## Fluid Mechanics and Hydraulic Machinery Miscellaneous

#### Fluid Mechanics and Hydraulic Machinery

1. Match the following pairs:
EquationPhysical Interpretation
P. ∇ × v̄ = 0I. Incompressible continuity equation
Q. ∇ . v̄ = 0II. Steady flow
R. ∆v̄/∆t = 0III. Irrotational flow
S. δv̄/δt = 0IV. Zero acceleration of fluid particle

1. P-III, Q-I, R-IV, S-II

##### Correct Option: C

P-III, Q-I, R-IV, S-II

1. For an in-compressible flow field, V, which one of the following conditions must be satisfied?

1. In compressible flow condition
∇.v̄ = 0

##### Correct Option: A

In compressible flow condition
∇.v̄ = 0

1. For the continuity equation given by∆ . V̄ = 0 to be valid, where V̄ is the velocity vector, which one of the following is a necessary condition?

1. The basic equation of continuity for fluid flow is given by

 ρu ρv ρw ρ u y z t

Now if ρ remains constant, then only we can write
∇.v̄ = 0
 i.e. δu + δv + δw = 0 δx δy δz

hence incompressible flow

##### Correct Option: D

The basic equation of continuity for fluid flow is given by

 ρu ρv ρw ρ u y z t

Now if ρ remains constant, then only we can write
∇.v̄ = 0
 i.e. δu + δv + δw = 0 δx δy δz

hence incompressible flow

1. Which combination of the following statements about steady incompressible forced vortex flow is correct?
P: Shear stress is zero at all points in the flow.
Q: Vorticity is zero at all points in the flow.
R: Velocity is directly proportional to the radius from the centre of the vortex.
S: Total mechanical energy per unit mass is constant in the entire flow field.
Select the correct answer using the codes given below:

1. Clearly zero shear stress and vortex.

##### Correct Option: B

Clearly zero shear stress and vortex.

1. The 2-D flow with, velocity
V̄ = (x + 2y + 2) î + (4 – y) ĵ is

1. v̄ = (x + 2y +2)î + (4 - y)ĵ
u =x + 2y + 2, v = 4 – y

 ∴ δv = 1, δv = - 1 δx δy

 ∴ δv + 1, δv = 0 δx δy

hence in compressible.
 Again, ω = 1 δv - δv 2 δx δy

 = 1 (0 - 2) = - 1. 2

hence not irrotational.

##### Correct Option: D

v̄ = (x + 2y +2)î + (4 - y)ĵ
u =x + 2y + 2, v = 4 – y

 ∴ δv = 1, δv = - 1 δx δy

 ∴ δv + 1, δv = 0 δx δy

hence in compressible.
 Again, ω = 1 δv - δv 2 δx δy

 = 1 (0 - 2) = - 1. 2

hence not irrotational.