Refresher - HYDRAULICS Quiz 1
PROBLEM 1-3:
A jet of water 25 mm in diameter and having a velocity of 7.5 m/s strikes against a
plate at right angles.
➀ Determine the force on the plate if the plate is fixed.
➁ Determine the force on the plate if the plate is moving in the same direction
as the jet with a uniform velocity of 3 m/s.
V’=0
Solution:
➀ Force on the plate if the plate is fixed: V1=7.5 m/s
Q = AV
π
Q = (0.025)2 (7.5) = 0.00368 m3 / s
4
QW 0.00368(9810)
F= (V1 - V2 ) = (7.5 - 0)
g 9.81
F = 27.6 N
➁ Force on the plate if the plate is moving in the same direction as the jet with a
uniform velocity of 3 m/s.
U = V1 - V1
V =3.0
’
U = 7.5 - 3 = 4.5 m/s
Q = AU
V1=7.5 m/s
π
Q' = (0.025)2 (4.5) = 0.0022 m3 / s
4
QW 0.0022(9810)
F= (V1 - V1 ) = (7.5 - 3)
g 9.81
F = 9.94 N
, Refresher - HYDRAULICS Quiz 1
PROBLEM 3-5:
A water tank has a sloping inclined at 45˚ with the horizontal. The total depth of water in the tank is 6.7 m.
A water jet issues from an orifice located on the inclined side of the tank under a hydrostatic head of 4 m.
or that orifice is located 2.7 m. vertically above the bottom of the tank. Coefficient of velocity is 1.0.
Negcting air resistance on the jet.
➀ Determine the maximum height of the issuing jet arises above the level of the center of orifice in
meters.
➁ Determine the time it takes for this particle of the jet to sit the ground that is 1.20 m. below the
bottom of the tank in seconds.
③ Determine the horizontal distance on the ground traveled by the jet from the center of the orifice in
meters.
Solution:
➀ Max. height the jet arises above the center of orifice:
V = Cv 2gH
w.s.
V1
V = 1.0 2(9.81)(4) V 2 =0
V
4m V
V = 8.86 m/s h 45˚
45˚
V1 = 8.86 Sin45˚= 6.26 2.7m V3
45˚ 45˚
V2 2 = V1 2 – 2gh
0 = (6.26)2 – 2(9.81)h 1.2m
h = 2 m.
➁ Time for the particle to sit the ground: w.s.
V2 = V1 – gt1 V1 V 2 =0
V
0 = 6.26 – 9.81t1 4m
B
t1 = 0.64 sec. A
45˚
Time to travel from A to B is t1 = 0.64 sec. 2.7m 45˚
V3
5.9
Time to travel from B to C: 1.2m
5.9 = ½ gt2 2 C
5.9 = ½ (9.81)t2 2
t2 = 1.1 sec.
Total time for the particle to sit the ground = 0.64 + 1.1 = 1.74 sec.
③ Horizontal distance:
V3 = 8.86 Cos 45˚ = 6.26 m/s
D = V3 t
D = 6.26(1.74) = 10.89 m.
PROBLEM 1-3:
A jet of water 25 mm in diameter and having a velocity of 7.5 m/s strikes against a
plate at right angles.
➀ Determine the force on the plate if the plate is fixed.
➁ Determine the force on the plate if the plate is moving in the same direction
as the jet with a uniform velocity of 3 m/s.
V’=0
Solution:
➀ Force on the plate if the plate is fixed: V1=7.5 m/s
Q = AV
π
Q = (0.025)2 (7.5) = 0.00368 m3 / s
4
QW 0.00368(9810)
F= (V1 - V2 ) = (7.5 - 0)
g 9.81
F = 27.6 N
➁ Force on the plate if the plate is moving in the same direction as the jet with a
uniform velocity of 3 m/s.
U = V1 - V1
V =3.0
’
U = 7.5 - 3 = 4.5 m/s
Q = AU
V1=7.5 m/s
π
Q' = (0.025)2 (4.5) = 0.0022 m3 / s
4
QW 0.0022(9810)
F= (V1 - V1 ) = (7.5 - 3)
g 9.81
F = 9.94 N
, Refresher - HYDRAULICS Quiz 1
PROBLEM 3-5:
A water tank has a sloping inclined at 45˚ with the horizontal. The total depth of water in the tank is 6.7 m.
A water jet issues from an orifice located on the inclined side of the tank under a hydrostatic head of 4 m.
or that orifice is located 2.7 m. vertically above the bottom of the tank. Coefficient of velocity is 1.0.
Negcting air resistance on the jet.
➀ Determine the maximum height of the issuing jet arises above the level of the center of orifice in
meters.
➁ Determine the time it takes for this particle of the jet to sit the ground that is 1.20 m. below the
bottom of the tank in seconds.
③ Determine the horizontal distance on the ground traveled by the jet from the center of the orifice in
meters.
Solution:
➀ Max. height the jet arises above the center of orifice:
V = Cv 2gH
w.s.
V1
V = 1.0 2(9.81)(4) V 2 =0
V
4m V
V = 8.86 m/s h 45˚
45˚
V1 = 8.86 Sin45˚= 6.26 2.7m V3
45˚ 45˚
V2 2 = V1 2 – 2gh
0 = (6.26)2 – 2(9.81)h 1.2m
h = 2 m.
➁ Time for the particle to sit the ground: w.s.
V2 = V1 – gt1 V1 V 2 =0
V
0 = 6.26 – 9.81t1 4m
B
t1 = 0.64 sec. A
45˚
Time to travel from A to B is t1 = 0.64 sec. 2.7m 45˚
V3
5.9
Time to travel from B to C: 1.2m
5.9 = ½ gt2 2 C
5.9 = ½ (9.81)t2 2
t2 = 1.1 sec.
Total time for the particle to sit the ground = 0.64 + 1.1 = 1.74 sec.
③ Horizontal distance:
V3 = 8.86 Cos 45˚ = 6.26 m/s
D = V3 t
D = 6.26(1.74) = 10.89 m.
