Test: Fluid Kinematics Level - 3 - Mechanical Engineering MCQ

dr: rdθ = Vr: Vθ or dr / rdθ = Vr / Vθ

Substituting for Vr and Vθ ; we obtain

dr / rdθ = cos θ / r 2 / sin θ / r 2 = cos θ / sin θ

dr / r = cos θ / sin θ dθ

Upon integration:

oge r = loge sin θ + loge C

loge r = loge (C sin θ)

r = C sin θ

This is the equation for the family of streamlines describing the flow.

For particular solution the value of constant C can be found from the condition that the streamline passes through

r = 2 and θ = π 2

∴ 2 = C sin π/ 2; C = 2

Thus the equation of streamline is, r = 2 sin θ

a = u ∂u /∂x (Now u = Q/A)

∴ a = Q² / A ∂ / ∂x ( 1 / A )

Also A(x) = π[r(x)]2

a(x = L) = −2Q²(r₂ − r₁) / π² L r^5/2

dx / u = dy / v = dz / w

Substituting for u, v and w, we obtain

dx / 2kx = dy / 2ky = −dz / 4kz

(a) (b) ©

Consider the expressions (a) and (b) and integrate

∫dx / x = ∫dy/ y

or loge x = loge y + constant which is equivalent to y = C1 x where C1 is a constant.

Likewise expressions (a) and (c) yield,

2∫dx / x = − ∫dz /- y

or loge x² = − loge z + constant

which is equivalent to z = C₂ / x²

where C2 is another constant. All the streamlines in the given flow field are described by equations

y = C₁ x and z = C₂ / x² with different values of

C₁ and C₂

For the streamline passing through the given point (1, 0, 1) the constants C₁ and C₂ are to be such that y = 0 and z = 1 at x = 1. ∴ C₁ = 0 and C₂ = 1

Thus the streamline passing through (1, 0, 1) is y = 0 and z = 1 / x²

∴ U = Q(t) / A(x)

a = dU / dt

∴ a = ∂U / ∂t + U ∂U /∂x

a = 1 / A ∂Q / ∂t + Q 2 / A ∂/∂x (1 / A)

But A = πr2

Where a is the acceleration at section x (whose radius is r) at time instant t (when discharge is Q) Given that discharge increases from 100 Lt/sec to 200 Lt/sec in 5 seconds.

∴∂Q /∂t = (200 − 100) / 5 × 10−3 = 0.02 m3/sec2

Also Q(t = 2) = 0.1 + 0.02 × 2 = 0.14 m²/sec

As r = r1 + r2 − r1 / Lx

∴ r (x = L /2) = 0.1 +0.05 − 0.1 / 1 × 1 /2

r (x = L / 2) = 0.075 m

Also ∂r / ∂x =r2 − r1 / L = 0.05 − 0.1 / 1

∂r / ∂x = −0.05

= 1/π × (0.075)2 × 0.02 − 2 / π × (0.075)5 × 0.142 × (−0.05)

∴ a ( L / 2 , 2) = 264 m/sec²

At r = 1 m, u = 0.75 m⁄s and hence C = 0.75 Acceleration,

a = ∂V / ∂t + V ∂V / ∂r

or steady state flow ∂V / ∂t = 0

a = V ∂V / ∂r = C / r² × ∂/∂r (C / r² )

= −2C /r⁵

Substituting C = 0.75

a = −2 × (0.75)² /r 5 = −1.125 / r⁵

i) When r = 0.5 m

a = −1.125 / (0.5)⁵= −36 m⁄s²

(ii) When r = 1.5 m

a = −1.125 /(1.5)⁵ = −0.148 m/s²

The minus sign indicates that acceleration is directed towards the intake.

V(x) = Q /A(x)

A(x) = A2 − A1 /L x + A1

A(x) = 0.006 b − 0.01 b / 1 + 0.01 b

A(x) = (−0.004 bx + 0.01 b) ⋯ ①

Also from section ①,

Q = A1 V1

⇒ Q = 0.01 b × 1

⇒ Q = 0.01 b ⋯ ②

Thus from ① and ②

V(x) = 0.01 b / −0.004 bx + 0.01 b

⇒ V(x) = 1 / 1 − 0.4x

But, considering steady state flow, the above expression in the Eulerian approach is equivalent to particle description in Lagrangian approach.

⇒dx / dt = 1 / 1 − 0.4x

⇒ (1 − 0.4x)dx = dt

⇒ ∫ (1 − 0.4x) dx= ∫ dt

⇒ [x − 0.2x2]|10 = t

⇒ (1 − 0.2) = t

⇒ t = 0.8 sec

∴ U(x, t) = Q /−ax + b

where U is the flow velocity at section x when time instant is t. [Eulerian approach].

But actually it defines velocity of particle crossing section x at time instant. [Lagrangian approach]

∴ for particle U = dx / dt

∴ dx /dt =Q / (−ax + b)

(−ax + b)dx = Q dt

∂ / ∂r (r Vr) + ∂ /∂θ (Vθ) = 0

From the given velocity components,

∂/∂r (r Vr) = Vr + r ∂ /∂r (Vr)

= 2 r sin θ cos θ + r ∂ / ∂r (2r sin θ cos θ )

= 2r sin θ cos θ + 2r sin θ cos θ

= 4 r sin θ cos θ

and ∂ / ∂θ (Vθ) = ∂ / ∂θ (−2r sin2 θ)

= −2r × 2 sin θ cos θ = −4r sin θ cos θ

∴∂ / ∂r (r Vr) + ∂ /∂θ (Vθ)

= 4r sin θ cos θ − 4r sin θ cos θ

= 0

The continuity equation is satisfied so the flow is incompressible.

ince ∂p/ ∂x is also negative therefore, Ω will be positive. Hence according to convention of FM (+)ve sign indicates anticlockwise rotation.

Therefore the correct option is (C).

ar = Vr ∂Vr / ∂r + Vθ / r ∂Vr / ∂θ −Vθ2 /r

and acceleration in azimuthal direction is given by,

aθ = Vr ∂Vθ /∂r + Vθ / r ∂Vθ /∂θ + Vr Vθ /r

Now ∂Vr / ∂θ = u sin θ (1 − a 2 /r2)

aθ = 0 + 0 − 0 ⇒ aθ = 0

∴ Option (D) is correct.

u = ∂ψ / ∂y ; v = − ∂ψ / ∂x ⇒ u = 2x² ; v = −4xy

As (∇ ×V ) ≠ 0 ; therefore flow is rotational, and velocity potential function does not exist for rotational flow.

∴ The definition of vorticity i.e. circulation per unit area can also be used for calculating circulation.

Γ = Ωz × A where A = πa²

∴ Γ = −kπa²

Hence (A) is the correct answer.

Circulation, ΓABCD is defined as the line integral of tangential velocity along closed contour ABCD.

Based on shape of contour y = 0 for 1st integral x = 1 for 2nd integral y = 1 for 3rd integral x = 0 for 4th integral

i.e. vorticity is non-uniform and thus the method mentioned in question 14 cannot be used directly.