Fluid Mechanics and Hydraulic Machinery Miscellaneous
 Velocity vector of a flow field is given as = 2xyî  x²zĵ. The vorticity vector at (1, 1, 1) is

View Hint View Answer Discuss in Forum
v = 2xyi – x 2zj
Velocity of a vector = ∇ × v= î 0  δ (  x²z)  î 0  δ (2xy)  k̂ 0  δ (x²z)  δ (2xy) δz δz δx δy
= î(x²) + k̂(2 × z  2X)
At(1 , 1 , 1),
∇ × v = î + k̂ (2 2) = i  4kCorrect Option: D
v = 2xyi – x 2zj
Velocity of a vector = ∇ × v= î 0  δ (  x²z)  î 0  δ (2xy)  k̂ 0  δ (x²z)  δ (2xy) δz δz δx δy
= î(x²) + k̂(2 × z  2X)
At(1 , 1 , 1),
∇ × v = î + k̂ (2 2) = i  4k
 A Prandtl tube (Pitotstatic tube with C = 1) is used to measure the velocity of water. The differential manometer reading is 10 mm of liquid coiumn with a relative density of 10. Assuming g = 9.8 m/s², the velocity of water (in m/s) is _____ .

View Hint View Answer Discuss in Forum
Velocity as water = C_{v} √2gh
C_{v} = 1 (Given)h = x s_{g}  1 s_{0}
= 0.01 (10 – 1) = 0.09 m
∴ velocity of flow = √2 × 9.8 × 0.09 = 1.328 m/ sCorrect Option: C
Velocity as water = C_{v} √2gh
C_{v} = 1 (Given)h = x s_{g}  1 s_{0}
= 0.01 (10 – 1) = 0.09 m
∴ velocity of flow = √2 × 9.8 × 0.09 = 1.328 m/ s
 A streamline and an equipotential line in a flow field

View Hint View Answer Discuss in Forum
dy × dy =  1 dx _{φ} dx _{ψ}
Slope of equipotential Line × slope of stream function = – 1
They are orthogonal to each other.Correct Option: B
dy × dy =  1 dx _{φ} dx _{ψ}
Slope of equipotential Line × slope of stream function = – 1
They are orthogonal to each other.
 The arrangement shown in the figure measures the velocity V of a gas of density 1 kg/m³ flowing through a pipe. The acceleration due to gravity is 9.81 m/s². If the manometric fluid is water (density 1000 kg/m³) and the velocity V is 20 m/ s, the differential head h(in mm) between the two arms of the manometer is _________.

View Hint View Answer Discuss in Forum
Dynamic pressure of gas = (δgh)_{water}
1 δ_{gas}v_{2} = δ_{m} × 9.81 × h 2 1 × 1 × (20)² = 1000 × 9.81 × h 2
h = 0.02038 m of water
h = 20.38 mm of waterCorrect Option: D
Dynamic pressure of gas = (δgh)_{water}
1 δ_{gas}v_{2} = δ_{m} × 9.81 × h 2 1 × 1 × (20)² = 1000 × 9.81 × h 2
h = 0.02038 m of water
h = 20.38 mm of water
 The Newtonian fluid has the following velocity field:
= x²yî + 2xy²ĵ  yz³k̂
The rate shear deformation ε_{yz} at the point x = –2, y = –1 and z = 2 for the given flow is

View Hint View Answer Discuss in Forum
V = x⊃î + 2xy²zĵ  yz²̂k
Shear strain_{yz} = 1 δv + δw = 1 (2xy²  z³) 2 δz δy 2
at (–2π, 2), we get_{yz} = 1 ( 4  8)) =  6 2 Correct Option: A
V = x⊃î + 2xy²zĵ  yz²̂k
Shear strain_{yz} = 1 δv + δw = 1 (2xy²  z³) 2 δz δy 2
at (–2π, 2), we get_{yz} = 1 ( 4  8)) =  6 2