Mechanical Engineering
Fluid Mechanics and Machines
WORKBOOK
WORKBOOK
WORKBOOK
WORKBOOK
WORKBOOK
2017
Detailed Explanations of
Try Yourself
Questions

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T1 : Solution
V
W
θ
W
cos
θ
W
sin
θ
W
F
viscous
θ
N
Balancing forces along the inclined plane.
F
viscous
=
W
sin
θ
⇒
μ
=
W
sin
θ
⇒
V
=
θ
μ
=
−
−
×
×
×
×
×
= 0.5625 m/s
Fluid Properties
1

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3
Workbook
T2 : Solution
Power =
T
ω
Calculating torque,
Torque =
F
×
radius
F
=
μ
μ
= 2 × 10
–1
Ns/m
2
A
=
π
D
l
=
π ×
×
= 0.1414 m
2
v
=
π
×
=
× π ×
×
= 1.131 m/s
y
= 2.5 mm
= 2.5 × 10
–3
m
F
=
−
−
×
×
×
×
= 12.79 N
Torque =
F
×
radius
=
×
= 0.576 Nm
ω
=
× π ×
= 25.12 rad/s
P
= 0.576 × 25.12 = 14.47 Watt
14.5 Watt

© Copyright
T1 : Solution
A
ρ
= 0.12 kg/m
3
0.4 m
0.5 m
ρ
w
= 10 kg/m
3
3
ρ
Hg
= 13.6 × 10
kg/m
3
3
P
A
+ (0.12 × 9.81 × 0.9) – {13.6 × 10
3
× 9.81 × 0.5} + {10
3
× 9.81 × 0.5} – {13.6 × 10
3
× 9.81 × 0.5}
+ {10
3
× 9.81 × 0.5} – {13.6 × 10
3
× 9.81 × 0.5} + {10
3
× 9.81 × 0.5} – {13.6 × 10
3
× 9.81 × 0.5} = 0
P
A
+ 1.05948 – 66708 + 2 × {4905} – 2 × {66708} = 0
P
A
=
190312.94 Pascal
Pressure in terms of height of water column
=
×
= 19.39 m of H
2
O column
T2 : Solution
outside volume = 600 m
3
inside volume = (V– 600)
ρ
ice berg
= 915 kg/m
3
ρ
sea water
= 1025 kg/m
3
Fluid Statics
2

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5
Workbook
Let the total volume of iceberg be “
v
”.
Buoyancy force = Weight of iceberg
⇒
ρ
sea water
× (
V
– 600) × 9.81 =
ρ
iceberg
×
v
× 9.81
⇒
1025 (
v
– 600) = 915 V
⇒
1025
v
– 915
v
= 1025 × 600
⇒
v
=
×
−
= 5590.9 m
3
Weight of the iceberg
=
ρ
iceberg
×
v
iceberg
× 9.81
= 915 × 5590.9 × 9.81
=
50184757.04 N
= 50.185 MN
T3 : Solution
h
M
G
B
r
Specific gravity of the cylinder is “
S
”.
Given,
r
=
For floatation to be unstable
GM
< 0
∴
We have
GM
=
−
∀
I
Calculating height of cylinder inside water, let it be “
Y
”.
∴
Buoyancy force = Weight of cylinder
⇒
ρ
w
× Volume of cylinder inside ×
g
=
ρ
cylinder
× Volume of cylinder ×
g
ρ
w
×
π
r
2
×
Y
× 9.81 =
s
×
ρ
w
×
π
r
2
×
h
×
9.81
Y
=
sh
∀
I
=
(
)
π
π
×
=
π
×
π ×
×
×

6
Mechanical Engineering
•
Fluid Mechanics and Machines
© Copyright
=
×
=
×
×
=
BG
=
OG
–
OB
=
−
=
−
=
(
)
−
∴
−
∀
I
< 0
(
)
−
−
< 0
2 – 9
s
(1 –
s
) < 0
2 < 9s – 9s
2
or
9
s
2
– 9
s
+ 2 < 0
9
s
2
– 3
s
– 6
s
+ 2 < 0
3
s
(3
s
– 1)– 2 (3
s
– 1) < 0
(3
s
–1)(3
s
– 2) < 0
⎛
⎞ ⎛
⎞
−
−
⎜
⎟ ⎜
⎟
⎝
⎠ ⎝
⎠
< 0
+
+
0.333
0.667
+
∞
+
∞
–
∴
<
s
<
T4 : Solution
F
B
T + W
D
T
= 1.5 m
= 5.5 kN
F
buoyancy
= Tension
+ Weight
ρ
w
× Volume 5
g
= Tension + Weight,
Weight =
F
buoyancy
– Tension
=
[
]
⎡
⎤
ρ
×
× π ×
×
−
×
⎢
⎥
⎣
⎦
=
[
]
⎡
⎤
⎛
⎞
×
× π ×
×
−
×
⎢
⎥
⎜
⎟
⎝
⎠
⎣
⎦
= 17335.7 – 5500 = 11835.7 N = 11.8 kN

© Copyright
7
Workbook
T5 : Solution
30°
5 sin30°
5 sin30°
30°
F
h
=
ρ
×
g
×
A
x
= 1000 × 10 × 2 × 5 × sin30° × 1 × 5 × sin30°
= 125 kN/m
F
v
=
θ
⎛
⎞
× π
−
×
×
⎜
⎟
⎝
⎠
× 1000 × 10
=
( )
⎛
⎞
× π ×
−
×
×
° ×
°
⎜
⎟
⎝
⎠
× 1000 × 10
= 22.65 kN/m
F
R
=
+
=
+
= 127.03 kN/m

© Copyright
T1 : Solution
x
= 0
x
= 0.375 m
v
= 1.5 m/s
v
= 15 m/s
Let the velocity by given by
u
=
a
+
b
x
∴
At
x
= 0,
u
= 1.5
∴
a
= 1.5
At
x
= 0.375,
u
= 15
∴
b

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- Fall '17
- Ravi Murmu
- Mechanical Engineering, Fluid Dynamics, Fluid Mechanics, www.madeeasypublications.org