# CHAPTER 9: FLUID MECHANICS WORKSHEET SOLUTIONS

Mechanics

Oct 24, 2013 (4 years and 7 months ago)

67 views

MR. SURRETTE

VAN NUYS HIGH SCHOOL

1

|
P a g e

PHYSICS

CHAPTER 9
:
FLUID MECHANICS

WORKSHEET

SOLUTIONS

1
.
What is the total pressure at the bottom of a
8

m deep swimming pool? (Note the pressure
contribution from the atmosphere is 1.01 x 10
5

N/m
2
, the density of water is 10
3

kg/m
3
, and

g

= 9.8 m/s
2
)

1
A.

(
1)
p

=
p
a

+

gh

(2)
p

= (1.01 x 10
5

N/m
2
)+(1.0 x 10
3

kg/m
3
)(9.8 m/s
2
)(
8

m)

(3)
p

= 1.79

x 10
5

N/m
2

2
.
A hydraulic lift raises a
1750

kg automobile with

a
6
00 N force is applied to the smaller piston. If
the smaller piston has an area of
8

cm
2
, what is the cross
-
s
ectional area of the larger piston?

2
A.

(1) F
1
/A
1

= F
2
/A
2

(2) A
2

= (F
2
A
1
)/F
1

(3)
F
2

= w =
mg

(4) A
2

= (mgA
1
)/F
1

(5) A
2

= (
1750

kg
)(
9.8 m/s
2
)(
8
.0 x 10
-
4

m
2
) /
6
00 N

(6)
A
2

=
2.29

x 10
-
2

m

3.

An object of mass 0.
61

kg is suspended from a scale and sub
merged in a liquid. If the reading on the
scale is
4
.0
0

N, then the buoyant force that the fluid exerts on the object is:

3A.

(1)
w = mg

(2) w = (0.61

kg)(9.8 m/s
2
)

(3)
w = 5.98

N

(4)
F
B

= w

w’

(5) F
B

=
5.98

N

4.00

N

(6)
F
B

= 1.9
8

N

4
.
If the

flow rate of a liquid going through a 2.
5
0 cm radius pipe is measured at
1.5

x 10
-
3

m
3
/s, what is
the average fluid velocity in the pipe?

4
A.

(1)
Flow rate = Av

(2) v = Flow rate / A

(3) v = Flow rate /

r
2

(4) v =
1.5

x 10
-
3

m
3
/s /

(
2
.5

x 10
-
2

m
)
2

(
5)
v = 0.76

m/s

MR. SURRETTE

VAN NUYS HIGH SCHOOL

2

|
P a g e

PHYSICS

5
.
A liquid filled tank has a hole on its vertical surface just above the bottom edge. If
the surface of the
liquid is 0.7

m above the hole, at what speed will the stream of liquid emerge from the hole?

5
A.

(1) v = (2gh)
1/2

(2) v =
[(2)(9.8 m/s
2
)(0.
7

m)]
1/2

(3)
v =
3.7

m/s

6
.
Water (density = 1 x 10
3

kg/m
3
) is flowing thr
ough a pipe whose radius is 0.03

m with a speed of 1
2

m/s. This same pipe goes up to the second floor of the building, 3 m higher, and the pressure remain
s
unch
anged. What is the cross
-
sectional area

of the pipe on the second floor?

6
A.

(1)
First Equation:

p
1

+ ½

v
1
2

+

1

=
p
2

+ ½

v
2
2

+

2

(2) ½ v
1
2

+

gy
1

= ½ v
2
2

+ gy
2

(3) v
1
2

= v
2
2

+ 2gy
2

(4) v
2

= [(v
1
2
)

2gh]
1/2

(5
) v
2

= [(1
2

m/s)
2

2(9.8
m/s
2
)(3 m)]
1/2

(6
)
v
2

=
9.23

m/s

(7
)
Second Equation:

A
1
v
1

= A
2
v
2

(8
) A
2

= A
1
v
1

/ v
2

(9
)
Third Equation:

A
1

=

r
2

(10
) A
2

= (

r
2
)v
1

/ v
2

(11
) A
2

=

(0.03

m)
2
(
12 m/s) / (
9.23

m/s
)

(12
)

A
2

=
3.68

x 10
-
3

m
2

MR. SURRETTE

VAN NUYS HIGH SCHOOL

3

|
P a g e

PHYSICS

CHAPTERS 8
-

9
:

TEMPERATURE, HEAT, AND

FLUIDS

QUIZ SOLUTIONS

1.
Three moles of an ideal gas are confined to a 30 liter container at a pressure of 3 atmospheres. What
is the gas temperature? (R = 0.0821 L.atm/mole.K)

1A.

(1)
PV = nRT

(2)

T = (PV) / (nR)

(3)

T = [(3 atm)(30 liters) / [(3

mol)(0.0821 L.atm/mol.K)]

(4)

T = 365.4 K

2.
Suppose the ends of a 35
-
m
-
long steel rail are rigidly clamped at 5
o
C to prevent expansion. The rail
has a cross
-
sectional area of 15 cm
2
. What force does the rail exert when it is heated to 38
o
C?

(

STEEL

= 1.1 x 10
-
5

/
o
C, Y
STEEL

= 2 x 10
11

N/m
2
)

2A.

(1)
Y = (F/A) / (

L/L)

(2)

(

L/L) = (F/A) / Y

(3)

Y(

L/L) = (F/A)

(4)

(F/A) = Y(

L/L)

(5)

F = (AY

L
) / L

(6)

L = L

T
: F = (AY
L

T
) /
L

(7)

F = AY

T

(8)

F = (1.5 x 10
-
3

m
2
)(2 x 10
11

N/m
2
)(1.1

x 10
-
5

/
o
C)(33
o
C)

(9)
F =

1.09 x 10
5

N

3.
An unknown gas condenses into a liquid at approximately 105
o

Kelvin. What temperature, in degrees
Fahrenheit, does this correspond to?

3A.

(1)

T
F

= 9/5
T
C

+ 32

(2)

T
F

= 9/5(
T

273
) + 32

(3)

T
F

= 9/5(105

27
3) + 32

(4)

T
F

=
-

270.4
o

4.
An auditorium has dimensions 15 m x 25 m x 20 m. How many molecules of air are needed to fill the
auditorium at standard temperature and pressure?

4A.

(1)

(1 mol / 22.4 liter) (1 liter / 10
-
3

m
3
) (6.022 x 10
23

molecules / m
ol) (7500 m
3

/ auditorium)

(2)

2.02 x 10
29

molecules

MR. SURRETTE

VAN NUYS HIGH SCHOOL

4

|
P a g e

PHYSICS

5.
A helium
-
filled balloon has a volume of 1.5 m
3
. As it rises in the Earth’s atmosphere, its volume
expands. What will be its new volume (in cubic meters) if its original temperature and pressure
are 5
o
C
and 1 atm, and its final temperature and pressure are

40
o
C and 0.1 atm?

5A.

(1)

(P
1
V
1
) / T
1

= (P
2
V
2
) / T
2

(2)

(T
2
P
1
V
1
) / T
1

= P
2
V
2

(3)

P
2
V
2

= (T
2
P
1
V
1
) / T
1

(4)

V
2

= (T
2
P
1
V
1
) / (T
1
P
2
)

(5)

V
2

= [(233 K)(1 atm)(1.5 m
3
)] / [(278 K)(0.1 atm)]

(6)

V
2

= 12.57 m
3

6.
If the rms velocity of a helium atom at room temperature is 1350 m/s, what is the rms velocity of an
oxygen (O
2
) molecule (mass of O
2

= 32 grams, mass of He = 4 grams)?

6A.

(1) v
rms

= [(
3RT
)/M]
1/2

(2)

v
rms

O
2

is proportional to (
1
/32
gra
ms
)
1/2

(3) v
rms

He is proportional to (
1
/4
grams
)
1/2

(4) v
rms

(O
2

/ He) is proportional to (32)
1/2

/ (4)
1/2

(5) v
rms

(O
2

/ He) = 2.8

(6) 1350 m/s / 2.8

(7)
v
rms

= 477.3 m/s

7.
A silver bar of length 35 cm and cross
-
sectional area 1 cm
2

is used to tr
ansfer heat from a 100
o
C
reservoir to a 0
o
C reservoir. How much heat is transferred per second?

(k
SILVER

= 427 W/m
o
C)

7A.

(1) H = kA(

T) / L

(2) H= (427 W/m
o
C)(1 x 10
-
4

m
2
)(100
o
C) / (0.35 m)

(3)
H = 12.20 W

8.
A solar heating system has a 30% conve
rsion efficiency; the solar radiant incident on the panels is
500 W/m
2
. What is the increase in temperature of 30 kg of water in a one hour period by a 5.0 m
2

area
collector? (The specific heat of water is 4186 J.kg.
o
C)

8A.

(1)
: 5.0 m
2

(
500 W / 1.0 m
2
) =
2500 W

(2)
:
2500 W

x 30% = 750 W

(3) 750 W = 750 Joules / second

(4)
Determine Q
: 3600 s (750 J / s) =
2.70 x 10
6

J

(5) Q = mc

T

(6)

T = Q / mc

(7)

T = 2.70 x 10
6

J / (30 kg)(4186 J/kg.
o
C)

(8)

T = 21.5
o
C

MR. SURRETTE

VAN NUYS HIGH SCHOOL

5

|
P a g e

PHYSICS

9.
How much heat energy is required to melt a 30 gram block of silver at 10
o
C? The melting point of
silver is 960
o

C, its specific heat is 2.34 x 10
2

J/kg.
o
C, and its h
eat of fusion is 8.83 x 10
4

J/kg.

9A.

(1)
Determine specific heat
:
Q
C
= mc

T

(2) Q
C
= (0.030 kg)(2.34x10
2

J/kg.
o
C)(960
o
C

10
o
C)

(3)
Q
C

= 6669 J

(4)
Determine latent heat
:
Q
L

= mL

(5) Q
L

= (0.030 kg)(8.83 x 10
4

J/kg)

(6)
Q
L

= 2649 J

(7)
Determine

total heat
: Q = Q
C

+ Q
L

(8) Q =
6669 J

+
2649 J

(9)
Q = 9318 J

10.
An ideal gas is maintained at an average pressure of 75 N/m
2

during an isobaric process while its
volume decreases by 0.25 m
3
. What work is done by the system on its environment?

10A
.

(1) W = P

V

(2) W = (75 Pa)(
-

0.25 m
3
)

(3)
W =
-

18.8 J

11.
A boulder of mass 700 kg tumbles down a mountainside and stops 180 m below. If the temperature
of the boulder, mountain, and surrounding air are all at 300 K, what is the change in entropy of the
Univ
erse?

11A.

(1)

S =

Q / T

(2)

Q = Energy
:

Q = U
g

= mgh

(3)

Q = (700 kg)(9.8 m/s
2
)(180 m)

(4)

Q = 1.24 x 10
6

J

(5)

S =
1.24 x 10
6

J

/ 300 K

(6)

S = 4116 J/K

12.
A bottle containing an ideal gas has a volume of 3 m
3

and a pressure of 1 x 1
0
5

N/m
2

at a
temperature of 300 K. The bottle is placed against a metal block that is maintained at 900 K and the gas
expands as the pressure remains constant until the temperature of the gas reaches 900 K. How much
work is done by the gas?

12A.

(1)
Det
ermine

V
: (
p
1
V
1
) / T
1

= (
p
2
V
2
) / T
2

(2) V
2

= (
p
1
V
1
T
2
) / (T
1
p
2
)

(3) V
2

= [(1 x 10
5

Pa)(3 m
3
)(900 K)] / [(300 K)(1 x 10
5

Pa)]

(4) V
2

= 9 m
3

(5)

V = V
2

V
1

(6)

V

= 9 m
3

3 m
3

=
6 m
3

(7)
Determine the Work
: W =
p

V

(8) W = (1 x 10
5

Pa)(
6 m
3
)

(9)
W = 6

x 10
5

J

MR. SURRETTE

VAN NUYS HIGH SCHOOL

6

|
P a g e

PHYSICS

13.
A gasoline engine absorbs 2600 J of heat energy and performs 600 J of mechanical work in each
cycle. What is the efficiency of the engine?

13A.

(1)
e

= (Q
H

Q
C
) / Q
H

(2)
Q
H

Q
C

= Work

(3)
e

=
Work

/ Q
H

(4)
e

= 600 J / 2600 J

(5)
e

= 23.1%

14.
Calculate the entropy change when 2 mol (36 grams) of water at 100
o
C is converted into steam (The
heat of vaporization of water is 540 cal/g).

14A.

(1)
Determine

Q
:

Q = mL

(2)

Q = (36 g)(540 cal/g)

(3)

Q = 19,440 cal

(4)
Convert to Jo
ules
:
19,440 cal

(4.186 J/cal) =
8.138 x 10
4

J

(5)
Determine Entropy
:

S =

Q / T

(6)

S =
8.138 x 10
4

J

/ 373 K

(7)

S = 218.2 J/K

15.
What is the total pressure at the bottom of a

12 m deep swimming pool? (Note the pressure contribution from the
atmosphere is 1.01 x 10
5

N/m
2
, the
density of water is 10
3

kg/m
3
, and
g

= 9.8 m/s
2
)

15A.

(1)
p

=
p
a

+

gh

(2)
p

= (1.01 x 10
5

N/m
2
)+(1.0 x 10
3

kg/m
3
)(9.8 m/s
2
)(12 m)

(3)
p

= 2.19 x 10
5

N/m
2

16.
A hydraulic lift raises a 1325 kg automobile when a 445 N

force is applied to the smaller piston. If
the smaller piston has an area of 6 cm
2
, what is the cross
-
sectional area of the larger piston?

16A.

(1) F
1
/A
1

= F
2
/A
2

(2) A
2

= (F
2
A
1
)/F
1

(3)
F
2

= w =
mg

(4) A
2

= (mgA
1
)/F
1

(5) A
2

= (
1325 kg
)(
9.8 m/s
2
)(
6 x
10
-
4

m
2
) /445 N

(6)
A
2

= 2.29 x 10
-
2

m

MR. SURRETTE

VAN NUYS HIGH SCHOOL

7

|
P a g e

PHYSICS

17.
An object of mass 1.24 kg is suspended from a scale and submerged in a liquid. If the reading on
the scale is 3.00 N, then the buoyant force that the fluid exerts on the object is:

17A.

(1)
w = mg

(2) w =
(1.24 kg)(9.8 m/s
2
)

(3)
w = 12.15 N

(4)
F
B

= w

w’

(5) F
B

=
12.15 N

3.00 N

(6)
F
B

= 9.15 N

1
8.
A liquid filled tank has a hole on its vertical surface just above the bottom edge. If the surface of
the liquid is 1.4 m above the hole, at what speed

will the stream of liquid emerge from the hole?

18A.

(1) v = (2gh)
1/2

(2) v = [(2)(9.8 m/s
2
)(1.4 m)]
1/2

(3)
v = 5.24 m/s

19.
Water (density = 1 x 10
3

kg/m
3
) is flowing through a pipe whose radius is 0.02 m with a speed of 11
m/s. This same pipe go
es up to the second floor of the building, 2.5 m higher, and the pressure remains
unchanged. What is the cross
-
sectional area of the pipe on the second floor?

19A.

(1)
First Equation:

p
1

+ ½

v
1
2

+

1

=
p
2

+ ½

v
2
2

+

2

(2) ½ v
1
2

+
gy
1

= ½ v
2
2

+ gy
2

(3) v
1
2

= v
2
2

+ 2gy
2

(4) v
2

= [(v
1
2
)

2gh]
1/2

(5) v
2

= [(11 m/s)
2

2(9.8 m/s
2
)(2.5 m)]
1/2

(6)
v
2

= 8.49 m/s

(7)
Second Equation:

A
1
v
1

= A
2
v
2

(8) A
2

= A
1
v
1

/ v
2

(9)
Third Equation:

A
1

=

r
2

(10) A
2

= (

r
2
)v
1

/ v
2

(11) A
2

=

(0.02 m)
2
(1
1 m/s) / (
8.49 m/s
)

(12)

A
2

= 1.63 x 10
-
3

m
2