Exercise 9: Model of a Submarine

Transcription

1 Fluid Mechanics, SG4, HT3 October 4, 3 Eample : Submarine Eercise 9: Model of a Submarine The flow around a submarine moving at a 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 Submarine p Q -Q y z 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, U = 8 m/s, L = m and ρ = kg/m 3. Use a potential flow description ū = φ, u = φ, 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 i = φ = The equation is linear and thus superposition can be used. We have freestream plus 3D source plus sink φ = U z }{{} freestream + Q 4π r }{{} source + Q 4π r }{{} sink The first term is in cylindrical coordinates (R, θ, z and the two last are in two different spherical coordinate systems with origin in z a and z + a, respectively. Transform the two second terms to cylindrical coordinates φ = Uz + Q 4π (z + a + R + Q 4π (z a + R

2 Velocity ū = φ = φ R ēr + φ R θ ēθ + φ z ēz = { { } QR ē R 4π((z + a + R QR Q(z + a }+ē 3/ 4π((z a + R 3/ z U+ 4π((z + a + R Q(z a 3/ 4π((z a + R 3/ We need to know the distance, b, from the point source to the stagnation point on the submarine nose. Thus we need to know the length of the submarine. b How long is the submarine? Compute where u z = for R = U + Q ( 4π (z + a (z a = 4πU Q = (z a (z + a (z a (z + a = 4az (z a Solving this system gives z = ±43. m, z = ±36.99 m, where the second solution lies inside the submarine. The length is then m and b = 3. m. Now we continue to solve a Use Bernoulli equation to determine the pressure fluctuations at z = L b a, R = Evaluate ū noticing that u R = p + ρ ū = p + ρu ū(z = L b a, R = = ē z {U + Inserting the given values gives ū = m/s and we get } Q 4π(L + b + Q 4π((L + b + a p p = ρ(u ū 6.86 N/m.7 mbar 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 the stream function is defined as u r = ψ r sin θ θ u θ = ψ r sin θ r Transformation between cylindrical and spherical coordinates Our velocity field gives ψ r sin θ θ = sin θ Q 4π r sin θ { u r = u R sin θ + u z cos θ R = r sin θ, z = r cos θ } (r + a + ar cos θ + 3/ (r + a ar cos θ 3/

3 ( cos θ U + Q { } r cos θ + a 4π (r + a + ar cos θ r cos θ a 3/ (r + a ar cos θ 3/ = U 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 = U cos θ + Q 4π r = ψ r sin θ θ Determine C from the stagnation point Ψ = Ur sin θ Q 4π cos θ + C Since Ψ = on the body we get The stream function is then u r (θ = π, r = r = Q 4πU C = Q 4π. Ψ = UR sin θ Q 4π (cos θ + }{{} source The shape is given by Ψ =. As r, θ then r sin θ d. This gives Figure : Rankine body for submarine problem z U d 4 Q 4π = d = 4Q Uπ Q d = Uπ 3

4 There is a simple way of determining the radius as z directly. The flow from the source must take up an particular area in the flow at infinity. Since no fluid can cross the streamlines this area must be equal to that of the Rankine body: Q = Uπ d Q d = Uπ =.7 m We can use the computed stream function for a point source and displace it to z = a. In cylindrical coordinates Ψ = Q ( z + a 4π (z a + R Transform to spherical coordinates Ψ = Q ( r cos θ + a 4π (r cos θ + a + r sin θ = Q 4π ( r cos θ + a r + a + ar cos θ The stream function for the submarine is then Ψ = Ur sin θ + Q ( r cos θ + a 4π r + a + ar cos θ + r cos θ a r + a ar cos θ Figure 3: Submarine body for submarine problem z At r = R and θ = π/ we get Ψ = UR + Q ( a 4π R + a + a R + a For the body Ψ = and we get Multiply by Computing this R R + a aq πu = R R + a + aq πu (R 3 + (R a a Q π U = d = R =.6 m 4

5 The comple potential The lines with constant stream function 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 comple function F (z = φ(, y + i Ψ(, y z = + iy Figure 4: Comple potential for submarine problem, solid: Ψ, dotted: φ This is an analytical function since the Cauchy Riemann equation holds φ = Ψ y and φ y = Ψ The velocity is then w(z = df dz = φ + i Ψ = u iv This enables the use of comple analysis, in particular conformal mapping that can be used to compute the flow over airfoil shapes. 5

6 Eample : Half body over a wall A line source of strength Q is located at (, a above a flat plate that coincides with the -ais. A uniform stream with velocity U flows along the -ais. Calculate the irrotational flow field. Method of images. Put a line source of equal strength at (, a in order to fulfill the condition of no flow through the plate. Superposition of a uniform flow and the two line sources gives the comple potential Comple velocity W = U + Q π The velocity field now becomes F = Uz + Q π ln(z ia + Q ln(z + ia π W = df dz = U + Q ( π z ia + z + ia W = U + Q ( π + i(y a + + i(y + a W = U + Q ( i(y a i(y + a π + + (y a + (y + a [ ( + (y a + u = U + Q π v = Q ( π + (y + a i ( y a + (y a + + (y a + + (y + a y a + (y a + y + a + (y + a ] y + a + (y + a 6

7 Conformal Mapping Flow past a rotating cylinder centered at z = λ at an angle of attack α ] F (z = U [(z + λe iα (a + λ + (z + λ eiα iγ log(z + λ π ( Mapping by z = Z + 4 Z a gives an airfoil shape with the potential F(z. A correct flow is not achieved unless the Kutta Joukovski condition is satisfied requiring Γ = 4πU(a + λ sin α 4 Z=f(z z=f (Z Figure 5: Conformal mapping from circle to airfoil shape, (a = 3,λ =.5 Eercise 3: Flow past a Symmetric Airfoil a Use conformal mapping to calculate the irrotational flow field around a symmetric airfoil. Joukowski transformation ζ(z = z + c z

8 Equation for the circle Equation for the airfoil Comple potential in the z-plane z = λ + (a + λe iθ ζ = λ + (a + λe iθ + a λ + (a + λe iθ Comple velocity in the z-plane Comple velocity in the ζ-plane ω = df/dz dζ/dz = F = U(z + λe iα (a + λ + U (z + λ eiα + iγ ln(z + λ π W = df dz = (a + Ue iα λ iγ U (z + λ eiα + π(z + λ (Ue iα (a + λ U (z + λ eiα + iγ π(z + λ /( a The velocity can then be found by introducing the reversed transformation z = ζ/ + ζ / a into z u = Re{ω(z}, v = Im{ω(z} b Calculate the Joukowski condition for the airfoil. The flow field has a singular point at the trailing edge of the airfoil at ζ = a. Resolve the singularity by choosing the circulation Γ so the numerator vanishes at the trailing edge z = a Ue iα Ue iα + iγ π(a + λ = Γ k = 4π(a + λu eiα e iα i = 4π(a + λu sin α 8