In cylindrical coordinates, a flow field is described as Vr = cos θ /r 2
Vθ = sin θ / r 2
Find the equation of streamline passing through the point (r = 2, θ = π/2).
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
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
Consider steady state flow of an incompressible fluid (density ρ) through a conical diffuser of length L whose diameter varies from d1 at inlet to d2 at outlet. (d1 < d2 ). Find the acceleration in the flow at x = L.
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 r5/2
In a steady state flow, the velocity components are u = 2kx ; v = 2ky ; w = −4kz Find the equation of streamline passing through the point (1, 0, 1).
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
A conical flow passage converges uniformly from 0.2 m diameter to 0.1 m diameter over a length of 1 metre. Find the magnitude of total acceleration at the middle of the diffuser.
Consider the following case.
Rate of flow varies linearly from 100 litres/s to 200 litres/s in 5 seconds and the time of interest is t = 2 sec.
(Velocity at any cross-section may be assumed to be uniform and perpendicular to the center line of passage).
∴ 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
Velocity distribution at entry to the pump intake is inversely proportional to square of radial distance from inlet to suction pipe. If the velocity at a radial distance of 1 m from the pipe inlet is 0.75 m/s, make calculations for the acceleration of flow at 0.5 m and 1.5 m from the inlet. Consider the streamlines to be radial.
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.
From a flow net diagram it is observed that distances between two consecutive streamlines at two consecutive cross-sections are 1 cm and 0.6 cm respectively. If the two sections are 1 m apart and velocity at the first section is 1 m/sec, find out the time taken by a fluid particle to move from the first to second section.
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
For a flow passage described by A(x) = −ax + b [Where A is the cross-sectional area at a distance x from inlet and a, b are constants] predict the time taken by a fluid particle to move from inlet to outlet of the flow. The length of the flow passage is L. The flow through the passage is steady and free of compressibility and viscous effects.
Also it can be assumed that flow velocity at a section is uniform and parallel to the centerline of the passage. The flow rate through the passage is Q.
∴ 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
A two-dimensional flow in cylindrical polar coordinates is given by Vr = 2r sin θ cos θ Vθ = −2r sin2 θ Check whether these velocity components represent a physical possible flow field.
∂ / ∂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
The continuity equation is satisfied so the flow is incompressible.
Consider fully developed axisymmetric Poiseuille flow in a round pipe of radius R with a pressure gradient (∂p/ ∂x) driving the flow as shown in figure (∂p/ ∂x is uniform and negative). The flow is steady, incompressible and axisymmetric about the x-axis. The velocity components are given by
uz = 1 / 4μ ( ∂p / ∂x ) (r2 − R2 ); ur = 0; uθ = 0
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).
Find the acceleration of a fluid particle at the point r = 2a, θ = π/2 for a 2-D flow given by
Vr = −u (1 −a2 / r2) cos θ
Vθ = u (1 + a2 / r2 ) sin θ
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.
A spherical balloon is filled with 3536π cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 70π cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases 50 minutes after the leakage began is
A flow field is defined by the stream function ψ = 2x2 y, determining the corresponding velocity potential.
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.
Obtain the relationship between stream function ψ and the velocity components Vr and Vθ in cylindrical polar coordinates.
A two-dimensional flow field is prescribed by the velocity components: u = ky and v = 0 where k is an arbitrary constant. Find the circulation about a closed contour conforming to the equation x2 + y2 − 2ay = 0.
∴ 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.
If the velocity field is given by: u = x+y and v = x3 −y Find the circulation around a closed contour defined by x = 1 ; y = 0, y = 1 and x = 0.
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.