Fluid Mechanics and Hydraulic Machinery Miscellaneous
 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:

View Hint View Answer Discuss in Forum
Clearly zero shear stress and vortex.
Correct Option: B
Clearly zero shear stress and vortex.
 The 2D flow with, velocity
V̄ = (x + 2y + 2) î + (4 – y) ĵ is

View Hint View Answer Discuss in Forum
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.
 Existence of velocity potential implies that

View Hint View Answer Discuss in Forum
Fluid is irrotational
Correct Option: B
Fluid is irrotational
 A closed cylinder having a radius R and height H is filled with oil of density ρ. If the cylinder is rotated about its axis at an angular velocity of ω, the thrust at the bottom of the cylinder is

View Hint View Answer Discuss in Forum
We know that
δp = ρv^{2} = ρ.ω^{2}r^{2} = δω^{2}r δr r r
[∵ v = ω × r]^{p} δp = ^{r} ρω^{2}r dr _{0} _{0} p = ρ ω^{2}r^{2} 2
Area of circular ring = 2πrdr
Force on elementry ring
= Intensity of pressure × Area of ring= ρ ω^{2}r^{2}2πr dr 2
∴ Total force on the top of the cylinder= ^{R} ρ ρω^{2}r^{2}2π r dr _{0} 2 = ρ ρω^{2}r^{2}2π ^{R} r^{3} dr 2 _{0} ρ . ω^{2}2π R^{2} = ρ ω^{2} × πR^{4} 2 4 4
Thrust at the bottom of the cylinder
= Weight of water in cylinder + Total force on the top of cylinder= ρg × πR^{2} × H + ρ ω^{2} × πR^{4} 4 = πR^{2} ρω^{2}R^{2} + ρgH 4 Correct Option: D
We know that
δp = ρv^{2} = ρ.ω^{2}r^{2} = δω^{2}r δr r r
[∵ v = ω × r]^{p} δp = ^{r} ρω^{2}r dr _{0} _{0} p = ρ ω^{2}r^{2} 2
Area of circular ring = 2πrdr
Force on elementry ring
= Intensity of pressure × Area of ring= ρ ω^{2}r^{2}2πr dr 2
∴ Total force on the top of the cylinder= ^{R} ρ ρω^{2}r^{2}2π r dr _{0} 2 = ρ ρω^{2}r^{2}2π ^{R} r^{3} dr 2 _{0} ρ . ω^{2}2π R^{2} = ρ ω^{2} × πR^{4} 2 4 4
Thrust at the bottom of the cylinder
= Weight of water in cylinder + Total force on the top of cylinder= ρg × πR^{2} × H + ρ ω^{2} × πR^{4} 4 = πR^{2} ρω^{2}R^{2} + ρgH 4
 The horizontal and vertical hydro static forces F_{x} and F_{y} on the semicircular gate, having a width w into the plane of figure, are

View Hint View Answer Discuss in Forum
F_{x} = 2ρghrw
F_{y} = w_{2}  w_{1} πr^{2} . ρg × w = πr^{2}ρgw 2 2 Correct Option: D
F_{x} = 2ρghrw
F_{y} = w_{2}  w_{1} πr^{2} . ρg × w = πr^{2}ρgw 2 2