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The velocity distribution in a two-dimensional, steady, inviscid flow field in the xy plane is
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Chapter 6 Solutions
Fox And Mcdonald's Introduction To Fluid Mechanics
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- 4 = 3x2 – y represents a stream function in a two – dimensional flow. The velocity component in 'x' direction at the point (1, 3) is:arrow_forward1. For incompressible flows, their velocity field 2. In the case of axisymmetric 2D incompressible flows, where is Stokes' stream function, and u = VXS, S(r, z, t) = Uz = where {r, y, z} are the cylindrical coordinates in which the flow is independent on the coordinate and hence 1 Ꭷ r dr 1 dy r dz Show that in spherical coordinates {R, 0, 0} with the same z axis, this result reads Y(R, 0, t) R sin 0 S(R, 0, t) UR uo Y(r, z, t) r = = -eq, and Up = = 1 ay R2 sin Ꮎ ᎧᎾ 1 ƏY R sin Ꮎ ᎧR -eq 2 (1) (2) (3)arrow_forward4. The velocity vectors of three flow fileds are given as V, = axĩ + bx(1+1)}+ tk , V, = axyi + bx(1+t)j , and V3 = axyi – bzy(1+t)k where coefficients a and b have constant values. Is it correct to say that flow field 1 is one-, flow filed 2 is two-, and flow filed 3 is three-dimensional? Are these flow fields steady or unsteady?arrow_forward
- Flow through a converging nozzle can be approximated by a one-dimensional velocity distribution u=vo (1+2). For the nozzle shown below, assume that the velocity varies linearly from u = vo at the entrance to u = 3v, at the exit. Compute the acceleration at the entrance and exit if vo=10m/s and L = 1m. x=0 X u= :326 x=Larrow_forward1. Stagnation Points A steady incompressible three dimensional velocity field is given by: V = (2 – 3x + x²) î + (y² – 8y + 5)j + (5z² + 20z + 32)k Where the x-, y- and z- coordinates are in [m] and the magnitude of velocity is in [m/s]. a) Determine coordinates of possible stagnation points in the flow. b) Specify a region in the velocity flied containing at least one stagnation point. c) Find the magnitude and direction of the local velocity field at 4- different points that located at equal- distance from your specified stagnation point.arrow_forward1. For a two-dimensional, incompressible flow, the x-component of velocity is given by u = xy2 . Find the simplest y-component of the velocity that will satisfy the continuity equation. 2. Find the y-component of velocity of an incompressible two-dimensional flow if the x-component is given by u = 15 − 2xy. Along the x-axis, v = 0.arrow_forward
- 3.4 Consider a steady, incompressible, 2D velocity field for motion parallel to the X-axis with constant shear. The shear rate is du/dy Ay. Obtain an expression for the velocity field V. Calculate the rate of rotation. Evaluate the stream function %3D for this flow field. Ay Ay + В і, о, Ay + By+ C 6. Ans: V= 2arrow_forwardVelocity field of an incompressible flow is given by V = 6xi − 6yj (m/s) a) Find the pathlines in x-y plane. Make a sketch of pathlines for x ≥ 0 and y ≥ 0. b) Find the streamlines. Make a sketch of streamlines for x ≥ 0 and y ≥ 0. c) At time t = 0 s, the position of a rectangular fluid element ABCD is described by the corner points A(1,3), B(2,3), C(1,2) and D(2,2). Determine the new position of the fluid element at time t = 1/6 sarrow_forward.3 2. 1. If u = 3x*yt and v = -6x°y´t“ answer the following questions giving reasons, Is this flow or fluid: (a) Real (Satisfies Continuity Principle). (b) Steady or unsteady. (c) Uniform or non-uniform. (d) One, two, or three dimensional. (e) Compressible or incompressible. Also, Find the acceleration at point (1,1).arrow_forward
- 1. For a flow in the xy-plane, the y-component of velocity is given by v = y2 −2x+ 2y. Find a possible x-component for steady, incompressible flow. Is it also valid for unsteady, incompressible flow? Why? 2. The x-component of velocity in a steady, incompressible flow field in the xy-plane is u = A/x. Find the simplest y-component of velocity for this flow field.arrow_forwardThe velocity field for a fluid flow is given by following expression: =(0.2x² + 2y+2.5)î +(0.5x+2y² – 6) ĵ+(0.15x² + 3y° + z)k The strain tensor at (2,1,–1) will be: 0.8 1.25 0.30 a) | -1.25 -4 0.30 -1 (0.8 1.25 0.70 b) | 1.25 2 0.30 -2 1 0.8 1.25 0.30) c) | 1.25 4 -2 0.30 -2 1 0.8 1.25 0.30 d) | 1.25 8. -2 0.8 2 1arrow_forwardThe velocity distribution in a 0.02 m diameter horizontal pipe conveying carbon tetrachloride (specific gravity = 1.59, absolute viscosity = 9.6 x 10-6 Pa sec) is given by the parabolic equation: v(r)=0.01(0.12- r?), where v(r) is the velocity in (m/s) at a distance r in (m) from the pipe center. What is discharge? O a. 3.13 x-8 m3/s O b. None of the mentioned O c. 1.047 x 108 m3/sec O d. 4.97 x 109 m3/secarrow_forward
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