FLUID MECHANICS (PRE – FINAL S4) March 15, 2017 Name ____________________________________ 1. A piece of wood wood of S = 0.651 is 8 cm square and 150 cm cm long. How many kilograms of lead lead weighing 11,200 3 kg/m must be fastened at one end of the stick so that it will float upright with 30.5 cm out of water? S 0.651
wood 0.651(1000) 651
kg
m3 V1 0.08(0.08)(1.5) 0.96 m3
65 1) 624.96 kg 0.96(651 W2 11,200V2 BF1 (1.5 0.305 30 5)1000 1195 kg BF2 1000V2 W1 W2 BF1 BF2 624.96 11,200V2 1195 1000V2 W1
V2
1195 - 624.96 0.056 m 3 10,200
W2 625.93 kg 2. A barge is loaded with with 150 Metric tons of coal. coal. The weight of the empty barge in air is 35 Metric ton. If the the barge is 5.5 m wide, 16 m long and 3 m high, what is its draft. (Depth below the water surface)
BF W1 W2 Vs (150 35)(1000) 1000(5.5)(16)h h 2.1 m W
3. A prismatic object 20 20 cm thick by 20 cm wide by 40 cm long long is weighed in water water at a depth of 50 cm and found to weigh 50 N. What is its weight in air and its specific gravity?
W2 50 Newton V1 (0.20 )(0.20 )(0.4) 0.016 m3 W1
1
m1 V1
m1 V1 S 1000 W1
W1 W1 W2
m1g Newton
W1 gV1 1000
W1 W1 W2
W1 9810V1
W1 W1 W2
1 9810V1
1 W1 W2
W1 W2
9,810(0.016 ) W1 156.96 50 W1 206.96 Newton S
206.96 206.96 - 50
1.312
FLUID MECHANICS QUIZ NO. 3 1. A block of wood wood has a vertical vertical projection of 15.24 15.24 cm when placed in water and 10.2 cm when when placed in alcohol. If the specific gravity of alcohol is 0.82, find the specific gravity of wood gravity of wood.
In water W
BF 1000A(h - 0.1524) 1
In Alcohol W
BF 0.82(1000) A(h - 0.102) 2
Equating eq. 1 and eq. 2 (h - 0.1524) 0.82(h - 0.102) h - 0.82h 0.1524 0.82(0.102) h 0.382 m
1000(0.2296) A 3 W B A(0.382) A(0.38 2) 4 eq.3 eq. 4 1000(0.2296) 0.382B B 0.2296 0.601 SB W
1000
0.382
2. A cubical block of wood 10 cm cm on a side side floats at the interface between oil and water with with its lower surface 3 1.50 cm below the interface. The density of the oil is 790 kg/m . What is the gage pressure at the lower face of the block? What is the mass and density of the block?
3. Compartments A and B of the tank shown in the figure below are closed and filled with air and a liquid with S = 0.6. If the atmospheric pressure is 101 KPa (abs) and the pressure gage reads 3.5 KPa (gage), determine the manometer reading h in cm.
3.5 9.81h 0.6(9.81)h 0.6(9.81)0.02 13.6(9.81)(0.03) 0 h
3.5 0.6(9.81)0.02 13.6(9.81)(0.03) 9.81(1 0.6)
1.9 m
4. The gate shown is hinged at A and rests on a smooth floor at B. The gate is 3 m square. Oil (S = 0.80) stands on the left side of the gate to a height of 1.5 m above A. Above the oil surface is a gas under a gage pressure of -6.9 KPa. Determine the amount of the vertical force P applied at B that would be required to open the gate.
P = -6.9 KPag
Free Surface
1.5m
h
A F
P
· CG · CP
45°
m 3
h 1.5 1.5 sin 45° 2.56 m F 6.9 0.80(9.81)2.56(3)(3) 118.8 KN 5.
A spherical buoy 2 m in diameter floats half submerge in a liquid with S = 1.5. What is the weight of the lead anchor weighing 7000 kg/m3 will completely submerged the buoy in the liquid.
L 1.5(1000) 1500
m3
4 3 1 4 (1)3 2.09 m 3 2 3
Vsphere Vs
kg
R 3 4.18 m 3
volume submerge
BF1 L ( Vs ) W1 1500( 2.09 ) 3135 kg W1
from figure 2 : W1 W2
BF1 BF2 BF1 1500(Vsphere ) 1500(4.18) 6270 kg W1 3135 kg W2 7000V2 BF2 1500V2 3135 7000V2 6270 1500V2 V2 0.57 m 3 W2 3990 kg 6.
A glass tube 1.5 m long and 25 mm diameter with one end closed is inserted vertically with the open end down into a tank of water until the open end is submerged to a depth of 1.2 m. If the barometric pressure is 98 KPa, and neglecting vapor pressure, how high will water rise in the tube. (Assume isothermal conditions for air)
fOR isothermal : P1V1 P2 V2 V1
(0.025)2 (1.5) 0.00074 m3
4 P2 98 9.81(1.2 x ) V2
1
(0.025 )2 (1.5 x ) 2 4
98 9.81(1.2 x ) (0.025 )2 (1.5 x ) 4 98(1.5 ) 98 9.81(1.2 x )(1.5 x ) 147 (98 11.772 9.81x )(1.5 x ) 147 (109.772 9.81x )(1.5 x ) 147 164.658 109.772 x 14.715 x 9.81x 2 9.81x 2 124.487 x 17.658 0 x 2 12.7 x 1.8 0 12.7 12.98 x 98 (0.025)2 (1.5) 4
2 x 0.14 m
14 cm
1. A brass cube 152.4 mm on a side weighs 298.2 N. We want to hold this cube in equilibrium under water by attaching a light foam buoy to it. If the foam weighs 707.3 N/m 3, what is the minimum required volume of the buoy?
298.2 N BF1 (0.1524) 3 (9810) 34.72 N W2 707.3 V2 1 BF2 V2 (9810) 2 298.2 707.3 V2 34.72 V2 (9810) 298.2 34.72 V2 0.029 m3 9810 707.3 W1
2. A man dives into a lake and tries to lift a large rock weighing 170 kg. If the density of the granite rock is 2700 kg/m3, find the force that the man needs to apply to lift it from the bottom of the lake. Assume density of lake water to be 1000 kg/m 3.
W V
2700V 170 kg
170
0.063 m 3
2700 BF 1000(0.063) 63 kg
BF T T 170 - 63 107 kg W
3. A 1.2 m diameter steel pipe, 6 mm thick, carries oil with S = 0.822 under a head of 122 m of oil. Compute a. The stress in the steel in KPa b. The thickness of the steel pipe required to carry a pressure of 1724 KPa with an allowable stress of 124 MPa
PD 2t P 0.822(9.81)(122) 983.8 KPa S
S t
983.8(1.2) 2(0.006)
1724(1.2) 124,000(2)
98,379 KPa 0.0083 m 8.3 mm
Example 1 A large pipe called a penstock in hydraulic work is 1.5 m in diameter. Here it is composed of wooden staves bound together by steel hoops each 3.23 cm 2 in area, and is used to conduct water from a reservoir to a powerhouse. If the maximum tensile stress permitted in the hoops is 130 MPa, what is the maximum spacing between hoops under a head of 30.5 m. L
m 5 . 1
P 9.81(30.5) 299.205 KPa D 1.5 m On the vertical projection F ; A LD A F 299.205(1.5)L 448.81L KN P
F 224.4L KN 2 Tensile Stress on the hoop T
T ; A 3.23 cm2 A A 0.000323 m2
S
S 130000 KPa 224.4L 0.000323 L 0.187 m 18.7 cm
130000
L
Example 2 Compute the wall stress in a 1200 mm diameter steel pipe 6 mm thick under a pressure of 970 KPa. Given: D = 1.2 m t = 0.006 m P = 970 KPa
S
PD 2t
970(1.2) 2(0.006)
97,000 KPa
Example 3 What is the minimum allowable thickness of 600 mm diameter steel pipe under an internal pressure of 860 KPa with a working stress in the steel of 70,000KPa.
S
PD 2t
860(0.600) 2t t 0.004 m 4 mm
70,000
Example 4 A wood stave pipe is bound by steel rods which take the entire bursting stress. Find the proper spacing for 25 mm steel rods for a 1800 mm diameter wood stave pipe under a pressure of 590 KPa if the working stress in the steel is 105,000 KPa. Given Dr = 0.025 m D = 1.8 m P = 590 KPa S = 105,000 KPa
F 2T A DL For the rods P
S
T
4
0.0252
105,000
4T
0.0252
T
51.54 KN
L
L
10 cm
2(51.54) 590(1.8)
0.10 m
QUIZ NO. 4 Problem No. 1 The velocity of water in a 10 cm diameter pipe is 3 m/sec. At the end of the pipe is a nozzle whose velocity coefficient is 0.98. If the pressure in the pipe is 55 KPa, what is the velocity in the jet? What is the diameter of the jet? What is the rate of discharge? What is the head loss? d1
0.10 m
m sec C v 0.98 v1
3
P1
55 KPa
Applying Bernoulli' s equation P1
HL P1
2
v1 2g
z1
P2
2
v2 2g
z 2 HL
2 1 v 2 1 v 22 v2 2 1 2 1 0 . 041 2g 2 2g Cv 2g (0.98) 2
v1 2g
2
0 0
v2 2g
2
v2 P 1 0.041 1 2g
2
0 0.041
v2 2g
2
v1 2g
P1 v 12 2g m 10.75 v2 sec 1 0.041 m3 2 Q Av (0.10) (3) 0.024 2g
4
Q
2
d2 (10.75) 0.024
4 d2 0.053 m 5.3 cm
HL
sec
3
m sec
(10.75)2 0.24 m 0.041 2g
Problem No. 2 A centrifugal pump draws water from a well at the rate of 142 L/sec of water through a 203 mm ID suction line and a 152 mm ID discharge line. The suction gauge located on the pump centerline reads 254 mm Hg vacuum, while the discharge gauge is 6 m above the pump centerline. If the power input to the water is 75 KW, find the reading of the discharge gauge in KPa. Q 0.142
m3
sec d1 0.203 m ; A 1
0.032 m2 d2 0.152 m; A 2 0.018 m 2 A d2 4 Q v A
m ; v2 sec z1 0; z 2 6 m
WP ht
ht
4. 4
P1
101.325 -254 33.86 KPa 760
Qht 75 (0.142)(9.81) P2
7. 8
m sec
v1
53.84 m
P1 v 22 v12 z 2 z1 HL 2g
2 2 P v v1 ht 2 z 2 z1 HL 1 2g P2 414.45 KPa
P2
Problem No. 3 A 15 KW suction pump draws water from a suction line whose diameter is 200 mm and discharges through a line whose diameter is 150 mm. The velocity in 150 mm line is 3.6 m/sec. If the pressure at point A in the suction line is 34.5 KPa below the atmosphere where A is 1.8 m below that of B on the 150 mm line, Determine the maximum elevation above B to which water can be raised assuming a head loss of 3 m due to friction. FP
15
KW
d1
0.2
m
d2
0.15
m
v2
3.6
m/sec
P1
-34.5
KPa
P2
0
KPa
HL
3
m
A1
0.03
m^2
A2
0.02
m^2
Q
0.064
m^3/sec
v1
2.025
m/sec
SW
9.81
KN/m^3
P1/SW
-3.52
m
P2/SW
0.000
m
v1^2/2g
0.209
m
v2^2/2g
0.66
m
z1
0.0
m
z2
1.8 + h
ht
24.04
D(Phead)
3.52
D(Vhead)
0.45
D(Ehead)
(1.8 + h)
h
15.27
m
m
Problem No. 4 A power nozzle throws a jet of water that is 50 mm in diameter. The diameter of the base of the nozzle and of the approach pipe is150 mm. If the power of the nozzle jet is 42 HP and the pressure head at the base of the nozzle is 54 m, compute the head lost in the nozzle.
Problem No. 5 A fire pump delivers water through a 150 mm main to a hydrant to which is connected a 75 mm hose, terminating in a 25 mm nozzle. The nozzle is 1.5 m above the hydrant and 10 m above the pump. Assuming frictional losses of 3 m from the pump to the hydrant, 2 m in the hydrant and, and 12 m from the hydrant to the base of the nozzle, and a loss in the nozzle of 6% of the velocity head in the jet, to what vertical height can the jet be thrown if the gage pressure at the pump is 550KPa.
Problem No. 6 Water issues from a circular orifice under a head of 12 m. The diameter of the orifice is 10 cm. If the discharge is found to be 75 L/sec, what is the coefficient of discharge? If the diameter at the vena cotracta is measured to be 8 c m, what is the coefficient of contraction and what is the coefficient of velocity. v
2gh theoretica l velocity
v
2(9.81)(12) 15.34
Q' 0.075 Q
m sec
m3 sec
(0.10 )2 (15.34 ) 0.12
4 Q'
0.625 Q a d' 8 Cc A D 12 Cd Cc(Cv )
m3 sec
Cd
Cv
0.625 0.67
0.67
0.93
Problem No. 7 A jet discharges from an orifice in a vertical plane under a head of 3.65 m. The diameter of the orifice is 3.75 cm and the measured discharge is 6 L/sec. The coordinates of the centerline of the jet are 3.46 m horizontally from the vena contracta and 0.9 m below the center of the orifice. Find the coefficient of discharge, velocity and contraction.
Problem No. 8 The inside diameters of the suction and discharge pipes of a pump are 20 cm and 15 cm, respectively. The discharge pressure is read by a gage at a point 2 m above the centerline of the pump, and the suction pressure is read by a gage 1 m below the pump centerline. If the pressure gage reads 145 KPa and the suction gage reads a vacuum of 250 mm Hg when diesel fuel (S = 0.82) is pumped at the rate of 30 L/sec, Find the KW power of the driving motor if overall pump efficiency is 75%. d1
0.20
m
d2
0.15
m
z1
-1
m
z2
2
m
P1
-33.33
Kpa
P2
145
KPa
S
0.82
SW(water)
9.81
KN/m^3
SW
8.0442
KN/m^3
Q
0.03
m^3/sec
A1
0.031
m^2
A2
0.018
m^2
v1
0.955
m/sec
v2
1.70
m/sec
P1/SW
-4.14
m
P2/SW
18.03
m
v1^2/2g
0.05
m
v2^2/2g
0.15
m
D(Phead)
22.17
m
D(Vhead)
0.10
m
D(ElHead)
3.00
m
HL
0.00
m
ht
25.27
m
WP
6.10
KW
e
0.75
BP
8.13
KW
Problem No. 9 A jet of water 7.6 cm in diameter discharges through a nozzle whose velocity coefficient is 0.96. If the pressure in the pipe is 82.7 KPa and the pipe diameter is 20 cm and if it is assumed that there is no contraction of the jet, what is the velocity at the tip of the nozzle? What is the rate of discharge? Q A 1v 1 A 2 v 2 4
d 2 v 2 v 2 d1 4 2 d2 v 2 2 v1 2g d1 2g 2 1
P1
0.20
m
d2
0.076
m
Cv
0.96
P1
82.7
Kpa
P2
0
KPa
v1
1.80
m/sec
v2
12.467
m/sec
SW
9.81
KN/m^3
g
9.81
m/sec^2
v2
12.47
m/sec
A1
0.031
m^2
A2
0.005
m^2
Q
0.057
m^3/sec
z1
P2
2
v2 2g
z2 HL
1 v 2 z2 2 1 2 C v 2g 4 2 2 1 v 2 P1 d2 v 2 P v z1 2 2 z 2 2 1 2 d1 2g 2g Cv 2g 4 2 v 2 1 d2 P1 P2 1 1 ( z1 z2 ) 2g C v 2 d1 4 2 d P P v2 1 2 2 1 2 ( z1 z 2 ) 2g C v d1 P1 P2 ( z1 z2 ) 2 v2 4 2g 1 d 2 2 C v d1 P1
d1
2
v1 2g 2
v1 2g
z1
P2
2
v2 2g
SAMPLE PROBLEMS APPLICATION OF BERNOULLI’S EQUATION
ExampleNo. 1 The water in a 10 m diameter, 2 m high aboveground swimming pool is to be emptied by unplugging a 3 cm diameter, 25 m long horizontal pipe attached to the bottom of the pool. Determine the maximum discharge rate of water through the pipe.
1 h=2 2
P1
P1
2
v1
2g
Z1
v2
2
2g
Z2
Q Av
0; v 1 0; P2 0; Z 1 0; Z 2 -2 m v2
000 0 2 v2
P2
v2
Q
2
2g
4
(0.03)2 (6.23) 0.0044 m3 /sec
( 2)
2
2g
2(9.81)(2 ) 6.23 m/sec
Example No. 2 A large tank open to the atmosphere is filled with water to a height of 5 m from the outlet tap. A tap near the bottom of the tank is now opened and water flows out from the smooth and rounded outlet. Determine the water velocity at the outlet.
P1
1
P1
v1
2
2g
Z1
v2
2
2g
Z2
0; v 1 0; P2 0; Z 1 0; Z 2 - 5 m
000 0 5 v2
P2
v2
v2
2
2g
( 5)
2
2g
2(9.81)5 9.623 m/sec
Example No. 3 The water level of a tank on a building roof is 20 m above the ground. A hose leads from the tank bottom to the ground. The end of the hose has a nozzle, which is poi nted straight up. What is the maximum height to which the water could rise.
P1
P1
2
v1
2g
Z1
P2
v2
2
2g
Z2
0; P2 0; Z 1 0; Z 2 h; v 2 0 2
0
v1
h
v1
2g
0 00h
2
2g
h max imum height v 1 velocity at the tip of the nozzle
Example no. 4 Water flows through a horizontal pipe at the rate of 1 Gal./sec. The pipe consist of two sections of diameter 4 in. and 2 in with a smooth reducing section. The pressure difference between the two pipe sections is measured by mercury manometer . Neglecting frictional effects, determine the differential height of mercury between the two pipe sections. Q = 1 gal/sec = 0.0038 m3/sec D1 = 4 in. = 0.1016 m D2 = 2 in = 0.0508 m
Mercury S
2
P1 v 1
2g
z1
P2 v 2
2
2g
z 2 HL
Q A 1v 1 A 2 v 2 v v1
Q A
HL
(0.1)2
m v 2 (3.82)2 3.82 ; 0.744 m sec 2g 2(9.81)
0.8(9.81) 7.848 20 2
2g
KN m3
2.55 m
7.848
P1 v 1
m v 1 (0.424)2 0.424 ; 0.009 m sec 2g 2(9.81) 2
0.030
D
2
(0.3)2
4
4
0.030 4
v2
2
;A
Z1
P2 v 2
2
2g
Z 2 HL
Z1 Z 2 : (Z2 - Z1 ) 0 2
P2 P1 v 1 v 2
2
2g 2g
HL
200 7.848
0.009 0.744 2.55
P2 174.22 KPa Example No. 5 A 300 mm pipe is connected by a reducer to a 100 mm pipe. Points 1 and 2 are at the same elevation. The pressure at point 1 is 200 KPa. Q = 30 L/sec flowing from 1 to 2, and the energy lost between 1 and 2 is equivalent to 20 KPa. Compute the pressure at 2 if the liquid is oil with S = 0.80. (174.2 KPa) P1
9.81x
P2
P1
P2
P1
9.81h 13.6(9.81)h 9.81x
P2
9.81h 13.6(9.81)h 123.606
1
from 1 to 2 P1
v1
2
2g
γ
Q
v1
v2
Z1
4
0.0038 π
4 v1
2
2g v2
2g P1
(0.47) 2
(1.875) 2
0.0113 0
γ
P2
0.47 m/sec
1.875 m/sec
0.0113m
2(9.81)
(0.0508) 2
2(9.81)
2
Z2
(0.1016) 2
A2
2
2g
0.0038 π
Q
v2
γ
A1
P2
0.18 m P2
0.18 0
γ
P1
0.0113 0.18
9.81(0.1687)
0.1687
γ
P2
P1
Equating eq. 1 and eq. 2 - 123.606h -1.655 - 1.655
h
h
- 123.606 0.0134 meters
h
0.53 inches
1.655 KPa
2
Example No. 6 A venturi meter having a diameter of 150 mm at the throat is installed in a 300 mm water main. In a differential gage partly filled with mercury (the remainder of the tube being filled with water) and connected with the meter at the inlet and at the throat, what would be the difference in level of the mercury columns if the discharge is 150 L/sec? Neglect loss of head. (h=273 mm)
Q 150
1m 3
L
x sec 1000L S of mercury 13.6
0.150
13.6(9.81) 133.416
m3 sec
KN m3
0.30m d2 0.15m Q 0.150 m 3 / sec HL 0 d1
v
v1
Q A
4
v2
(0.30 )
Z1
2.122 m / sec; 2
0.150
(0.15 )
4 P1
2
0.150
8.488 m / sec; 2
2
v1
2g
Z1
P2
v2
v1
0.230 m
2g v2
2
2g
3.672 m
2
2g
Z 2
Z2 0
P1 P2
v2
2
2
v1
3.672 0.230 2g 2g P1 P2 3.442 m P1 P2 3.442(9.81) 33.766 KPa P1 9.81( x ) 9.81h 133.416h 9.81x
P2
P1 9.81h 133.416h P2 P1 P2
133.416h 9.81h P1 P2 h(133.416 9.81) P1 P2 33.766 h (133.416 9.81) (133.416 9.81) h 0.273 m h 273 mm
Example no. 7 A mechanical engineer of an industrial plant is required to install a centrifugal pump to lift 15 L/sec of water from a sump to a storage tank on a tower. The water is to be delivered into a 105 KPag tank and the water level in the tank is 20 m above the water level in the sump. Pump centerline is 4 m above the water level in the sump. The suction pipe is 100 mm diameter and discharge pipe is 65 mm diameter. Head loss at suction is 3 times the velocity head in the suction line and head loss at discharge is 20 times the velocity head in the discharge pipeline. Other data are as follows: p = 75 % m = 80 % E = 220 Volts Motor – 3-Phase Power Factor = 0.92 Requirements: a. Sketch of the problem b. Total dynamic head in m c. Water Power in KW d. Motor power in KW e. Line current drawn by the motor in amperes f. Total power cost per day for 10 hours a day continuous operation and a power costs of P 5.00/KW-hr 105 KPa 2
20 65 mm
100 mm
4m
1
Q Av Q A Q vs A s v
0.015
4
vd
Q A s
vs
2g vd
m sec
4.52
(0.065 ) 2
(1.91)2 2(9.81)
0.186 m
( 4.52)2 2(9.81)
1.041m
2
2g
(0.10 )2 0.015
4 2
1.91
v s 2 hLs 3 2g 0.56 m v d 2 hLd 20 2g 20.83 m
m sec
HL ht
0.56 20.83 21.39 m
P2
P1 v 2 2 v 12 Z 2 Z1 HL 2g
at 1 to 2, datum line through point 1
105 - 0
0 (20 0) 20.83 9.81 h t 52.091 m ht
WP 0.015(9.81)(52.091) 7.67 KW 7.67 10.22 KW 0.75 10.22 12.8 KW MP 0.75 BP
MP
3 EI(P.F.) 1000
3 (220)I(0.92) 1000 I 36.44 amperes
12.8
Cost 12.8(10 )(5) P639.00 Example No. 8 Figure below shows a siphon discharging oi l (sp gr 0.90). The siphon is composed of 8 cm. pipe from A to B followed by 10 cm. pipe from B to the open discharge at C. The head l osses are from 1 to 2, 0.34 m; from 2 to 3, 0.2 m; from 3 to 4;0.8 m. Compute the discharge in L/sec
4.5 m
3m
0
Example No. 9 The liquid in the figure has a specific gravity of 1.5. The gas pressure P A is 35 KPa and PBis -15 KPa. The orifice is 100 mm in diameter with Cd = Cv = 0.95. Determine the velocity in the jet a nd the discharge when h = 1.2. (9.025 m/sec; 0.071 m 3/sec) P A
Without considering head lost
·1
P1
h
PB
2
v1 2g
2
v2 2g
Z2
Datum at point 1 : Z1
0 ; Z 2 - h; v 1 0 2
·2
Z1
P2
v2 2g v2
P1 P2
Z1 Z 2
P P 2g 1 2 h theoretical velocity
1.5(9.81) 14.715 v2
KN m3
35 15 v t 2(9.81) 1.2 9.5 m/sec 14.715
Example no. 10 A pump draws water from reservoir A and lifts it to reservoir B as shown in the figure. The loss of head from A to 1 is 3 times the velocity head in the 150 mm pipe and the loss of head from 2 to B is 20 times the velocity head in the 100 mm pipe. Compute the horsepower output of the pump and the pressure heads at 1 and 2 when the discharge is 20 L/sec. (FP= 20.73 HP; 5.74 m ; 84.3 m) 2·
72 m
100 mm
1· A
6m
B
I-1
150 mm
I-2
V-1
V-2
pump
20 m3 0.020 1000 sec Q v A 0.020 1.13 m/sec vs 2 (0.150) 4
Q
vd
0.020
4
Water Power 0.020(9.81)(78.63) Water Power 15.43 KW 20.7 HP At 1 to A (Point 1 on datum)
P1
2.55 m / sec
(0.100)2
vs2 hLs 3 2g 0.20 m v d2 hLd 20 2g 6.63 m
2
000 PA
2
v1 P v Z1 A S ZB hLs 2g 2g PA
(1.13)2 6 0.20 2(9.81)
5.73 meters ; PA 56.2 KPa
At B to 2 (Point B on datum) PB
HL hLs hLd 6.83 m
2
2.55)2 0 0 0 78 6.63 2(9.81)
2
v P v 1 Z1 ht 2 2 Z2 HL 2g 2g
v1 and v2 are negligible P1 P2 0 gage (Atmospheric) ht Z2 Z1 HL ht 72 6.83 78.63 m
2
PB
At point 1 to 2 P1
2
v P v d ZB 2 2 Z2 hLd 2g 2g
PB
84.3 meters; PB 827 KPa
Example No. 11 The diameters of the suction and discharge pipe of a fuel pump for a day tank of a gasoline engine are 150 mm and 100 mm, respectively. The discharge pressure gauge located 10 m above the pump centerline reads 320 KPa and the suction pressure gauge which is 4 m below the pump centerline reads a vacuum of 60 KPa. Head losses due to pipe friction, turbulence and fittings amounts to 15 m. If gasoline with a relative density S = 0.75 is pumped at the rate of 35 L/sec, find a. The total dynamic head developed by the pump b. The fluid power in KW c. The brake or shaft power delivered to the fluid for a pump efficiency of 75% d. The brake torque if the pump speed is 1200 RPM e. The electrical power input to the pump motor for a motor efficiency of 92% f. The line current drawn by the motor if the motor is 3 – phase, 240 volts, and 0.9 Power Factor g. The cost of power for 5 hours operation, if electricity costs P 1.50/KW-hr
Q S SW(water) SW g d1 d2 P1 P2 z1 z2 HL FP A1 A2 v1 v2 v1^2/2g v2^2/2g P1/sw P2/sw D(Phead)
0.035 0.75 9.81 7.3575 9.81 0.150 0.100 -60.00 320.00 -4 10 15 75 0.018 0.008 2.0 4.5 0.20 1.01 -8.15 43.49 51.65
m^3/sec KN/m^3 KN/m^3 m/sec^2 m m KPa KPa m m m KW m^2 m^2 m/sec m/sec m m m m m
D(vhead)
0.81
m
D(Zhead) ht Water Power Pump Efficiency Brake Power Motor Efficiency Motor Power E (Volts) Power Factor Phase N(RPM) Torque(N-m) Line Current PowerCost Cost
14 81.46 20.98
m m KW
75.00
%
27.97
KW
92.00
%
30.40 220.00 0.90 3.00 1200.00 222.57 88.65 1.50 228.01
KW Volts Phase RPM N-m Amperes Pesos/KW-hr Pesos
Example No. 12 A jet of liquid is directly vertically upward. At A (Nozzle tip) its diameter is 75 mm and its velocity is 10 m/sec. Neglecting air friction, determine its diameter at a point 4 m above A. d1 v1 P1 P2 g A1 Q SW P1/SW P2/SW v1^2/2g z1
0.075 10 0 0 9.81 0.0044 0.0442 9.81 0 0 5.097 0
m m/sec Kpa KPa m/sec^2 m^2 m^3/sec KN/m^3 m m m m
z2 HL v2^2/2g v2 A2 d2 d2
4 0 9.097 13.36 0.0033 0.065 6.5
m m m m/sec m^2 m cm
Example No. 13 A closed vessel contains water up to a height of 2 m and over the water surface there is air having a pressure of 8.829 N/cm 2 above atmospheric pressure. At the bottom of the vessel there is an orifice of diameter 15 cm. Find the rate of flow of water from orifice if C d = 0.6. h
2
m
P1
88.29
Kpa
g
9.81
m/sec^2
SW
9.81
KN/m^3
P1/SW
9.0
m
d(orifice) A(Orifice) Theoretical Velocity
0.15
m
0.018
m^2
14.69
m/sec
Cd
0.60
Q(m^3/sec)
0.1558
m^3/sec
Q(L/sec)
155.8
L/sec
Example No. 14 The 600 mm pipe shown in the figure conducts water from a reservoir A to a pre ssure turbine, which discharges through another 600 mm pipe into tailrace B. The loss of head from A to 1 is 5 times the velocit y head in the pipe and the loss of head from 2 to B is 0.2 times the velocit y head in the pipe. If the discharge is 700 L/sec , what horsepower is being given up by the water to the turbine and what are the pressure heads at 1 and 2.(FP = 537.4 HP; 53.628 m; 4.75 m)
Q 700 L/sec 0.70 m 3 /sec v
v2 2g
Q A
4(0.70)
(0.6)2
( 2. 5 ) 2 2(9.81)
2.5 m/sec
0.32 m
5(0.320 1.6 m hL2-B 0.20(0.32) 0.064 m HL 1.6 0.064 1.664 meters hLA-1
At A to B P A
v A
2
2g
Z A
PB
vB
2
2g
Z B HL h
With datum line through poin B
0; Z A 60 m P A PB 0 gage v A VB 0 h ( Z A ZB ) HL h 60 1.664 58.336 meters WP Qh 0.70(9.81)(58.336) 400.6 KW 537 HP ZB
At A to 1 P A
v A
2
Z A
2g
P1
2
v1
2g
Z1 hLA 1
with datum through point A ZA
0 ; Z1 -55.5 m
000 P1
P1
P1
0.32 55.5
55.5 0.32 55.18 meters 55.18(9.81) 541.3 KPa
At 2 to B, with datum through point 2 Z2 P2
P2
P2
P2
0 ; ZB -4.5 m
v2
2
2g
Z2
PB
vB
2
2g
ZB hL 2B
0.32 0 0 0 4.5 0.064 4.5 0.32 0.064 4.756 meters -4.756(9.81) -46.7 KPa 46.7 KPa vacuum
FLUID MECHANICS (PRE – FINAL S4) March 15, 2017 Name ____________________________________ 4. A piece of wood of S = 0.651 is 8 cm square and 150 cm long. How many kilograms of lead weighing 11,200 kg/m3 must be fastened at one end of the stick so that it will float upright with 30.5 cm out of water? S 0.651
wood 0.651(1000) 651
kg
m3 V1 0.08(0.08 )(1.5) 0.96 m3
0.96(651) 624.96 kg W2 11,200V2 BF1 (1.5 0.305)1000 1195 kg BF2 1000V2 W1 W2 BF1 BF2 624.96 11,200V2 1195 1000V2 W1
V2
1195 - 624.96 0.056 m 3 10,200
W2 625.93 kg 5. A barge is loaded with 150 Metric tons of coal. The weight of the empty barge in air is 35 Metric ton. If the barge is 5.5 m wide, 16 m long and 3 m high, what is its draft. (Depth below the water surface)
BF W1 W2 Vs (150 35)(1000) 1000(5.5)(16)h h 2.1 m W
6. A prismatic object 20 cm thick by 20 cm wide by 40 cm long is weighed in water at a depth of 50 cm and found to weigh 50 N. What is its weight in air and its specific gravity?
W2 50 Newton V1 (0.20 )(0.20 )(0.4) 0.016 m3 W1
1
m1 V1
m1 V1 S 1000 W1
W1 W1 W2
m1g Newton
W1 gV1 1000
W1 W1 W2
W1 9810V1
W1 W1 W2
1 9810V1
1 W1 W2
W1 W2
9,810(0.016 ) W1 156.96 50 W1 206.96 Newton S
206.96 206.96 - 50
1.312
In a hydroelectric power plant, the water surface on the crest of the dam is at elevation 75.3 m while the water surface just at the outlet of the head gate is at elevation 70.4 m. T he head gate has 5 gates of 0.91 m x 0.91 m leading to the penstock and are fully opened. Assume 61% as coefficient of discharge, determine a. The quantity of water that enters the hydraulic turbine in m3/sec b. The KW power that the turbine will developed, assuming eturbine = 90% efficiency and the turbine is 122 m below the entrance of the penstock c. The brake torque for a speed N = 1800 rpm d. The number of generator poles if f = 60 Hertz e. The electrical power developed by the generator if electrical and windage loses amounts to 18% h 75.4 70.4 5 m head producing the flow
2gh theoretical velocity Q 5(Av) theoretical flow for 5 gates Q' CdQ Actual flow v
Q' 0.61(5)(0.91x0.91) 2(9.81)5
25.016
m3 sec
total dynamic head h t 122 5 127 m BP eTurbine Q' h t 25.016(9.81)(127 )(0.90 ) 28,050.02 KW 2TN BP ht
60,000
148,810 N m GP 122, 024.16 KW T
FINAL EXAM (March 25, 2017) SET 2 NAME _____________________________________
1. An object weighs 25.95 N when submerged in kerosene (S = 0.81) and weighs 26.6 N when submerged in gasoline (S = 0.68). Determine the specific weight of the object.
6.23(9.81) 61.116
KN m3
2. Determine the water power and mechanical efficiency of a centrifugal pump which has an input of 3 KW. If the pump has a 203 mm diameter suction line and a 152 mm diameter discharge line and handles 10 L/sec of water at 66°C ( =980 kg/m3; = 9.6 KN/m3). The suction line gauge shows 102 mm Hg vacuum and the discharge gauge shows 180 KPa. The Discharge gauge is located 61 cm above the center of the discharge pipeline and the pump inlet and discharge lines are at the same elevation. BP D1 D2 Q A1 A2 v1 v2 g v1^2/2g v2^2/2g SW P1 P2 z2 z1 (P2-P1)/sw (v2^2v1^2)/2g (z2-z1) HL Ht WP
3 0.203 0.152 0.01 0.032 0.018 0.309 0.551 9.810 0.005 0.015 9.6 -13.60 180 0.61 0 20.17
KW m m m^3/sec m^2 m^2 m/sec m/sec m/sec^2 m m KN/m^3 Kpa KPa m m m
0.01
m
0.61 0.00 20.79 1.996
m m m KW
em
66.52
%
3. A piece of wood of S = 0.651 is 8 cm square and 150 cm long. How many kilograms of lead weighing 11,200 kg/m3 must be fastened at one end of the stick so that it will float upright with 30.5 cm out of water? S 0.651
wood 0.651(1000) 651
kg
m3 V1 0.08(0.08 )(1.5) 0.96 m 3
0.96(651) 624.96 kg W2 11,200V2 BF1 (1.5 0.305)1000 1195 kg BF2 1000V2 W1 W2 BF1 BF2 624.96 11,200V2 1195 1000V2 W1
V2 W2
1195 - 624.96 0.056 m 3 10,200
625.93 kg
4. At one point in a pipeline the water speed is 3 m/sec and the gage pressure is 50 KPa. Find the gage pressure at a second point in the line, 11 m lower than the f irst, if the pipe diameter at the second point is twice at the first.
FLUID MECHANICS (PRE – FINAL S2) March 04, 2017 Name ____________________________________ 1. The diameters of the suction and discharge pipe of a fuel pump for a day tank of a gasoline engine are 150 mm and 100 mm, respectively. The discharge pressure gauge located 10 m above the pump centerline reads 320 KPa and the suction pressure gauge which i s 4 m below the pump centerline reads a vacuum of 60 KPa. Head losses due to pipe friction, turbulence and fittings amounts to 15 m. If gasoline with a relative density S = 0.75 is pumped at the rate of 35 L/sec, find a. b. c. d. e. f. g.
The total dynamic head developed by the pump The fluid power in KW The brake or shaft power delivered to the fluid for a pump efficiency of 75% The brake torque if the pump speed is 1200 RPM The electrical power input to the pump motor for a motor efficiency of 92% The line current drawn by the motor if the motor is 3 – phase, 240 volts, and 0.9 Power Factor The cost of power for 5 hours operation, if electricity costs P 1.50/KW-hr
Q S SW(water) SW g d1 d2 P1 P2 z1 z2 HL FP A1 A2 v1 v2 v1^2/2g v2^2/2g P1/sw P2/sw D(Phead) D(vhead)
0.035 0.75 9.81 7.3575 9.81 0.150 0.100 -60.00 320.00 -4 10 15 75 0.018 0.008 2.0 4.5 0.20 1.01 -8.15 43.49 51.65 0.81
m^3/sec KN/m^3 KN/m^3 m/sec^2 m m KPa KPa m m m KW m^2 m^2 m/sec m/sec m m m m m m
D(Zhead) ht Water Power Pump Efficiency Brake Power Motor Efficiency Motor Power E (Volts) Power Factor Phase N(RPM) Torque(N-m) Line Current PowerCost Cost
14 81.46 20.98
m m KW
75.00
%
27.97
KW
92.00
%
30.40 220.00 0.90 3.00 1200.00 222.57 88.65 1.50 228.01
KW Volts Phase RPM N-m Amperes Pesos/KW-hr Pesos
2. A jet of liquid is directly vertically upward. At A (Nozzle tip) its diameter is 75 mm and its velocity is 10 m/sec. Neglecting air friction, determine its diameter at a point 4 m above A.
d1 v1 P1 P2 g A1 Q SW P1/SW P2/SW v1^2/2g z1 z2 HL v2^2/2g v2 A2 d2 d2
0.075 10 0 0 9.81 0.0044 0.0442 9.81 0 0 5.097 0 4 0 9.097 13.36 0.0033 0.065 6.5
m m/sec Kpa KPa m/sec^2 m^2 m^3/sec KN/m^3 m m m m m m m m/sec m^2 m cm
3. A closed vessel contains water up to a height of 2 m and over the water surface there is air having a pressure of 8.829 N/cm 2 above atmospheric pressure. At the bottom of the vessel there is an orifice of diameter 15 cm. Find the rate of flow of water from orifice if Cd = 0.6. h P1 g SW P1/SW d(orifice) A(Orifice) Theoretical Velocity Cd Q(m^3/sec) Q(L/sec)
2 88.29 9.81 9.81 9.0 0.15 0.018
m Kpa m/sec^2 KN/m^3 m m m^2
14.69
m/sec
0.60 0.1558 155.8
m^3/sec L/sec
Example No. 15 A Francis turbine is installed with a vertical draft tube. The pressure gauge located at the penstock leading to the turbine casing reads 372.6 KPa and velocity of water at inlet is 6 m/sec. The discharge is 2.5 m 3/sec. The hydraulic efficiency is 85%, and the overall efficiency is 82%. The top of the draft tube is 1.5 m below the centerline of the spiral casing, while the tailrace level is 2.5 m from the top of the draft tube. There is no velocity of whirl at the top or bottom of the draft tube and leakage losses are negligible. Calculate, a) the net effective head in meters (43.817 m) b) the brake power in kw. (881.2 kw) c) the plant output for a generator efficiency of 92%. (810.7 kw) d) the mechanical efficiency (96.550)
GIVEN: P1 = 372.6 KPa v = 6 m/sec Q = 2.5 m 3/sec eh = 0.85 e = 0.82 ZB = (1.5 + 2.5) = 4 m at point b to 2 (datum line through point 2)
P A
2
v A 2g
Z A
P2
2
v2 2g
Z2 hL A 2 h
0 v2 0
P2
Z2 0
0 (negligible)
hL A 2
2
h
P A
h
372.6 9.81
v A 2g
Z A ( 6) 2 4 2(9.81)
h 43.817 m WP Qh WP 2.5(9.81)( 43.817 ) 1074.6 kw BP 1074.6(0.82) 881.2 KW PLANT OUTPUT
881.2(0.92) 810.7 KW
e em eh e v 0.82 em (0.85)(1) em
0.82 0.85
0.965 96.5%
Example no. 16 A 4 m3/hr pump delivers water to a pressure tank. At the start, the gauge reads 138 KPa until it reads 276 KPa and then the pump was shut off. The volume of the tank is 160 Liters. At 276 KPa, the water occupied 2/3 of the tank volume. a) Determine the volume of water that can be taken out until the gauge reads 138 KPa. b) If 1 m 3/hr of water is constantly used, in how many minutes from 138 KPa will the pump run until the gauge reads 276 KPa.
138 101.325 239.325 KPa P2 276 101.325 377.325 KPa P1
1 3 P1V1 P2 V2 V2
(0.160) 0.0533 m3
0.084 m3 Vw@138 (0.160 - 0.084) 0.076 76 Liters V1
2 (0.16) 0.1067 106.7 Liters 3 Vtaken out 0.1067 0.076 0.0307 30.7 Liters Vw @ 276
4t 1t t
0.0307
0.01023 hrs 0.614 min 36.84 sec
