter: W
Question
Water moves through a constricted pipe in steady, ideal flow. At the lower point shown in the figure below, the pressure is 1.85 105 Pa and the pipe radius is 2.70 cm. At the higher point located at y = 2.50 m, the pressure is 1.29 105 Pa and the pipe radius is 1.30 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.
________________m3/s
http://www.webassign.net/sercp9/9-p-052.gif
Explanation / Answer
(a)
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) - ) ]
= [ 2((1.68×10Pa - 1.28×10Pa)/1000kg/m³ + 9.81(-2.5m)) / ((3/1.3) - ) ]
= 1.064m/s
(b)
v = (R/R)²v
= (3/1.3)² 1.064m/s
= 5.666m/s
(c)
dV/dt = Av = R²v
= (0.03m)² 1.064m/s
= 3.01×10³m³/s
= 3.01L/s
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