Quick and dirty first principles modelling
- 1. Stochastic ApS Quick and dirty First principles modelling Scaling, Energy and Symmetry Methods
- 2. Stochastic ApS Buckingham P theorem Given M variables in N fundamental dimensions (m, s, kg, etc.), there are (usually) M-N independent dimensionless combinations of these variables (dimensionless numbers) Drag on a sphere Buckingham’s theorem says every one of these dimensionless variables is a function of the others and nothing else Drag D kgms-2 Flow speed U ms-1 Radius a m Viscosity m kgm-1s-1 Density r kgm-3 2 2 D U ar 5 variables, 3 dimensions 2 dimensionless parameters Re Uar m = ( )2 2 Re D U a r =
- 3. Stochastic ApS Silencing the whistling boiler Hot air Water / steam pipes Vortex shedding or shear layer instability Unsteady flow forcing Resonant acoustic response Acoustic response synchronizes flow unsteadiness to resonance
- 4. Stochastic ApS Silencing the whistling boiler Hot air Water / steam pipes Vortex shedding or shear layer instability Unsteady flow forcing Damped acoustic response Micro-perforated plate
- 5. Stochastic ApS Silencing the whistling boiler Hot air Water / steam pipes Vortex shedding or shear layer instability Unsteady flow forcing Damped acoustic response Micro-perforated plate acoustic velocity acoustic pressure Viscous laminar flow gives pressure differential and loss
- 6. Stochastic ApS Silencing the whistling boiler Resistivity: • Too large, pressure waves reflect and cavity just splits into two • Too small and no loss • Just right when Hot boiler How does R change with temperature and frequency? acoustic velocity acoustic pressure Viscous laminar flow gives pressure differential and loss p u R p u= R cr d 2a
- 7. Stochastic ApS Silencing the whistling boiler Resistivity: • Too large, pressure waves reflect and cavity just splits into two • Too small and no loss • Just right when Hot boiler How does R change with temperature and frequency? acoustic velocity acoustic pressure Viscous laminar flow gives pressure differential and loss p u R p u= R cr Pressure difference p kgm-1s-2 velocity w ms-1 hole size a m plate thickness d m density r kgm-3 viscosity n m2s-1 frequency w s-1 7 variables, 3 dimensions 4 dimensionless parameters a d ( ) 2 p d a wnr ( ) 2 2 p d a wr 2 a v w
- 8. Stochastic ApS Assume • Small Reynolds number • Thickness only enters in pressure drop, not in pressure gradient • Note that p is actually a Fourier amplitude and so is even in its real part (doesn’t care if frequency is positive or negative) and odd in its imaginary part (has to be real when transformed back) Silencing the whistling boiler Resistivity: • Too large, pressure waves reflect and cavity just splits into two • Too small and no loss • Just right when Hot boiler How does R change with temperature and frequency? R p u= R cr Pressure difference p kgm-1s-2 velocity w ms-1 hole size a m plate thickness d m density r kgm-3 viscosity n m2s-1 frequency w s-1 7 variables, 3 dimensions 4 dimensionless parameters a d ( ) 2 p d a wnr ( ) 2 2 p d a wr 2 a v w ( ) 2 2 , p d a a a w d w nr n = ( ) 2 2 p d a a w w nr n = ( ) 22 2 2 0 1 2 p d a a a k ik k w v v w w nr = + + +
- 9. Stochastic ApS Energy methods Statics Potential energy in a deflection = work done making deflection Need an expression for the potential energy in terms of the deflection shape (bit tedious) Approximate the deflection shape (satisfy boundary conditions) – usually polynomials or sometimes trig functions Differentiate the potential energy with respect to deflection to get the force. Equate to get the deflection Dynamics Average kinetic energy = average potential energy Can use to find resonant frequencies by approximating the deflection shape Much more useful for looking at modified systems M
- 10. Stochastic ApS Energy methods Without added mass. Kinetic energy On average Average potential energy is the same by Virial theorem With added mass Kinetic energy On average Average potential energy is unchanged. Apply virial theorem once more Dynamics Average kinetic energy = average potential energy Can use to find resonant frequencies by approximating the deflection shape Much more useful for looking at modified systems M 2 2 2 2 0 1 d sin 2 2 L u mL T m x C t t w w = = ( ) ( ), cosu x t y x tw= 2 2 4 mL T C w = 2 2 2 2 2 21 sin sin 2 2 M M M M mL T C t M t w w w w = + ( ) 2 2 4 M T CmL M w = + 2 M CmL CmL M w w = +
- 11. Stochastic ApS Symmetry methods Symmetries are transformations that leave mathematical objects unchanged (invariant) Finding critical points using symmetry The curve is unchanged by the reflection Moreover, the minimum is a fixed point of this transformation ( )2y x x= − ' 2x x= − ( )( ) ( )2 ' 2 ' 2 ' ' 2y x x x x= − − − = − min min min min min2 1x x x x x = = − =
- 12. Stochastic ApS Symmetry methods Multistaging Reduce flow noise by staging pressure drop over several stages Flow accelerated (isentropically) to high speed in the valve. Non-dimensionalize sound power with stream power Power law relation to Mach number Compressible form of Bernoulli’s equation gives and so 0p 1p 2 maxQu 2 0 max 1 1 m p u p − 0 0 1 1 1 m m p p M p p − 2 max n aW Qu M= 1 2 2 0 0 1 1 1 n mn m a p p W p p + −
- 13. Stochastic ApS Symmetry methods Multistaging Reduce flow noise by staging pressure drop over several stages Flow accelerated (isentropically) to high speed in the valve. Non-dimensionalize sound power with stream power Power law relation to Mach number Compressible form of Bernoulli’s equation gives and so 0p 1p 2 maxQu 2 0 max 1 1 m p u p − 0 0 1 1 1 m m p p M p p − 2 max n aW Qu M= 1 2 2 0 0 1 1 1 n mn m a p p W p p + − 0p 1p 2p 1Np − Np We want to minimize but we have to keep the total pressure drop fixed Can solve by multivariate calculus with Lagrange multipliers Or you can notice that both the sound power and the constraint are invariant under permutations of qi. For example, let and for all the other i. The order of the sum and product change, but not the result. The minimum is a fixed point so for all possible permutations, which is only possible if all the pressure ratios are equal. 1 1 22 0 1 mnnN m a i i i W q q − + = − 1 0 0 N i iN p p q − = = 1 Nq q = 1i iq q − = 1 2 1 2min min , , , , , ,N Nq q q q q q =