Solutions Manual for
Fluid Mechanics Seventh Edition in SI Units
Frank M. White Chapter 4 Differential Relations for Fluid Flow
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P4.1 An idealized velocity field is given by the formula
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 2 z dt ∂t ∂x ∂y ∂z or:
dV = (4x +16t 2 x)i + (−4ty + 4t 4 y)j + (16txz +16x 2 z)k dt
at (x, y, z) = (−1, +1, 0), we obtain (d) At (–1, +1, 0) there are many unit vectors normal to dV/dt. One obvious one is k. Ans. P4.2 A three-dimensional flow field has the velocity given by V = 4x 2 i + 3xyj + 5t(x + y)zk
Find the local and total accelerations in terms of x, y, z, and t. Solution
Velocity V = (4x 2 , 3xy, 5t(x + y)z) ∂V Local acceleration = (0, 0, 5(x + y)z) ∂t
Convective acceleration
ax = u
∂u ∂u ∂u +v +w ∂x ∂y ∂z
= (4x 2 )(8x) + (3xy)(0) + 5t(x + y)z(0) = 32x 3
2
ay = u
∂v ∂u ∂u +v +w ∂x ∂y ∂z
= (4x 2 )(3y) + (3xy)(3x) + 5t(x + y)z(0) = 12x 2 y + 9x 2 y = 21x 2 y ∂w ∂w ∂w az = u +v +w ∂x ∂y ∂z = (4x 2 )(5tz) + (3xy)(5tz) + 5t(x + y)z(5t)(x + y) = 20tx 2 z +15txyz + 25t 2 (x + y)2 z
Total acceleration = Local accel + convective accel ∂v = (32x 3 , 21x 2 y, 20tx 2 z + 15txyz + 25t 2 (x + y)2 z + 5(x + y)z) ∂t P4.3 Flow through the converging nozzle in Fig. P4.3 can be approximated by the onedimensional velocity distribution
(a) Find a general expression for the fluid acceleration in the nozzle. (b) For the specific case Vo = 3 m/s and L = 15 cm, compute the acceleration, in g’s, at the entrance and at the exit.
Fig. P4.3
Solution: Here we have only the single ‘one-dimensional’ convective acceleration:
For L = 0.15 m and Vo = 3 m/s,
du 2(3)2 2x = 1+ = 120(1+13.3x), with x in m dt 0.15 0.15
At x = 0, du/dt = 120 m/s2 (12 g’s); at x = L = 0.15 m, du/dt = 360 m/s2 (37 g’s). Ans. (b)
3
P4.4 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)
P4.5
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
4
These may be plotted for various values of the dimensionless constant C, as shown:
The streamlines converge and the velocity increases to the right.
Ans.(a)
(b) The accelerations are calculated from Eq. (4.2):
ax = u
∂u ∂u U2 x + v = [U o (1+ x / L)](U o / L) + 0 = o (1+ ) ∂x ∂y L L
ay = u
∂v ∂v U o2 y + v = 0 + (−U o y / L)(−U o / L) = ∂x ∂y L L
Ans.(b)
(c) Find the resultant of ax and ay from Ans.(b) above and introduce y/L from Ans.(a):
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.
P4.6 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?
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Solution: (a) For two-dimensional steady flow, the acceleration components are du ∂u ∂ u x U y U2 = u + v = U o o + −U o (0) = 2o x dt ∂x ∂ y L L L L dv ∂v ∂v x y U U 2 = u + v = U o (0) + −U o − o = 2o y dt ∂x ∂y L L L L
Therefore the resultant a = (U 2o /L2 )(xi + yj) = (U 2o /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
m U 2o U 2o m |r| = = 2 m, solve for U o = 6.3 2 2 2 s s L (1.5 m)
Ans. (b)
P4.7 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) + (8x)(2) = 16 x ∂x ∂y ∂v ∂v ay = u + v = (2 y)(8) + (8x)(0) = 16 y ∂x ∂y a = i ax + ja y = 16 i x+16j y = 16(i x+ j y) = 16r ax = u
Ans.(a) Radial, yes. Ans.(b)
(c) Find the streamlines in the manner suggested in Chapter 1, Eq. (1.41):
dx dy dx dy = , or : = , u v 2 y 8x
2 y dy = 8x dx
Integrate :
, or : y 2 = 4 x 2 + const
∫ 2 y dy = ∫ 8x dx
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)
P4.8 Consider a sphere of radius R immersed in a uniform stream Uo, as shown in Fig. P4.8. According to the theory of Chap. 8, the fluid velocity along streamline AB is given by R3 V = ui = U o 1+ 3 i x
6
Fig. P4.8
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 = U o (1+ R 3 /x 3 ), or: dt
−R
∫ −4R
dx = 1+ R 3 /x 3
t
∫ U o dt, 0
−R
R (x + R)2 R −1 2x − R or: U o t = x − ln 2 + tan =∞ R 3 −4 R 6 x − Rx + R 2 3
It takes an infinite time to actually reach the stagnation point, where the velocity is zero. Ans. (b) P4.9 When a valve is opened, fluid flows in the expansion duct of Fig. P4.9 according to the approximation x Ut V = iU 1− tanh L 2L
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. P4.9
7
Solution: This is a one-dimensional unsteady flow. The acceleration is ax =
Ut ∂u ∂u x U x U 2 Ut +u = U 1− sech − U 1− tanh ∂t ∂x 2L L L 2L 2L L
=
Ut 1 Ut U2 x (1− )[sech 2 − tanh 2 ] L 2L L 2 L
At (x, t) = (L, L/U), ax = (U2/L)(1/2)[sech2(1) – 0.5tanh2 (1)] ≈ 0.0650 U2/L Ans. (a) The acceleration becomes zero when Ut 1 Ut 2Ut 1 sech 2 − tanh 2 , or sinh 2 = 2, 2 L 2 L L or:
Ut ≈ 1.146 L
Ans. (b)
The acceleration starts off positive, then goes through zero and turns negative as the negative convective acceleration overtakes the decaying positive local acceleration.
P4.10 An idealized incompressible flow has the proposed three-dimensional velocity distribution V = 4xy2i + f(y)j – zy2k Find the appropriate form of the function f(y) which satisfies the continuity relation. Solution: Simply substitute the given velocity components into the incompressible continuity equation:
∂u ∂ v ∂ w ∂ ∂f ∂ df + + = (4xy 2 ) + + (−zy 2 ) = 4y 2 + − y 2 = 0 ∂ x ∂ y ∂z ∂ x ∂ y ∂z dy or:
df = −3y 2 . Integrate: f (y) = dy
∫ (−3y2 )dy = −y3 + constant
Ans.
P4.11 Show that Eq. (4.6) can be written as dρ + ρ∇ ⋅ V = 0 dt
where the d/dt is the total time derivative. Note that dρ/dt = 0 is a necessary and sufficient condition for a flow being incompressible. Solution: ∂π + ∇ ⋅ (ρ v) = 0 ∂t By product rule ∇ ⋅ (ρ v) = ρ∇ ⋅ v + (v ⋅ ∇)ρ Eq. (4.6)
8
∴
∂ρ ∂ρ + ∇ ⋅ (ρ v) ≡ + (v ⋅ ∇)ρ + ρ∇ ⋅ v ∂t ∂t ρ d = + ρ∇⋅v dt
Hence, the equation can be written as dρ + ρ∇ ⋅ v = 0 dt
dρ =0 dt ⇔∇⋅v= 0
Incompressible flow ⇔
P4.12 Check whether continuity is satisfied by the following flow fields:
(a) u = x, v = –y (b) u = x2y + y3, v = x3 – xy2 (c) u = sin2(xy), v = cos2(xy) x x (d) u = 2 , v= 2 2 x +y x + y2 (e) vr = r sin θ cos θ, vθ = –r sin2 θ Solution: Continuity eqn ∇ ⋅ v = 0 ∂u ∂v ∂w ∂ x + ∂ y + ∂ z = 0 (Cartesian coord) 1 ∂ (rυ ) + 1 ∂υθ + ∂υz = 0 (cylindrical coord) r ∂ ∂ r ∂ ∂θ ∂ z
∂u ∂v + = 1+ (−1) ≡ 0 (OK) ∂x ∂y
a.
u = x, v = −y,
b.
u = x 2 y + y 3 , v = x 3 − xy 2 ∂u ∂v + = (2xy) + (−2xy) ≡ 0 ∂x ∂y
(OK)
u = sin 2 xy v = cos 2 xy ∂u ∂v + = (2 sin xy)(cos xy)(y) + (2 cos xy)(−sin xy)(x) ∂x ∂y ≡/ 0 (not OK) x y d. u = 2 , v= 2 2 x +y x + y2
c.
∂ u ∂ v (x 2 + y 2 ) − x(2x) (x 2 + y 2 ) − y(2y) + = + ≡ 0 (OK) ∂x ∂y (x 2 + y 2 )2 (x 2 + y 2 )2 e.
υr = r sin θ cosθ , υθ = −r sin 2 θ 1∂ 1 ∂υθ 1 1 (rυr ) + = (2r sin θ cosθ ) + (−2r sin θ cosθ ) ≡ 0 (OK) r ∂r r ∂θ r r
9
P4.13 Consider the simple incompressible plane flow pattern u = U, v = V, and w = 0, where U and V are constants. (a) Convert these velocities into polar coordinate components, vr and vθ. [HINT: Make a sketch of the velocity components.] (b) Determine whether these new components satisfy the continuity equation in polar coordinates. Solution: This is harder than it looks. Make a sketch of each separate cartesian component: y
V
y
(a)
(b) U
θ
θ
x
x
(a) We can resolve each figure into radial and circumferential components. For Figure (a), U has a radial component U cosθ and a circumferential component (-U sinθ). For Figure (b), V has a radial component V sinθ and a circumferential component V cosθ. Combine these into the result vr = U cosθ + V sin θ
;
vθ = −U sin θ + V cosθ
Ans.(a)
(b) The original (cartesian) distribution, being constant velocity, obviously satisfied continuity. The new version, in polar coordinates, requires some effort. From Eq. (4.9) for incompressible flow,
1 ∂ 1 ∂ (r vr )+ (v ) = r ∂r r ∂θ θ 1 = (U cosθ +V sin θ ) + r
1 ∂ 1 ∂ [ r (U cosθ +V sin θ )] + (−U sin θ +V cosθ ) r ∂r r ∂θ 1 (−U cosθ − V sin θ ) = 0 Yes, satisfied. Ans.(b) r
P4.14 Derive Eq. (4.12b) for cylindrical coordinates by considering the flux of an incompressible fluid in and out of the elemental control volume in Fig. 4.2. Solution: For the differential CV shown,
∂ρ out − ∑dm in = 0 dυol + ∑dm ∂t
Fig. 4.2
10
∂ρ dr ∂ r + dθ dr dz + ρ vr r dz dθ + (ρ vr )dr(r + dr)dz dθ + ρ vθ dz dr ∂t 2 ∂r dr dr ∂ ∂ + (ρ vθ )dθ dz dr + ρ vz r + dθ dr + (ρ vz ) r + dθ dr ∂θ 2 ∂z 2 dr − ρ vr r dz dθ − ρ vθ dz dr − ρ vz r + dθ dr = 0 2 Cancel (dθ drdz) and higher-order (4th-order) differentials such as (dr dθ dz dr) and, finally, divide by r to obtain the final result:
∂ρ 1 ∂ 1 ∂ ∂ + (ρ rv r ) + (ρ vθ ) + (ρ v z ) = 0 Ans. ∂t r ∂r r ∂θ ∂z
P4.15 Spherical polar coordinates (r, θ, φ) are defined in Fig. P4.15. The cartesian transformations are x = r sinθ cosφ y = r sinθ sinφ z = r cosθ Do not show that the cartesian incompressible continuity relation (4.12a) can be transformed to the spherical polar form
Fig. P4.15
1 ∂ 2 1 ∂ 1 ∂ (r υr ) + (υθ sin θ ) + (υφ ) = 0 2 rsin θ ∂θ rsin θ ∂φ r ∂r What is the most general form of υr when the flow is purely radial, that is, υθ and υφ are zero? Solution: Note to instructors: Do not assign the derivation of this continuity relation, it takes years to achieve, the writer can’t do it successfully. The problem is only meant to acquaint students with spherical coordinates.
If υθ = υφ = 0, then
1 ∂ 2 1 (r υr ) = 0, so, in general, υ r = 2 fcn(θ , φ ) Ans. 2 r ∂r r
11
P4.16 For an incompressible plane flow in polar coordinates, we are given
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 cosθ + r 2 sin θ )] + , r ∂r r ∂θ r ∂r r ∂θ 1 ∂vθ or : = −4 r 2 cosθ − 3r sin θ r ∂θ Integrate : vθ = − 4 r 3 sin θ + 3r 2 cosθ + f (r)
Ans.
We can’t determine the form of the “constant of integration” f(r) without further information.
P4.17 For incompressible polar-coordinate flow, what is the most general form of a purely circulatory motion, υθ = υθ (r, θ, t) and υr = 0, which satisfies continuity? Solution: If vr = 0, the plane polar coordinate continuity equation reduces to: 1 ∂ vθ = 0, or: vθ = fcn(r) only r ∂θ
Ans.
P4.18 What is the most general form of a purely radial polar-coordinate incompressibleflow pattern, υr = υr(r, θ, t) and υθ = 0, which satisfies continuity? Solution: If vθ = 0, the plane polar coordinate continuity equation reduces to: 1∂ 1 (rvr ) = 0, or: vr = fcn(θ ) only r ∂r r
P4.19
Ans.
Consider the plane polar coordinate velocity distribution vr =
C r
vθ =
K r
vz = 0
where C and K are constants. (a) Determine if the incompressible equation of continuity is satisfied. (b) By sketching some velocity vector directions, plot a single streamline for C = K. What might this flow field simulate? Solution: (a) Evaluate the incompressible continuity equation (4.12b) in polar coordinates: 1 ∂ 1 ∂ 1 ∂ C 1 ∂ K (r vr ) + (vθ ) = (r ) + ( )= 0 + 0 = 0 r ∂r r ∂θ r ∂r r r ∂θ r
Ans.(a)
Incompressible continuity is indeed satisfied. (b) For C = K, we can plot a representative streamline by putting in some velocity vectors and sketching a line parallel to them:
12
The streamlines are logarithmic spirals moving out from the origin. [They have axisymmetry about O.] This simple distribution is often used to simulate a swirling flow such as a tornado.
P4.20 An excellent approximation for the two-dimensional incompressible laminar boundary layer on the flat surface in Fig. P4.20 is u ≈ U (2
y y3 y4 −2 + ) for y ≤δ , where δ = C x1/ 2 , C = constant 3 4 δ δ δ
Fig. P4.20
(a) Assuming a no-slip condition at the wall, find an expression for the velocity component v(x, y) for y ≤ δ. (b) Then find the maximum value of v at the station x = 1 m, for the particular case of airflow, when U = 3 m/s and δ = 1.1 cm. Solution: (a) With u known, use the two-dimensional equation of continuity to find v: ∂v ∂u 2 y dδ 6 y 3 dδ 4 y 4 dδ = − = −U (− + − ) , ∂y ∂x δ 2 dx δ 4 dx δ 5 dx dδ or : v = 2U dx
y
∫( 0
y
δ2
−
3y 3 2 y 4 dδ y 2 3y 4 2 y5 + ) dy = 2U ( − + ) dx 2δ 2 4δ 4 5δ 5 δ 4 δ5
Ans.(a)
13
(b) First evaluate C from the given data at x = 1 m:
δ = 0.011m = C (1m)1/ 2 , hence dδ 1 Or, alternately , = C x −1/ 2 = dx 2
C = 0.011 m1/ 2 1 δ δ ( ) x −1/ 2 = 1/ 2 2 x 2x
Substitute this into Ans.(a) above and note that v rises monotonically with y to a maximum at the outer edge of the boundary layer, y = δ . The maximum velocity v is thus
vmax ≈ 2U
dδ 1 3 2 m 0.011m 3 m ( − + ) = 2(3 )[ ]( ) ≈ 0.0050 Ans.(b) dx 2 4 5 s 2(1m) 20 s
This is slightly smaller than the exact value of vmax from laminar boundary theory (Chap. 7).
P4.21 A piston compresses gas in a cylinder by moving at constant speed V, as in Fig. P4.21. Let the gas density and length at t = 0 be ρo and Lo, respectively. Let the gas velocity vary linearly from u = V at the piston face to u = 0 at x = L. If the gas density varies only with time, find an expression for ρ(t).
Fig. P4.21
Solution: The one-dimensional unsteady continuity equation reduces to
x ∂ρ ∂ dρ ∂u + (ρ u) = + ρ , where u = V 1 − , L = L o − Vt, ρ = ρ (t) only L ∂ t ∂x dt ∂x ρ t ∂u V dρ dt Enter =− and separate variables: ∫ =V∫ ∂x L ρ L o − Vt ρ o o
Lo The solution is ln(ρ/ρ o ) = −ln(1 − Vt/L o ), or: ρ = ρ o Ans. Lo − Vt
P4.22 An incompressible flow field has the cylindrical velocity components υθ = 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: 1∂ 1 ∂ vθ ∂ vz (rvr ) + + = 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.
14
P4.23 A two-dimensional incompressible velocity field has u = K(1 – e–ay), for x ≤ L and 0 ≤ y ≤ ∞. What is the most general form of v(x, y) for which continuity is satisfied and v = vo at y = 0? What are the proper dimensions for constants K and a? Solution: We can find the appropriate velocity v from two-dimensional continuity:
∂v ∂u ∂ = − = − [K(1 − e−ay )] = 0, or: v = fcn(x) only ∂y ∂x ∂x Since v = vo
at y = 0 for all x, then it must be that v = vo = const
Ans.
The dimensions of K are {K} = {L/T} and the dimensions of a are {L–1}. Ans.
P4.24 Air flows under steady, approximately one-dimensional conditions through the conical nozzle in Fig. P4.24. If the speed of sound is approximately 340 m/s, what is the minimum nozzle-diameter ratio De/Do for which we can safely neglect compressibility effects if Vo = (a) 10 m/s and (b) 30 m/s? Solution: If we apply one-dimensional continuity to this duct,
Fig. P4.24
π 2 π D o = ρ eVe D 2e , or Vo ≈ Ve(D e /D o )2 if ρ o ≈ ρ e 4 4 To avoid compressibility corrections, we require (Eq. 4.18) that Ma ≤ 0.3 or, in this case, the highest velocity (at the exit) should be Ve ≤ 0.3(340) = 102 m/s. Then we compute ρ oVo
(D e /D o )min = (Vo /Ve )1/2 = (Vo /102)1/2 = 0.31 if Vo = 10 m/s Ans. (a)
= 0.54 if Vo = 30 m/s Ans. (b)
P4.25 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 vθ, the equation of continuity (4.9) becomes ∂v 1 ∂ 1 ∂ ∂ (r vr ) + z = 0 = (r vr ) + (Bz) , r ∂r ∂z r ∂r ∂z ∂ B or : (r vr ) = − B r ; Integrate : r vr = − r 2 + f (z) ∂r 2 B f (z) Finally, vr = − r + Ans. 2 r
The “function of integration”, f(z), is arbitrary, at least until boundary conditions are set. __________________________________________________________________________
15
P4.26 A tank volume υ contains gas at conditions (ρo, po, To). At time t = 0 it is punctured by a small hole of area A. According to the theory of Chap. 9, the mass flow out of such a hole is approximately proportional to A and to the tank pressure. If the tank temperature is assumed constant and the gas is ideal, find an expression for the variation of density within the tank. Solution: This problem is a realistic approximation of the “blowdown” of a high-pressure tank, where the exit mass flow is choked and thus proportional to tank pressure. For a control volume enclosing the tank and cutting through the exit jet, the mass relation is d d exit = 0, or: exit = −C p A, where C = constant (m tank ) + m ( ρυ ) = −m dt dt p(t) p dp CRTo A t Introduce ρ = and separate variables: ∫ =− ∫ dt RTo p υ p o o
The solution is an exponential decay of tank density: p = po exp(–CRToAvt/ υ ). Ans. P4.27 For incompressible laminar flow between parallel plates (see Fig. 4.12b), the flow is two-dimensional (v ≠ 0) if the walls are porous. A special case solution is , where A and B are constants. (a) Find a general formula for velocity v if v = 0 at y = 0. (b) What is the value of the constant B if v = vw at y = +h? Solution: (a) Use the equation of continuity to find the velocity v:
∂v ∂u = − = −(−B)(h2 − y 2 ) ∂y ∂x Integrate : v = B ∫ (h2 − y 2 )dy = B(h2 y −
y3 ) + f (x) 3
y3 ) Ans.(a) 3 (b) Just simply introduce this boundary condition into the answer to part (a): If v =0 at y =0, then f (x) = 0. ∴ v = B(h2 y −
3vw h3 v( y = h) = vw = B(h − ) , hence B = 3 2 h3 3
Ans.(b)
P4.28 An incompressible flow in polar coordinates is given by b vr = K cosθ 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. P4.28
16
Solution: Substitute into plane polar coordinate continuity:
b 1 ∂ 1∂ 1 ∂ vθ b ?1 ∂ (rvr ) + = 0= K cosθ r − + −K sin θ 1+ 2 = 0 Satisfied r ∂r r ∂θ r ∂r r r ∂θ r 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).
P4.29 Show that, in Eq. (4.38), the gravity terms can be absorbed into the pressure gradient terms through the introduction of an effective pressure. Solution: In Eq. (4.38), the gravity and pressure gradient terms are ρ g − ∇p . Define p* = − ρ g ⋅ x + p = −ρ (g x x + g y y + g z z) + P then − ∇p* = ρ g − ∇p
or −
∂p x ∂p = ρ gx − ∂x ∂x
−
∂p v ∂p = ρg y − ∂y ∂y
−
∂p v ∂p = ρg y − ∂z ∂z
P4.30 Curvilinear, or streamline, coordinates are defined in Fig. P4.30, where n is normal to the streamline in the plane of the radius of curvature R. Show that Euler’s frictionless momentum equation (4.36) in streamline coordinates becomes ∂V / ∂t + V (∂V / ∂s) = −(1 / ρ )(∂p / ∂s) + gs
−V
∂θ V 2 1 ∂p − =− + gn ∂t R ρ ∂n
(1)
(2)
Fig. P4.30
Further show that the integral of Eq. (1) with respect to s is none other than our old friend Bernoulli’s equation (3.76).
17
Solution: This is a laborious derivation, really, the problem is only meant to acquaint the student with streamline coordinates. The second part is not too hard, though. Multiply the streamwise momentum equation by ds and integrate:
∂V dp dp dp ds + VdV = − + g s ds = − −g sin θ ds = − −g dz ρ ρ ρ2 ∂t 2
Integrate from 1 to 2:
∫ 1
V 2 − V12 ∂V ds + 2 + ∂t 2
2
∫ 1
dp + g(z 2 − z1 ) = 0 (Bernoulli) Ans. ρ
P4.31 A frictionless, incompressible steady-flow field is given by V = 2xyi – y2j in arbitrary units. Let the density be ρo = constant and neglect gravity. Find an expression for the pressure gradient in the x direction. Solution: For this (gravity-free) velocity, the momentum equation is
∂V ∂V 2 ρ u +v = −∇p, or: ρ o [(2xy)(2yi) + (−y )(2xi − 2yj)] = −∇p ∂y ∂x Solve for ∇p = −ρ o (2xy 2 i + 2y 3 j), or:
∂p = −ρo 2xy 2 ∂x
Ans.
P4.32 Consider the incompressible flow field of Prob. P4.7, 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. P4.7 we found the accelerations, so we can proceed to Navier-Stokes:
∂u ∂u ∂p ∂p +v ) = ρ[0+(8x)(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 ∇ 2 v =− +0+0; ∂x ∂y ∂y ∂y
ρ (u
∂p =−16ρ 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 = ∂x ∂y
∫ −16ρ x dx + ∫ −16ρ y dy = − 8ρ (x 2 + y 2 ) + const
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.
18
P4.33 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 = −2y + 2y + 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: ∂y ∂z ∂x ∂z ∂x ∂x ∂y
ρ[−2xy(−2y) + (y 2 − x 2 )(−2x)] = −
∂p ∂p + µ (0 + 0 + 0), or: = −2 ρ (xy 2 + x 3 ) ∂x ∂x
and, similarly for the y-momentum relation, ∂ 2v ∂ 2v ∂ 2v ∂v ∂v ∂v ∂p ρ u + v + w = − + µ 2 + 2 + 2 , or: ∂y ∂z ∂y ∂x ∂ x ∂ y ∂z
ρ[−2xy(−2x) + (y 2 − x 2 )(2y)] = −
∂p ∂p + µ (−2 + 2 + 0), or: = −2 ρ (x 2 y + y 3 ) ∂y ∂y
The two gradients ∂ p/∂ x and ∂ p/∂ y may be integrated to find p(x, y): p=
∫
x 2 y2 x 4 ∂p dx|y =Const = −2 ρ + + f (y), then differentiate: ∂x 4 2
∂p df df ρ = −2 ρ (x 2 y) + = −2 ρ (x 2 y + y 3 ), whence = −2 ρ y 3 , or: f(y) = − y 4 + C ∂y dy dy 2 ρ Thus: p = − (2x 2 y 2 + x 4 + y 4 ) + C = pa 2
at (x, y) = (0, 0), or: C = p a
Finally, the pressure field for this flow is given by 1 p = p a − ρ (2x 2 y 2 + x 4 + y 4 ) Ans. (b) 2
P4.34 For the velocity distribution of Prob. P4.5, determine if (a) the equation of continuity and (b) the Navier-Stokes equation are satisfied. (c) If the latter is true, find the pressure distribution p(x,y) when the pressure at the origin equals po. Neglect gravity. Solution: Recall that we were given u = Uo(1+x/L) and v = -Uo y/L. (a) Test continuity:
∂u ∂v ∂ x ∂ y U U + = [U o (1+ )] + (−U o ) = o − o ≡ 0 ∂x ∂y ∂x L ∂y L L L
OK,satisfied. Ans.(a)
19
(b) Now substitute these velocities into the x- and y- Navier-Stokes equations:
∂u ∂u x U y 1 ∂p 1 ∂p +v = U o (1+ ) o + (−U o )(0) = − + ν ∇ 2u = − +0 ∂x ∂y L L L ρ ∂x ρ ∂x ∂v ∂v x y U 1 ∂p 1 ∂p u +v = U o (1+ )(0) + (−U o )(− o ) = − + ν ∇2v = − +0 ∂x ∂y L L L ρ ∂y ρ ∂y u
Solve for the two pressure gradients and cross-differentiate to see if they agree:
∂p ρU o2 x = − (1+ ) ∂x L L
∂p ρU o2 y = − ∂y L L
∂2 p Check = 0 for both ∂x∂y
Thus, before finding p(x,y), we know this is an exact solution to Navier-Stokes. Ans.(b) (c) Integrate the two pressure gradients to find the pressure distribution:
p=
∫
∂p ρU 2 x2 ∂p df ρU 2 y 2 dx = − o (x+ ) + f (y) ; Then = , f = − o + const ∂x L 2L ∂y dy L 2L p = −
ρU o2 x 2 y2 (x+ + ) + po L 2L 2L
Ans.(c)
This is the same as Bernoulli’s equation, but that is a bit hard to see.
P4.35 According to potential theory (Chap. 8) for the flow approaching a rounded twodimensional body, as in Fig. P4.35, the velocity approaching the stagnation point is given by u = U(1 – a2/x2), where a is the nose radius and U is the velocity far upstream. Compute the value and position of the maximum viscous normal stress along this streamline. Is this also the position
Fig. P4.35
of maximum fluid deceleration? Evaluate the maximum viscous normal stress if the fluid is SAE 30 oil at 20°C, with U = 2 m/s and a = 6 cm. Solution: (a) Along this line of symmetry the convective deceleration is one-dimensional: ax = u
a 2 2a 2 a2 a4 ∂u = U 1− 2 U 3 = 2U 2 3 − 5 ∂x x x x x
This has a maximum deceleration at
da x = 0, or at x = − √ (5 / 3) a = −1.29a dx
Ans. (a)
20
The value of maximum deceleration at this point is (b) The viscous normal stress along this line is given by
τ xx = 2µ
2a 2 U ∂u 4µ U = 2µ 3 with a maximum τ max = at x = −a Ans. (b) ∂x a x
Thus maximum stress does not occur at the same position as maximum deceleration. For SAE 30 oil at 20°C, we obtain the numerical result
SAE 30 oil, ρ = 917
kg kg 4(0.29)(2.0) , µ = 0.29 , τ max = ≈ 39 Pa Ans. (b) 3 m⋅s (0.06 m) m
P4.36 The answer to Prob. 4.17 is υθ = f(r) only. Do not reveal this to your friends if they are still working on Prob. 4.17. Show that this flow field is an exact solution to the NavierStokes equations (4.38) for only two special cases of the function f(r). Neglect gravity. Interpret these two cases physically. Solution: Given vθ = f(r) and vr = vz = 0, we need only satisfy the θ-momentum relation: 1 ∂ ∂ v 1 ∂ 2 v ∂v v ∂v 1 ∂p vθ θ ρ vr θ + θ θ = − + µ − r θ + 2 , 2 r ∂θ r ∂θ r2 ∂r r ∂ r ∂ r r ∂θ
1 d df f 1 1 or: ρ (0 + 0) = −0 + µ r + 0 − 2 , or: f ′′ + f ′ − 2 f = 0 r dr dr r r r This is the ‘equidimensional’ ODE and always has a solution in the form of a power-law, f = Crn. The two relevant solutions for these particular coefficients are n = ±1: f1 = C1r (solid-body rotation); f2 = C2/r (irrotational vortex) Ans.
P4.37
Consider incompressible flow at a volume rate Q toward
θ = π/4
a drain at the vertex of a 45° wedge of width b, as in Fig. P4.37. Neglect gravity and friction and assume purely radial
r
inflow. (a) Find an expression for vr(r). (b) Show that the viscous term in the r-momentum equation is zero.
θ Drain
Fig. P4.37
(c) Find the pressure distribution p(r) if p = po at r = R. Solution: (a) Assume one-dimensional, steady, radial inflow. Then, at any radius r,
vr =
−Q −Q −C 4Q = = , where C = area ( π / 4)r b r πb
Ans.(a)
Q
21
The velocity is negative because the flow is inward. (b) The r-momentum equation is not written out in Chapter 4; it is Eq. (D.5) of Appendix D. The viscous term is
2 ∂vθ 1 ∂ ∂vr vr 2 ∂vθ ) = ν[ (r )− − ]= r ∂r ∂r r 2 r 2 ∂θ r 2 r 2 ∂θ 1 ∂ ∂ −C (−C / r) 1 ∂ C C C C = ν[ (r ( )) − − 0] = ν [ ( )+ ] = ν (− + ) = 0 r ∂r ∂r r r ∂r r r 3 r2 r3 r3
ν (∇ 2 vr −
vr
−
Ans.(b)
(c) With the viscous term zero, the r-momentum equation reduces to
ρ (vr
∂vr −C C ∂p ∂p ρ C2 ) = ρ ( )( ) = − , or : = ∂r r r2 ∂r ∂r r3
Integrate : p = − Finally,
ρ C2 2r2
+ C1 ; at r = R, p = po = −
p = po +
ρ C2 2 R2
+ C1 , C1 = po +
ρ C2 C2 4Q ( − ) , where C = 2 2 2 R πb r
ρ C2 2 R2
Ans.(c)
The two terms in parentheses are the velocities-squared at r = R and r = r, respectively. In other words, it integrates to Bernoulli’s equation because the viscous term is zero (irrotational flow).
P4.38 A proposed three-dimensional incompressible flow field has the following vector form: V = Kxi + Kyj – 2Kzk (a) Determine if this field is a valid solution to continuity and Navier-Stokes. (b) If g = –gk, find the pressure field p(x, y, z). (c) Is the flow irrotational? Solution: (a) Substitute this field into the three-dimensional incompressible continuity equation:
∂u ∂ v ∂ w ∂ ∂ ∂ + + = (Kx) + (Ky) + ( −2Kz) ∂ x ∂ y ∂z ∂ x ∂y ∂z = K + K − 2K = 0 Yes, satisfied. Ans. (a)
(b) Substitute into the full incompressible Navier-Stokes equation (4.38). The laborious results are: ∂p x − momentum: ρ (K 2 x + 0 + 0) = − + µ (0 + 0 + 0) ∂x
y − momentum: ρ (0 + K 2 y + 0) = −
∂p + µ (0 + 0 + 0) ∂y
z − momentum: ρ {0 + 0 + (−2Kz)(−2K )} = −
∂p + ρ (−g) + µ (0 + 0 + 0) ∂z
22
Integrate each equation for the pressure and collect terms. The result is p = p(0,0,0) – ρ gz – (ρ /2)K2(x2 + y2 + 4z2) Ans. (b) Note that the last term is identical to (ρ/2)(u2 + v2 + w2), in other words, Bernoulli’s equation. (c) For irrotational flow, the curl of the velocity field must be zero: ∇ × V = i(0 – 0) + j(0 – 0) + k(0 – 0) = 0 Yes, irrotational. Ans. (c)
P4.39 From the Navier-Stokes equations for incompressible flow in polar coordinates (App. E for cylindrical coordinates), find the most general case of purely circulating motion υθ (r), υr = υz = 0, for flow with no slip between two fixed concentric cylinders, as in Fig. P4.39. Solution: The preliminary work for this !""(r) " r No slip
r!a r!b
Fig. P4.39
problem is identical to Prob. 4.36 on an earlier page. That is, there are two possible solutions for purely circulating motion υθ(r), hence vθ = C1r +
C2 , subject to vθ (a) = 0 = C1a + C2 / a and vθ (b) = 0 = C1b + C2 /b r
This requires C1 = C2 = 0, or vθ = 0 (no steady motion possible between fixed walls) Ans. P4.40 A constant-thickness film of viscous liquid flows in laminar motion down a plate inclined at angle θ, as in Fig. P4.40. The velocity profile is u = Cy(2h – y) v = w = 0 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. y
g
h (y)
! x
Fig. P4.40
23
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 ρ u + v = − + ρg x + µ∇ 2 u, or: 0 = 0 + ρg sinθ + µ (−2C) ∂y ∂x ∂x ρ g sinθ Solve for C = Ans. (a) 2µ The flow rate per unit width is found by integrating the velocity profile and using C: h
Q=
∫ 0
h 2 ρ gh3 sinθ u dy = ∫ Cy(2h − y)dy = Ch 3 = per unit width 3 3 µ 0
Ans. (b)
P4.41 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 Navier-Stokes is relevant:
Fig. P4.41
ρ
dw d 2w ρg = 0 = ρ g + µ 2 , or: w′′ = − , 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=
ρg 2 (h − x 2 ) Ans. 2µ
P4.42 Show that the incompressible flow distribution, in cylindrical coordinates, where C is a constant, (a) satisfies the Navier-Stokes equation for only two values of n. vr = 0
vθ = C r n
vz = 0
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?
24
Solution: (a) The important direction here is the θ-momentum equation, Eq. (D.6):
1 1 ∂p v 2 ∂v ∂vθ + (V • ∇)vθ + vr vθ = − + ν (∇ 2 vθ − θ2 + 2 θ ) , or : r r ∂θ ∂t ρ r ∂θ r vθ 1 ∂ ∂ Cr n n (r (Cr ))− 2 ] , or : 0 + 0 + 0 = 0 + ν (∇ vθ − 2 + 0) = ν [ r ∂r ∂r r r 1 ∂ ν[ (rnCr n−1 )−Cr n−2 ] = ν (Cn 2 r n−2 − Cr n−2 ) = 0 r ∂r 2
Cancel C and ν and rn-2. These terms equal zero only if n2 = 1, or n = ±1. Ans.(a). (b) Find the respective pressure distributions for n = 1 and n = -1. Use Eq. (D.5), which reduces simply to ∂p/∂r = ρvθ2/r. Try this for each distribution, n = ±1: Case1, n = 1 :
∫
Case2, n = −1 :
p po
∫
dp = p po
dp =
∫
ρ (C 2 r 2 ) dr ; or : R r r
∫
p = po +
ρC 2 2 (r − R 2 ) 2
Ans.(b1 )
ρ (C 2 / r 2 ) ρC 2 1 1 dr ; or : p = po + ( − ) R 2 R2 r 2 r r
Ans.(b2 )
Case 1, vθ = Cr, is solid-body rotation. Case 2, vθ = C/r, is an irrotational potential vortex.
P4.43 Attempt this question if you have learnt how to classify second-order partial differential equations. For one-dimensional flow u = u(y, t) under simple shear, and in the absence of pressure gradient and gravity, show that the Navier-Stokes equations reduce to ∂2 u ∂u =v 2 ∂t ∂y where v = ρ / µ is the kinematic viscosity of the fluid. Can you identify what type of partial differential equation it is? What is the physics represented by this equation, and the role played by the parameter v? Is it dissipative or conservative? Solution: Full Navier-Stokes eqn ∂u 2 ∂2u ∂2u ∂u ∂u ∂u ∂u ∂p ρ + u + v + w = − + µ 2 + 2 + 2 + gx ∂x ∂y ∂z ∂x ∂y ∂z ∂t ∂x || 0 as u is indep of x
Hence ρ
|| 0 as v=0
|| 0 as w=0
|| 0 no press grad
|| 0 as u is indep of x
|| 0 as u is indep of z
|| 0 no gravity
∂u ∂2 u ∂u ∂2 u = µ 2 or = v 2 (v = µ / ρ ) ∂t ∂t ∂y ∂y
This is a parabolic type of partial differential equation, also known as diffusion equation. The N-S equation reduces to a momentum diffusion equation, where the kinematic viscosity plays the role of momentum diffusivity. Diffusion is dissipative. Momentum or energy is dissipated because of viscosity.
25
P4.44 Reconsider the angular-momentum balance of Fig. 4.5 by adding a concentrated body couple Cz about the z axis [6]. Determine a relation between the body couple and shear stress for equilibrium. What are the proper dimensions for Cz? (Body couples are important in continuous media with microstructure, such as granular materials.) Solution: The couple Cz has to be per unit volume to make physical sense in Eq. (4.39):
Fig. 4.5
1 ∂τ xy 1 ∂τ yx 1 d 2θ dx − dy dx dy dz + Cz dx dy dz = ρ dx dy dz(dx 2 + dy 2 ) 2 τ xy − τ yx + 2 ∂x 2 ∂y 12 dt Reduce to third order terms and cancel (dx dy dz): τ yx − τ xy = Cz Ans. The concentrated couple allows the stress tensor to have unsymmetrical shear stress terms. P4.45 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
ρ cp u or :
∂T ∂2T ∂u d 2T du =0=k + µ ( )2 = k + µ ( )2 , 2 2 ∂x ∂y dy ∂y dy
d 2T dy 2
µ 2Uy 2 4µ U 2 2 dT 4µ U 2 y 3 = − ( ) = −( ) y ; Integrate : = −( ) +C1 k h2 dy k h4 k h4 3
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 h4 µU 2 ) + C2 ; at y = h : T = Tw = −( ) + C2 , ∴ C2 = Tw + 3k k h4 12 k h4 12
The final solution for T(y) is, like Ex. 4.6, a quartic polynomial:
µU 2 y4 T ( y) = Tw + (1 − ) 3k h4
Ans.
26
P4.46 As mentioned in Sec. 4.10, the velocity profile for laminar flow between two plates, as in Fig. P4.46, is 4u y(h − y) u = max 2 υ=w=0 h If the wall temperature is Tw at both walls, use
Fig. P4.46
the incompressible-flow energy equation (4.75) to solve for the temperature distribution T(y) between the walls for steady flow. Solution: Assume T = T(y) and use the energy equation with the known u(y): 2 4u 2 du dT d 2T d 2T max ρc p = k 2 + µ , or: ρc p (0) = k 2 + µ 2 (h − 2y) , or: dt dy dy dy h
d2 T 16µu 2max 2 dT −16µu 2max 2 4y 3 2 2 = − (h − 4hy + 4y ), Integrate: = h y − 2hy + + C 1 dy 3 dy 2 kh 4 kh 4
Before integrating again, note that dT/dy = 0 at y = h/2 (the symmetry condition), so C1 = –h3/6. Now integrate once more: 16µu 2max 2 y 2 y3 y4 T=− h − 2h + + C y + C2 1 3 3 kh 4 2
If T = Tw at y = 0 and at y = h, then C2 = Tw. The final solution is: T = Tw +
8µ u 2max y y 2 4y 3 2y 4 + − − Ans. k 3h h 2 3h 3 3h 4
This is exactly the same solution as Problem P4.45, except that, here, the coordinate y is measured from the boo tom wall rather than the centerline. P4.47 Suppose that we wish to analyze the rotating, partly-full cylinder of Fig. 2.23 as a spin-up problem, starting from rest and continuing until solid-body-rotation is achieved. What are the appropriate boundary and initial conditions for this problem? Solution: Let V = V(r, z, t). The initial condition is: V(r, z, 0) = 0. The boundary conditions are Along the side walls: vθ (R, z, t) = RΩ, vr(R, z, t) = 0, vz(R, z, t) = 0. At the bottom, z = 0: vθ (r, 0, t) = rΩ, vr(r, 0, t) = 0, vz(r, 0, t) = 0. At the free surface, z = η : p = patm, τrz = τθz = 0.
27
P4.48 For the draining liquid film of Fig. P4.40, what are the appropriate boundary conditions (a) at the bottom y = 0 and (b) at the surface y = h?
Fig. P4.40
Solution: The physically realistic conditions at the upper and lower surfaces are: (a) at the bottom, y = 0, no-slip: u(0) = 0 Ans. (a)
(b) At the surface, y = h, no shear stress, µ
∂u ∂u = 0, or (h) = 0 Ans. (b) ∂y ∂y
P4.49 Suppose that we wish to analyze the sudden pipe-expansion flow of Fig. P3.64, 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 flat-faced surface between the two.
Fig. P3.64
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. 4.38.] 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.
28
P4.50 For the sluice gate problem of Example 3.10, list all the boundary conditions needed to solve this flow exactly by, say, Computational Fluid Dynamics (CFD). 2 3 1 2
4
3
Solution: There are four different kinds of boundary conditions needed, as labeled. (1) Known velocity V1 upstream, and of course the depth y1 must be known. (2) Known pressure patm at both the upstream and downstream free surfaces. (3) No-slip (V = 0) all along the bottom and on the gate inner wall. (4) The downstream flow is complicated because we don’t know V2 or y2 and therefore cannot specify them. What CFD modelers do is to have an adjustable upper boundary and specify that the exit flow is “smooth”, or “zero gradient”, that is, ∂V/∂x = 0.
P4.51 Fluid from a large reservoir at temperature To flows into a circular pipe of radius R. The pipe walls are wound with an electric-resistance coil which delivers heat to the fluid at a rate qw (energy per unit wall area). If we wish to analyze this problem by using the full continuity, Navier-Stokes, and energy equations, what are the proper boundary conditions for the analysis? Solution: Letting z = 0 be the pipe entrance, we can state inlet conditions: typically uz(r, 0) = U (a uniform inlet profile), ur(r, 0) = 0, and T(r, 0) = To, also uniform. At the wall, r = R, the no-slip and known-heat-flux conditions hold: uz(R, z) = ur(R, z) = 0 and k(∂ T/∂ r) = qw at (R, z) (assuming that qw is positive for heat flow in). At the exit, z = L, we would probably assume ‘free outflow’: ∂ uz/∂ z = ∂ T/∂ z = 0. Finally, we would need to know the pressure at one point, probably the inlet, z = 0.
P4.52 What conditions regarding velocity, shear stress, and velocity gradient can be specified on a plane surface about which the flow is symmetrical? Solution: On a surface about which the flow is symmetrical, the normal velocity, the tangential velocity gradient (and hence the shear stress) are equal to zero.
29
P4.53 Given the incompressible flow V = 3yi + 2xj. Does this flow satisfy continuity? If so, find the stream function ψ(x, y) and plot a few streamlines, with arrows. Solution: With u = 3y and v = 2x, we may check ∂ u/∂ x + ∂ v/∂ y = 0 + 0 = 0, OK. Find the streamlines from u = ∂ ψ/∂ y = 3y and v = –∂ψ/∂x = 2x. Integrate to find
ψ=
3 2 y − x2 2
Ans.
Set ψ = 0, ±1, ±2, etc. and plot some streamlines at right: flow around corners of half-angles 39° and 51°. y
51° 39°
x
Fig. P4.53
P4.54 Show that, for two-dimensional steady creeping flow where the inertia can be ignored, the Navier-Stokes equations in terms of the stream function reduce to the biharmonic equation ∂4ψ ∂4ψ ∂4ψ ∇ 4ψ = 4 + 2 2 2 + 4 = 0 ∂x ∂x ∂y ∂y Compare this equation with the original Navier-Stokes equations. Solution: 2-D steady creeping flow
0=−
1 ∂p ∂2u ∂2u + v 2 + 2 ρ ∂x ∂y ∂x
0=−
1 ∂p ∂2 v ∂2 v + v + ρ ∂y ∂x 2 ∂y 2
Introduce a stream function ψ( x, y) ∂ψ ∂ψ such that u= , v=− ∂y ∂x Hence, substituting into the eqns above
0=−
1 ∂p ∂3ψ ∂3ψ + v 2 + 3 —① ρ ∂x ∂x ∂y ∂y
0=−
∂3ψ ∂3ψ 1 ∂p − v 3 + —② ρ ∂y ∂x∂y 2 ∂x
30
Eliminating p by
∂ ∂ ①− ② ∂y ∂x get
∂4ψ ∂4ψ ∂4ψ + 2 + =0 ∂x 4 ∂x 2∂y 2 ∂y 4
or
∇ 4ψ = 0
or
∇ 2 (∇ 2ψ) = 0
Comparing with the original N-S equations
Biharmonic 4th Order one variable (ψ )
N-S 2nd Order 3 variables (µ, v, p)
no parameter
viscosity as the parameter
P4.55 Consider the following two-dimensional incompressible flow, which clearly satisfies continuity: u = Uo = constant, v = Vo = constant Find the stream function ψ(r, θ) of this flow, that is, using polar coordinates. Solution: In cartesian coordinates the stream function is quite easy: u = ∂ψ/∂ y = Uo and v = –∂ψ/∂x = Vo or: ψ = Uoy – Vox + constant But, in polar coordinates, y = rsinθ and x = rcosθ. Therefore the desired result is
ψ (r, θ ) = Uor sinθ – Vor cosθ + constant Ans. P4.56 Investigate the stream function = K(x2 – y2), K = constant. Plot the streamlines in the full xy plane, find any stagnation points, and interpret what the flow could represent. Solution: The velocities are given by
u=
∂ψ ∂ψ = −2Ky; v = − = −2Kx ∂y ∂x
This is also stagnation flow, with the stream-lines turned 45° from Prob. 4.55.
Fig. P4.56
31
P4.57 A two-dimensional incompressible flow field for x > 0 is described by y2 V(x, y), = y ln x,− 2x
(a) Show that mass conservation is satisfied. (b) Find the stream function. (c) Determine the rate of flow through the passage between (1, 1) and (2, 2). (d) What are the slopes of the streamlines passing through the points (1, 1) and (2, 2)? (e) Calculate the x-component of the pressure gradient at (1, 1) if the fluid is inviscid and gravity can be neglected. Solution:
y2 V = ylnx,− 2x ∂u ∂v y 2y Δ ⋅V = + = + − ≡ 0 ∂x ∂y x 2x u=
∂ψ ⇒ψ = ∂y
v=−
∫ u dy = ∫ ylnx dy =
∂ψ ⇒ψ = ∂x
∫ −v dx = ∫
Ans. y2 lnx + C1 (x) 2
y2 y2 dx = lnx + C2 (y) 2x 2
2 On matching, ψ = y lnx + Constant 2 0 for simplicity
12 ln1 = 0 2 22 ψ (2,2) = ln2 = 2ln2 2 Flow rate betn (1,1) and (2,2) = ψ (2,2) − ψ (1,1) = 2ln2 1 At (1,1) u = 1ln1 = 0, v = − 2 22 At (2,2) u = 2ln2 v = − 2 = −1 2
ψ (1,1) =
Slope of streamline
dy v = dx u
∴ at (1,1) slope is ∞ (ie vertical) 1 (2,2) slope is − 2ln2
32
∂u ∂p ∂u = ρ u + v (for gx = 0,u = 0) ∂x ∂y ∂x y y2 = ρ (ylnx)( ) + (− )(lnx) x 2x −
=ρ
y2 lnx 2x
P4.58 In 1851, George Stokes (of Navier-Stokes fame) solved the problem of steady incompressible low-Reynolds-number flow past a sphere, using spherical polar coordinates (r, θ) – [Ref. 5, page 168]. In these coordinates, the equation of continuity is ∂ 2 ∂ (r vr sin θ ) + (r vθ sin θ ) = 0 ∂r ∂θ
(a) Does a stream function exist for these coordinates? (b) If so, find its form. Solution: Two velocity components and two continuity terms. Yes, ψ exists! (b) The stream function should be defined such that continuity takes the form
∂2ψ ∂2ψ + ≡ 0 , or : ∂r ∂θ ∂θ ∂r 1 ∂ψ 1 ∂ψ vθ = ; vr = − 2 r sin θ ∂r r sin θ ∂θ
Ans.(a)
−
Ans.(b)
P4.59 The velocity profile for incompressible pressure-driven laminar flow between parallel plates (see Fig. 4.12b) has the form u = C(h2 – y2), where C is a constant. (a) Determine if a stream function exists. (b) If so, determine a formula for the stream function, Solution: (a) A stream function exists, for a single velocity component u, if ∂u/∂x = 0, which it certainly is, since u is a function only of y. Yes, ψ exists. Ans.(a) (b) Finding the stream function is just a matter of direct integration:
v= − u=
∂ψ = 0 , hence ψ is a function only of y ∂x
∂ψ y3 = C(h2 − y 2 ) ; Integrate : ψ = C(h2 y − ) + constant ∂y 3
Ans.(b)
33
P4.60 A two-dimensional, incompressible, frictionless fluid is guided by wedge-shaped walls into a small slot at the origin, as in Fig. P4.60. The width into the paper is b, and the volume flow rate is Q. At any given distance r from the slot, the flow is radial inward, with constant velocity. Find an expression for the polar-coordinate stream function of this flow.
Fig. P4.60
Solution: We can find velocity from continuity:
vr = −
Q Q 1 ∂ψ =− = A (π /4)rb r ∂θ
ψ=−
from Eq. (4.101). Then
4Q θ + constant !b
Ans.
This is equivalent to the stream function for a line sink, Eq. (4.131).
P4.61 For the fully developed laminar-pipe-flow solution of Eq. (4.137), find the axisymmetric stream function ψ(r, z). Use this result to determine the average velocity V = Q/A in the pipe as a ratio of umax. Solution: The given velocity distribution, vz = umax(1 – r2/R2), vr = 0, satisfies continuity, so a stream function does exist and is found as follows: vz = u max (1 − r 2 /R 2 ) =
r2 1 ∂ψ r4 , solve for ψ = u max − + f(z), now use in 2 r ∂r 2 4R
r2 1 ∂ψ df r4 vr = 0 = − = 0 + , thus f(z) = const, ψ = umax − Ans. 2 r ∂z dz 2 4R
We can find the flow rate and average velocity from the text for polar coordinates:
π 2 R2 R4 Q1-2 = 2π (ψ2 − ψ1 ), or: Q 0-R = 2π u max − − u (0 − 0) = R u max max 2 2 4R 2 1 Then Vavg = Q/A pipe = [(π /2)R 2 u max /(π R 2 )] = umax 2
Ans.
34
P4.62 An incompressible stream function is defined by
ψ (x, y) =
U (3x 2 y − y 3 ) 2 L
where U and L are (positive) constants. Where in this chapter are the streamlines of this flow plotted? Use this stream function to find the volume flow Q passing through the
Fig. E4.7
rectangular surface whose corners are defined by (x, y, z) = (2L, 0, 0), (2L, 0, b), (0, L, b), and (0, L, 0). Show the direction of Q. Solution: This flow, with velocities u = ∂ψ/∂ y = 3U/L2(x2 – y2), and v = –∂ψ/∂ x = –6xyU/L2, is identical to Example 4.7 of the text, with “a” = 3U/L2. The streamlines are plotted in Fig. E4.7. The volume flow per unit width between the points (2L, 0) and (0, L) is Q/b = ψ (2L, 0) − ψ (0, L) =
U U (0 − 0) − 2 [3(0)2 L − L3 ] = UL, or: Q = ULb Ans. 2 L L
Since ψ at the lower point (2L, 0) is larger than at the upper point (0, L), the flow through this diagonal plane is to the left, as per Fig. 4.9 of the text.
P4.63 For the incompressible plane flow of Prob. P4.7, 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) + (8x) = 0 + 0 = 0 Yes, Ψ exists. ∂x ∂y ∂x ∂y
Ans.(a)
(b) Find the stream function by relating velocities to derivatives of ψ :
u = 2y =
∂ψ ∂ψ ; v = 8x = − ; Integrate : ψ = y 2 − 4x 2 + constant Ans.(b) ∂y ∂x
35
(c) Plot a few streamlines, that is, plot
y2 = 4x2 + C for various C. Here are the results:
We are showing only the upper half plane, which is the mirror image of the lower half. P4.64
Review Sect. 4.8, and devise a right-hand rule which can help indicate the sense of rotation of a
two-dimensional element in the (x, y) plane.
Solution: Right hand rule #1 x = thumb y = pointer z = middle finger ∴ z is pointing out of the plane of the paper Right hand rule #2 z = thumb, pointing out of the paper all other fingers curve up pointing counterclockwise
Hence, rotation in the (x, y) plane is positive counter-clockwise
36
P4.65 Investigate the velocity potentialφ = Kxy, K = constant. Sketch the potential lines in the full xy plane, find any stagnation points, and sketch in by eye the orthogonal streamlines. What could the flow represent? Solution: The potential lines, φ = constant, are hyperbolas, as shown. The streamlines,
Fig. P4.65
sketched in as normal to the φ lines, are also hyperbolas. The pattern represents plane stagnation flow (Prob. 4.56) turned at 45°. P4.66 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 L ∂x ∂y
To find the stream function ψ, use the definitions of u and v and integrate: u=
xy y 2 x y ∂ψ = 2V − , ∴ ψ = 2V − + f (x) L L ∂y L 2L
∂ψ 2Vy df 2Vy = + = −v = ∂x L dx L 2 2xy y df = 0 and ψ = V − + const dx L L Evaluate
Thus
Ans.
37
P4.67 Check whether the following functions can represent the velocity potential of an irrotational flow.
(a) φ = x + y (b) φ = x 2 − y 2 (c) φ = In(x + y) (d) φ = sin x + cos y 1 (e) φ = r + cosθ r Solution: Check if ∇ 2φ = 0
∂ 2φ ∂ 2φ 2 + 2 = 0 (cartesian coord) ∂x ∂y 1 ∂ ∂φ 1 ∂ 2φ (r ) + r ∂r ∂r r 2 ∂θ 2 = 0 (cylindrical coord) a. φ = x + y,
∂ 2φ ∂ 2φ + = 0+0 ≡ 0 ∂x 2 ∂y 2
(O.K.)
∂ 2φ ∂ 2φ b. φ = x − y , + =2−2≡0 ∂x 2 ∂y 2 2
2
c. φ = ln(x + y),
(O.K.)
∂ 2φ ∂ 2φ 1 1 + 2 =− − ≡/ 0 2 2 ∂x ∂y (x + y) (x + y)2
d. φ = sin x + cos y,
∂ 2φ ∂ 2φ + = −sin x − cos y ≡/ 0 ∂x 2 ∂y 2
(not O.K.) (not O.K.)
1 e. φ = r + cosθ r 1 ∂ ∂φ 1 ∂2φ 1 1 1 1 r + 2 2 = 1+ 2 cosθ + 2 r + (−cosθ ) ≡ 0 r ∂r ∂r r ∂θ r r r r
(O.K.)
P4.68 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.
38
P4.69 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 ∂ y2
Yes
(b) Now use φ to find u and v and then integrate to find ψ.
u=
∂φ ∂ψ y2 = y + 2x = , hence ψ = + 2xy + f (x) ∂x ∂y 2
v=
∂φ ∂ψ df x2 = x − 2y = − = −2y − , hence f (x) = − + const ∂y ∂x dx 2
The final stream function is thus ψ =
1 2 (y − x 2 ) + 2xy + 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 ) + 2xy = 2 2
Ans. (c)
P4.70 Liquid drains from a small hole in a tank, as shown in Fig. P4.70, such that the velocity field set up is given by υr ≈ 0, υz ≈ 0, υθ = 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. Solution: From Appendix D, the angular velocity is z patm
r z=H
z=0 r=R
Fig. P4.70
ωz =
1∂ 1 ∂ (rvθ ) − (v ) = 0 (IRROTATIONAL) r ∂r r ∂θ θ
39
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: 1 1 2 2 patm + ρ Vr=∞ + ρ gH = patm + ρ Vr=R + ρ gz c 2 2
Introduce Vr=∞ = 0 and Vr=R
ω 2 R2 = ω R to obtain zC = H − 2g
Ans.
P4.71 For the incompressible plane flow of Prob. P4.7, with velocity components u = 2y, v = 8x, w = 0, determine (a) if a velocity potential exists. (b) If so, determine the form of the velocity potential, and (c) plot a few representative potential lines. Solution: (a) A velocity potential exists if the vorticity is zero. Here, for plane flow in (x, y) coordinates, we need only evaluate rotation around the z axis:
ς z = 2ωz =
∂v ∂u − = 8 − 2 = +6 ≠ 0 ∂x ∂y
Rotational, φ does not exist. Ans.(a)
(b, c) There is no velocity potential – no plot, no formula. The flow has constant vorticity.
P4.72 Show that the linear Couette flow between plates in Fig. 1.6 has a stream function but no velocity potential. Why is this so? Solution: Given u = Vy/h, v = 0, check continuity:
Fig. 1.6
∂u ∂v ? + = 0 = 0 + 0 (Satisfied therefore ψ exists). Find ψ from ∂x ∂y u=
Vy ∂ψ ∂ψ V 2 = , v = 0 = − , solve for ψ = y + const h ∂y ∂x 2h
Ans.
Now check irrotationality:
2ωz =
∂v ∂u ? V − = 0 = 0 − ≠ 0! (Rotational, φ does not exist.) Ans. ∂x ∂y h
40
P4.73 Find the two-dimensional velocity potential φ(r,θ) for the polar-coordinate flow pattern υr = Q/r, υθ = K/r, where Q and K are constants. Solution: Relate these velocity components to the polar-coordinate definition of φ : vr =
Q ∂φ K 1 ∂φ = , vθ = = ; solve for φ = Qln(r) + Kθ + const r ∂r r r ∂θ
Ans.
P4.74 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. P4.75 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. P4.75
Solution: Evaluate the angular velocity:
2ωz =
∂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=
y ∂φ Ky ∂φ Kx =− 2 ; v= = 2 , solve for φ = K tan−1 = Kθ 2 2 x ∂x ∂y x + y x +y
The φ lines are plotted above. They represent a counterclockwise line vortex.
Ans.
41
P4.76 A plane polar-coordinate velocity potential is defined by
φ=
K cosθ r
K = const
Find the stream function for this flow, sketch some streamlines and potential lines, and interpret the flow pattern. Solution: Evaluate the velocities and thence find the stream function:
Fig. P4.76
vr =
∂φ Kcosθ 1 ∂ψ 1 ∂φ Ksinθ ∂ψ =− 2 = ; vθ = =− 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. P4.77 A stream function for a plane, irrotational, polar-coordinate flow is
ψ = Cθ − K lnr C and K = const Find the velocity potential for this flow. Sketch some streamlines and potential lines, and interpret the flow pattern. Solution: If this problem is given early enough (before Section 4.10 of the text), the
Fig. 4.77
students will discover this pattern for themselves. It is a line source plus a line vortex, a tornado-like flow, Eq. (4.134) and Fig. 4.14 of the text. Find the velocity potential: vr =
1 ∂ψ C ∂φ ∂ψ K 1 ∂φ = = ; vθ = − = = , solve φ = C ln(r) + Kθ r ∂θ r ∂ r ∂ r r r ∂θ
The streamlines and potential lines are plotted above for negative C (a line sink).
Ans.
42
P4.78 For the velocity distribution of Prob. P4.5, (a) determine if a velocity potential exists and, if it does, (b) find an expression for φ(x,y) and sketch the potential line which passes through the point (x, y) = (L/2, L/2). Solution: Recall the given flow, u = Uo(1+x/L) and v = −Uo(y/L). (a) Calculate if the flow is irrotational. For plane flow, only one term of the curl(V) is needed:
2ω z =
∂v ∂u − = 0 − 0 = 0 ; Yes, curl(V) = 0 ∂x ∂y
Therefore a velocity potential does exist. Ans.(a) (b) To find φ, integrate from u and v: ∂φ = u ; Thus φ = ∂x
∫ u dx =
∂φ df y = v = 0+ = −U o , ∂y dy L Thus
x x2 U (1+ ) dx = U (x+ ) + f ( y) ∫ o L o 2L or : f = −U o
φ = (U o L)(
y2 + constant 2L
x x2 − y2 + ) + const L 2L2
Ans.(b)
For a potential line to pass through (L/2, L/2), we must have φ /(Uo/L) = [1/2 + {(1/2)2 (1/2)2}/2] = ½. For convenience let the const = 0. Thus we are to plot this potential line:
ϕ 1 x x 2 − y2 = = + Uo L 2 L 2L2 The result is plotted (red) in the graph below, along with the (blue) ψ line, which has the analytic form ψ = Uo(y + xy/L) = 3UoL/4.
43
P4.79 A steady, two-dimensional flow has the following polar-coordinate velocity potential:
ϕ = C r cosθ + K ln r where C and K are constants. Determine the stream function ψ(r, θ) for this flow. For extra credit, let C be a velocity scale U, let K = UL, and sketch what the flow might represent.
∂ϕ K 1 ∂ψ = C cosθ + = , hence ψ = C r sin θ + Kθ + f (r) ∂r r r ∂θ 1 ∂ϕ ∂ψ vθ = = −C sin θ + K(0) = − , hence ψ = C r sin θ + Kθ + constant r ∂θ ∂r vr =
Ans.
Solution: Write out the ψ and φ expressions for polar-coordinate velocities: Extra credit: Plot a typical streamline for C = U and K = UL:
All the streamlines are logarithmic spirals coming out from the origin in every direction.
P4.80 Consider the following stream functions of two-dimensional horizontal flow fields (i) ψ = A(y – x) (ii) ψ = Ay2 (iii) ψ = Ar2 (iv) ψ = Qθ/(2π) where A and Q are positive constants, and (x, y) and (r, θ) are the Cartesian and polar coordinates, respectively. For each of these flow fields, (a) find the velocity components; (b) identify the flow field and draw some streamlines to show the flow pattern; (c) determine if the flow is irrotational or not; (d) find the velocity potential if it exists; and (e) find the pressure gradients. You are given that the gradient and Laplacian operators in polar coordinates are ∂ 1 ∂ 1 ∂ ∂ 1 ∂2 2 ∇≡ , . , ∇ ≡ r + r ∂r ∂r r 2 ∂θ 2 ∂r r ∂θ
44
Solution: (a) For cartesian coord. (u,v) = (∂ψ / ∂y, − ∂ψ / ∂x) 1 For polar coord. (ur , uθ ) = ∂ψ / ∂θ , − ∂ψ / ∂r r ψ = A( y − x) ⇒ (u,v) = ( A, A)
ψ = Ay 2 ⇒ (u,v) = (2 Ay, 0) ψ = Ar 2 ⇒ (ur , uθ ) = (0, − 2 Ar) ψ = Qθ / 2π ⇒ (ur , uθ ) = (Q / 2π r, 0) (b)
∂ 2ψ ∂ 2ψ (c) i) ∇ ψ = 2 + 2 = 0 (irrotational) ∂x ∂y 2
ii) ∇ 2ψ = 2A ≠ 0 (rotational) iii) ∇ 2ψ =
1 ∂ ∂ψ 1 ∂2ψ = 4A ≠ 0 (rotational) r + r ∂r ∂r r 2 ∂θ 2
iv) ∇ 2ψ = 0 (irrotational) (d) i) φ =
∫ u dx = ∫ v dy = A(x + y) + const
0
0 Q iv) φ = ∫ ur dr = ∫ ruθ dθ = lnr + const 2π
45
∂p ∂p = =0 ∂x ∂y ∂p ∂p ii) couette flow = =0 ∂x ∂y ∂p iii) vortex = ρuθ2 / r = 4 ρ A 2 r, ∂r
(e) i) uniform flow
∂p =0 ∂θ
Q 2 1 ∂p ∂ur iv) source = −ρ u r = ρ 3 , 2π r ∂r ∂r
∂p =0 ∂θ
P4.81 A CFD model of steady two-dimensional y =1.1 m
incompressible flow has printed out the values of
ψ = 1.9552 m2/s
2.0206
V? α?
stream function ψ(x, y), in m2/s, at each of the four corners of a small 10cm-by-10cm cell, as y = 1.0 m
shown in Fig. P4.81. Use these numbers to
1.7308 m2/s
x = 1.5 m
estimate the resultant velocity in the center of
1.7978 x = 1.6 m
Fig. P4.81
the cell and its angle α with respect to the x axis. Solution: Quick analysis: the ψ values are higher on the top than the bottom, therefore u is to the right. The ψ values are higher on the right than the left, therefore v is down. There are several ways to estimate the center velocities. One simple way is to compute average values of ψ on the sides:
1.9879 0
v u 1.8430
1.9092
1.7643
Then ucenter ≈ Δψ/Δy = (1.9879-1.7643 m2/s)/(0.1m) = 2.236 m/s to the right. And vcenter ≈ Δψ/Δx = (1.9092-1.8430 m2/s)/(0.1m) = 0.662 m/s down. The resultant and its angle are
46
V ≈
(2.236)2 + (0.662)2 = 2.332 m / s ; α = tan −1 (
0.662 ) = 16.5 o down Ans. 2.236
The ψ values in this problem are in fact taken
α
V
from an exact solution, V = 2.3315 m/s, α = 16.505°.
P4.82 Consider the following two-dimensional function f(x, y):
f = A x 3 + B x y2 + 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 = 6Ax + Bx + 2C = 0 if B = − 3A and C = 0
ϕ = A x 3 − 3A xy 2 + D
The velocity potential is
Ans.(a)
(b) To find ψ, use φ to get u and v and work backwards to get the stream function:
∂ϕ ∂ψ = 3A x 2 − 3A y 2 = , ∴ ψ = 3A x 2 y − A y 3 + f (x) ∂x ∂y ∂ϕ ∂ψ df v= = −6Axy = − = −6Axy − , ∴ f = const ∂y ∂x dx
u =
Finally,
ψ = 3A x 2 y − A y 3 + const
Ans.(b)
P4.83 Water flows through a two-dimensional narrowing wedge at 0.04 m3/min per meter of
Q
r
Drain
width into the paper. If this inward flow is purely radial, find an expression, for (a) the stream function, and (b) the velocity potential of the flow. Assume one-dimensional flow. The included angle of the wedge is 45°.
Fig. P4.83
47
Solution: The wedge angle equals π/4 radians. At any given position r, the inward flow equals
vr = −
Q Q 4Q / ( π b) = − = − A ( π / 4)rb r
where 4(Q / b) / π =
(
)(
4 0.04m 3 /min/m 1 min/60 s
π
)
= 0.00085
m3 s−m
We have already been advised that vθ = 0. (a) Work from radial velocity to stream function:
vr = −
0.00085m 3 /s − m 1 ∂ψ = ; r ∂θ r
Solve
ψ = − 0.00085 θ
Ans.(a)
Note that r must be in meters. (b) Work from radial velocity to obtain velocity potential:
vr = −
0.00085m 3s − m ∂φ = ; ∂r r
Solve
φ = −0.00085 ln(r)
Ans.(b)
P4.84 A CFD model of steady two-dimensional incompressible flow has printed out the values of velocity potential φ(x, y), in m2/s at each of the four corners of a small 10cm-by10cm cell, as shown in Fig. P4.84. Use these numbers to estimate the resultant velocity in the center of the cell and its angle α with respect to the x axis.
Solution: Quick analysis: the φ values are lower on the left than the right, therefore u is to the right. The φ values are lower on the top than the bottom, therefore v is down. There are several ways to estimate the center velocities. One simple way is to compute average values of φ on the sides:
48
Then ucenter ≈ Δφ/Δx = (5.0923-4.8688 m2/s)/(0.1m) = 2.235 m/s to the right. And vcenter ≈ |Δφ/Δy| = (5.0137-4.9474 m2/s)/(0.1m) = 0.663 m/s down. The resultant and its angle are V ≈
(2.235)2 + (0.663)2 = 2.331 m / s ; α = tan −1 (
0.663 ) = 16.5 o Ans. 2.235
The φ values in this problem are in fact taken
α
V
from an exact solution, V = 2.3315 m/s, α = 16.505° down.
P4.85
Consider the two-dimensional incompressible polar-coordinate velocity potential
ϕ
= 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? (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 = − Bsin θ − r ∂θ r
At first it doesn’t look as if we can find a stagnation point, but indeed there is one: r = L , θ = 180 : vθ = 0 , vr = B −
BL = 0 L
Ans.(b)
As discussed later in Chap. 8, this is the velocity potential of a Rankine half-body.
49
(c) With the velocities known, check the continuity equation: 1 ∂ 1 ∂vθ B cosθ B cosθ (r vr ) + = 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 =
P4.86
1 ∂ψ ∂ψ BL = B cosθ ; vθ = − = − Bsin θ − r ∂θ ∂r r Integrate : ψ = B r sin θ + B L ln r + const
Ans.(c)
Given the following steady axisymmetric stream function:
ψ =
B 2 r4 (r − 2 ) , where B and R are constants 2 2R
valid in the region 0 ≤ r ≤ R and 0 ≤ z ≤ L. (a) What are the dimensions of the constant B? (b) Show whether this flow possesses a velocity potential and, if so, find it. (c) What might this flow represent? [HINT: Examine the axial velocity vz.] Solution: (a) From the definition of ψ(r, z) in Eqs. (4.105), the dimensions of ψ are {L3/T}. Thus B has velocity dimensions, {B} = {L/T}. Ans.(a) (b) To test for irrotationality, first find the velocity components from Eqs. (4.106):
vr = −
1 ∂ψ 1 ∂ψ 1 B 4r 3 r2 = 0 ; vz = = (2r − 2 ) = B(1− 2 ) r ∂z r ∂r r2 2R R
Now evaluate the curl of the velocity, which has only one possible non-zero component. From Appendix D, Eq. (D.11), ∂v ∂v 2Br 2ωθ = r − z = 0 − 2 ≠ 0 Rotational , ϕ does not exist. Ans.(b) ∂z ∂r R (c) The interpretation of the flow follows immediately from the velocity components. The velocity profile is a paraboloid of revolution and represents Poiseuille pipe flow, Eq. (4.137). Ans.(c)
*P4.87
A two-dimensional incompressible flow has the velocity potential
ϕ = K(x 2 − y 2 ) + C ln(x 2 + y 2 ) where K and C are constants. In this discussion, avoid the origin, which is a singularity (infinite velocity). (a) Find the sole stagnation point of this flow, which is somewhere in the upper half plane. (b) Prove that a stream function exists and then find ψ(x, y), using the hint that
∫ dx / (a 2 + x 2 ) = (1 / a) tan −1 (x / a).
50
Solution: (a) Find the velocity components and see where they both equal zero:
u =
∂ϕ 2Cx ∂ϕ 2Cy = 2Kx + 2 ; v = = −2Ky + 2 2 ∂x x +y ∂x x + y2
For positive K and C, u cannot be zero anywhere except at x = 0. Then v = 0 if
2Ky =
2C , or : y
stagnation at
x =0 and y =
C K
Ans.(a)
(b) First check the velocities to see if continuity is satisfied:
∂u ∂v 2C 4Cx 2 2C 4Cy 2 + = [2K + 2 − ] + [−2K + 2 − ]= 0 ∂x ∂y x + y 2 (x 2 + y 2 )2 x + y 2 (x 2 + y 2 )2 The algebra is messy but, indeed, continuity is satisfied, ψ exists. Ans.(b) – part 1. Now integrate the velocity components to find the stream function : u=
∂ψ 2Cx = 2Kx + 2 ∂y x + y2
and v = −
∂ψ 2Cy = −2Ky + 2 ∂x x + y2
y Integrate to obtain ψ = 2Kxy + 2C tan −1 ( ) + const x
Ans.(b)
P4.88 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 = 40 km, the velocity is 45 m/s and the pressure 99 kPa, calculate the velocity and pressure at r = 160 km. Solution: (a) The velocities are calculated from φ, as requested in Prob. P4.68:
vr =
∂ (U R θ ) = 0 ∂r o
; 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 = 4R = 160 km, vθ = Bernoulli : p1 +
U o R U o 45 m = = = 11.25 4R 4 4 s
ρ 2 ρ V1 = p2 + V22 , or : 2 2
99,000 + (1.225 / 2)(45)2 = p2 + (1.225 / 2)(11.25)2 , Solve p100 mi = 100,163 Pa Ans.(b)
The pressure far from the storm is approximately sea-level standard pressure.
51
P4.89 An incompressible, irrotational, two-dimensional flow has the following stream function in polar coordinates:
ψ =
Ar n sin(nθ ) ,
where A and n are constants.
Find an expression for the velocity potential of this flow. Solution: Use ψ to find the velocity components, then integrate back to find φ.
P4.90 Study the combined effect of the two viscous flows in Fig. 4.12. That is, find u(y) when the upper plate moves at speed V and there is also a constant pressure gradient (dp/dx). Is superposition possible? If so, explain why. Plot representative velocity profiles for (a) zero, (b) positive, and(c) negative pressure gradients for the same upper-wall speed V.
Fig. 4.12
Solution: The combined solution is u=
V y h 2 dp y 2 1+ + − 1 − 2 h 2µ dx h 2
The superposition is quite valid because the convective acceleration is zero, hence what remains is linear: ∇p = µ∇2V. Three representative velocity profiles are plotted at right for various (dp/dx).
Fig. P4.90
52
P4.91 An oil film drains steadily down the side of a vertical wall, as shown. After an initial development at the top of the wall, the film becomes independent of z and of constant thickness. Assume that w = w(x) only that the atmosphere offers no shear resistance to the film. (a) Solve Navier-Stokes
for w(x). (b) Suppose that film thickness and [∂w/∂x] at the wall are measured. Find an expression which relates µ to this slope [∂w/∂x]. Solution: First, there is no pressure gradient ∂p/∂z because of the constant-pressure atmosphere. The Navier-Stokes z-component is µ(d2w/dx2) = ρg, and the solution requires w = 0 at x = 0 and (dw/dx) = 0 (no shear at the film edge) at x = δ. The solution is:
w=
ρ gx (x − 2δ ) Ans. (a) NOTE: w is negative (down) 2µ
The wall slope is dw/dx |wall = −ρ gδ / µ , rearrange: µ = − ρ gδ / [dw / dx |wall ]
P4.92 Modify the analysis of Fig. 4.13 to find the velocity vθ when the inner cylinder is fixed and the outer cylinder rotates at angular velocity Ωo. May this solution be added to Eq. (4.140) to represent the flow caused when both inner and outer cylinders rotate? Explain your conclusion. Solution: We apply new boundary conditions to Eq. (4.139) of the text:
Ans. (b)
Fixed
ro "$
# r ri Fluid: !, µ
Fig. 4.13
vθ = C1r + C2 /r; At r = ri , vθ = 0 = C1ri + C2 /ri
At r = ro , vθ = Ωo ro = C1ro + C2 /ro r / ri − ri / r Solve for C1 and C2 . The final result: vθ =Ωo ro Ans. ro / ri − ri / ro
This solution may indeed be added to the inner-rotation solution, Eq. (4.140), because the convective acceleration is zero and hence the Navier-Stokes equation is linear.
53
P4.93 A solid circular cylinder of radius R rotates at angular velocity Ω in a viscous incompressible fluid which is at rest far from the cylinder, as in Fig. P4.93. Make simplifying assumptions and derive the governing differential equation and boundary conditions for the velocity field vθ in the fluid. Do not solve unless you are obsessed with this problem. What is the steady-state flow field for this problem?
Fig. P4.93
Solution: We assume purely circulating motion: vz = vr = 0 and ∂/∂θ = 0. Thus the remaining variables are vθ = fcn(r, t) and p = fcn(r, t). Continuity is satisfied identically, and the θ-momentum equation reduces to a partial differential equation for vθ:
∂ vθ µ 1 ∂ ∂ vθ vθ = r − subject to vθ (R, t) = ΩR and vθ (∞, t) = 0 Ans. ∂ t ρ r ∂ r ∂ r r2 I am not obsessed with this problem so will not attempt to find a solution. However, at large times, or t = ∞, the steady state solution is vθ = ΩR2/r. Ans.
P4.94 The flow pattern in bearing lubrication can be illustrated by Fig. P4.94, where a viscous oil (ρ, µ) is forced into the gap h(x) between a fixed slipper block and a wall moving at velocity U. If the gap is thin, it can be shown that the pressure and velocity distributions are of the form p = p(x), u = u(y), υ = w = 0. Neglecting gravity, reduce the Navier-Stokes equations (4.38) to a single differential equation for u(y). What are the proper boundary conditions? Integrate and show that
u=
y 1 dp 2 (y − yh) +U 1 − h 2µ dx
where h = h(x) may be an arbitrary slowly varying gap width. (For further information on lubrication theory, see Ref. 16.)
Fig. P4.94
54
Solution: With u = u(y) and p = p(x) only in the gap, the x-momentum equation becomes
With C1 and C2 evaluated, the solution is exactly as listed in the problem statement:
P4.95 Consider a viscous film of liquid draining uniformly down the side of a vertical rod of radius a, as in Fig. P4.95. At some distance down the rod the film will approach a terminal or fully developed draining flow of constant outer radius b, with υz = υz(r), υθ = υr = 0. Assume that the atmosphere offers no shear resistance to the film motion. Derive a differential equation for υz, state the proper boundary conditions, and solve for the film velocity distribution. How does the film radius b relate to the total film volume flow rate Q?
Fig. P4.95
Solution: With vz = fcn(r) only, the Navier-Stokes z-momentum relation is
ρ or:
dvz ∂p = 0 = − + ρg + µ∇ 2 vz , dt ∂z
1 d dvz ρg ρgr 2 + C1 ln(r) + C2 r = − , Integrate twice: vz = − r dr dr µ 4µ
The proper B.C. are: u(a) = 0 (no-slip) and µ
The final solution is v z =
ρ gb 2 r ρ g 2 ln − (r − a 2 ) Ans. a 4µ 2µ
b
The flow rate is Q =
∫ a
∂ vz (b) = 0 (no free-surface shear stress) ∂r
vz 2π r dr =
πρ ga 4 (−3σ 4 − 1 + 4σ 2 + 4σ 4 lnσ ), 8µ
where σ =
b a
Ans.
55
P4.96 A flat plate of essentially infinite width and breadth oscillates sinusoidally in its own plane beneath a viscous fluid, as in Fig. P4.96. The fluid is at rest far above the plate. Making as many simplifying assumptions as you can, set up the governing differential equation and boundary conditions for finding the velocity field u in the fluid. Do not solve (if you can solve it immediately, you might be able to get exempted from the balance of this course with credit).
Fig. P4.96
Solution: Assume u = u(y, t) and ∂p/∂x = 0. The x-momentum relation is
P4.97 SAE 10 oil at 20°C flows between parallel plates 8 mm apart, as in Fig. P4.97. 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. P4.97 2
Vavg =
dp h 2 u max = , 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 NOTE: The Reynolds number, based upon plate half-width, is 16, laminar.
m3 s⋅m
Ans.
56
P4.98 SAE 30W oil at 20°C flows through the 9-cm-diameter pipe in Fig. P4.98 at an average velocity of 4.3 m/s. (a) Verify that the flow is laminar. (b) Determine the volume flow rate in m3/h. (c) Calculate the expected reading h of the mercury manometer, in cm.
D = 9 cm
V
SAE 30W Oil
h Hg
Solution: (a) Check the Reynolds number. For SAE 30W oil, from Appendix A.3, ρ = 891 kg/m3 and µ = 0.29 kg/(m⋅s). Then
2.5 m Fig. P4.98
Red = ρVd/µ = (891 kg/m3)(4.3 m/s)(0.09 m)/[0.29 kg/(m⋅s)] = 1190 < 2000 Laminar Ans. (a) (b) With average velocity known, the volume flow follows easily: Q = AV = [(π /4)(0.09 m)2](4.3 m/s)(3600 s/h) = 98.5 m3/h Ans. (b) (c) The manometer measures the pressure drop over a 2.5 m length of pipe. From Eq. (4.147),
V = 4.3
m Δp R 2 Δp (0.045 m)2 = = , solve for Δp = 12320 Pa s L 8µ 2.5 m 8(0.29 kg/m⋅s)
Δpmano = 12320 = (ρ merc − ρ oil )gh = (13550 − 891)(9.81)h, Solve h = 0.099 m Ans. (c)
P4.99 The viscous oil in Fig. P4.99 is set into steady motion by a concentric inner cylinder moving axially at velocity U inside a fixed outer cylinder. Assuming constant pressure and density and a purely axial fluid motion, solve Eqs. (4.38) for the fluid velocity distribution vz(r). What are the proper boundary conditions? Fig. P4.99
Solution: If vz = fcn(r) only, the z-momentum equation (Appendix E) reduces to:
ρ
dvz ∂p µ d dvz = − + ρg z + µ ∇ 2 vz , or: 0 = 0 + 0 + r dt ∂z r dr dr
The solution is vz = C1 ln(r) + C2, subject to vz(a) = U and vz(b) = 0 Solve for C1 = U/ln(a/b) and C2 = –C1 ln(b) The final solution is: v z = U
ln(r / b) ln(a / b)
Ans.
P4.100 Reconsider Prob. P4.99, and show that the flow approximates to a plane Couette flow as the clearance becomes very small: δ ≡ b – a b. You may use in your proof the approximation ln(1 + α) ≈ α as |α| 1.
57
Solution: Prob. 4.88 has the solution Vz = Uln(r / b) / ln(a / b) For small gap δ ≡ b − a b define y = b − r, then r = b − y
a = b −δ ln(r / b) = ln(1 − y / b) ≈ −y / b ln(a / b) = ln(1 − δ / b) ≈ −δ / b Hence Vz ≈ U(−y / b) / (−δ / b) = Uy / δ which is plane Couette flow
P4.101 A 10-cm-diameter rod is pulled steadily at 0.2 m/s through a 1.5-m long fixed cylinder whose clearance of 1 mm is filled with a lubricating oil of density ρ = 870 kg/m3 and viscosity µ = 0.1 Pa · s, as shown in Fig. P4.101. Estimate the force required to pull the rod, using the velocity distribution deduced in Prob. P4.99. Re-do the problem on assuming that the flow approximates to a plane Couette flow. Compare the results by the two approaches.
Fig. P4.101
Solution: By Prob. 4.88 Vz = Uln(r / b) / ln(a / b) Force = shear stress × surface area ⇒ F = τ / r = a × 2 π aL ∂Vz U = −µ / ln(a / b) where τ = −µ ∂r r U ∴ F = −µ / ln(a / b) (2 π aL) = −2 πµUL / ln(a / b) a Now a = 0.05 m, b = 0.051 m, L = 1.5 m, U = 0.2 m/s, µ = 0.1 Pa ⋅ s F = −2 π × 0.1× 0.2 ×1.5 / ln(0.05 / 0.051) = 9.52 N If the flow is assumed to be Couette flow τ = µV / δ δ = 0.001 m F = (µV / δ )(2 π aL) = 2 πµUL(a / δ ) = 2 π × 0.1× 0.2 ×1.5 × 0.05 / 0.001 = 9.42 N which is 1% smaller than the exact value.
58
P4.102 Modify Prob. 4.99 so that the outer cylinder also moves to the left at constant speed V. Find the velocity distribution υz(r). For what ratio V/U will the wall shear stress be the same at both cylinder surfaces? Solution: We merely modify the boundary conditions for the known solution in 4.99: vz = C1 ln(r) + C2, subject to vz(a) = U and vz(b) = –V Solve for C1 = (U + V)/ln(a/b) and C2 = U – (U + V)ln(a)/ln(a/b) The final solution is
Ans.
The shear stress τ = µ(U + V)/[r ln(a/b)] and is never equal at both walls for any ratio of V/U unless the clearance is vanishingly small, that is, unless a ≈ b. Ans.
P4.103 It is desired to pump ethanol at 20°C through 25 meters of straight smooth tubing under laminar-flow conditions, Red = ρVd/µ < 2300. The available pressure drop is 10 kPa. (a) What is the maximum possible mass flow, in kg/h? (b) What is the appropriate diameter? Solution: For ethanol at 20°C, ρ = 789 kg/m3 and µ = 0.0012 kg/m-s. From Eq. (4.138), Qlaminar =
π R 4 Δp π d 4 Δp = 8µ L 128µ L
Clearly, flow increases with diameter, so maximum mass flow requires the maximum diameter consistent with the maximum Reynolds number. The Reynolds number may be written out:
ρVd 4ρQ 4ρ π d 4 Δp ρ d 3 Δp Re d = = = ( )( ) = < 2300 µ π dµ π d µ 128µ L 32µ 2 L 2300(32)µ 2 L 2300(32)(0.0012)2 (25) Or : d = = = 3.36E − 7 m3 ρ Δp (789)(10,000) 3
Solve for
dmax = 0.00695m ≈ 7 mm
Ans.(b)
The maximum mass flow is m max = ρQmax = (789
kg m3
)[
π (0.00695m)4 10,000Pa kg kg ]( ) = 0.0151 = 54 Ans.(a) 128(0.0012kg / m • s) 25m s h
Light liquids like ethanol stay laminar only for tiny diameters. To work the same problem with, say, SAE 30W oil, µ = 0.29 kg/m-s, would result in dmax = 26 cm, or 37 times larger. The maximum oil mass flow would be nearly nine thousand times larger.
59
P4.104 Consider 2-D incompressible steady Couette flow between parallel plates with the upper plate moving at speed V, as in Fig. 4.12a. Let the fluid be nonnewtonian, with stress given by
τ xx
∂u c = a ∂x
τ yy
∂ v c = a ∂y
τ xy = τ yx
c 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:
Many terms drop out because v = 0 and τxx and ∂u/∂x = 0 (because u does not vary with x). Thus we only have c ∂τ xy d a du du = = constant, u = C1 y + C2 = 0, or: ∂ y dy 2 dy dy
The boundary conditions are no-slip at both walls: u(y = –h) = 0 = C1(–h) + C2; u(y = +h) = V = C1(+h) + C2, solve C1 =
V V , 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)
P4.105 A tank of area Ao is draining in laminar flow through a pipe of diameter D and length L, as shown in Fig. P4.105. Neglecting the exit-jet kinetic energy and assuming the pipe flow is driven by the hydrostatic pressure at its entrance, derive a formula for the tank level h(t) if its initial level is ho.
60
Fig. P4.105
Solution: For laminar flow, the flow rate out is given by Eq. (4.147). A control volume mass balance shows that this flow out is balanced by a tank level decrease:
Qout =
π D 4 Δp dh = − Ao where Δp ≈ ρ gh(t) 128µ L dt
Thus we can separate the variables and integrate to find the tank level change: h
∫
ho
t π D4ρ g dh π D 4 ρg =−∫ dt, or: h = ho exp − t Ans. h 128µ LAo 128µ LAo 0
P4.106 A number of straight 25-cm-long microtubes, of diameter d, are bundled together into a “honeycomb” whose total cross-sectional area is 0.0006 m2. The pressure drop from entrance to exit is 1.5 kPa. It is desired that the total volume flow rate be 1 m3/h of water at 20°C. (a) What is the appropriate microtube diameter? (b) How many microtubes are in the bundle? (c) What is the Reynolds number of each microtube? Solution: For water at 20°C, ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. Each microtube of diameter D sees the same pressure drop. If there are N tubes,
Q=
1 m3 π D 4 Δp π D 4 (1500 Pa) = NQtube = N =N = 1.47E5 N D 4 3600 s 128 µ L 128(0.001kg/m⋅s)(0.25 m) At the same time, N = Abundle /Atube =
0.0006 m 2 (π /4)D 2
Combine to find D2 = 2.47E−6 m2 or D = 0.00157 m and N = 310 Ans.(a, b) With D known, compute V = Q/Abundle = Qtube/Atube = 0.462 m/s and ReD = ρVD/µ = (998)(0.462)(0.00157)/(0.001) = 724 (laminar) Ans. (c)
61
P4.107 A long solid cylinder rotates steadily
r
in a very viscous fluid, as in Fig. P4.107.
ρ, µ
θ
Assuming laminar flow, solve the Navier-Stokes
R equation in polar coordinates to determine the
Ω
resulting velocity distribution. The fluid is at rest
Fig. P4.107
far from the cylinder. [HINT: the cylinder does not induce any radial motion.] Solution: We already have the useful hint that vr = 0. Continuity then tells us that (1/r)∂vθ/∂θ = 0, hence vθ does not vary with θ. Navier-Stokes then yields the flow. From Eq. D.6, the tangential momentum relation, with ∂p/∂θ = 0 and vθ = f(r), we obtain Eq. (4.139):
1 d dvθ v C (r ) = θ2 , Solution : vθ = C1 r + 2 r dr dr r r As r →∞, vθ → 0 , hence C1 = 0 C2 ΩR 2 2 At r = R, vθ = ΩR = ; C2 = ΩR ; Finally, vθ = R r
Ans.
Rotating a cylinder in a large expanse of fluid sets up (eventually) a potential vortex flow. ________________________________________________________________________ *P4.108
Two immiscible liquids of
V equal thickness h are being sheared
h
y
ρ2, µ2
between a fixed and a moving plate, as in Fig. P4.108. Gravity is neglected,
h
x
ρ1, µ1 Fixed
and there is no variation with x.
Fig. P4.108 Find an expression for (a) the velocity at the interface; and (b) the shear stress in each fluid. Assume steady laminar flow.
62
Solution: Treat this as a Ch. 4 problem (not Ch. 1), use continuity and Navier-Stokes:
Continuity :
∂u ∂v ∂v + = 0+ = 0 ; thus v = const = 0 for no − slip at the walls ∂x ∂y ∂y
This tells us that there is no velocity v, hence we need only consider u(y) in Navier-Stokes: The velocity profiles are linear in y but have a different slope in each layer. Let uI be the
ρ1,2 (u
∂u ∂u ∂p ∂2 u ∂2 u d 2u + v ) = − + µ1,2 ( 2 + 2 ) or : 0 + 0 = 0 + µ1,2 (0 + 2 ) ∂x ∂y ∂x ∂x ∂y dy Thus u = a + b y
velocity at the interface. (a) The shear stress is the same in each layer:
τ = µ1
uI V − uI µ2 = µ2 Solve for u I = V h h µ1 + µ 2
Ans.(a)
(b) In terms of the upper plate velocity, V, the shear stress is
τ = (
µ1 µ 2 V ) µ1 + µ 2 h
Ans.(b)
________________________________________________________________________ P4.109 Consider lubricated Poiseuille flow through a channel bounded by two wide and long parallel plates, as is shown in Fig. P4.109. The flow is driven by a known pressure gradient K = –dp/dx. The fluid I in the core part of the channel has density ρ1 and dynamic viscosity µ1, while the lubricating fluid II has density ρ2 and dynamic viscosity µ2, which are much smaller in magnitude than those of fluid I. Fluid II is a thin sheath layer near each wall (b – a a). Gravity may be ignored, and the flow is assumed to be fully developed.
Fig. P4.109
Solution:
x − momentum equation: ⇒ u(y) = −k
d 2u 1 =− k 2 µ dy
y2 + C1 y + C2 2µ y2 + C1 y + C2 , 2µ1
Fluid I (0 < y < a)
u1 (y) = −k
Fluid II (a < y < b)
y2 u2 (y) = −k + C3 y + C4 , 2µ 2
u1′ = −k
y + C1 µ1
u2′ = −k
y + C3 µ2
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B.C. symmetry ⇒ u1′ (0) = 0 ⇒ C1 = 0 @mid-plane continuity of stress ⇒ µ1u1′ (a) = µ 2u2′ (a) @interface no slip @ wall ⇒
µ 2 (b) = 0
⇒
continuity of velocity @ interface ⇒ ∴
µ1 (y) =
C4 = k
⇒
C3 = 0
b2 2µ 2
⇒
k (b 2 − y 2 ) 2µ 2 ka 2 µ1 b 2 C2 = 1+ −1 2µ1 µ 2 a 2
µ2 =
µ1 (a) = µ 2 (a) ⇒
ka 2 µ1 b 2 y 2 1+ −1 − 2µ1 µ 2 a 2 a 2 a
Flow rate Q1 = 2 ∫ µ1dy 0
=
ka 3 2 µ1 b 2 + −1 µ1 3 µ 2 a 2
The lubrication due to fluid II is significant when
µ1 b 2 2 −1 ≥ µ2 a 2 3
P4.110 Reconsider Prob. P1.49 and calculate (a) the inner shear stress and (b) the power required, if the exact laminar-flow formula, Eq. (4.140) is used. (c) Determine whether this flow pattern is stable. [HINT: The shear stress in (r, θ) coordinates is not like plane flow.] Solution: The exact laminar-flow velocity is Eq. (4.140), and the shear stress is Eq. (D.9):
vθ = Ωi ri [
τ rθ = µ (
(ro / r)− ( r / ro ) ] (ro / ri )−(ri / ro )
dvθ vθ Ωi ri 2r − ) = µ[ ] o dr r (ro / ri )−(ri / ro ) r 2
Recall the data from Prob. P1.49: ri = 5 cm, ro = 6 cm, L = 120 cm, µ = 0.86 kg/m-s (SAE 50W oil), and Ωi = 900 rev/min = 94.25 rad/s. At the inner cylinder,
τ inner = µ [
Ωi ri 2r 94.25(0.05) 2(0.06) ] o = (0.86)[ ] = 531 Pa Ans.(a) (ro / ri )−(ri / ro ) r 2 0.06 / 0.05 − 0.05 / 0.06 (0.05)2 i
The moment and power required are M = τ i 2π ri2 L = (531) 2π (0.05)2 (1.20) = 10.0 N − m Power = Ωi M = (94.25rad / s)(10.0 N − m) = 943 watts
Ans.(b)
The shear stress, moment, and power are all 31% larger than the approximate linear-profile analysis of Prob. 1.49.
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(c) The stability of this flow is determined by Taylor’s criterion, Eq. (4.141):
Ta =
ri (ro − ri )3 Ω2i (µ / ρ ) 2
=
(0.05)(0.06 − 0.05)3 (94.25)2 (0.86 / 902)2
= 490 < 1700 STABLE Ans.(c)
To finish this, we had to look up the density of SAE 50W oil, ρ = 902 kg/m3. _______________________________________________________________________ P4.111 Reconsider the pressure-driven flow between two parallel plates of Fig. 4.12(b) for the case of slip flow at both walls. The boundary conditions are
u = λ
du dy
at y = ±h,
where λ ≥ 0 is a constant, known as the slip length. (a) Solve the x-momentum equation to obtain the following axial velocity profile: u( y) = K
h2 y 2 λ 1− +2 2µ h2 h
where K = –dp/dx is the axial pressure gradient. Hint: use either one of the slip boundary conditions given above, and the symmetry condition du/dy = 0 at y = 0. (b) Sketch the velocity profile. (c) Deduce a relation between the flow rate Q (per width of the channel) and the pressure gradient K. (d) Owing to the boundary slip, the pressure gradient required to drive the same flow rate through the channel can be reduced. Show that the relative reduction in the pressure gradient is given by
K no-slip − K Q
K no-slip
=
3λ h + 3λ
where Kno-slip is the pressure gradient when there is no slip at the boundaries (i.e., λ = 0). Solution:
x − momentum eqn
d 2u dy
2
=−
1 K µ
⇒ u( y) = −K By symmetry,
du = 0 at y = 0 dy
⇒
y2 + C1 y + C2 2µ
C1 = 0
du at y = h dy h 2 2λ ⇒ C2 = K 1+ 2µ h 2 h 2 y 2λ ∴ u( y) = K 1− + 2µ h h Partial slip @ wall,
u = −λ
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h
Flow-rate
Q = 2 ∫ u dy 0
=K
h3 2 2 λ + µ 3 h
For the same flow-rate Q = K no-slip ⇒ ∴
Kslip K no-slip
h3 2 h3 2 2 λ = Kslip + µ 3 µ 3 h =
1 1+ 3λ / h
K no-slip − Kslip K no-slip
= 1−
1 3λ = 1+ 3λ / h h + 3λ
_______________________________________________________________________ FUNDAMENTALS OF ENGINEERING EXAM PROBLEMS: Answers Chapter 4 is not a favorite of the people who prepare the FE Exam. Probably not a single problem from this chapter will appear on the exam, but if some did, they might be like these: FE4.1 Given the steady, incompressible velocity distribution V = 3xi + Cyj + 0k, where C is a constant, if conservation of mass is satisfied, the value of C should be (a) 3 (b) 3/2 (c) 0 (d) –3/2 (e) –3 FE4.2 Given the steady velocity distribution V = 3xi + 0j + Cyk, where C is a constant, if the flow is irrotational, the value of C should be (a) 3 (b) 3/2 (c) 0 (d) –3/2 (e) –3 FE4.3 Given the steady, incompressible velocity distribution V = 3xi + Cyj + 0k, where C is a constant, the shear stress τxx at the point (x, y, z) is given by (a) 3µ (b) (3x + Cy)µ (c) 0 (d) Cµ (e) (3 + C)µ FE4.4 Given the steady incompressible velocity distribution u = Ax, v = By, and w = Cxy, where (A, B, C) are constants. This flow satisfies the equation of continuity if A equals (a) B , FE4.5 is
(b) B + C ,
(c) B – C ,
(d) – B ,
(e) –(B + C)
For the velocity field in Prob. FE4.4, the convective acceleration in the x direction (a) A x2 ,
(b) A2 x ,
(c) B2 y ,
(d) B y2 ,
(e) C x2 y
FE4.6 If, for laminar flow in a smooth straight tube, the tube diameter and length both double, while everything else remains the same, the volume flow rate will increase by a factor of (a)
2 ,
(b) 4 ,
(c) 8 ,
(d)
12 ,
(e) 16
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COMPREHENSIVE PROBLEMS C4.1 In a certain medical application, water at room temperature and pressure flows through a rectangular channel of length L = 10 cm, width s = 1 cm, and gap thickness b = 0.3 mm. The volume flow is sinusoidal, with amplitude Qo = 0.5 ml/s and frequency f = 20 Hz, that is, Q = Qosin(2π f t). (a) Calculate the maximum Reynolds number Re = Vb/ν, based on maximum average velocity and gap thickness. Channel flow remains laminar for Re < 2000, otherwise it will be turbulent. Is this flow laminar or turbulent? (b) Assume quasi-steady flow, that is, solve as if the flow were steady at any given Q(t). Find an expression for streamwise velocity u as a function of y, µ, dp/dx, and b, where dp/dx is the pressure gradient required to drive the flow through the channel at flow rate Q. Also estimate the maximum magnitude of velocity component u. (c) Find an analytic expression for flow rate Q(t) as a function of dp/dx. (d) Estimate the wall shear stress τw as a function of Q, f, µ, b, s, and time t. (e) Finally, use the given numbers to estimate the wall shear amplitude, τwo, in Pa. Solution: (a) Maximum flow rate is the amplitude, Qo = 0.5 ml/s, hence average velocity V = Q/A: Q 0.5E−6 m 3/s V= = = 0.167 m /s bs (0.0003m)(0.01m)
Vb (0.167)(0.0003) = ν (0.001 / 998) = 50 (laminar) Ans. (a)
Re max =
(b, c) The quasi-steady analysis is just like Eqs. (4.142−144) of the text, with “h” = b/2: u=
−1 dp b 2 −1 dp b 2 2 −sb 3 dp 2 , Qmax = umax bs = − y , umax = 2µ dx 4 2µ dx 4 3 12µ dx
(d) Wallshear: τ wall = µ
du b dp 6µ Q 6µ Qo = = = sin(2 πf t) Ans. (d) dy wall 2 dx sb 2 sb 2
(e) For our given numerical values, the amplitude of wall shear stress is:
τ wo =
6µQo 6(0.001)(0.5E−6) = = 3.3 Pa Ans. (e) sb 2 (0.01)(0.0003)2
Ans. (b, c)
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C4.2 A belt moves upward at velocity V, dragging a film of viscous liquid of thickness h, as in Fig. C4.2. Near the belt, the film moves upward due to no-slip. At its outer edge, the film moves downward due to gravity. Assuming that the only non-zero velocity is v(x), with zero shear stress at the outer film edge, derive a formula for (a) v(x); (b) the average velocity Vavg in the film; and (c) the wall velocity VC for which there is no net flow either up or down. (d) Sketch v(x) for case (c). Solution: (a) The assumption of parallel flow, u = w = 0 and v = v(x), satisfies continuity and makes the x- and z-momentum equations irrelevant. We are left with the y-momentum equation:
Fig. C4.2
There is no convective acceleration, and the pressure gradient is negligible due to the free surface. We are left with a second-order linear differential equation for v(x):
At the free surface, x = h, τ = µ(dv/dx) = 0, hence C1 = –ρgh/µ. At the wall, v = V = C2. The solution is
(b) The average velocity is found by integrating the distribution v(x) across the film:
(c) Since hvavg ≡ Q per unit depth into the paper, there is no net up-or-down flow when
68
(d) A graph of case (c) is shown below. Ans. (d)
