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Unformatted text preview: Problem 5.1 [Difﬁcultyz 1] 5.1 Which of the following sets of equations represent pos
sible twodimensional incompressible ﬂow cases?
(a) u, = 2:2 +192 —*r1}t; v = 3:3 +  4}»)
(b’) u=2xy—x2y; v=2ry—yz+x2
(c) u = f: +2y; v = x1? —yt
(d) u a (2x + dﬁxz; v a —3(x + ﬂy: Given: The list of velocity ﬁelds provided above
F ind: Which of these ﬁelds possibly represent twodimensional, incompressible ﬂow
Solution: We will check these ﬂow ﬁelds against the continuity equation Governing 6
Equations: ——
‘ 8x (1) Incompressible' ﬂow (p is constant)
(2) Two dimensional ﬂow (velocity is not a ﬁmction of z) (pu) + — + — + a—’0 = 0 (Continuity equation) Assumptions: 1 Based on the two assumptions listed above, the continuity equation reduces to: This is the criterion against which we will check all of the ﬂow ﬁelds. a) u(x,y,t> = 2ox2+y2—x2~y v(x,y,t) = x3 +x(y2—4y) 9u<x,y,t) = 4x— 2xy Elwyn) = way—4)
Hence 9u + a—vssﬂ
6x 6y
b 2 2 . 2 a a
) u(x,y,t) = My — x y v(x,y,t) = 2X'y y + X —u(x,y,t) = 2'y  2Xy —v(x,y,t) = 2'x 2y
' ﬁx 6y
Hence 53—u + £9—v at 0
6X 6y
2 2 a a
c) u(xsy:t) = X 't+ V(XJY>t) = X't _Y't _u(xzy,t) = 2't'x _V(X:YJt) = _t
Hence 59—u + 6—v ¢ 0
6x 6y
d) u(x,y,t) = (2'X+4'Y)'X't v(x,y,t) = 3‘(x+ y)y't a—u(x,y,t) = t(Zx+ 4'y) + 2tX a—v(x,y,t) = —t(3x+ 3y) — 3'ty
6x 53' Hence ' 6—u + 8—v eé 0 3X ay NOTINCOMPRESSIBLE Problem 5.2 5 Given: Velocity ﬁelds Find:
Solution: Governing
Equation: Assumption: This is the criterion against which we will check all of the flow 1e a) u(x,y,z,t) = 2'y2 + 2x.z
a—u(x,y,z,t) = 22
ﬁx
Hence b) u(X,y,z,t) = xyzt a—u(x,y,z,t) = tyz
6X Hence 0) u(x,y,z,t) = x2 + 2~y + 22 a—u(x,y,z,t) = 2): Hence 5.12 Which of Efollowing sets oflequations representvpose
:sible'threeadimemional incompressible ﬂow leases‘2.
(a) u = 2}? + 2x2; 1) = —2yz +’ ﬁfyz; w = 23x26 +55!“ (b), n==xzz1§v =>xyzt“z; w‘= 22613—3”). a 
(C) u =2? + 2y + z“; u=" x  2y .‘i’ Z; W =' ail+1? +2.5 Which are 3D incompressible v(x,y,z,t) = —2'y~z + 6x2yz 9—V(x,y,z,t) = 6'X2'Z — 22 6y 2u+6—v+Qw:r&0 6X By 62 V(x,y,z,t) = —xyzt2 2—v(x,y,z,t) = —t2xz 6y iu+a—V+1W¢0 6x By 62 Iv(x,y,z,t) = x — 2~y + z a—v(x,y,z,t) = —2
5y a—u+a—v+a—W=0 6x 6y az .INCOMPRE‘SSIBLE We will check these ﬂow ﬁelds against the continuity equation w(x,y,z,t) = 3)(222 + x3vy4 —a—w(x,y,z,t) = 6x2z W(x,y,z,t) = —2X~z + y2 + 22 a—w(x,y,z,t) = 2 — 2x
Bz [Difﬁculty: 2] Problem 5.4 [Difﬁculty: 1] 5:4 The three components of'velocity in aveloeity ﬁeld are
given by u.‘=A:::.+ By+ Cz, vi: Dxrl— Ey+ Fz, and w=Gx +
Hy +15, 'Detenninejthe relationship among the ,gzcoeﬁ‘idents through I. that is necessary if. this is robe va pmsib‘le
incompressible, ﬂow ﬁeld, Given: The velocity ﬁeld provided above
Find: The conditions under which this ﬁelds could represent incompressible ﬂow Solution: We will check this ﬂow ﬁeld against the continuity equation Governing a a 6 5p . E i t' : — + — + — + — = 0 (Continuity eq ation
qua Ions 64m) é1y(pv) 6 (PW) _ u ) Assumptions: (1)VIncompressible ﬂow (p is constant) 5 Problem 5.6 Lnifﬁcuuy: 2] ﬂ\ Y 5.671.110 x vc'ompom’mt of. velOcity in a steade incompressible
ﬂow ﬁeld in the .ryplane is u =A/x, where A. =2"m2/s, andx
isxmeasured in meters. Find the simplest y Component of
velocity fer th's ﬂow: ﬁeld. Given: The x—component of velocity in a steady, incompressible ﬂow ﬁeld
Find: The simplest y—component of velocity for this ﬂow ﬁeld
Solution: We will check this ﬂow ﬁeld against the continuity equation Governing
a a a a . . .
Equations: — + — (pv) + + —’0 = 0 (Contlnmty equat1on)
6x 62 62‘
Assumptions: Incompressible ﬂow (p is constant) @ a
, (2) Two dimensional ﬂow (velocity is not a function of z) '
\ ’ a‘u av _ ' \‘____
Based on the' two assumptions listed above, the continuity equation reduces to: — — _ 6a A — = —— There re~from continuity, we hav The partial of u with respect to X i:
6x )52 A Ay
—d +fx=—+fx
2 y () x2 () Integrating this expression will yield the ycomponent of velocity:
' x The simplest version of this velocity component would result when f(x) = 0: Problem 5.7 [Difﬁculty: 2]
' A 5.7“The y. component ofveloc'ity in asteadyanoxnpxwsible
ﬂow ﬁeld; in the .ry plane 'is v=Axy(xz—yz), were A =
3 11145“ and x and y are measured. in meters. Find the
‘ simplest .tf component 'of‘velocity for this ﬂowﬁ‘eld. Given: y component of velocity Find: x component for incompressible ﬂow; Simplest x components? Solution: Basic a_(p.u) + a_(p.v) + 2mm + ip = 0
equation: 5" 53’ 62 at
Assumptions: Incompressible ﬂow (p is constant) @ Flow is only in the x—y plane Hence 3
x — 3xy2) dx = —iA~x4 + EAxzyz + f(y) Integrating This basic equation is valid for steady and unsteady ﬂow (t is not explicit) There are an inﬁnite number of solutions, since f(y) can be any function of y. The simplest is f(y) = 0 3 1
u(x,y) = EvAxzy2 — ZA‘x4 ...
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This note was uploaded on 09/05/2011 for the course EGN 3353C taught by Professor Lear during the Spring '07 term at University of Florida.
 Spring '07
 Lear
 Fluid Mechanics

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