Exercise 9, Ex. 6.3 ( submarine The flow around a submarine moving at at velocity V can be described by the flow caused by a source and a sink with strength Q at a distance a from each other. V x Submarine p Q -Q y L a Figure : Coordinate system for submarine problem a If one wants to construct a pressure sensor that will register an approaching submarine at a distance L, what sensitivity is needed for the sensor? Assume an ideal fluid and that a = 8 m, Q = 95 m 3 /s, V = 8 m/s, L = m and ρ = kg/m 3 roughly corresponding to a submarine with length 8.5 meters and 6 m radius. Use a potential flow description ū = φ, u = φ x, v = φ y The flow is always irrotational due to the definition of the velocity potential, ω = ū = φ =, curl(grad= For incompressibility we get, ū = φ = ( φ i x i = φ = The equation is linear and thus superposition can be used. We have freestream plus 3D source plus sink, φ = V freestream + Q 4π r source + Q = V + 4π r sink Q 4π ( + a + x + y + Switch to cylindrical coordinates and notice that x + y = r. This gives Q 4π ( a + x + y + ū = φ = φ r ēr + φ r θ ēθ + φ ē = Qr ē r 4π(( + a + r Qr Q( + a +ē 3/ 4π(( a + r 3/ V + 4π(( + a + r Q( a 3/ 4π(( a + r 3/
Use Bernoullis equation to determine the pressure fluctuations at = L a, r =, Evaluate ū noticing that u r =, p ρ + ū = constant = p ρ + V ū( = L a, r = = ē V + Inserting the given values gives ū = 7.999 and we get, Q 4π(L + Q 4π((L + a p p = ρ(v ū 7. N/m.7 mbar b How long is the submarine? Compute where u = for r =, V + Q ( 4π ( + a ( a = 4πV Q = ( a ( + a ( a ( + a = 4a ( a Solving this system gives = 43., = 36.99, where the second solution lies inside the submarine. The length is then 43. 86 m. c How wide is the submarine? To get this we need to compute the shape of the submarine. The stream-function is constant along streamlines and is useful for this. In spherical coordinates we get, ū ē R = u R = ū ē R ψ R sin θ θ = u R = ūē R ψ R sin θ R = u θ = ūē θ ē r = ē θ cos θ + ē R sin θ notice that u r (ē θ cos θ + ē R sin θ + u (ē R cos θ ē θ sin θ ē = ē R cos θ ē θ sin θ ē R = u r sin θ + u cos θ u R = u r sin θ + u cos θ = r = R sin θ, = R cos θ = sin θ Q 4π R sin θ (R + a + ar cos θ + 3/ (R + a ar cos θ 3/ ( cos θ V + Q R cos θ + a 4π (R + a + ar cos θ R cos θ a 3/ (R + a ar cos θ 3/ = V cos θ + Q ( R + a cos θ 4π (R + a + ar cos θ R a cos θ 3/ (R + a ar cos θ 3/ This is difficult to integrate. Simplify to a Rankine body by neglecting the sink and say that a =, u R = V cos θ + Q 4π R = ψ R sin θ θ Determine C from the stagnation point, Ψ = V R sin θ Q 4π cos θ + C u R (θ = π, R = R = Q 4πV
Since Ψ = on the body we get, C = Q 4π. The streamfunction is then, Ψ = V R sin θ Q (cos θ + 4π source In cartesian coordinates we get the streamfunction for a source, Ψ = Q ( 4π x + + The shape is given by Ψ =, As R, θ then R sin θ d. This gives, 5 x 5 5 5 5 5 5 Figure : Rankine body for submarine problem V d 4 Q 4π = d = 4Q V π r = 4Q V π There is a simple way of determining the radius as directly. The flow from the source must take up an particular area in the flow an infinity. Since no fluid can cross the streamlines this area must be equal to that of the Rankine body: Q Q = V πr r = V π = 6.338 m 3
We can use the computed streamfunction and displace it a distance, Ψ = Q ( 4π x + ( + Notice that R = in spherical coordinates and change coordinate system, Ψ = Q ( R cos θ R + = 4π R Q ( R cos θ R sin θ + (R cos θ R 4π R + R R R cos θ + The streamfunction for the submarine is then, Ψ = V R sin θ + Q ( R cos θ + a 4π R + a + Ra cos θ + R cos θ a R + a Ra cos θ x 8 6 4 4 6 8 Figure 3: Submarine body for submarine problem At = or θ = π/ we get, Ψ = V R + Q ( a 4π R + a + a R + a For the body Ψ = and we get, Multiply by, Computing this, R R + a aq πv = R R + a + aq πv (R 3 + (R a a Q π V = R = 6.3 m 4
The complex potential The lines with constant streamfunction Ψ are the streamlines. They are orthogonal to the lines of constant velocity potential φ which are equipotential lines. Since both of them satisfy Laplace s equation we can define a complex function, F ( = φ(x, y + i Ψ(x, y = x + iy 8 6 4 8 6 4 5 45 4 35 3 5 Figure 4: Complex potential for submarine problem, solid: Ψ, dotted: φ This is an analytical function since the Cauchy Riemann equation holds, φ x = Ψ y and φ y = Ψ x The velocity is then, w( = df d = φ x + i Ψ x = u iv This enables the use of complex analysis, in particular conformal mapping that can be used to compute the flow over airfoil shapes. 5
Example: Flow past a rotating cylinder centered at = λ at an angle of attack α, ] F ( = U [( + λe iα (a + λ + ( + λ eiα iγ log( + λ π ( Mapping by = Z + 4 Z a gives an airfoil shape with the potential F(. A correct flow is not achieved unless the Kutta Joukovski condition is satisfied requiring, Γ = 4πU(a + λ sin α 4 Z=f( 3 3 5 5 4 4 =f (Z Figure 5: Conformal mapping from circle to airfoil shape, (a = 3,λ =.5 6