Chapter 4 • Differential Relations for Fluid Flow P4.1 An idealized velocity field is given by the formula
V = 4txi − 2t 2 yj + 4 xzk Is this flow field steady or unsteady? Is it two- or three-dimensional? At the point (x, y, z) = (–1, +1, 0), compute (a) the acceleration vector and (b) any unit vector normal to the acceleration. Solution: (a) The flow is unsteady because time t appears explicitly in the components. (b) The flow is three-dimensional because all three velocity components are nonzero. (c) Evaluate, by laborious differentiation, the acceleration vector at (x, y, z) = (−1, +1, 0). du ∂ u ∂u ∂u ∂u = +u +v +w = 4x + 4tx(4t) − 2t 2 y(0) + 4xz(0) = 4x + 16t 2 x dt ∂ t ∂x ∂y ∂z
dv ∂ v ∂v ∂v ∂v = +u +v +w = −4ty + 4tx(0) − 2t 2 y( −2t 2 ) + 4xz(0) = −4ty + 4t 4 y dt ∂ t ∂x ∂y ∂z dw ∂ w ∂w ∂w ∂w = +u +v +w = 0 + 4tx(4z) − 2t 2 y(0) + 4xz(4x) = 16txz + 16x 2z dt ∂t ∂x ∂y ∂z dV or: = (4x + 16t 2 x)i + (−4ty + 4t 4 y) j + (16txz + 16x 2z)k dt dV at (x, y, z) = (−1, +1, 0), we obtain = −4(1 + 4t 2 )i − 4t(1 − t 3 ) j + 0 k Ans. (c) dt (d) At (–1, +1, 0) there are many unit vectors normal to dV/dt. One obvious one is k. Ans. P4.2 Flow through the converging nozzle in Fig. P4.2 can be approximated by the one-dimensional velocity distribution
⎛ 2 x ⎞ u ≈ Vo ⎜ 1 + ⎟ υ ≈ 0 w ≈ 0 ⎝ L ⎠ (a) Find a general expression for the fluid acceleration in the nozzle. (b) For the specific case Vo = 10 ft/s and L = 6 in, compute the acceleration, in g’s, at the entrance and at the exit.
Fig. P4.2
Solution: Here we have only the single ‘one-dimensional’ convective acceleration:
du ∂ u ⎡ ⎛ 2 x ⎞ ⎤ 2Vo 2Vo2 ⎛ 2x ⎞ =u = Vo 1 + ⎟ ⎥ = ⎜ 1 + ⎟ Ans. (a) dt ∂ x ⎢⎣ ⎜⎝ L ⎠ ⎦ L L ⎝ L ⎠ ft du 2(10)2 ⎛ 2 x ⎞ For L = 6ʹ′ʹ′ and Vo = 10 , = ⎜⎝1 + ⎟ = 400(1 + 4x), with x in feet s dt 6 /12 6 /12 ⎠ At x = 0, du/dt = 400 ft/s2 (12 g’s); at x = L = 0.5 ft, du/dt = 1200 ft/s2 (37 g’s). Ans. (b)
P4.3 A two-dimensional velocity field is given by V = (x2 – y2 + x)i – (2xy + y)j in arbitrary units. At (x, y) = (1, 2), compute (a) the accelerations ax and ay, (b) the velocity component in the direction θ = 40°, (c) the direction of maximum velocity, and (d) the direction of maximum acceleration. Solution: (a) Do each component of acceleration: du ∂u ∂u =u +v = (x 2 − y 2 + x)(2x + 1) + (−2xy − y)(−2y) = a x dt ∂x ∂y
dv ∂v ∂v =u +v = (x 2 − y 2 + x)( −2y) + (−2xy − y)(−2x − 1) = a y dt ∂x ∂y At (x, y) = (1, 2), we obtain ax = 18i and ay = 26j Ans. (a) (b) At (x, y) = (1, 2), V = –2i – 6j. A unit vector along a 40° line would be n = cos40°i + sin40°j. Then the velocity component along a 40° line is
V40° = V⋅n40° = (−2i − 6 j) ⋅ (cos 40°i + sin 40° j) ≈ 5.39 units Ans. (b) (c) The maximum acceleration is amax = [182 + 262]1/2 = 31.6 units at ∠55.3° Ans. (c, d) _______________________________________________________________________ P4.4
A simple flow model for a two-dimensional converging nozzle is the distribution
x u = U o (1 + ) L
v = −U o
y L
w=0
(a) Sketch a few streamlines in the region 0
(c) Where is the largest resultant acceleration and its numerical value?
Solution: The streamlines are in the x-y plane and are found from the velocities:
dx dy = u v
or integrate :
dx
dy
∫ U o (1 + x / L) = − ∫ U o y / L
Cancel U o
L ln(1 + x / L) = − L ln( y / L) + const , or : ln[( y / L)(1 + x / L)] = constant Finally the streamlines :
y = L
C 1 + x/L
Ans.( a )
These may be plotted for various values of the dimensionless constant C, as shown:
1.0
C=1
0.9 0.8
y/L
C = 0.75 C = .5
0.7 0.6 0.5
C = 0.25
0.4 0.3 0.2 0.1 0.0 0
0.2
0.4
x/L
0.6
0.8
1
The streamlines converge and the velocity increases to the right. (b) The accelerations are calculated from Eq. (4.2):
Ans.(a)
ax = u
U2 ∂u ∂u x + v = [U o (1 + x / L)](U o / L) + 0 = o (1 + ) ∂x ∂y L L
U o2 y ∂v ∂v + v = 0 + (−U o y / L)(−U o / L) = Ans.(b) ∂x ∂y L L (c) Find the resultant of ax and ay from Ans.(b) above and introduce y/L from Ans.(a): ay = u
a = ax2 + a 2y =
1 + 2η +η 2 + C 2 /(1 + η )2
,
where η = x / L
Ans.(c)
We observe that the resultant acceleration increases with x and is greatest at x = L, where its numerical value is (Uo2/L) [4 + C2/4]1/2.
P4.5 The velocity field near a stagnation point (see Example 1.10) may be written in the form U x −U o y u= o v= U o and L are constants L L (a) Show that the acceleration vector is purely radial. (b) For the particular case L = 1.5 m, if the acceleration at (x, y) = (1 m, 1 m) is 25 m/s2, what is the value of Uo? Solution: (a) For two-dimensional steady flow, the acceleration components are
Uo2 du ∂u ∂ u ⎛ x ⎞ ⎛ Uo ⎞ ⎛ y ⎞ =u +v = U + −U (0) = 2 x dt ∂x ∂ y ⎜⎝ o L ⎟⎠ ⎜⎝ L ⎟⎠ ⎜⎝ o L ⎟⎠ L dv ∂v ∂ v ⎛ x ⎞ y ⎞ ⎛ Uo ⎞ Uo2 ⎛ =u +v = U (0) + ⎜ − U o ⎟ ⎜ − y ⎟ = ⎝ dt ∂x ∂ y ⎜⎝ o L ⎟⎠ L ⎠ ⎝ L ⎠ L2 Therefore the resultant a = (U2o /L2 )(xi + yj) = (U2o /L2 )r (purely radial) Ans. (a) (b) For the given resultant acceleration of 25 m/s2 at (x, y) = (1 m, 1 m), we obtain
|a| = 25
U 2o m U 2o m r| = | = 2 m, solve for U = 6.3 o s s2 L2 (1.5 m)2
Ans. (b)
P4.6 An incompressible plane flow has the velocity components u = 2y, v = 8x, w = 0. (a) Find the acceleration components. (b) Determine if the vector acceleration is radial. (c) Which two streamlines of this flow are straight lines?
Solution: (a, b) With no z activity, we can stick to steady two-dimensional formulas:
∂u ∂u +v = (2 y )(0) + (8 x)(2) = 16 x ∂x ∂y ∂v ∂v ay = u +v = (2 y )(8) + (8 x)(0) = 16 y ∂x ∂y a = i ax + j a y = 16 i x + 16 j y = 16(i x + j y ) = 16 r ax = u
Ans.(a) Radial , yes. Ans.(b)
(c) Find the streamlines in the manner suggested in Chapter 1, Eq. (1.41):
dx dy = , u v Integrate :
or :
dx dy = , 2y 8x
∫ 2 ydy = ∫ 8x dx
2 y dy = 8 x dx
, or : y 2 = 4 x 2 + const
The streamlines are all curved except when the constant of integration is zero, for which there are two straight streamlines: y = ± 2 x . Ans.(c)
P4.7 Consider a sphere of radius R immersed in a uniform stream Uo, as shown in Fig. P4.7. According to the theory of Chap. 8, the fluid velocity along streamline AB is given by
⎛ R3 ⎞ V = ui = Uo ⎜ 1 + 3 ⎟ i x ⎠ ⎝
Fig. P4.7
Find (a) the position of maximum fluid acceleration along AB and (b) the time required for a fluid particle to travel from A to B. Note that x is negative along line AB. Solution: (a) Along this streamline, the fluid acceleration is one-dimensional:
du ∂u =u = U o (1 + R 3 /x 3 )( −3U o R 3 /x 4 ) = −3U o R 3 (x −4 + R 3x −7 ) for x ≤ −R dt ∂x The maximum occurs where d(ax)/dx = 0, or at x = –(7R3/4)1/3 ≈ –1.205R Ans. (a) (b) The time required to move along this path from A to B is computed from
u=
dx = Uo (1 + R3 /x3 ), or: dt
−R
∫
−4R
t
dx = Uo dt, 1 + R3 /x3 ∫0 −R
⎡ R (x + R)2 R −1 ⎛ 2x − R ⎞ ⎤ or: U o t = ⎢ x − ln 2 + tan =∞ ⎜⎝ ⎟⎠ ⎥ 2 6 x − Rx + R 3 R 3 ⎣ ⎦ −4R It takes an infinite time to actually reach the stagnation point, where the velocity is zero. Ans. (b)
P4.8 When a valve is opened, fluid flows in the expansion duct of Fig. P4.8 according to the approximation
x ⎞ Ut ⎛ V = iU ⎜ 1 − ⎟ tanh L ⎝ 2 L ⎠
Find (a) the fluid acceleration at (x, t) = (L, L/U) and (b) the time for which the fluid acceleration at x = L is zero. Why does the fluid acceleration become negative after condition (b)?
Fig. P4.8
Solution: This is a one-dimensional unsteady flow. The acceleration is
ax =
∂u ∂u x ⎞ U x ⎞ ⎛ U ⎞ ⎛ ⎛ Ut ⎞ ⎛ ⎛ Ut ⎞ +u = U ⎜ 1 − ⎟ sech 2 ⎜ ⎟ − U ⎜ 1 − ⎟ ⎜ ⎟ tanh⎜ ⎟ ⎝ 2L ⎠ L ⎝ L ⎠ ⎝ 2L ⎠ ⎝ 2L ⎠ ⎝ L ⎠ ∂t ∂x =
U 2 ⎛ x ⎞ ⎡ ⎛ Ut ⎞ ⎤ 2 ⎛ Ut ⎞ 1 ⎜ 1 − ⎟ ⎢sech ⎜ ⎟ − tanh ⎜ ⎟ ⎥ L ⎝ 2L ⎠ ⎣ ⎝ L ⎠ 2 ⎝ L ⎠ ⎦
At (x, t) = (L, L/U), ax = (U2/L)(1/2)[sech2(1) – 0.5tanh(1)] ≈ 0.0196 U2/L Ans. (a) The acceleration becomes zero when
1 ⎛ Ut ⎞ 1 ⎛ Ut ⎞ ⎛ 2Ut ⎞ sech 2 ⎜ ⎟ = tanh ⎜ ⎟ , or sinh ⎜ = 2, ⎝ L ⎠ 2 ⎝ L ⎠ ⎝ L ⎟⎠ 2 Ut or: ≈ 1.048 Ans. (b) L The acceleration starts off positive, then goes through zero and turns negative as the negative convective acceleration overtakes the decaying positive local acceleration.
P4.13
For an incompressible plane flow in polar coordinates, we are given
vr = r 3 cos θ + r 2 sin θ
Find the appropriate form of circumferential velocity for which continuity is satisfied.
Solution: Substitute into continuity, Eq. (4.9), for incompressible flow:
1 ∂ 1 ∂ 1 ∂ 1 ∂vθ (r vr ) + (vθ ) = [r (r 3 cos θ + r 2 sin θ )] + , r ∂r r ∂θ r ∂r r ∂θ 1 ∂vθ or : = − 4 r 2 cos θ − 3 r sin θ r ∂θ Integrate : vθ = − 4 r 3 sin θ + 3 r 2 cos θ + f (r )
Ans.
We can’t determine the form of the “constant of integration” f(r) without further information.
P4.19 An incompressible flow field has the cylindrical velocity components υ = Cr, υz = K(R2 – r2), υr = 0, where C and K are constants and r ≤ R, z ≤ L. Does this flow satisfy continuity? What might it represent physically? Solution: We check the incompressible continuity relation in cylindrical coordinates: θ
1∂ 1 ∂ vθ ∂ vz (rv r ) + + = 0 = 0 + 0 + 0 satisfied identically r ∂r r ∂θ ∂z
Ans.
This flow also satisfies (cylindrical) momentum and could represent laminar flow inside a tube of radius R whose outer wall (r = R) is rotating at uniform angular velocity.
P4.22 In an axisymmetric flow, nothing varies with θ ; the only nonzero velocities are vr and vz (see Fig. 4.2 of the text). If the flow is steady and incompressible and vz = Bz, where B is constant, find the most general form of vr which satisfies continuity.
Solution: With no θ variation and no v , the equation of continuity (4.9) becomes θ
∂vz 1 ∂ 1 ∂ (r vr ) + = 0 = (r vr ) + r ∂r ∂z r ∂r ∂ or : (r vr ) = − B r ; Integrate : r vr ∂r B Finally, vr = − r + 2
∂ ( Bz ) , ∂z B = − r 2 + f ( z) 2 f ( z) Ans. r
The “function of integration”, f(z), is arbitrary, at least until boundary conditions are set. __________________________________________________________________________ P4.25 An incompressible flow in polar coordinates is given by b ⎞ ⎛ vr = K cos θ ⎜ 1 − 2 ⎟ ⎝ r ⎠ b ⎞ ⎛ vθ = − K sin θ ⎜ 1 + 2 ⎟ ⎝ r ⎠ Does this field satisfy continuity? For consistency, what should the dimensions of constants K and b be? Sketch the surface where vr = 0 and interpret. Fig. P4.25 Solution: Substitute into plane polar coordinate continuity:
1∂ 1 ∂ vθ ? 1 ∂ ⎡ ⎛ b ⎞ ⎤ 1 ∂ (rvr ) + = 0= K cos θ ⎜ r − ⎟ ⎥ + ⎢ ⎝ r ∂r r ∂θ r ∂ r ⎣ r ⎠ ⎦ r ∂θ
b ⎞ ⎤ ⎡ ⎛ ⎢ −K sin θ ⎜⎝1 + r 2 ⎟⎠ ⎥ = 0 Satisfied ⎣ ⎦
The dimensions of K must be velocity, {K} = {L/T}, and b must be area, {b} = {L2}. The surfaces where vr = 0 are the y-axis and the circle r = √b, as shown above. The pattern represents inviscid flow of a uniform stream past a circular cylinder (Chap. 8).
P4.28 Consider the incompressible flow field of Prob. P4.6, with velocity components u = 2y, v = 8x, w = 0. Neglect gravity and assume constant viscosity. (a) Determine whether this flow satisfies the Navier-Stokes equations. (b) If so, find the pressure distribution p(x, y) if the pressure at the origin is po. Solution: In Prob. P4.6 we found the accelerations, so we can proceed to Navier-Stokes:
∂u ∂u ∂p ∂p + v ) = ρ[ 0 + (8 x)(2)] = − + ρ g x + µ ∇ 2u = − + 0 + 0; ∂x ∂y ∂x ∂x ∂v ∂v ∂p ∂p ρ (u + v ) = ρ[(2 y )(8) + 0] = − + ρ g y + µ ∇2v = − + 0 + 0; ∂x ∂y ∂y ∂y
ρ (u
∂p = − 16 ρ x ∂x ∂p = − 16 ρ y ∂y
Noting that
∂ 2 p /(∂x∂y) = 0 in both cases, we conclude Yes, satisfies Navier - Stokes. Ans.(a) (b) The pressure gradients are simple, so we may easily integrate:
dp =
∂p ∂p dx + dy , or : p = ∫ −16 ρ x dx + ∫ −16 ρ y dy = − 8ρ ( x 2 + y 2 ) + const ∂x ∂y If p(0, 0) = po , then p = po − 8ρ ( x 2 + y 2 )
Ans.(b)
This is an exact solution, but it is not Bernoulli’s equation. The flow is rotational. P4.29 Consider a steady, two-dimensional, incompressible flow of a newtonian fluid with the velocity field u = –2xy, v = y2 – x2, and w = 0. (a) Does this flow satisfy conservation of mass? (b) Find the pressure field p(x, y) if the pressure at point (x = 0, y = 0) is equal to pa. Solution: Evaluate and check the incompressible continuity equation:
∂u ∂ v ∂ w + + = 0 = −2 y + 2 y + 0 ≡ 0 Yes! Ans. (a ) ∂ x ∂ y ∂z (b) Find the pressure gradients from the Navier-Stokes x- and y-relations:
⎛ ∂ 2u ∂ 2u ∂ 2u ⎞ ⎛ ∂ u ∂u ∂ u ⎞ ∂p ρ ⎜ u + v + w ⎟ = − + µ ⎜ 2 + 2 + 2 ⎟ , or: ⎝ ∂ x ∂y ∂ z ⎠ ∂x ∂ y ∂ z ⎠ ⎝ ∂ x ∂p ∂p ρ[ −2 xy(−2 y) + ( y2 − x 2 )(−2 x)] = − + µ(0 + 0 + 0), or: = −2 ρ( xy 2 + x 3 ) ∂x ∂x and, similarly for the y-momentum relation,
⎛ ∂ 2 v ∂ 2 v ∂ 2 v ⎞ ⎛ ∂ v ∂v ∂ v ⎞ ∂p + v + w ⎟ = − + µ ⎜ 2 + 2 + 2 ⎟ , or: ⎝ ∂ x ∂y ∂ z ⎠ ∂y ∂ y ∂ z ⎠ ⎝ ∂ x ∂p ∂p ρ[ −2 xy(−2 x) + ( y2 − x 2 )(2 y)] = − + µ(−2 + 2 + 0), or: = −2 ρ( x 2 y + y3 ) ∂y ∂y
ρ ⎜ u
The two gradients ∂ p/∂ x and ∂ p/∂ y may be integrated to find p(x, y):
⎛ x 2 y 2 x 4 ⎞ ∂p p=∫ dx| = −2ρ ⎜ + ⎟ + f ( y), then differentiate: ∂ x y = Const 4 ⎠ ⎝ 2 ∂p df df ρ = −2ρ( x 2 y) + = −2ρ( x 2 y + y3 ), whence = −2ρ y3 , or: f ( y) = − y 4 + C ∂y dy dy 2 ρ 2 2 4 4
Thus: p = − (2 x y + x + y ) + C = pa at ( x, y) = (0,0), or: C = pa 2 Finally, the pressure field for this flow is given by p = pa −
1 ρ (2x 2 y 2 + x 4 + y 4 ) Ans. (b) 2
y
g
P4.36 A constant-thickness film of viscous liquid flows in laminar motion down a plate inclined at angle θ, as in Fig. P4.36. The velocity profile is
h (y)
u = Cy(2h – y) v = w = 0 x
Find the constant C in terms of the specific weight and viscosity and the angle θ. Find the volume flux Q per unit width in terms of these parameters.
Fig. P4.36
Solution: There is atmospheric pressure all along the surface at y = h, hence ∂p/∂x = 0. The x-momentum equation can easily be evaluated from the known velocity profile:
⎛ ∂ u ∂ u ⎞ ∂p + v ⎟ = − + ρgx + µ∇2 u, or: 0 = 0 + ρgsinθ + µ(−2C) ⎝ ∂ x ∂ y ⎠ ∂x ρ g sinθ Solve for C = Ans. (a) 2µ
ρ ⎜ u
The flow rate per unit width is found by integrating the velocity profile and using C:
h
h
2 ρgh3 sinθ Q = ∫ u dy =∫ Cy(2h − y)dy = Ch3 = per unit width Ans. (b) 3 3 µ 0 0
P4.37 A viscous liquid of constant density and viscosity falls due to gravity between two parallel plates a distance 2h apart, as in the figure. The flow is fully developed, that is, w = w(x) only. There are no pressure gradients, only gravity. Set up and solve the Navier-Stokes equation for the velocity profile w(x). Solution: Only the z-component of NavierStokes is relevant:
ρ
Fig. P4.37
dw d 2w ρg = 0 = ρg + µ 2 , or: w ʹ′ʹ′ = − , w( − h) = w( + h) = 0 (no-slip) dt µ dx
The solution is very similar to Eqs. (4.142) to (4.143) of the text:
w=
P4.38
ρg 2 (h − x 2 ) Ans. 2µ
Show that the incompressible flow distribution, in cylindrical coordinates,
vr = 0 vθ = C r n vz = 0 where C is a constant, (a) satisfies the Navier-Stokes equation for only two values of n. Neglect gravity. (b) Knowing that p = p(r) only, find the pressure distribution for each case, assuming that the pressure at r = R is po. What might these two cases represent?
Solution: (a) The important direction here is the θ-momentum equation, Eq. (D.6):
∂vθ v 1 1 ∂p 2 ∂v + (V • ∇)vθ + v r vθ = − + ν (∇ 2 vθ − θ2 + 2 θ ) , or : ∂t r ρr ∂θ r r ∂θ 0 + 0 + 0 = 0 + ν (∇ 2 vθ −
vθ
r2
+ 0) = ν [
1 ∂ ∂ Cr n (r (Cr n )) − 2 ] , or : r ∂r ∂r r
1 ∂ (rnCr n−1 ) − Cr n−2 ] = ν (Cn 2 r n−2 − Cr n−2 ) = 0 r ∂r Cancel C and ν and rn-2. These terms equal zero only if n2 = 1, or n = ±1. Ans.(a).
ν[
(b)
Find the respective pressure distributions for n = 1 and n = -1. Use Eq. (D.5), which
reduces simply to ∂p/∂r = ρv 2/r. Try this for each distribution, n = ±1: θ
p
Case1, n = 1 : ∫ dp po
p
=
Case 2, n = −1 : ∫ dp = po
r
∫R
ρ (C 2 r 2 ) r r
∫R
dr ; or :
ρ (C 2 / r 2 ) r
p=
dr ; or :
po +
p = po +
ρC 2 2
(r 2 − R 2 )
ρC 2 2
(
1 R
2
−
1 r2
Ans.(b1 ) )
Ans.(b2 )
Case 1, v = Cr, is solid-body rotation. Case 2, v = C/r, is an irrotational potential vortex. θ
θ
P4.40 For pressure-driven laminar flow between parallel plates (see Fig. 4.12b), the velocity components are u = U(1– y2/ h2), v = 0, and w = 0, where U is the centerline velocity. In the spirit of Ex. 4.6, find the temperature distribution T(y) for a constant wall temperature Tw.
Solution: There are no variations with x or z, so the energy equation (4.53) reduces to
∂T ∂ 2T ∂u 2 d 2T du ρ cp u = 0 = k 2 + µ ( ) = k 2 + µ ( )2 , ∂x ∂y dy ∂y dy or :
d 2T dy 2
= −
µ 2Uy k
(
h2
2
) = −(
4µ U 2 k h4
4µ U 2 y 3 dT ) y ; Integrate : = −( ) + C1 dy k h4 3 2
The condition T = Tw at ±h is equivalent to dT/dy = 0 at y = 0. Thus C1 = 0. Integrate again:
T = −(
4µ U 2 y 4 4µ U 2 h 4 µU 2 ) + C ; at y = h : T = T = − ( ) + C , ∴ C = T + 2 w 2 2 w 3k k h4 12 k h4 12
The final solution for T(y) is, like Ex. 4.6, a quartic polynomial:
T ( y ) = Tw +
µU 2 3k
(1 −
y4 h4
)
Ans.
P4.44 Suppose that we wish to analyze the sudden pipe-expansion flow of Fig. P3.59, using the full continuity and Navier-Stokes equations. What are the proper boundary conditions to handle this problem? Solution: First, at all walls, one would impose the no-slip condition: ur = uz = 0 at all solid surfaces: at r = r1 in the small pipe, at r = r2 in the large pipe, and also on the flatfaced surface between the two.
Fig. P3.59
Second, at some position upstream in the small pipe, the complete velocity distribution must be known: u1 = u1(r) at z = z1. [Possibly the paraboloid of Prob. 4.34.] Third, to be strictly correct, at some position downstream in the large pipe, the complete velocity distribution must be known: u2 = u2(r) at z = z2. In numerical (computer) studies, this is often simplified by using a “free outflow” condition, ∂ u/∂ z = 0. Finally, the pressure must be specified at either the inlet or the outlet section of the flow, usually at the upstream section: p = p1(r) at z = z1.
P4.55 For the incompressible plane flow of Prob. P4.6, with velocity components u = 2y, v = 8x, w = 0, determine (a) if a stream function exists. (b) If so, determine the form of the stream function, and (c) plot a few representative streamlines. Solution: (a) Check to see is two-dimensional continuity is satisfied:
∂u ∂v ∂ ∂ + = (2 y) + (8 x) = 0 + 0 = 0 ∂x ∂y ∂x ∂y
Yes, ψ exists.
Ans.(a)
(b) Find the stream function by relating velocities to derivatives of ψ :
u = 2y =
∂ψ ∂ψ ; v = 8x = − ; Integrate : ψ = y 2 − 4 x 2 + constant Ans.(b) ∂y ∂x
(c) Plot a few streamlines, that is, plot results:
y2 = 4x2 + C for various C. Here are the
6
y 2 = 4x 2 + C
5 4 3 2 1 0 -3
-2
-1
0
1
2
3
We are showing only the upper half plane, which is the mirror image of the lower half. P4.57 A two-dimensional incompressible flow field is defined by the velocity components
⎛ x y ⎞ u = 2V ⎜ − ⎟ ⎝ L L ⎠
v = −2V
y L
where V and L are constants. If they exist, find the stream function and velocity potential. Solution: First check continuity and irrotationality:
∂ u ∂ v 2V 2V + = − = 0 ψ exists; ∂x ∂y L L ⎛ ∂ v ∂ u ⎞ 2V ⎞ ⎛ ∇ xV = k ⎜ − ⎟ = k ⎜ 0 + ⎟ ≠ 0 φ does not exist ⎝ ⎝ ∂ x ∂ y ⎠ L ⎠ To find the stream function ψ, use the definitions of u and v and integrate:
u=
⎛ xy y2 ⎞ ∂ψ ⎛ x y ⎞ = 2V ⎜ − ⎟ , ∴ ψ = 2V ⎜ − ⎟ + f ( x ) ∂y L 2 L ⎝ L L ⎠ ⎝ ⎠ ∂ψ 2Vy df 2Vy Evaluate = + = −v = ∂x L dx L
Thus
⎛ 2 xy y2 ⎞ df = 0 and ψ = V ⎜ − ⎟ + const dx L ⎠ ⎝ L
Ans.
P4.58 Show that the incompressible velocity potential in plane polar coordinates φ(r,θ) is such that
υr =
∂φ ∂r
υθ =
1 ∂φ r ∂θ
Finally show that φ as defined satisfies Laplace’s equation in polar coordinates for incompressible flow. Solution: Both of these things are quite true and easy to show from the definition of the gradient vector in polar coordinates. Ans.
P4.59 Consider the two-dimensional incompressible velocity potential φ = xy + x2 – y2. (a) Is it true that ∇2φ = 0, and, if so, what does this mean? (b) If it exists, find the stream function ψ(x, y) of this flow. (c) Find the equation of the streamline which passes through (x, y) = (2, 1). Solution: (a) First check that ∇2φ = 0, which means that incompressible continuity is satisfied.
∇ 2φ =
∂ 2φ ∂ 2φ + = 0+2−2 = 0 ∂ x2 ∂ y 2
Yes
(b) Now use φ to find u and v and then integrate to find ψ.
u=
∂φ ∂ψ y2 = y + 2x = , hence ψ = + 2 xy + f ( x ) ∂x ∂y 2
v=
∂φ ∂ψ df x2 = x − 2y = − = −2 y − , hence f ( x ) = − + const ∂y ∂x dx 2
The final stream function is thus ψ =
1 2 ( y − x 2 ) + 2 xy + const 2
Ans. (b)
(c) The streamline which passes through (x, y) = (2, 1) is found by setting ψ = a constant:
1 3 5 At ( x, y) = (2, 1), ψ = (12 − 2 2 ) + 2(2)(1) = − + 4 = 2 2 2 Thus the proper streamline is ψ =
1 2 5 ( y − x 2 ) + 2 xy = 2 2
Ans. (c)
P4.60 Liquid drains from a small hole in a tank, as shown in Fig. P4.60, such that the velocity field set up is given by υr ≈ 0, υz ≈ 0, υ = ωR2/r, where z = H is the depth of the water far from the hole. Is this flow pattern rotational or irrotational? Find the depth zc of the water at the radius r = R.
z patm
r z=H
θ
z=0 r=R
Fig. P4.60
Solution: From Appendix D, the angular velocity is
ωz =
1∂ 1 ∂ (rvθ ) − (vθ ) = 0 (IRROTATIONAL) r ∂r r ∂θ
Incompressible continuity is valid for this flow, hence Bernoulli’s equation holds at the surface, where p = patm, both at infinity and at r = R:
patm +
1 1 ρ Vr2=∞ + ρ gH = patm + ρ Vr2=R + ρ gzc 2 2
Introduce Vr =∞ = 0 and Vr=R = ω R to obtain
zC = H −
ω 2 R2
Ans.
2g
P4.64 Show that the velocity potential φ(r, z) in axisymmetric cylindrical coordinates (see Fig. 4.2 of the text) is defined by the formulas:
υr =
∂φ ∂r
υz =
∂φ ∂z
Further show that for incompressible flow this potential satisfies Laplace’s equation in (r, z) coordinates. Solution: Both of these things are quite true and are easy to show from their definitions. Ans.
P4.65 A two-dimensional incompressible flow is defined by
u=−
Ky x + y2 2
υ=
Kx x + y2 2
where K = constant. Is this flow irrotational? If so, find its velocity potential, sketch a few potential lines, and interpret the flow pattern.
Fig. P4.65
Solution: Evaluate the angular velocity:
2ω z =
∂v ∂u K 2Kx 2 K 2Ky 2 − = 2 − + − = 0 (Irrotational) Ans. ∂ x ∂ y x + y 2 (x 2 + y 2 )2 x 2 + y 2 (x 2 + y 2 )2
Introduce the definition of velocity potential and integrate to get φ(x, y):
u=
∂φ Ky ∂φ Kx ⎛ y ⎞ =− 2 ; v= = 2 , solve for φ = K tan−1 ⎜ ⎟ = Kθ 2 2 ⎝ x ⎠ ∂x ∂y x + y x +y
Ans.
The φ lines are plotted above. They represent a counterclockwise line vortex.
P4.66 A plane polar-coordinate velocity potential is defined by
K cos θ K = const r Find the stream function for this flow, sketch some streamlines and potential lines, and interpret the flow pattern.
φ=
Solution: Evaluate the velocities and thence find the stream function:
Fig. P4.66
vr =
∂φ Kcosθ 1 ∂ψ 1 ∂φ Ksinθ ∂ψ =− = ; vθ = =− 2 =− , 2 ∂r r ∂θ r ∂θ ∂r r r solve ψ = −
Ksinθ r
Ans.
The streamlines and potential lines are shown above. This pattern is a line doublet.
P4.71 Consider the following two-dimensional function f(x, y):
f = A x3 + B x y 2 + C x 2 + D ,
where A > 0
(a) Under what conditions, if any, on (A,B,C,D) can this function f be a steady, plane-flow velocity potential? (b) If you find a φ(x, y) to satisfy part (a), also find the associated stream function ψ(x, y), if any, for this flow. Solution: (a) If f is to be a plane-flow velocity potential, it must satisfy Laplace’s equation:
∇ 2 f = 6 Ax + Bx + 2C = 0 The velocity potential is
if
B = − 3 A and
C=0
φ = A x 3 − 3 A xy 2 + D
Ans.( a)
∂φ ∂ψ = 3A x 2 − 3A y 2 = , ∴ ψ = 3 A x 2 y − A y 3 + f ( x) ∂x ∂y ∂φ ∂ψ df v= = − 6 Axy = − = − 6 Axy − , ∴ f = const ∂y ∂x dx
u =
Finally, ψ = 3 A x 2 y − A y 3 + const Ans.(b) (b) To find ψ, use φ to get u and v and work backwards to get the stream function:
P4.74
Consider the two-dimensional incompressible polar-coordinate velocity potential
= B r cosθ + B L θ where B is a constant and L is a constant length scale. (a) What are the dimensions of B?
φ
(b) Locate the only stagnation point in this flow field. (c) Prove that a stream function exists and then find the function ψ(r, θ).
Solution: (a) To give φ its correct dimensions of {L2/T}, the constant B must have the dimensions of velocity, or {L/T}. Ans.(a)
(b) Calculate velocities in polar coordinates:
vr =
∂φ = B cosθ ∂r
;
vθ =
1 ∂φ BL = − B sin θ − r ∂θ r
At first it doesn’t look as if we can find a stagnation point, but indeed there is one:
BL = 0 Ans.(b) L As discussed later in Chap. 8, this is the velocity potential of a Rankine half-body. r = L , θ = 180! : vθ = 0 , vr = B −
(c) With the velocities known, check the continuity equation:
B cosθ B cosθ 1 ∂ 1 ∂vθ (r v r ) + = 0 = − = 0 r ∂r r ∂θ r r
Yes, satisfied
Continuity is satisfied. Find the stream function from the definition of ψ(r, θ):
vr =
1 ∂ψ = B cosθ r ∂θ Integrate :
∂ψ BL = − B sin θ − ∂r r ψ = B r sin θ + B L ln r + const ;
vθ = −
Ans.(c)
P4.77 Outside an inner, intense-activity circle of radius R, a tropical storm can be simulated by a polar-coordinate velocity potential φ(r, θ) = Uo R θ, where Uo is the wind velocity at radius R. (a) Determine the velocity components outside r = R. (b) If, at R = 25 mi, the velocity is 100 mi/h and the pressure 99 kPa, calculate the velocity and pressure at r = 100 mi. Solution: (a) First, convert Uo = 100 mi/h = 44.7 m/s and R = 25 mi = 40,200 m. The velocities are calculated from φ, as requested in Prob. P4.58:
vr =
∂ (U o R θ ) = 0 ∂r
;
vθ =
U R 1 ∂ (U o R θ ) = o r ∂θ r
Ans.(a )
Outside the “intense” region, the wind is simulated as a circulating “potential vortex” whose velocity drops off inversely as the radius. (b) The flow is irrotational, otherwise φ would not exist. Thus Bernoulli’s equation applies outside r = R, with no elevation change at the ocean surface. Take surface air density to be sea-level standard, ρ = 1.225 kg/m3.
At r = 4 R = 100 mi, vθ =
ρ
Bernoulli : p1 +
2
Uo R U 44.7 m = o = = 11.2 4R 4 4 s
V12 = p2 +
ρ 2
V22 , or :
99,000 + (1.225 / 2)(44.7)2 = p2 + (1.225 / 2)(11.2)2 , Solve p100 mi = 100,150 Pa Ans.(b) The pressure far from the storm is approximately sea-level standard pressure.
P4.86 SAE 10 oil at 20°C flows between parallel plates 8 mm apart, as in Fig. P4.86. A mercury manometer, with wall pressure taps 1 m apart, registers a 6-cm height, as shown. Estimate the flow rate of oil for this condition. Solution: Assuming laminar flow, this geometry fits Eqs. (4.143, 144) of the text:
Fig. P4.86
Vavg =
2 2 ⎛ dp ⎞ h umax = ⎜ ⎟ , where h = plate half-width = 4 mm ⎝ dx ⎠ 3µ 3
For SAE 10W oil, take ρ = 870 kg/m3 and µ = 0.104 kg/m⋅s. The manometer reads Δp = (ρHg – ρoil)gΔh = (13550 – 870)(9.81)(0.06) ≈ 7463 Pa for Δx = L = 1m
Then V =
Δp h 2 ⎛ 7463 Pa ⎞ (0.004)2 m = ⎜ ≈ 0.383 ⎟ Δx 3µ ⎝ 1 m ⎠ 3(0.104) s
The flow rate per unit width is Q = VA = (0.383)(0.008) ≈ 0.00306
m3 s⋅ m
Ans.
NOTE: The Reynolds number, based upon plate half-width, is 16, laminar.
P4.91 Consider 2-D incompressible steady Couette flow between parallel plates with the upper plate moving at speed V, as in Fig. 4.16a. Let the fluid be nonnewtonian, with stress given by
τ xx
⎛ ∂ u ⎞ = a ⎜ ⎟ ⎝ ∂ x ⎠
c
τ yy
⎛ ∂ v ⎞ = a ⎜ ⎟ ⎝ ∂ y ⎠
c
c
τ xy = τ yx
a ⎛ ∂ u ∂ v ⎞ = ⎜ + ⎟ , a and c are constants 2 ⎝ ∂ y ∂ x ⎠
Make all the same assumptions as in the derivation of Eq. (4.140). (a) Find the velocity profile u(y). (b) How does the velocity profile for this case compare to that of a newtonian fluid? Solution: (a) Neglect gravity and pressure gradient. If u = u(y) and v = 0 at both walls, then continuity specifies that v = 0 everywhere. Start with the x-momentum equation:
⎛ ∂ u ∂ u ⎞ ∂ p ∂τ xx ∂τ xy + v ⎟ = ρgx − + + ⎝ ∂ x ∂ y ⎠ ∂x ∂x ∂y
ρ ⎜ u
Many terms drop out because v = 0 and τxx and ∂u/∂x = 0 (because u does not vary with x). Thus we only have ∂τ xy d ⎡ a ⎛ du ⎞ c ⎤ du = = constant , u = C1 y + C2 ⎢ ⎜ ⎟ ⎥ = 0, or: ∂ y dy ⎢⎣ 2 ⎝ dy ⎠ ⎥⎦ dy The boundary conditions are no-slip at both walls: V V u(y = –h) = 0 = C1(–h) + C2; u(y = +h) = V = C1(+h) + C2, solve C1 = , C2 = 2h 2 The final solution for the velocity profile is:
u( y) =
V V y+ 2h 2
Ans. (a)
This is exactly the same as Eq. (4.140) for the newtonian fluid! Ans. (b)
