Download Fluids Assignment 2 – Hydrostatics and more Quizzes Fluid Mechanics in PDF only on Docsity!
IN FIG. P2.1, IN WHICH FLUID WILL A PRESSURE OF 650 kPa WILL BE ACHIEVED?
SOLUTION:
REQUIRED:
Fluid where 650 kPa will be achieved.
SOLUTION:
P
1
= P
A
- Σγh ; Up to bottom of Ethyl Alcohol
P
1
= 90 , 000 Pa + ( 773. 3 kg/m
3
)( 9. 81 m/s
2
)( 60 m)
P
1
= 545 , 164. 38 Pa or 545. 164 kPa
- 164 kPa < 650 kPa
P
2
= P
A
- Σγh ; Up to bottom of Oil
P
2
= P
1
oil
h
oil
P
2
= 545 , 164. 38 Pa + ( 899. 6 kg/m
3
)( 9. 81 m/s
2
)( 10 m)
P
2
= 633 , 415. 14 Pa or 633. 415 kPa
- 415 kPa < 650 kPa
P
3
= P
A
- Σγh ; Up to bottom of Water
P
3
= P
2
water
h
water
P
3
= 633 , 415. 14 Pa + ( 9810 N/m
3
)( 5 m)
P
3
= 682 , 465. 14 Pa or 682. 465 kPa
- 465 kPa > 650 kPa
CONCLUSION:
Since the pressure at the bottom of the oil is less than 650 kPa and the pressure at the
bottom of the water is greater than 650 kPa. The fluid where 650 kPa will be achieved is
WATER.
FOR A GAGE READING - 17.1 kPa , DETERMINE THE ELEVATIONS OF THE LIQUIDS IN THE OPEN
PIEZOMETER COLUMNS E, F AND G AND THE DEFLECTION (H) OF THE MERCURY IN THE U-TUBE
MANOMETER. (SEE FIG. P2.2)
SOLUTION:
GIVEN:
P
gage reading
= − 17. 1 kPa
REQUIRED:
Elevation of the fluids − Elev
E
, Elev
F
, Elev
G
, h
SOLUTION:
P
E
= P
gage
0 = − 17 , 100 Pa + ( 0. 7 )( 9810 N/m
3
)(h)
h = 2. 490 m
Elev
E
= 15 m − h Elev
E
= 12. 510 m
P
1
= P
gage
- Σγh = − 17 , 100 Pa + ( 0. 7 )( 9810 N/m
3
)( 3 m)
P
1
= 3501 Pa P
1
= P
F
F
3501 Pa = ( 9810 N/m
3
)(h) h = 0. 357 m
Elev
F
= 12 m + h Elev
F
= 12. 357 m
P
2
= P
1
- Σγh = 3501 Pa + ( 9810 N/m
3
)( 4 m)
P
2
= 42 , 741 Pa P
2
= P
G
G
42 , 741 Pa = ( 1. 6 )( 9810 N/m
3
)(h) h = 2. 723 m
Elev
G
= 8 m + h Elev
G
= 10. 723 m
0 = P
2
3
)( 8 m − 4 m) − ( 13. 6 )(( 9810 N/m
3
)(h
mercury
0 = 42741 + ( 9810 N/m
3
)( 8 m − 4 m) − ( 13. 6 )(( 9810 N/m
3
)(h
mercury
h
mercury
= 0. 614 m
THE HYDRAULIC JACK SHOWN IN FIG 2.4 IS FILLED WITH OIL AT 55lb/ft
3
. NEGLECTING THE
WEIGHT OF THE TWO PISTONS, WHAT FORCE F (lb) ON THE HANDLE IS REQUIRED TO SUPPORT
THE 220 - lb WEIGHT?
SOLUTION:
GIVEN:
W = 2200 lb
SOLUTION:
P
1
= P
2
Let F
2
be the force applied in 1 in ∅
F
1
A
1
F
2
A
2
2200 lb
π ×
( 3 in)
2
F
2
π ×
( 1 in)
2
F
2
= 244. 444 lb
ΣM
A
0 = −F
2
1 in
F = 14. 379 lb
CALCULATE THE PRESSURE DIFFERENCE BETWEEN POINTS A AND B. (SEE FIG 2.5)
SOLUTION:
REQUIRED:
P
A
− P
B
SOLUTION:
Let x be the distance between the 24 in and 12 in measurements
P
A
= P
B
- 4 lb ft
3
24 in + x
1 ft
12 in
- 4 lb ft
3
12 in
1 ft
12 in
− ( 0. 92 )( 62. 4 lb ft
3
⁄ )( 12 in) (
1 ft
12 in
− ( 0. 92 )( 62. 4 lb ft
3
⁄ )( 24 in + x) (
1 ft
12 in
3
)( 24 in) (
1 ft
12 in
P
A
− P
B
= 906. 048 lb/ft
2
760 mmHg IS EQUIVALENT TO WHAT PRESSURE HEAD OF WATER (m)?
SOLUTION:
GIVEN:
P = 760 mmHg
SOLUTION:
h =
P
γ
h =
760 mmHg (
101325 Pa
760 mmHg
( 9810 N/m
3
h = 10. 329 m
