MIT Department of Mechanical Engineering 2.25 Advanced Fluid Mechanics Kundu & Cohen 6.14 This problem is from “Fluid Mechanics” by P. K. Kundu and I. M. Cohen In a two-dimensional constant density potential ﬂow, a source of strength m is located a meters above an inﬁnite plane. Find the velocity on the plane, the pressure on the plane, and the reaction force on the plane. 1 Solution: Here we use the method of images and note that in the complex plane the two point sources are located at z = ±ia, where i2 = −1. Now the complex potential for this ﬂow is given by w= m m m ln(z − ia) + ln(z + ia) − ln a2 2π 2π 2π (1) where we have included the term ln a2 to make the arguments of the logarithms dimensionless (recall that ln z − ln a = ln az ). The inclusion of this term, however, does not aﬀect the velocity ﬁeld, since it is eliminated when we take the gradient of w. The above result in Eq.(1) is similar, but not identical to Eq. (6.50) in Kundu. Why? The equation expands to w= m m ln a2 ln(x2 − y 2 + a2 + i2xy) − 2π 2π (2) Since the logarithm of a complex quantity, ζ = |ζ|eiθ , is ln ζ = ln |ζ| + iθ, for the imaginary part of Eq.(2), we have ψ= since here θ = tan−1 ( 2xy x2 −y 2 +a2 m tan−1 2π 2xy x2 − y 2 + a2 (3) ) . We obtain the x and y-components of the velocity, ux and uy respectively from ux = ∂ψ m x(x2 + y 2 + a2 ) = ∂y π a4 + 2a2 x2 − 2a2 y 2 + x4 + 2x2 y 2 + y 4 (4) and uy = − ∂ψ m y(x2 + y 2 − a2 ) = 4 2 2 ∂x π a + 2a x − 2a2 y 2 + x4 + 2x2 y 2 + y 4 (5) To determine the velocity along the horizontal plane, we solve for ux and uy at y = 0. Accordingly, ux,p = m x(x2 + a2 ) m x = 4 2 2 4 2 π x + a2 π a + 2a x + x (6) whereas uy,p = 0 and hence the condition of no penetration along the plane is satisﬁed. 2 (7) We can normalize this result to obtain Up ≡ 1 xa ux,p a = m π 1 + xa22 (8) 0.2 Up 0.1 0 −0.1 −0.2 −15 −10 −5 0 x/a 5 10 15 2 The pressure along the plane is given by the Bernoulli equation P∞ = Pp − 12 ρux,p , where P∞ is taken to be given and is the pressure very far away from the plane where the velocity is essentially zero, but it also happens to equal the pressure at the stagnation point. So the pressure at the plane is x2 1 1 m2 Pp = P∞ − ρu2x,p = P∞ − ρ 2 2 2 2 π (x + a2 )2 (9) This result can also be non-dimensionalized to Pp − P∞ 1 =− 2 1 m2 π ρ 2 a2 3 x2 a2 1+ x2 a2 2 (10) 0 1 m2 (P p − P ∞ )/ ρ 2 2 a −0.005 −0.01 −0.015 −0.02 −0.025 −0.03 −15 −10 −5 0 x/a 5 10 15 The net upward (i.e. positive y-direction) force per unit depth, F ' , on the plane is simply the integral of the gage pressure acting on it. Here we assume that the pressure on the bottom side of the plane is everywhere P∞ . F' = ∞ −∞ −ΔP n̂dx = ∞ −∞ −(Pp − P∞ )dx = ∞ −∞ 1 m2 x2 ρ 2 2 dx 2 π (x + a2 )2 (11) The right hand side of Eq. (11) is 1 m2 1 tan−1 F = ρ 2 2 π 2a ' x a x − 2 2(a + x2 ) ∞ = −∞ 1 m2 π − ρ 4a π 2 2 − π 2 (12) So the net force acting per unit depth of the plane is F' = 1 m2 ρ 4π a (13) D 4