HW#12b
Note: numbers used in solution steps are different from your WebAssign values.
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Note: numbers used in solution steps are different from your WebAssign values.
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First, The thinner the pipe the higher the speed. A
1
v
1
=A
2
v
2 ,
due to the continuity, the total flow rate is the same.
When pipe becomes wide at a location, flow speed reduces. When pipe becomes narrower flow speed is high.
Pipe area is proportional to diameter square. A=3.14*D
2
/4, here D is diameter.
A
1
v
1
=A
2
v
2 ;
So,
A
1
v
1
/A
2
= v
2
, So,
D
1
2
v
1
/
D
2
2
= v
2
So, at narrower location: v
2
= (0.02
2
/0.018
2
)*0.85 =1.05 m/s
The change of the speed is due to the change of pipe diameter.
Once we know the speed and height at the new location we can use the Bernoulli’s equation
P
1
+1/2
ρ
v
1
2
+
ρ
g y
1
= P
2
+1/2
ρ
v
2
2
+
ρ
g y
2
To find out the pressure P
2
at the new location.
165*10
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 Spring '09
 JamesM.Lockhart
 Physics, Frequency, Hertz, International System of Units

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