This document contains diagrams and questions related to fluid mechanics: 1) It shows diagrams of different devices moving in fluid and asks whether each will move in the positive or negative x-direction assuming equal pressure at the entrance and exit. 2) It shows a diagram of a sprinkler and asks to determine the torque required to prevent its rotation given the fluid velocity and distance from the point of rotation. 3) It shows a diagram of a vane moving through fluid and asks to determine the force on the vane given its angle, velocity through the fluid, and the fluid velocity striking it. 4) It asks which factors the pressure at the summit of a siphon depends on out of liquid density

2. L
Oil, depth “h”
Moving bell, width “b”
V

3. A B
Mercury
Air, 180 Pa abs.
Water
h
80 cm

4. C
D
Plan View
Water
B
4 cm
3 cm
B.C
D

5. 25 m/ S 25 m/ S
F
Plate
d2 = 4 cm
d1 = 6 cm
C.V

6. W. S
R
B
C
D
A

7. y2
2 m Orifice “1”
Orifice “2”
W. S
10 m

8. Z = 25
m
Z = 0
L = 70 m
Valve
L = 100 m
10 m
30 m
Water
A B
C

9. (a) (b) (c) (d)
If the motion of devices, shown in the figure, is restricted to x-direction
alone, state whether each will move in +x or –x direction. Assume
pressure at entrance and exit to be essentially same.

10. Determine the torque about point “O” required to prevent the rotation
of sprinkler, shown in the figure. (2 A  𝑈2 ∙ 𝑥)
𝜔
x
x
U
U

11. A vane with a turning angle of 60𝑜 moves with a constant velocity of
6.0 m/s as shown in the figure. Water flowing out from a nozzle 60 mm
diameter strikes the vane tangentially at 30m/s. Determine the force
on the vane?
(1624.425 N @ -60o with x-axis). `
600
6.0 m/s
30 m/s

12. If all the losses are neglected, the pressure at the summit of a siphon
is:
a) Independent of liquid density
b) Is independent of Q in siphon
c) Depends on height of summit above reservoir level, velocity
and mass density.
If head over the triangular notch is doubled, discharge will increase to:
a) 2 Q
b) 2.828 Q
c) 5.657 Q
d) 4 Q.

13. A body floats in between the plane of separation of two fluids of unit
weights 𝛾1 and 𝛾2 and if 𝑉1 and 𝑉2 are the volumes of the body in the
two fluids, the buoyant force is given by:
a) FB = γ1 ∙ V1 − γ2 ∙ V2
b) FB = γ1 ∙ V1 /γ2 ∙ V2
c) FB = γ1 ∙ V1 + γ2 ∙ V2
d) FB = γ1 ∙ V2 +γ2 ∙ V1