baum (lmb2768) – HW09 – distler – (56295)
1
This
printout
should
have
22
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
For the siphon problem, note that the pres
sure in the hose must be everywhere non
negative
001
10.0 points
The picture below shows water forced out of
a fire extinguisher by compressed air.
The
water jets out at speed 11 m
/
s from a nozzle
located at height 1
.
78 m above the water level
inside the extinguisher.
h
v
Calculate the
gauge
pressure of the com
pressed air inside the extinguisher; that is,
the pressure di
ff
erence
P
gauge
air
=
P
inside
air

P
outside
air
.
The acceleration of gravity is 9
.
8 m
/
s
2
.
Water
density is 1000 kg
/
m
3
and the extinguisher’s
body is much wider than the water jet.
Correct answer: 77
.
944 kPa.
Explanation:
Let :
v
jet
= 11 m
/
s
,
h
= 1
.
78 m
,
g
= 9
.
8 m
/
s
2
,
and
ρ
= 1000 kg
/
m
3
.
This problem is a straightforward application
of Bernoulli’s equation
P
+
ρ
g y
+
ρ
2
v
2
= const
.
The flow begins at the water level inside the
extinguisher and continues all the way to the
jet emerging from the nozzle.
At the inside
water level, the pressure of the water equals
the pressure of the compressed air inside the
extinguisher.
In the jet emerging from the
nozzle, however, the water pressure equals the
atmospheric pressure of the inside air. Thus,
P
inside
air
+
ρ
2
v
2
init
=
P
outside
air
+
ρ
2
v
2
jet
+
ρ
g h
so that
P
gauge
air
=
P
inside
air

P
outside
air
=
ρ
g h
+
ρ
2
(
v
2
jet

v
2
init
)
.
Since the extinguisher’s body is much wider
than the jet, the initial speed of the water
flow inside the extinguisher is much slower
than the jet (
v
init
v
jet
) so
v
2
jet

v
2
init
≈
v
2
jet
,
and consequently
P
gauge
air
≈
ρ
g h
+
ρ
2
v
2
jet
= (1000 kg
/
m
3
) (9
.
8 m
/
s
2
) (1
.
78 m)
+
1000 kg
/
m
3
2
(11 m
/
s)
2
=
77944 Pa
.
002
10.0 points
A large water tank is emptying by having
water spurt out two holes at the bottom of
the tank, where hole 1 is twice the diameter
of hole 2.
Which of the following statements is cor
rect?
1.
The water speed out of hole 1 is twice
that of the water speed out of hole 2, while
the mass flow rate is twice that of hole 2.
2.
The water speed out of hole 1 is equal to
the water speed out of hole 2, as is the mass
flow rate out of hole 2.
3.
The water speed out of hole 1 is equal to
that of the water speed out of hole 2, while
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
baum (lmb2768) – HW09 – distler – (56295)
2
the mass flow rate is four times that of hole 2.
correct
4.
The water speed out of hole 1 is equal to
the water speed out of hole 2, while the mass
flow rate is twice that of hole 2.
5.
The water speed out of hole 1 is twice
that of the water speed out of hole 2, while
the mass flow rate is four times that of hole
2.
Explanation:
The openings at the top of the tank and at
both holes are at atmospheric pressure,
p
a
.
If
we approximate the water speed at the top,
y
=
y
t
, of the large tank to be
v
t
= 0, and
take
y
1
=
y
2
= 0 at the holes, then Bernoulli’s
equation for flow between the top of the tank
and hole 1 is
p
a
+
ρ
g y
t
=
p
a
+
1
2
ρ
v
2
1
,
and between the top of the tank and hole 2 is
p
a
+
ρ
g y
t
=
p
a
+
1
2
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '09
 IRENEPOLYCARPOU
 Physics, Correct Answer

Click to edit the document details