Test: Fluid Kinematics Level - 1 - Mechanical Engineering MCQ

u = 0.5 + 0.8x = 0

∴ x = −0.5 / 0.8 = −0.625

v = 1.5 − 0.8y = 0

∴ y = 1.5 / 0.8 = 1.875

i.e there is only one stagnation point at (−0.625, 1.875)

As per definition of stream function

u = ∂ψ / ∂y and v = −∂ψ / ∂x

∴ u = −3 & v = −4 So V- = −3î − 4ĵ

and magnitude of resultant velocity is given |⃗V⃗ | = √u2 + v2 = 5

Similarly for constant flow rate through a straight, uniform diameter pipe, there is no change of velocity, (neither magnitude nor direction) so total acceleration is zero.

For a flow through bend where the discharge is not constant, the velocity changes both due to geometry of pipe and with time, so both local and convective acceleration exist.

For variable discharge through a straight pipe, the velocity will change with time, but it will be same across the whole pipe at any particular time instant, i.e the flow is unsteady but uniform, only local acceleration exists.

∂2ψ / ∂x2 + ∂2ψ / ∂y2 = 0

Checking all the given options, we find option (A) satisfies the equation.

we know u = −∂ϕ / ∂x and v = −∂ϕ / ∂y

∴ u = −(2x), v = −(−2y)

⇒ u = −2x, v = 2y

Thus at (1, 1),⃗V⃗ = −2î + 2ĵ |⃗V⃗ | = √4 + 4 = 2 × 21/2

For incompressible fluid, A₁V₁ = A₂V₂ + A₃V₃ + A₄V₄

1000 = 10 V₂ + 450 + 200

V₂ = 35 cm/s

u = ∂ψ / ∂y and v = −∂ψ / ∂x

So u = 3x, v = −3y

at point (2, 3)

⃗V = 6î − 9ĵ and |⃗V⃗ | = √62 + 92 = 10.816

≈ 10.82 Units

Ωz = 0

Checking option (A)

Ωz = ∂v / ∂x −∂u / ∂y =3/2 − 1 ≠ 0

Checking option (B)

Ωz = ∂v / ∂x −∂u / ∂y = 2xy − 2xy = 0

If ρ = C

i. e. ρ ≠ ρ(x, y, z, t)

Hence equation ① is reduced to V ∙⃗V = 0 for incompressible flow.

ax = (2x2 + y2)(4x) + (−4xy)(2y) + 0 + 0

∴ at point (1, 2) ax

= (2 + 4)(4) − (8)(4) ax

= 24 − 32 ax

= −8 units

i. e u = x, v = −y

For a streamline in 2D flow,

dy / dx =v / u ⟹ dy / dx =−y /x

⇒ ln y = ln 1 / x + C

at (1, 2) ln 2 = ln 1 + C

C = ln 2

So, ln y = ln 2 / x

or xy = 2 or xy − 2 = 0

ωz = 1/[∂v / ∂x − ∂u / ∂y ]

Therefore in the given case ε̇ xy = 2 ∂u / ∂y and ωz = 0

Thus along with translation, elements will deform but shall not rotate.