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A 2D flow defined by velocity field ⃗V⃗ = (x + 2y + 2)î + (4 − y)ĵ is
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 nonzero.
A stream function is given by (x2 − y2). The velocity potential function of the flow will be
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.
The velocity potential function of a flow field is given by x2 − y2 + constant. Its stream function will be given by
∂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.
The area of a 2 m long tapered duct decreases as ! = (0.5 − 0.2x) where ‘x’ is the distance in meters. At a given instant a discharge of 0.5 m3⁄s is flowing in the duct and is found to increase at a rate of 0.2 m^{3} /s^{2}. The local acceleration (in m⁄s2) at x = 0 will be
∂Q / ∂t = 0.2 m3/s^{2}
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/s^{2}
Water flows into an open drum from the top at a constant mass flow rate. Water exits through a pipe near the base with a mass flow rate proportional to the height of the liquid. If the drum is initially empty, which of the following gives the plot of liquid height with time?
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 hT curve will decrease.
A stream function is given by ψ = 4x2y + (2 + t) y2 Find the flow rates across the faces of the triangular prism OMN (Fig.) having a thickness of 5 m in the zdirection at the time instant t = 2 seconds. (in m3/s)
The coordinates of point N are (0, 2)
∴ ψN = 0 + (2 + 2) × 22 = 16 The coordinates 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
For a twodimensional potential flow, the velocity potential function is given by ϕ = 4x(3y − 4). The numerical value of stream function at the point (2, 3) is
∴ 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
In a stream line steady state flow, two points A and B on a stream line are 1 m apart and the velocity of a particle varies linearly from 2 m/s to 5 m/s as the particle moves from A to B. What is the acceleration of fluid at B?
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
Statement I: When flow is unsteady, both normal and tangential components of acceleration will occur.
Statement II: During unsteady flow, in addition to the change of velocity along the path, the velocity will also change with time.
Also consider flow of water at an increasing or decreasing flow rate through a straight pipe of uniform crosssection. In this case normal acceleration is zero, even though tangential acceleration is nonzero and a function of time, making the flow unsteady.
Statement I: Nonviscous flow between two plates held parallel with a very small spacing between them is an example of irrotational flow.
Statement II: Forced vortex implies irrotational flow.
A forced vortex represents a rotational flow. Hence statement (II) is false.
The stream function for a flow field is ψ = 3x2 y + (2 + t)y2. The velocity at a point P for position vector r = î + 2ĵ − 3k̂ at time t = 2 will be
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
In a twodimensional incompressible steady flow, the velocity component u = !ex is obtained. What is the other component v of velocity?
∂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)
The instantaneous velocity and density fields for flow through a diffuser are given as u = u0 e−2x/L and ρ = ρ0 e−x/L Find the rate of change of density for the same time instant at x = L.
∂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 coordinate, 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}
State whether the following flow field is physically possible? u = 3xy^{2} + 2x + y^{2} and v = x^{2} − 2y − y^{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 nonzero.
Thus, flow is rotational.
Which of the following statements is correct in respect of velocity potential function & stream function?
1. Stream function is defined only for an incompressible irrotational flow
2. Velocity Potential function is defined only for an incompressible irrotational flow
3. If the stream function exists it shall definitely satisfies the Laplace equation
4. If the velocity potential function exists it shall definitely satisfies the Laplace equation
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.
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