P13.32. A 1.0cm. diameter pipe widens to 2.0cm, then narrows to 0.50cm diameter. The speed in the first section is 4.0
m/s, what is the speed in the other two sections?
The equation of continuity applies as long as the pipe is full. Note that A1, A2, and A3 and v1, v2, and v3 are the crosssectional areas and the speeds in the first, second, and third segments of the pipe.
Solve: (a) The equation of continuity is
A1 v1
A2 v2
r 21 v1
A3 v3
2
(0.005 m) ( 4.0 m/s)
v2
0.005 m
0.01 m
r 22 v2
r 23 v3
2
(0.01 m) v 2
r 21 v1
r 23 v3
2
(0.0025 m ) v3
2
(4.0 m /s )
r 22 v2
1.0 m/s
0.005 m
v3
2
(4.0 m /s)
0.0 025 m
16 m /s
(b) The volume flow rate through the pipe is
Q
A1v1
(0.005 m) 2(4.0 m/s) 3.1 10 4 m3/s
P13.33. What does the second pressure gauge read?
Treat the oil as an ideal fluid obeying Bernoulli’s equation.
Consider the path connecting point 1 in the lower pipe with
point 2 in the upper pipe a streamline.
Solve: Bernoulli’s equation is
1 2
p2
v2
2
Using p1 200 kPa
105 Pa 110 kPa.
2.00
gy2
105 Pa,
p1
1 2
v1
2
gy1
900 kg/m3, y2 – y1
p2
p1
1
2
10.0 m, v1
v 21
v22
g(y1
2.0 m/s, and v2
y 2)
3.0 m/s, we get p2
1.1
P13.35. A rectangular trough of height 0.45m width 0.3m and length 1.2m has a small drain hole at the bottom of one
end. What is the velocity of the water stream coming out of the hole?
Treat the water as an ideal fluid
obeying Bernoulli’s equation.
g(y2
y1 ) (p1
p2 )
1
2
2
v1
2
v2
Consider a point at the surface and a point in the hole. The pressure at the two points is the same (Patm), so p1 – p2 0.
Further assumptions are that the area of the trough is so large that (1) it doesn't matter, and (2) the speed of the water at the
top is zero (v1 0).
Call y2 – y1 h 0.45 m.
Solve: Since the pressures are equal we have
1
gh
v21 v22
2
Now set v1
0 and cancel .
gh
1 2
v2
2
Solve for v2.
v2
2gh
2(9.8 m/s2 )(0.45 m)
3.0 m/s
Assess: The result is independent of the area of the trough, as long as it is big enough that we can assume v1
0. The
result v2 = 2gh is known as Torricelli’s theorem.
P13.53.
The pictorial representation below gives the relevant diameters of a 2ml syringe. A nurse uses the syringe to
inject medicine into a patient with a blood pressure of 140/100. (a) How much force must he exert on the plunger with his
thumb? (b) if the syringe is emptied in 2 seconds, what is the speed of flow of the medicine through the needle?
Solve: (a) Because the patient’s blood pressure is 140/100, the minimum fluid pressure needs to be 100 mm of Hg above
atmospheric pressure. Since 760 mm of Hg is equivalent to 1 atm and 1 atm is equivalent to 1.013 105 Pa, the minimum
pressure is 100 mm 1.333 104 Pa. The excess pressure in the fluid is due to force F pushing on the internal 6.0-mmdiameter piston that presses against the liquid. Thus, the minimum force the nurse needs to apply to the syringe is
F
(b) The flow rate is Q
fluid pressure
area of plunger
(1.333 10 4 Pa)[ (0.003 m)2 ] 0.38 N
vA, where v is the flow speed of the medicine and A is the cross-sectional area of the needle. Thus,
v
Q
A
2.0 10 6 m3 /2.0 s
(0.125 10 3 m)2
20 m/s
P13.54. A water pistol with a 1.0 mm exit hole is fired horizontally 70cm above the ground. The water stream hits the
ground 1.2m away from the exit hole. What is the volume flow rate during the squirt?
The kinematic equation y = ½ g t2 can be used to find the velocity of the water as it leaves the pistol, and the volume flow
rate is just the velocity times the area of the exit hole.
Solve:
floor.
The water has no vertical component of initial velocity as it leaves the pistol. It takes a time t
The
horizontal
distance
it
travels
in
this
time
is x (vx ) i t. Solving
for
the
initial
the water, (vx)i = x/t:
(vx )i
x
g
2y
(1.2 m)
(9.80 m/s2 )
2(0.70 m)
3.2 m/s
The volume flow rate as it leaves the pistol is
Q
vA
v r2
2.5 10 6 m3/s
P13.57.
For the situation shown in the diagram, (a) what is the
water pressure as it exits the tube? and (b) calculate the height, h, of
the water in the tube.
We will treat the water as an ideal fluid obeying Bernoulli’s equation.
Let point 1 be beneath the standing column and point 2 be where the
water exits the pipe.
Solve:
(a) The pressure of the water as it exits into the air is p2
patmos.
(b) Bernoulli’s equation, Equation 13.14, relates the pressure, water
speed, and heights at points 1 and 2:
1 2
1 2
p1
v1
gy1 p2
v2
gy2
p1 p2
2
2
From the continuity equation,
v1A1
v2 A2
1
2
v 22
v21
(4m/s)(5 10 4 m2 ) v1(1 0 10 4 m2 ) 20 10 4 m3/s
g(y 2
v1
y1)
2m/s
Substituting into Bernoulli’s equation,
1
p1 p2 p1 pa tm os
(1000 kg/m 3 )[( 4 m/s)2 (2 m/s)2 ] (1000 k g/m 3 )(9.8 m/s)(4.0 m)
2
6000 Pa 39,200 Pa 45,200 Pa
But p1
p2
gh, where h is the height of the standing water column. Thus
h
45,200Pa
4.6m
(1000 kg/m3 )(9.8m/s2 )
2 y /g to hit the
velocity of
P13.58.
Assuming the air flowing through the nozzle
diagrammed at right behaves as an ideal, incompressible fluid.
(a) Calculate the flow speeds at points 1 and 2. (b) Calculate
the volume flow rate.
The ideal fluid (that is, air) obeys Bernoulli’s equation. The air
speeds at points 1 and 2 are v1 and v2, and the cross-sectional
areas of the pipes at these points are A1 and A2. Points 1 and 2
are at the same height, so y1 y2.
Solve:
(a) The height of the mercury is 10 cm. So, the pressure at point 2 is larger than at point 1 by
g(0.10m) (13,600 kg/m3 )(9.8m/s2)(0.10 m) 13,328 Pa
Hg
p2 p1 13,328 Pa
Using Bernoulli’s equation,
p1
1
2
ai r
v 21
air
gy1
p2
1
2
v12
v22
a ir
v22
ai r
2( p2
gy2
p1 )
ai r
1
2
2(13,3 28 Pa)
(1.28 kg/m3 )
p2
p1
ai r
Substituting v1
A2 v2
v1
A2
v
A1 2
(0.005m) 2
v 25 v2
(0.001m)2 2
25v2 in the Bernoulli equation, we get
(25v2) 2 v22 20,825 m2 /s2
Thus v1 25v2 145 m/s 150 m/s.
(0.005 m) 2 (5.78 m/s)
(b) The volume flow rate A2 v2
v2 5.78 m/s 5.8 m/s
4.5 10 3 m3/s.
v 22
20,82 5 m2 /s 2
From the continuity equation, we can obtain another equation connecting v1 and v2:
A1v1
v 21