Test: Fluid Kinematics Level - 2 - Mechanical Engineering MCQ

Substituting the given velocity components in LHS.

∂u /∂x + ∂v / ∂y = 1 − 1 = 0

Hence flow is incompressible.

For the flow to be irrotational, Ωz must be zero. i. e ∂v /∂x − ∂u /∂y = 0

Substituting velocity components in LHS

∂v /∂x − ∂u /∂y = 0 − (2) = −2

Hence flow is rotational or not irrotational because vorticity is non-zero.

From definition of stream function (ψ)

∂ψ / ∂y = u and ∂ψ / ∂x = −v

hence u = −2y, v = −2x We also know that ∂ϕ /∂x = −u and ∂ϕ /∂y = −v

i. e ∂ϕ / ∂x = 2y and ∂ϕ / ∂y = 2x

Now dϕ = ∂ϕ / ∂x dx + ∂ϕ ∂y / dy

⟹ dϕ = 2ydx + 2xdy

ϕ = 2yx + 2xy + C

ϕ = 4(xy) + C or

ϕ = 2xy + K

where K is constant.

∂ϕ / ∂x = −u and ∂ϕ / ∂y = −v

hence u = −2x and v = +2y

Also

∂ψ / ∂y = u ⟹ ∂ψ / ∂y = −2x ⋯ ①

and

∂ψ / ∂x = −v ⟹ ∂ψ / ∂x = −2y ⋯ ②

Partially integrating both the equations ① and ② we get

ψ = −2xy + f(x) + C1 ⋯ ③ and

ψ = −2xy + g(y) + C2 ⋯ ④

respectively.

Since both the equations ③ and ④ are same hence f(x) and g(y) will be zero and C1 = C2

∴ ψ = −2xy + constant.

∂Q / ∂t = 0.2 m3/s²

Now we know that Q = A × V or V = Q / A

he local acceleration aL = ∂V / ∂t = ∂(Q⁄A) / ∂t

Since area is not dependent on time, we can write

aL = 1 / (0.5 − 0) × 0.2 = 0.4 m/s²

The height (h) = Volume of fluid / Area of tank

so dh / dt = Rate of change of volume / Area of tank

Assuming Area to be constant

dh / dt = Qin − Qout / Area

dh / dt = Qin − kh / Area

Where, Qin, k and !rea are constant and at t = 0, h = 0

So clearly as h will increase, dh / dt will decrease i.e as h increases, slope of h-T curve will decrease.

The coordinates of point N are (0, 2)

∴ ψN = 0 + (2 + 2) × 22 = 16 The co-ordinates of point M are (3, 0)

∴ ψM = 0 + 0 = 0 The coordinates of point O are (0, 0)

∴ ψO = 0 + 0 = 0

Flow rate across face NO

= 5 (ψN − ψO)

= 5 (16 − 0) = 80 m3⁄s

Flow rate across face MO

= 5 (ψM − ψO)

= 5(0 − 0) = 0 m3⁄s

Flow rate across face NM

= 5(ψN − ψM)

= 5 (16 − 0)

= 80 m3⁄s

∴ u = 16 − 12y & v = −12x

Also dψ = ∂ψ / ∂x dx + ∂ψ /∂y dy

and ∂ψ ∂x = −v and ∂ψ ∂y = u

∴ dψ = 12xdx + (16 − 12y)dy

ψ = 12x2 2 + 16y − 12y2 2

ψ = 6x2 + 16y − 6y2

Hence at point (2, 3) ψ = 6(2)2 + 16(3) − 6(3)2 ψ = 18 units

For a flow a = U ∂U / ∂x +∂U /∂t

As flow is steady

∂U / ∂t = 0

Also it is given, that as particle moves from A to B its velocity increases linearly,

therefore U = ax + b And

hence

∂U/∂x =UB − UA / xB − xA = 5 − 2 / 1 − 0 = 3/sec

Thus, a(x) = U(x) × 3 m/sec Hence a at B

a(xB) = U(xB) ∂U /∂x

a(xB) = 5 × 3

a(xB) = 15 m/sec

Also consider flow of water at an increasing or decreasing flow rate through a straight pipe of uniform cross-section. In this case normal acceleration is zero, even though tangential acceleration is non-zero and a function of time, making the flow unsteady.

A forced vortex represents a rotational flow. Hence statement (II) is false.

By definition of ψ ; ∂ψ /∂y = u and ∂ψ/ ∂x = −v

Therefore u = 3x2 + 2(2 + t)y and v = −6yx

Thus the velocity vector is

⃗V = uî + vĵ

⃗V = [3x2 + 2(2 + t)y]î − 6xyĵ

at r = î + 2ĵ − 3k̂ , i. e. (1, 2, −3)

and t = 2

V = [3(1)2 + 4(2 + 2)]î − 6(1 × 2)ĵ

V = 19î − 12ĵ

∂u/ ∂x + ∂v / ∂y = 0 ⋯ ①

Given u = Ae^x Therefore ∂u / ∂x = Ae^x

Substituting the same in equation ①

Ae^x + ∂v / ∂y = 0

⇒∂v / ∂y = −Ae^x

⇒ v = −Ae^x y + g(x)

∂ρ / ∂t + ∂(ρu) / ∂x = 0

∴ ∂ρ / ∂t = −∂(ρu) / ∂x

Consider the RHS of the above equation.

As it is a partial derivative with respect to space co-ordinate, time terms will be treated as constants. Hence substituting instantaneous density field and velocity field in RHS we get instantaneous rate of change of density.

∂ρ /∂t = −∂ / ∂x (ρ0e−x/L u0e−2x / L)

∂ρ / ∂t = −∂ / ∂x (ρ0 u0e −3x /L )

Therefore at x = L

∂ρ / ∂t = 3 / L ρ0 u0 e^-3

⇒∂u / ∂x + ∂v / ∂y = 3y2 + 2 − 2 − 3y2 = 0 The flow satisfies incompressible continuity equation, so it is incompressible.

Since u and v are not functions of time, flow is steady. Also

∂v / ∂x − ∂u / ∂y = 2x − 6xy − 2y

i.e vorticity is non-zero.

Thus, flow is rotational.

A velocity potential function is defined only for irrotational flow whether it is compressible or incompressible. However it satisfies the Laplace equation only when flow is incompressible.