MATH Homework #1 Solution - by Ben Ong

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Bertha Mae Joseph
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1 MATH 46 - Homewok #1 Solution - by Ben Ong Deivation of the Eule Equations We pesent a fist pinciples deivation of the Eule Equations fo two-dimensional fluid flow in thee-dimensional cylndical cooodinates (, θ, z). We assume no vetical flow (u z = 0) and no vetical vaiations ( / z 0). We define u (, θ, t) and u θ (, θ, t) to be flow components in the ê and ê θ diections espectively. Consevation of mass tells us that the change of mass inside a contol volume has to equal the net flux of mass though the contol volume. The change in mass due to time vaiation in density and volume is given by ] ρ ρ(, θ, t + t) ρ(, θ, t) ( θ) z t ( θ) z (1) t Denoting the flux though sufaces with ê (, θ, z) and ê ( +, θ, z) nomal vectos as flux = ρ(, θ, t) ( u (, θ, t) t ) ( θ) z flux + = ρ( +, θ, t) ( u ( +, θ, t) t )( ( + ) θ ) z We see that the net flux in the ê is Similaly (net flux) (ρu ) t θ z () Thus and flux θ = ρ(, θ, t) ( u θ (, θ, t) t ) z flux θ+ θ = ρ(, θ + θ, t) ( u θ (, θ + θ, t) t ) z (net flux) θ (ρu θ) θ t z (3) θ (net flux) z 0 (4) Combining the fou above equations give the consevation of mass condition ρ t + 1 (ρu ) + 1 (ρu θ ) θ = 0 (5) 1

2 Newton s Law tells us that the net change in momentum is equal to the net impulse. The net change in momentum is affected by (i) momentum flux leaving though the contol volume, and though (ii) time vaiation in density and volume of the contol box. The momentum is given by the poduct of the momentum density and the volume. m(t) = ρ(, θ, t) u (, θ, t)ê (θ) + u θ (, θ, t)ê θ (θ) ]( ( θ) z ) m(t + t) = ρ(, θ, t + t) u (, θ, t + t)ê (θ) + u θ (, θ, t + t)ê θ (θ) ]( ( θ) z ) thus the contibution to the net change fom (ii) is m t { ρ ] } ( ) u ê + u θ ê θ t ( θ) z t (6) Denoting the flux though sufaces with ê (, θ, z) and ê ( +, θ, z) nomal vectos as m() = ρ(, θ, t) u (, θ, t)ê (θ) + u θ (, θ, t)ê θ (θ) ]( (u (, θ, t) t)( θ) z ) m( + ) = ρ( +, θ, t) u ( +, θ, t)ê (θ) + u θ ( +, θ, t)ê θ (θ) ] ( (u ( +, θ, t) t )( ( + ) θ ) z) Thus the net flux out of the two sufaces is appoximated by taking Similaly m( + ) m() lim 0 = m { ρu u ê + u θ ê θ ] } ( t θ z) (7) m(θ) = ρ(, θ, t) u (, θ, t)ê (θ) + u θ (, θ, t)ê θ (θ) ]( (u θ (, θ, t) t) z ) m(θ + θ) = ρ(, θ + θ, t) u (, θ + θ, t)ê (θ + θ) + u θ (, θ + θ, t)ê θ (θ + θ) ] ( (uθ (, θ + θ, t) t ) z) Fom vecto calculus, Substituting into m(θ + θ) gives ê (θ + θ) = ê (θ) + θê θ (θ) (8) ê θ (θ + θ) = ê θ (θ) θê (θ) (9) m(θ + θ) = ρ(, θ + θ, t) u (, θ + θ, t)(ê (θ) + θê θ (θ)) + u θ (, θ + θ, t)(ê θ (θ) θê (θ)) ] ( (uθ (, θ + θ, t) t ) z)

3 Thus m(θ + θ) m(θ) t z = ρ(, θ + θ, t)u (, θ + θ, t)u θ (, θ + θ, t) ρ(, θ, t)u (, θ, t)u θ ((, θ, t) ] ê + ρ(, θ + θ, t)u θ (, θ + θ, t)u θ (, θ + θ, t) ρ(, θ, t)u θ (, θ, t)u θ ((, θ, t) ] ê θ + ρ(, θ + θ, t)u (, θ + θ, t)u θ (, θ + θ, t) ] θê θ ρ(, θ + θ, t)u θ (, θ + θ, t)u θ (, θ + θ, t) ] θê m(θ+ θ) m(θ) Dividing both sides by θ and taking lim θ 0 gives θ { m θ ρuθ (u ê + u θ ê θ ) } ] + ρu θ (u ê θ u θ ê ) t z θ (10) θ Combining equations (6), (7) and (10) gives the net change in momentum t {ρu } + {ρu u } + ] θ {ρu u θ } ρu θ ( θ z t)ê + t {ρu θ} + {ρu θu } + θ ] {ρu θu θ } + ρu u θ ( θ z t)ê θ (11) If you expand the deivatives, and impose consevation of mass (equation (5)), the net change in momentum simplifies to (djm: the two exta tems come fom the CV geomety (8) & (9)) ρ u t u + ρu + 1 ρu θ + ρ u θ t + ρu u θ ρu θ ] ( θ z t)ê u θ + 1 ρu u θ θ θ + ρu u θ ] ( θ z t)ê θ (1) We now need to calculate the net impulse on the system in ode to deive the emaining Eule Equations. Thee ae two contibutions; (i) Fom a given body foce density F (, θ, t) = F (, θ, t)ê + F θ (, θ, t)ê θ + F z (, θ, t)ê z and (ii) fom the intenal pessue. The impulse contibution fom (i) is simply the foce density*volume* t. F (, θ, t) = (F (, θ, t)ê + F θ (, θ, t)ê θ + F z (, θ, t)ê z )( θ z) t (13) 3

4 The foce = (pessue*aea) exeted on the sufaces with ê (, θ, z) and ê ( +, θ, z) nomal vectos is F P = P (, θ, t) ( θ) z ] ê (θ) F + P = P ( +, θ, t) ( + ) θ z ] ê (θ) Thus, the net foce contibution fom the two sufaces Similaly Using equation (9) gives {F P } {P } θ z]ê (θ) P ] + P θ z]ê (θ) (14) F θ P = P (, θ, t) z ] ê θ (θ) F θ+ θ P = P (, θ + θ, t) z ] ê θ (θ + θ) F θ+ θ P = P (, θ + θ, t) z ] (ê θ (θ) θê (θ)) Thus, the contibution fom the two sufaces gives ] P {F P } θ θ êθ(θ) P ê (θ) z θ (15) And the contibution fom sufaces with ê z as nomals is {F P } z 0 (16) Thus the net impulse is (djm: note the amazing cancellation of the non-gadient P tem) F ê + F θ ê θ + F z ê z + P ê + 1 P ] θ êθ θ z t (17) Combining equations (1) and (17) gives us Newton s Law u (ê ) t + u u + u θ u θ u θ = 1 P ] ρ + F (18) u θ (ê θ ) t + u u θ + u θ u θ θ + u u θ = 1 1 ] P ρ θ + F θ (19) (ê z ) 0 = F z (imposes condition on body foce) (0) 4

5 Rotational Flows Given u = ( Ωy, Ωx, 0), find the pessue which poduces a flow solution to the incompessible Eule Equations with F = ρgẑ Incompessible Eule Equations u = 0 (1) u t + ( u ) u P = We notice that equation (1) is automatically satisfied. Expanding () Substituting u = ( Ωy, Ωx, 0), we get Solving (3) gives u t + (uu x + vu y + wu z ) = P x v t + (uv x + vv y + wv z ) = P y w t + (uw x + vw y + ww z ) = P z g + F () Ω x = P x (3) Ω y = P y (4) P z = g (5) P = Ω x + f(y, z) (6) Diffeentiating (6) and compaing with (4) gives Solving (7) gives P y = f (y, z) = Ω y (7) P = Ω (x + y ) + g(z) (8) Diffeentiating (8) and compaing with (5) and solving gives P = Ω (x + y ) gz + constant (9) 5

6 djm:] Since the fluid has voticity, u = (0, 0, Ω) 0, the Benoulli theoem fo iotational flow does not apply. Howeve, the flow is steady and so the Benouilli theoem fo steamlines does apply, but the Benouilli function can have diffeent constant values on diffeent steamlines (cicles aound the axis of otation) and cannot be used to infe the suface geomety. An astonome could make a liquid mio telescope by spinning mecuy on a a paabolic suface. Actually, this is being done at a UBC eseach station in maple Ridge; they have a 6m diamete mio! They need the coect angula velocity to get a unifom coating of the mecuy 6

Classical Mechanics Homework set 7, due Nov 8th: Solutions

Classical Mechanics Homework set 7, due Nov 8th: Solutions Classical Mechanics Homewok set 7, due Nov 8th: Solutions 1. Do deivation 8.. It has been asked what effect does a total deivative as a function of q i, t have on the Hamiltonian. Thus, lets us begin with