Soal-Jawab Tentang Fluida
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Transcript of Soal-Jawab Tentang Fluida
Mid-term Exam1. Newtonian Fluid, Non-Newtonian Fluid, Ideal Fluid2. Viscosity, Kinematic Viscosity3. Viscosity4. Bulk Modulus of Elasticity5. Surface Tension, Capillarity 6. Vapor Pressure (Cavitation)7. Perfect Gas Law8. Pressure9. Pressure variation10. Monometer11. Pressure Variation and Center of Pressure12. Buoyancy13. Continuity Equation and Bernoulli Equation14. Force from Fluid
Viscosity (Absolute Viscosity)
Kinematic Viscosity
Physical Properties of Water
7. Capillarity
0. Newtonian Fluid
Shear Stress
Ideal Fluid = 0
Newtonian Fluid = constant
Non-Newtonian Fluid = not constant
Ideal Plastic = constant
> yield stress
du
dy
du
dy
1. Viscosity, kinematic viscosity
a. What is the viscosity of gasoline at 25 ºC ?
b. Determine the kinematic viscosity of benzen at 27 ºC ?
(Use the figure)
(Use the figure)
c. A fluid has a viscosity 0.6 Pa·s and a relative density of 0.7. Determine its kinematic viscosity.
1. Answer
a.42.8 10 [ ]Pa s
b.
Viscosity of gasoline at 25 ºC :
Viscosity of benzen at 27 ºC : 7 27.0 10 [ / ]m s
Kinematic Viscosity :
4 23
0.6[ ]8.6 10 [ / ]
0.7 1000[ / ]
Pa sm s
kg m
c.
2. Viscosity
Determine the viscosity of fluid between shaft and sleeve in the figure.
V = 0.1 m/s
200 mm
100 N
74 mm
0.07 mm
2. Answer
AUF
t
3
3 3 2
100[ ] (0.07 10 )[ ]
(74 10 ) (200 10 )[ ] 0.1[ / ]
1.5[ ]
Ft
AU
N m
m m s
Pa s
3
3 3 2
100[ ]
0.1[ / ]
0.07 10 [ ]
(74 10 ) (200 10 )[ ]
F N
U m s
t m
A m
3. Viscosity
A 25 mm diameter steel cylinder 300 mm long falls, because of its own gravity force at an uniform rate of 0.1 m/s inside a tube of slightly larger diameter. A castor-oil film of constant thickness is between the cylinder and the tube. Determine the clearance between the tube and the cylinder. The temperature is 38 ºC. Relative density of steel = 7.85.
3. Answer
AUF
t
233
1
3 3 2
25 10(300 10 ) 7.85 1000 [ ]
2
2.6 10 [ ]( )
0.1[ / ]
(25 10 ) (300 10 )[ ]
?[ ]
F ma g N
Pa s fromthe figure
U m s
A m
t m
3 31
233
4
(25 10 ) ( 300 10 ) 0.12.6 10
25 10( 300 10 ) 7.85 1000
2
2.1 10 [ ]
AUt
Fg
m
4. Viscosity
A piston of diameter 50.00 mm moves within a cylinder of 50.10 mm. Determine the percent decrease in force necessary to move the piston when the lubricant warms up from 0 to 120 ºC. (use viscosity of crude oil)
constant
89 % decrease
UAF
t
2 2
1 1
1 1 1 0.111 0.89F
F
4. Answer
Viscosity at 0ºC
1 = 1.8 10-2 2 = 2.0 × 10-3
Viscosity at 120ºC
5. Bulk Modulus of Elasticity
dp : Pressure change, ex. (p2 – p1)
dV : Volume change, ex. (V2 – V1)
Unit of K: [Pa]dp = 0.1 MPa
1 m3 of Water at 20 ºC
dV = ?
/
dpK
dV V
Determine the change of volume.(use the table)
5. Answer
6
9
5 3
0.1 10 1
2.2 101
4.5 10 [ ]22000
dp VdV
K
m
/
dpK
dV V Water at 20 ºC 2.20 GPa (from the table)
6. Bulk Modulus of Elasticity
For K = 2.2 GPa for bulk modulus of elasticity of water, what pressure is required to reduce its volume by 0.5 % ?
6. Answer
/
dpK
dV V
92.2 10 [ ]
/ 0.5 /100
?
K Pa
dV V
dp
9
7
/ 2.2 10 0.5 /100
1.1 10 [ ]
dp K dV V
Pa
7. Capillarity
Determine the capillary rise for distilled water at 40 ºC in a circular 6 mm diameter glass tube.(Use the figure)
7. Answer
From the figure,2.5 mm of capillarity rise
8. Vapor Pressure
A vertical cylinder 300 mm in diameter is fitted (at the top) with a tight but frictionless piston and is completely filled with water at 70 ºC. The outside of the piston is exposed to an atmospheric pressure of 100 kPa. Calculate the minimum force applied to the piston that will cause the water to boil ? (use the table to determine the vapor pressure)
8. Answer
d = 300mm
70 ºC
100 kPa
Force ?From the table, vapor pressure of waterat 70 ºC : 31.4 kPa
Deference between Atmosphere and Vapor Pressures: 100 – 31.4 [kPa]
The force required :
233
3
300 10(100 31.4) 10
2
4.85 10 [ ]
F p A
N
9. Cavitation
At What pressure can cavitation be expected at the inlet of a pump that is handling water at 20 ºC ?
Pump
9. Answer
From the table,cavitation is expected at 2.5 kPa of pressure
10. Perfect Gas Law
A gas with relative molecular mass of 44 is at a pressure of 0.9 MPa and a temperature of 20 ºC. Determine its density.
10. Answer
P = RT
61 30.9 10
1.6 10 [ / ]8312
(273 20)44
Pkg m
RT
11. Absolute Pressure
h = 380 mm
Atmospheric Pressure = ??
A B
Mercury (s = 13.6)
11. Answer
A Bp p
3
4
13.6 1000 9.80665 380 10
5.07 10 [ ]
A B HGp p gh
Pa
Pressure at A:
Pressure at B:
B HGp gh
Taking the points A and B,??Ap
12. Manometer
A
h2 = 3 mP0= 101.3 kPa
h1 = 380 mm
Mercury (s = 13.6)
Oil(s = 0.9)
PA (gage) = ??, PA (abs) = ??
12. AnswerA
h2 =
3 m
1
P0
2
h1 = 380 m
m
Mercury (s = 13.6)
Oil(s = 0.9)
PA (gage) = ??PA (abs) = ??
B C
B Cp p
Pressure at C:
1C HGp gh
Taking the points B and C,
Pressure at B:
1B C HGp p gh
Pressure at the point A:
2 1 2
1 2
3
4
9.80665 13.6 1000 380 10 0.9 1000 3
2.4 10 [ ]( )
A B oil HG oil
HG oil
p p gh gh gh
g h h
Pa Gage pressure
Absolute pressure at the point A:
, 0
4 3
5
2.4 10 101.3 10
1.3 10 [ ]( )
A abs Ap p p
Pa Absolute pressure
13. Manometer
Water
250 mm
Oil (s = 0.90)
PA
PB
1625 mm
500 mm
PA – PB = ??
13. Answer
Water
h2 = 250 mm
Oil (s = 0.90)
PA
PB
h 1 = 1
625
mm
h 3 = 5
00 m
m
PA – PB = ??
C D
2C D w oilp p gh
Pressures at A and B:
1 2
1 2 3
A C w
B D w
p p g h h
P p g h h h
Taking the points C and D,
Pressure difference at the points A and B:
1 2 1 2 3
1 2 1 2 3
2 3
2 3 2
3
9.80665 1000 0.250 0.500 0.9 1000 0.250
5.1 10
A B
C w D w
C D w
w oil w
w oil
p p
p g h h p g h h h
p p g h h h h h
gh gh
g h h gh
Pa
B
Oil (s = 0.90)
14. Manometer
A
150 mm
250mm150 mm
150 mm
200mm
Mercury (s=13.6)
Water
PA – PB = ??
14. Answer
B
Oil (s = 0.90)A
h1 = 150 mm
Mercury (s=13.6)
Water
C D
E F
G Hh2 = 150 mm
h 3 = 2
50 m
m
h 4 = 2
00 m
m
h 5 = 1
50 m
m
2 1
3
4
5
5 4 3 2 1
5 4 2 3 1
9.80665 1000 0.9 0.150 13.6 1000 0.200 0.150 1000 0.250 0.150
A C Hg w
C D
D E w
E F
F G Hg
G H
H B oil
A B oil Hg w Hg w
A B oil Hg w
p p gh gh
p p
p p gh
p p
p p gh
p p
p p gh
p p gh gh gh gh gh
p p g h h h h h
34.4 10 Pa
Water
15. Pressure variation (Inclined Surface)
22.2 kN
600
2.4 m
4.5 m
fixed
fixed
d = ?
Determine the depth d when the gate opens.
Answer:D >= 2.75m
15. Answer
Water
22.2 kN
= 600
b = 2.4 mh 1
= 4
.5 m
d = ?
Centroid
Center of Pressure
h 2
2
2
hy
y C
Moment balance on the plate:
2
2 22
32
22 2 2 2 2 2 2
2 2
1 1 2 2
2 2 2
2 22 2
32
3
2
sin
122
2 2 2 6 32 2
sin
2 2sin
3 3
2sin
9
2sin
9 sin
9sin 60
2
G G G
G
GC
C
c C w C C
w
w
w
w
d h
I k A k bh
bhI
k h bh h bh h h hy y
h hy
F h F h y
P A h y gy bh h y
h hg bh h
g bh
dg b
dgb
31 1
3 3
39
4 22.2 10 4.52 1000 9.80665 2.4
2.4
F h
d m
F 1
F 2
16. Pressure Variation
Water
d = ??
Atmospheric Pressure P0
1.3 m
1.2 m
G
y
b = 4 m
h = 3 m
Determine depth d when water gateopens automatically.
d > 6.45
16. Answer
Water
d = ??
a2 = 1.3 m
a1 = 1.2 m
G
y
h =
3 m
b = 4 m
2
2
1
1
1 2
2
1 1 2
1
2
1 2 1
122
2
,
122
2
2 2 12 2
GC
C
Equation for position of center of pressure
hk h
y y d ay h
d a
y d a a
Combining two equations above
hh
d a d a ah
d a
h h h hd a a d a
2 21 2
2 1 2
21 2
1 2
2
2
2 2 4 12 2
2 3 2
2
3.0 1.2 3.0 1.3 3.01.2 1.3
2 3 23.0
1.32
6.45
ha a hh h hd a a a
ha a hha a
dh
a
d
d
3 322 22
22
3
22 2 2
1
3 12
1212
2
hh
G hh
GG G G G
Conditions
y bhI y dA y b dy b
bhI h
I k A k bh kbh bh
hy d a
17. Buoyancy
350 kN
12 m
6 m
6 m
15 m
2.4 m
400 kN
d (draft) = ??
Water
17. Answer (1/2)
F2 = 350 kN
a1 = 12 m
a4 = 6 m
6 m
a2 = 15 m
a3 = 2.4 m
F1 = 400 kN
d (draft) = ??
Water
17. Answer (2/2)2
2 14 1 1 2
3
22 1 11
3 4
2 2 1 11 1
3 4
2 1
3
32
.
( )
2
( )0
2
( )4
2
( )2
2
(15 12) 750 1012 12 4
2 2.4 1000 9.80665 6
(15 12
w
w
w
Buoyancy equals gravity force
a a dga a d F F
a
a a F Fd a d
a ga
a a F Fa a
a gad
a a
a
2)
2 2.4
12512 144 2.5
12 13.269.80665 1.01, 20.21.25 1.25
0 1.01d d
3 3 31 2
1 2 13
1 4
41 1 2 1
3
2 1 41
3
22 1
4 1
400 10 350 10 750 10
( )
1
2
( )2
( )2
2
( )
2
w
w
w
w
w
Gravity Force
F F F N
Eqation between x and d
dx a a a
a
Buoyancy
F gV
g a x d a
dadg a a a a
a
a a d dag a
a
a a dga a d
3a
18. Bernoulli Eq. (Pitot Tube)
V1 1 = 300 mm 2 = 200 mm V2
Flow rate = 0.3 m3
h = ??
Mercury (s = 13.6)
water
18. Answer
1 2
2
22
2
22
2
2
1
Pr
,
4
2
4 1
2
1000 4 0.3 1
2 13.6 1000 1000 9.806653.1416 0.2
3.69 10
Hg w
wHg w
w
Hg w
essure Difference
p p gh
Combining two equations above
Qgh
Qh
g
Pa
1 1 2 2
2 2
1 21 2
2 2 22 22
2 21 1 2 2
1 2
1 1 2
222
1 2 22
,2 2
4
2
2 2
0,
4
2 2
w w
w
Continuity Equation
Q AV A V
A A
Q Q QV
A
Bernoulli Equation
V p V pgz gz
V z z
V Qp p
19. Bernoulli Eq. (Venturi Tube)
V1
d2 = 150 mm
Flow rate: Q = ???
Mercury (s = 13.6)h=480 mm
V2
d1 = 300 mm
z2
z1
1.5 m
oil = 0.800
(1)
(2)
19. Answer (1/2)
V1
d2 = 150 mm
Flow rate: Q = ???
Mercury (s = 13.6)h1=480 mm
V2
d1 = 300 mm
z2
z1
h2=1.5 m
oil = 0.800
(1)
(2)
(3)
19. Answer (2/2)
1 2 1 2
1 3 1 3 2 2
21 2
2 22 2
2 1
Pr int
,
,
1 1
2
2 2
Hg oil oil
Hg oil oil
Hg oil oil
oil
essure Diffrencebetween po s Aand B
p p gh gh
p p gh p p gh
Combining twoequations above
gh gh Q
d d
2
1 2 2 2 4 42 1
2 4 4
3
16 1 12 1
1000 13.6 16 1 12 1 9.80665 0.48
800 3.1416 0.150 0.300
0.224 /
Hg
oil
gh
Q gh gh ghd d
m s
1 1 2 2
2 2
1 21 2
2 21 1 2 2
1 2
2 21 22 1 2 1
22 21 2 2
2 2 221
2 21 2 2 2
22 1
,2 2
2 2
1
2
1
2
1 1
2
oil oil
oil
oil
oil
Continuity Equation
Q AV A V
d dA A
Bernoulli Equation
V p V pgz gz
p pV V g z z
p p AV V gh
A
p p A V
A A
22
21 2
22 22 1
1 1
2oil
gh
p p Qgh
A A
20. Force From Fluid
= ??F = ??
(1)
(2)
V1
V2
600
d1 = 300 mm
d2 = 200 mm
200 L/s
(Overhead View)
P1 = 150 kPa
Fluid : Water (Incompressible) = constant)
21. Force from Fluid
H = ??
V
V
Water Jet
1 = 150 mm
V
300
3 m
= 125 mm
How much is H at least to support the plate ?