= Unit outward normal to the interface contained within the cross section ( to
|
|
- Ὑπατια Λιάπης
- 7 χρόνια πριν
- Προβολές:
Transcript
1 Two-Phase Flow Equations n z c a n n c Note: Walls are drawn such that n z = Unit vector in the flow direction n = Unit outward normal vector to the phase surface A x = constant ut this doesn t have to hold true n c = Unit outward normal to the interface contained within the cross section ( to z n ) 1
2 Transverse View a = Cross sectional area of phase c = Contour etween phases i c = Contour etween phase and the wall Limitin forms of Leinitz Rule F dc 1) FrtdS (, ) = ds Fv i n t t n n and Gauss s Theorem a a c c ) dc BdS = B n ds B n z n n z a a c i c c c Similar to that shown for sinle phase flow a n n B = n = = dc z 3) z z c n nc Bein with the eneral alance equation written for phase 4) ( v ) t ψ ψ = ψ ψ s Interate the eneral alance equation over the area a 5) ds ( v) ds sds ds t ψ ψ = ψ ψ a a a a and apply Gauss s Theorem and Leinitz Rule
3 6) dc ψds ψv nzds ψ( v vi) n = t z n n a a ci c dc ψ nds ψ n ψ ds z n n s z s c a ci c ac c Note: ( v v dc i) n ψ = 0 such that n n c c 7) dc ψds ψv nzds ψ( v vi) n = t z n n a a ci dc ψ nds ψ n ψ ds z n n s z s c a ci c ac c Mass Conservation ψ = ψ s = 0 ψ = 0 dc 1) ds v nzds ( v vi) n = 0 t z n n a a ci c Also a a1 a a Under conditions such as su-cooled nucleate oilin, we could have a situation such that at some interfaces we would have evaporation and at others condensation. To model such a situation, we would need mass, enery and momentum equations for each species of a phase, as well as terms to descrie how one species transitioned to the other, i.e. a 1 a 3
4 Define Γ ( v vi) n which represents mass exchane at the interface due to evaporation and condensation dc ) ds v nzds Γ = 0 t z n n a a ci In terms of our averain notation dc 3) < > a < vz > a Γ = 0 t z n n Let a =< α > A x ci c c δ ci Γ dc n n c 4) Ax < > < α > < v > < α > Ax δ = 0 t z Momentum Conservation ψ v ψ ψ = = s = T = ( PI σ ) 5) n vds vv nds z v( v vi) dc= t z c n i nc a a n n PI nzds PI dc σ nzds σ dc ds z n n z n n c c a ci c a ci c a Proect alon the flow direction y dottin aainst the unit vector n z 6) n vds vvds v( v vi) dc= t z c n i nc a a n nz n n PdS P dc σzzds σ dc zds z n n z n n c c a ci c a ci c a 4
5 7) Let: ci c c ci dc Ax < v > < α > < vv > < α > Ax Γ v = t z n nc ci nz n < P >< α > Ax P dc < σzz >< α > Ax z n nc z c n n n n σ dc σ dc< > A < α > c z z x z n nc n nc ci Γ dc v vˆ n n =δ c n n n n a P dc =< P > dc = < P > = < P > < α > A z z x n nc n nc z z c n n σ dc = < τ > P z w 1 w n nc n n σ dc =< τ > P z i 1 i n nc where P i is the interfacial perimeter. Sustitutin into the area averaed momentum equation ives A ˆ x < v > < α > < vv > < α > Ax δ v = t z < α > A < P > < P > < P > < α > A < σ > < α > A z z z < τ > P < τ > P< > A < α > 8) { } x 1 x zz x w 1 w i 1 i x z Internal Enery 9) ( u ) ( uv) = P v q t Aain, the internal enery equation does not satisfy the eneral alance equation, ut we can still employ our area averain techniques. Interate over the area a, applyin Leinitz Rule and Gauss s Theorem 5
6 n 10) uds ( uv ) nds z ( u)( v vi) dc t z n nc a a ci = P v ds q nzds n n q dc q dc z n n n n a c c a c ci 11) dc Ax < u > < α > < uv > < α > Ax Γ v = t z n n P v ds q P q P a w w i i ci c Let a P v ds =< P > v ds a Note, the weihtin is different thus the pressure is different than the other area averaed pressure. n =< > =< > dc n =< P > v nzds ( v vi) n vi dc z n n c n n c a c c d < P > dc n =< P > < v > < α > Ax Γ < P > vi dc dz n n n n P vds P vds P v nzds v dc z n nc a a a c ci c c c Recall Leinitz Rule vi n FrtdS (, ) FrtdS (, ) Frt (, ) dc t = t n n a a c v n < P > dc = < P > ds < P > ds i n nc t t c a a = < P > a a < P > t t c a t =< P > 6
7 Such that the internal enery equation is 1) A u α u v α A u < P > x < >< > < > < > x Γ = t z n nc c < P > < v > < α > Ax < P > < α > Ax q wpw q ipi z t dc Jump Conditions Jump conditions provide the couplin of the phasic equations across the interface. The eneral form of the ump condition is v v n = v v n 13) [ ψ ( ) ψ ] [ ψ ( ) ψ ] a a i sa a i s Continuity (Mass) ψ = ψ = 0 s ( v v ) n = ( v v ) n a a i a i Γ = Γ a Note: na nac = n nc c dc dc Γ a = Γ n n n n δ = δ a a ac c c 7
8 Momentum ψ = v ψ = PI σ s v ( v v ) PI σ n = v ( v v ) PI σ n a a a i a a a i Proect alon flow direction y dottin aainst n z ava( va vi) PI a nz σa n z n a= v( v vi) PI nz σ n z n Γ v Pn n σ n n = Γ v Pn n σ n n a a a z a a z a z a z dc dc Γ v Pn n σ n n = Γ v Pn n σ n n a a a z a a z a a a z z na nac n nc dc dc dc dc δ ˆ ˆ ava Pn a z na σanz na = δ v Pn z n σnz n n n n n n n n n a ac c ac c c ci ci ci ci ci Assumin that pressure is continuous on the interface δ < > z a ˆ ava τia 1 Pi Pa n a n ac ci δ vˆ < τ > P = δ vˆ < τ > P a a ia 1 i i 1 i n n nz n dc = δ ˆ v< τi> 1 Pi P n n c i a ac dc Total Enery ( ) ( ) e ( v v ) q PI σ v n = e ( v v ) q PI σ v n a a a i a a a a a i Note v v e = u 8
9 a aua( va vi) va va( va vi) q a Pa( va vi) Pv a i σa va na = u ( v v ) v v ( v v ) q P( v v ) Pv σ v n i i i i a aha( va vi) va va( va vi) q a Pv a i σa va na = h ( v v ) v v ( v v ) q Pv σ v n i i i Assume the inetic enery and viscous terms alance a v v ( v v ) σ v = v v ( v v ) σ v such that a a a i a a i h v v q Pv n [ ( ) ] a a a i a a i a = h v v q Pv n [ ( ) ] i i Multiply oth sides y dc n n and interate over the interfacial contour c dc dc dc Γ h q n Pv n = a a a a a i a na nac na nac na nac ci ci ci dc dc dc Γ h q n Pv n i n nc n nc n nc ci ci ci Assume pressure is continuous on the interface, such that ˆ dc δ aha q iapi Pv a i na n n ci a ac ˆ dc = δ h q ipi Pv i n n n ci c δ hˆ q P = δ hˆ q P a a ia i i i 9
10 Two Phase Equation Summary Apply the area averaed two-phase equations to a liquid vapor system. We will aain assume all averae pressures are the same, the averae of products is the product of averaes and we can drop the averae notation. Liquid Mass A x α α v Ax = δ t z Vapor Mass A α α v A = δ t z x x Liquid Internal Enery ˆ Ax αu αu v Ax δ h P αv Ax P αax q w Pw q = ipi t z z t Vapor Internal Enery ˆ Ax α u α uvax δ h = P αvax P αax q wpw q ipi t z z t Liquid Momentum A ( α v ) ( α v v ) A = vˆ A P δ α τ P τ P αa t z z x x x w w i i x z Vapor Momentum P A ( α v ) ( α v v ) A = δ vˆ α A τ P τ P α A t z z x x x w w i i x z Jump Conditions δ = δ δ hˆ q P = δ hˆ q P i i i i δ vˆ τ P = δ vˆ τ P i i i i 10
11 The area averaed two-phase equations constitute the Six Equation model for modelin two phase systems as it contains six conservation equations, one for each phase. Assumin the interfacial and wall interaction terms can e expressed in terms of the followin fundamental variales, we have as unnowns Fundamental Variales, u, v, P, α, T 1 Equations: Conservation Equations 6 State Equations: = ( u, P) T = T ( u, P ) Volume Constraint α = 1 1 An additional equation is require that relates the phase pressures P = P( P ) 1 which closes the system. A common assumption is that the phase pressures are equal, i.e. P = P = P which leads to the six equation, sinle pressure model that is the asis for most desin codes. Liquid Mass A x α α v Ax = δ t z Vapor Mass A α α v A = δ t z x x Liquid Internal Enery ˆ Ax αu αuvax δ h P αvax P αax q w Pw qp = i i t z z t 11
12 Vapor Internal Enery ˆ Ax α u α uvax δ h = P αvax P αax q wpw q ipi t z z t Liquid Momentum A ( α v ) ( α v v ) A = δ vˆ α A P τ P τ P αa t z z x x x w w i i x z Vapor Momentum A ( α v ) ( α v v ) A = δ vˆ α A P τ P τ P α A t z z x x x w w i i x z Jump Conditions δ = δ δ hˆ q P = δ hˆ q P i i i i δ vˆ τ P = δ vˆ τ P i i i i 1
13 Mixture Equations The Mixture Equations are a convenient form of the two-phase equations which are otained y addin the phasic equations. The Mixture Equations provide loal alances on mass, enery and momentum as opposed to alances on each phase. Mixture Mass Ax ( α α ) ( αv α v) Ax δ δ = 0 t z 0 Mixture Momentum P A ( α v αv ) ( αv v α vv) A = A τ P τ P ( α α) A t z z δ vˆ τ ˆ i Pi δ v τipi x x x w w wl wl x z 0 Mixture Internal Enery A ( α u α u ) ( α u v α u v ) A P ( α v α v ) A q = P t z z q P δ hˆ q ˆ i Pi δ h q ipi x x x w w wl wl 0 Mixture Variale Definitions α α v αv αv = G u αu αu v v τ P τ P τ P w w w w wl wl qp w w = q wpw q w Pw The mixture equations then simplify to 13
14 Mixture Mass Ax vax = 0 t z Mixture Momentum A v ( α v v α v v ) A A P = τ P A t z z x x x w w x z Mixture Internal Enery A u ( α u v α u v ) A P ( α v α v ) A q = P t z z x x x w w The convective terms in the momentum and internal enery equations are all of the form αvφ αvφ. We wish to express these convective terms as mixture quantities plus correction terms, i.e. α vφ α v φ = φ v Correction Terms Define φ α φ α φ Note: This is already consistent with our definitions of u and v aove. φ v= v α φ α φ v φ v = α φ αφ αv αv φ v = [ αφ αφ] α α φ v = [ αφv αφ v ] [ αφv αφ v] α α φ v = [ α φ v α φv α φ( v v )] [ α φ v α φv α φ ( v v )] 14
15 φ v = ( α α ) αα αα [ αφv αφ ] ( ) v φ v v φ( v v ) αα φ v= αφv α φv ( φ φ)( v v ) Internal Enery Equation a) Convective Enery Term αvφ αvφ = αvu αvu φ = u φ = u φ = u : ) Pressure Wor Term αvφ αvφ = αv αv φ = υ φ = υ φ = υ : d αα A ( u) ( uva ) = P ( va ) q P ( u u)( v v) A t z dz z x x x w w x αα P ( υ υ )( v v) A z x Momentum Equation Momentum Flux Term αvφ αvφ = αvv αvv φ = v φ = v φ = v : P αα A ( v) ( vva ) = A τ P A ( v v ) A t z z z x x x w w x z x The Mixture Equations are not linearly independent from the phasic equations, and as such can e used to replace a phasic equation, ut can not e used in addition to the phasic equations. Since the mixture equations reduce to the sinle phase equations for α = 0, they are convenient for handlin phase appearance and disappearance. Cominations of Mixture Equations and phasic equations can e produce various simplifications to the Six Equation Models. 15
16 5 Equation Models Five Equation Models are composed of 3 mixture and phasic equations. Two common Five Equation models are: a) Phasic Equations = Mass Momentum Since we only have the mixture enery equation, information reardin enery distriution amon the phases has een lost. Assume least massive phase to e at saturation (could e liquid or vapor u ( P, T) = u ( P) and ( Pu, ) = ( P) s ) Phasic Equations = Mass Enery s Since we have only the mixture momentum equation, information reardin relative phase velocity has een lost. The eneral approach it to correlate the relative velocity in terms of the other system variales, i.e. vr = v v = vr( α,, u, P) Correlation Drift Flux Models are a common example of this approach. Four Equation Models Four Equation Models are composed of three mixture equations and one phasic equation. The most common of the four equation models consist of the three mixture equations and the liquid phase mass equation. The liquid phase mass equation is A x α α v Ax = δ t z Since liquid phase velocity alone does not appear in the mixture equations, we can eliminate liquid phase velocity in favor of mixture velocity and relative velocity y notin v α v v = = α v α v α v α v α v v = v = ( α α ) v α ( v v ) 16
17 v v α = ( v v ) Sustitutin into the liquid phase mass equation ives αα A α αva = δ ( v v) A t z z x x x which is of the same eneral form as the mixture equations. As we have only the mixture momentum equation, a correlation is required for relative velocity. In addition, since we have only the mixture enery equation, the vapor phase is usually taen to e at saturation. The four equation model does allow for the liquid phase to e sucooled. Three Equation Models Three equation models are ased solely on the mixture equations. A correlation is required for relative velocity and when two phases are present, they are oth assumed to e at saturation. In either the 3, 4 or 5 equation models homoeneous flow is otained y simply settin v r = 0. 17
18 Drift Flux Models The Drift Flux is defined in terms of the relative velocity as We can also define a Drift Velocity such that = α α ( v v ) = α α v r V = α ( v v ) = α v r For ravity dominated flows in the asence of wall shear, an equation which has een found to correlate a wide variety of data ives n = αα v where v is the terminal rise velocity of a sinle ule in an infinite fluid. The relative velocity, Drift Velocity and Drift Flux are then all proportional to the terminal rise velocity. One model for terminal rise velocity can e otained from the followin force alances. Buoyancy Force: V ( ) Surface Tension: Dra Force: Pσ C A d v Balancin the Buoyancy and Surface Tension forces ives V ( ) = Pσ V P = σ d ( ) ˆ σ ( ) dˆ Balancin the uoyancy and dra forces ives v V( ) = CdA 18
19 19 and solvin for v 1 ˆ ( ) d Cd V v C A = ( ) ( ) ( ) d d v C C C C σ σ = = 1/4 ( ) v C σ =
20 Numerical Solution of the Three Equation Model Numerical solution of the two phase equations is ased on the assumption of Gloal Compressiility, i.e. the spatial pressure distriution is unimportant and system parameters can e evaluated at the loal system pressure. This eliminates the need for a spatially discretized momentum equation. Examine the open channel illustrated aove. Fluid enters sucooled and can leave sucooled, a two phase mixture or superheated. We assume the 3 equation model is valid where the relative velocity vr = v v is availale y an appropriate correlation. Consistent with the three equation model, oth phases are at equilirium when two phases are present. The oundary conditions are inlet velocity, pressure, density and internal enery. We also assume a now exit pressure. Equations: Mixture Mass A x x = t z ( uva ) 0 Mixture Internal Enery d αα A ( u) ( uva ) = P ( va ) q P ( u u)( v v) A t z dz z αα P ( υ υ )( v v) Ax z x x x w w x 0
21 Interate the mass and enery equations over a node centered at and ounded y ± 1/ Mass d V ( vax) 1/ ( vax) 1/ 0 dt = Internal Enery d V ( u) ( uvax) 1/ ( uvax) 1/ = P[( vax) 1/ ( vax) 1/ ] q dt αα αα υ υ 1/ 1/ ( u u)( v v) Ax P ( )( v v) Ax 1/ 1/ S The term S contains the two phase correction terms and is zero for sinle phase flow. Semi-Implicit Time Discretization W assume a Semi-Implicit time discretization, where velocities are evaluated at new time, and convected properties are evaluated at old time Mass V v A v A Δt tδt t t tδ t t tδt 1/ 1/ 1/ 1/ 1/ 1/ = 0 Internal Enery ( u) ( u) V u v A u v A P v A v A q S Δt tδt t t tδ t t tδ t t tδ t tδt t t [( ) x] 1/ [( ) x] 1/ = 1/ x 1/ 1/ x 1/ Consistent with our treatment of convected properties in the sinle phase equations, we assume any oundary valued property Ψ can e represented in terms of the cell centered properties y v Ψ = Ψ Ψ Ψ Ψ tδt t 1 t t 1 t t 1/ 1 tδt 1 v 1/ 1/ { } { } 1
22 State Equations a) Sinle Phase = ( u, P) ( u) = ( u, P) u ) Two Phase = α α = ( P) α [ ( P) ( P)] f f ( u) = αu αu = f ( Pu ) f ( P) α f ( Pu ) ( P) f ( Pu ) f ( P) These state equations can e expressed in eneral as = ψ (, ) P ( u) = u( ψ, P) For Sinle Phase ψ = u and Two Phase ψ = α The Mass and Internal Enery Equations are linear in the new time values. The equations are nonlinear as a result of the state equations. Linearize the equations via the Newton-Raphson technique Mass V v A v A Δt 1 t t 1 t 1 1/ 1/ 1/ 1/ 1/ 1/ = 0 Internal Enery V u v A u v A P v A v A q S Δt 1 t ( u) ( u) t 1 t 1 t 1 1 t t [( ) x] 1/ [( ) x] 1/ = 1/ x 1/ 1/ x 1/
23 State = ( ψ, P ) ( ψ ψ ) ( P P ) P ψ u u u = u( ψ, P ) ( ψ ψ ) ( P P ) P ψ u where the state equation derivatives are a) Sinle Phase = ( up, ) ψ = u = ψ u P = P P u u = ( u, P) u u = ( up, ) u ψ u P u P = u P u ) Two Phase ( α, P) = ( P) α [ ( P) ( P)] ψ α f f = ψ = f f = α α P P P 3
24 u( α, P) = ( P) u ( P) α [ ( P) u ( P) ( P) u ( P)] f f f f u = ( Pu ) ( P) f ( Pu ) f ( P) ψ u f uf u = α uf f α u P P P P P The linearized equations can e written in the form Mass 1 t 1 t 1 t a1v 1/ a v 1/ = S1 Internal Enery ( ) t t t 1 1/ 1/ = u v v S State i) ii) 1 = δψ δp ψ P 1 ( ) u u u = u δψ δp ψ P Sustitute i) into the Mass Equation, and ii) into the Internal Enery Equation p t 1 t 1 t P a1v 1/ a v 1/ = S1 ψ P δψ δ u u t 1 t 1 t δψ δp 1v 1/ v 1/ = S u ψ P Divide the Mass Equation y eliminateδψ, divide the Internal Enery Equation y ψ u ψ and sutract to 4
25 iii) av v cδ P= ξ 1 1 1/ 1/ For J nodes, the numer of unnowns are J velocities for inlet velocity nown and P (or δ P ). We have the J equations implied y iii) aove. One more equation is required. We assume a simplified Momentum Boundary Condition of the form ( v ) P Pexit K Δ = t t t J 1/ J 1/ which can e linearized to ive P P K ( v v ) v t 1 J 1/ 1 = exit J 1/ J 1/ J 1/ δ P= P P = dv C 1 1 J 1/ J The linear system of equations may e written in matrix form, where the matrix structure is illustrated elow where vn = v n 1/ a1 c1 v1 ξ1 v1/ a c v ξ 3 a3 c 3 v ξ 3 3 = d 1 δ P C J While not tridiaonal, solution alorithms can e developed for this structure that are similar to the Thomas Alorithm for tridiaonal systems. 5
26 Recirculatin Systems The flow paths within U-Tue steam enerators and Boilin Water Reactors may e approximated as the recirculatin system illustrated elow. For the one dimensional sements contained etween the inlet manifold (node 1) and the exit manifold (node J) the mass and internal enery equations are of the same form as the previous open channel prolem av v δ P= ξ [, J 1] 1/ 1/ For the exit manifold, the mass and internal enery equations reduce to av v cv δ P= ξ J exit J 1 J J and for the inlet manifold (node 1) av v cv δ P= ξ 3 1 In addition, we have the momentum oundary condition at the steam outlet d v δ P= C ex exit J Assumin the velocity v is nown, the unnowns are: v 3, v ( [, J 1] ), v exit, v, P δ J 6
27 The numer of equation is J1, the last equation is an interated momentum equation around the entire loop. The mixture momentum equation (valid for oth sinle and two phase conditions) is 1 P f v v v [ vva ] = φ δ( z z ) ψ t A z z D x or in non conservative form x 0 e 1 αα sin θ ( v v ) Ax Ax z v v P f v v v = φ δ( z z ψ t z z D e 0 ) 1 αα sin θ ( ) v v Ax Ax z z (interate from center of one node to center of the next) z 1 Results in an equation for the new iterate values of velocity of the form B v 1/ = e which closes the system of equations. 7
2 Composition. Invertible Mappings
Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan Composition. Invertible Mappings In this section we discuss two procedures for creating new mappings from old ones, namely,
Διαβάστε περισσότεραCHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS
CHAPTER 5 SOLVING EQUATIONS BY ITERATIVE METHODS EXERCISE 104 Page 8 1. Find the positive root of the equation x + 3x 5 = 0, correct to 3 significant figures, using the method of bisection. Let f(x) =
Διαβάστε περισσότεραPartial Differential Equations in Biology The boundary element method. March 26, 2013
The boundary element method March 26, 203 Introduction and notation The problem: u = f in D R d u = ϕ in Γ D u n = g on Γ N, where D = Γ D Γ N, Γ D Γ N = (possibly, Γ D = [Neumann problem] or Γ N = [Dirichlet
Διαβάστε περισσότεραEvery set of first-order formulas is equivalent to an independent set
Every set of first-order formulas is equivalent to an independent set May 6, 2008 Abstract A set of first-order formulas, whatever the cardinality of the set of symbols, is equivalent to an independent
Διαβάστε περισσότεραHomework 8 Model Solution Section
MATH 004 Homework Solution Homework 8 Model Solution Section 14.5 14.6. 14.5. Use the Chain Rule to find dz where z cosx + 4y), x 5t 4, y 1 t. dz dx + dy y sinx + 4y)0t + 4) sinx + 4y) 1t ) 0t + 4t ) sinx
Διαβάστε περισσότεραderivation of the Laplacian from rectangular to spherical coordinates
derivation of the Laplacian from rectangular to spherical coordinates swapnizzle 03-03- :5:43 We begin by recognizing the familiar conversion from rectangular to spherical coordinates (note that φ is used
Διαβάστε περισσότεραSection 8.3 Trigonometric Equations
99 Section 8. Trigonometric Equations Objective 1: Solve Equations Involving One Trigonometric Function. In this section and the next, we will exple how to solving equations involving trigonometric functions.
Διαβάστε περισσότερα[1] P Q. Fig. 3.1
1 (a) Define resistance....... [1] (b) The smallest conductor within a computer processing chip can be represented as a rectangular block that is one atom high, four atoms wide and twenty atoms long. One
Διαβάστε περισσότεραPhys460.nb Solution for the t-dependent Schrodinger s equation How did we find the solution? (not required)
Phys460.nb 81 ψ n (t) is still the (same) eigenstate of H But for tdependent H. The answer is NO. 5.5.5. Solution for the tdependent Schrodinger s equation If we assume that at time t 0, the electron starts
Διαβάστε περισσότεραExample Sheet 3 Solutions
Example Sheet 3 Solutions. i Regular Sturm-Liouville. ii Singular Sturm-Liouville mixed boundary conditions. iii Not Sturm-Liouville ODE is not in Sturm-Liouville form. iv Regular Sturm-Liouville note
Διαβάστε περισσότεραLecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3
Lecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3 1 State vector space and the dual space Space of wavefunctions The space of wavefunctions is the set of all
Διαβάστε περισσότεραSrednicki Chapter 55
Srednicki Chapter 55 QFT Problems & Solutions A. George August 3, 03 Srednicki 55.. Use equations 55.3-55.0 and A i, A j ] = Π i, Π j ] = 0 (at equal times) to verify equations 55.-55.3. This is our third
Διαβάστε περισσότεραChapter 6: Systems of Linear Differential. be continuous functions on the interval
Chapter 6: Systems of Linear Differential Equations Let a (t), a 2 (t),..., a nn (t), b (t), b 2 (t),..., b n (t) be continuous functions on the interval I. The system of n first-order differential equations
Διαβάστε περισσότερα3.4 SUM AND DIFFERENCE FORMULAS. NOTE: cos(α+β) cos α + cos β cos(α-β) cos α -cos β
3.4 SUM AND DIFFERENCE FORMULAS Page Theorem cos(αβ cos α cos β -sin α cos(α-β cos α cos β sin α NOTE: cos(αβ cos α cos β cos(α-β cos α -cos β Proof of cos(α-β cos α cos β sin α Let s use a unit circle
Διαβάστε περισσότερα( y) Partial Differential Equations
Partial Dierential Equations Linear P.D.Es. contains no owers roducts o the deendent variables / an o its derivatives can occasionall be solved. Consider eamle ( ) a (sometimes written as a ) we can integrate
Διαβάστε περισσότεραEE512: Error Control Coding
EE512: Error Control Coding Solution for Assignment on Finite Fields February 16, 2007 1. (a) Addition and Multiplication tables for GF (5) and GF (7) are shown in Tables 1 and 2. + 0 1 2 3 4 0 0 1 2 3
Διαβάστε περισσότεραHOMEWORK 4 = G. In order to plot the stress versus the stretch we define a normalized stretch:
HOMEWORK 4 Problem a For the fast loading case, we want to derive the relationship between P zz and λ z. We know that the nominal stress is expressed as: P zz = ψ λ z where λ z = λ λ z. Therefore, applying
Διαβάστε περισσότεραParametrized Surfaces
Parametrized Surfaces Recall from our unit on vector-valued functions at the beginning of the semester that an R 3 -valued function c(t) in one parameter is a mapping of the form c : I R 3 where I is some
Διαβάστε περισσότεραwave energy Superposition of linear plane progressive waves Marine Hydrodynamics Lecture Oblique Plane Waves:
3.0 Marine Hydrodynamics, Fall 004 Lecture 0 Copyriht c 004 MIT - Department of Ocean Enineerin, All rihts reserved. 3.0 - Marine Hydrodynamics Lecture 0 Free-surface waves: wave enery linear superposition,
Διαβάστε περισσότερα6.3 Forecasting ARMA processes
122 CHAPTER 6. ARMA MODELS 6.3 Forecasting ARMA processes The purpose of forecasting is to predict future values of a TS based on the data collected to the present. In this section we will discuss a linear
Διαβάστε περισσότεραΑπόκριση σε Μοναδιαία Ωστική Δύναμη (Unit Impulse) Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο. Απόστολος Σ.
Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο The time integral of a force is referred to as impulse, is determined by and is obtained from: Newton s 2 nd Law of motion states that the action
Διαβάστε περισσότεραHomework 3 Solutions
Homework 3 Solutions Igor Yanovsky (Math 151A TA) Problem 1: Compute the absolute error and relative error in approximations of p by p. (Use calculator!) a) p π, p 22/7; b) p π, p 3.141. Solution: For
Διαβάστε περισσότεραOther Test Constructions: Likelihood Ratio & Bayes Tests
Other Test Constructions: Likelihood Ratio & Bayes Tests Side-Note: So far we have seen a few approaches for creating tests such as Neyman-Pearson Lemma ( most powerful tests of H 0 : θ = θ 0 vs H 1 :
Διαβάστε περισσότεραReminders: linear functions
Reminders: linear functions Let U and V be vector spaces over the same field F. Definition A function f : U V is linear if for every u 1, u 2 U, f (u 1 + u 2 ) = f (u 1 ) + f (u 2 ), and for every u U
Διαβάστε περισσότεραFinite Field Problems: Solutions
Finite Field Problems: Solutions 1. Let f = x 2 +1 Z 11 [x] and let F = Z 11 [x]/(f), a field. Let Solution: F =11 2 = 121, so F = 121 1 = 120. The possible orders are the divisors of 120. Solution: The
Διαβάστε περισσότεραFourier Series. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics
Fourier Series MATH 211, Calculus II J. Robert Buchanan Department of Mathematics Spring 2018 Introduction Not all functions can be represented by Taylor series. f (k) (c) A Taylor series f (x) = (x c)
Διαβάστε περισσότεραJesse Maassen and Mark Lundstrom Purdue University November 25, 2013
Notes on Average Scattering imes and Hall Factors Jesse Maassen and Mar Lundstrom Purdue University November 5, 13 I. Introduction 1 II. Solution of the BE 1 III. Exercises: Woring out average scattering
Διαβάστε περισσότεραES440/ES911: CFD. Chapter 5. Solution of Linear Equation Systems
ES440/ES911: CFD Chapter 5. Solution of Linear Equation Systems Dr Yongmann M. Chung http://www.eng.warwick.ac.uk/staff/ymc/es440.html Y.M.Chung@warwick.ac.uk School of Engineering & Centre for Scientific
Διαβάστε περισσότεραDESIGN OF MACHINERY SOLUTION MANUAL h in h 4 0.
DESIGN OF MACHINERY SOLUTION MANUAL -7-1! PROBLEM -7 Statement: Design a double-dwell cam to move a follower from to 25 6, dwell for 12, fall 25 and dwell for the remader The total cycle must take 4 sec
Διαβάστε περισσότεραAreas and Lengths in Polar Coordinates
Kiryl Tsishchanka Areas and Lengths in Polar Coordinates In this section we develop the formula for the area of a region whose boundary is given by a polar equation. We need to use the formula for the
Διαβάστε περισσότεραProblem Set 9 Solutions. θ + 1. θ 2 + cotθ ( ) sinθ e iφ is an eigenfunction of the ˆ L 2 operator. / θ 2. φ 2. sin 2 θ φ 2. ( ) = e iφ. = e iφ cosθ.
Chemistry 362 Dr Jean M Standard Problem Set 9 Solutions The ˆ L 2 operator is defined as Verify that the angular wavefunction Y θ,φ) Also verify that the eigenvalue is given by 2! 2 & L ˆ 2! 2 2 θ 2 +
Διαβάστε περισσότεραApproximation of distance between locations on earth given by latitude and longitude
Approximation of distance between locations on earth given by latitude and longitude Jan Behrens 2012-12-31 In this paper we shall provide a method to approximate distances between two points on earth
Διαβάστε περισσότεραD Alembert s Solution to the Wave Equation
D Alembert s Solution to the Wave Equation MATH 467 Partial Differential Equations J. Robert Buchanan Department of Mathematics Fall 2018 Objectives In this lesson we will learn: a change of variable technique
Διαβάστε περισσότεραAreas and Lengths in Polar Coordinates
Kiryl Tsishchanka Areas and Lengths in Polar Coordinates In this section we develop the formula for the area of a region whose boundary is given by a polar equation. We need to use the formula for the
Διαβάστε περισσότεραSolutions to Exercise Sheet 5
Solutions to Eercise Sheet 5 jacques@ucsd.edu. Let X and Y be random variables with joint pdf f(, y) = 3y( + y) where and y. Determine each of the following probabilities. Solutions. a. P (X ). b. P (X
Διαβάστε περισσότεραTMA4115 Matematikk 3
TMA4115 Matematikk 3 Andrew Stacey Norges Teknisk-Naturvitenskapelige Universitet Trondheim Spring 2010 Lecture 12: Mathematics Marvellous Matrices Andrew Stacey Norges Teknisk-Naturvitenskapelige Universitet
Διαβάστε περισσότεραC.S. 430 Assignment 6, Sample Solutions
C.S. 430 Assignment 6, Sample Solutions Paul Liu November 15, 2007 Note that these are sample solutions only; in many cases there were many acceptable answers. 1 Reynolds Problem 10.1 1.1 Normal-order
Διαβάστε περισσότερα1 String with massive end-points
1 String with massive end-points Πρόβλημα 5.11:Θεωρείστε μια χορδή μήκους, τάσης T, με δύο σημειακά σωματίδια στα άκρα της, το ένα μάζας m, και το άλλο μάζας m. α) Μελετώντας την κίνηση των άκρων βρείτε
Διαβάστε περισσότεραNumerical Analysis FMN011
Numerical Analysis FMN011 Carmen Arévalo Lund University carmen@maths.lth.se Lecture 12 Periodic data A function g has period P if g(x + P ) = g(x) Model: Trigonometric polynomial of order M T M (x) =
Διαβάστε περισσότερα4.6 Autoregressive Moving Average Model ARMA(1,1)
84 CHAPTER 4. STATIONARY TS MODELS 4.6 Autoregressive Moving Average Model ARMA(,) This section is an introduction to a wide class of models ARMA(p,q) which we will consider in more detail later in this
Διαβάστε περισσότεραOrdinal Arithmetic: Addition, Multiplication, Exponentiation and Limit
Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit Ting Zhang Stanford May 11, 2001 Stanford, 5/11/2001 1 Outline Ordinal Classification Ordinal Addition Ordinal Multiplication Ordinal
Διαβάστε περισσότεραInstruction Execution Times
1 C Execution Times InThisAppendix... Introduction DL330 Execution Times DL330P Execution Times DL340 Execution Times C-2 Execution Times Introduction Data Registers This appendix contains several tables
Διαβάστε περισσότεραANSWERSHEET (TOPIC = DIFFERENTIAL CALCULUS) COLLECTION #2. h 0 h h 0 h h 0 ( ) g k = g 0 + g 1 + g g 2009 =?
Teko Classes IITJEE/AIEEE Maths by SUHAAG SIR, Bhopal, Ph (0755) 3 00 000 www.tekoclasses.com ANSWERSHEET (TOPIC DIFFERENTIAL CALCULUS) COLLECTION # Question Type A.Single Correct Type Q. (A) Sol least
Διαβάστε περισσότεραChapter 6: Systems of Linear Differential. be continuous functions on the interval
Chapter 6: Systems of Linear Differential Equations Let a (t), a 2 (t),..., a nn (t), b (t), b 2 (t),..., b n (t) be continuous functions on the interval I. The system of n first-order differential equations
Διαβάστε περισσότεραLecture 26: Circular domains
Introductory lecture notes on Partial Differential Equations - c Anthony Peirce. Not to be copied, used, or revised without eplicit written permission from the copyright owner. 1 Lecture 6: Circular domains
Διαβάστε περισσότεραRadiation Stress Concerned with the force (or momentum flux) exerted on the right hand side of a plane by water on the left hand side of the plane.
upplement on Radiation tress and Wave etup/et down Radiation tress oncerned wit te force (or momentum flu) eerted on te rit and side of a plane water on te left and side of te plane. plane z "Radiation
Διαβάστε περισσότεραMath 6 SL Probability Distributions Practice Test Mark Scheme
Math 6 SL Probability Distributions Practice Test Mark Scheme. (a) Note: Award A for vertical line to right of mean, A for shading to right of their vertical line. AA N (b) evidence of recognizing symmetry
Διαβάστε περισσότεραEcon 2110: Fall 2008 Suggested Solutions to Problem Set 8 questions or comments to Dan Fetter 1
Eon : Fall 8 Suggested Solutions to Problem Set 8 Email questions or omments to Dan Fetter Problem. Let X be a salar with density f(x, θ) (θx + θ) [ x ] with θ. (a) Find the most powerful level α test
Διαβάστε περισσότεραSection 7.6 Double and Half Angle Formulas
09 Section 7. Double and Half Angle Fmulas To derive the double-angles fmulas, we will use the sum of two angles fmulas that we developed in the last section. We will let α θ and β θ: cos(θ) cos(θ + θ)
Διαβάστε περισσότεραMatrices and Determinants
Matrices and Determinants SUBJECTIVE PROBLEMS: Q 1. For what value of k do the following system of equations possess a non-trivial (i.e., not all zero) solution over the set of rationals Q? x + ky + 3z
Διαβάστε περισσότεραforms This gives Remark 1. How to remember the above formulas: Substituting these into the equation we obtain with
Week 03: C lassification of S econd- Order L inear Equations In last week s lectures we have illustrated how to obtain the general solutions of first order PDEs using the method of characteristics. We
Διαβάστε περισσότερα6.1. Dirac Equation. Hamiltonian. Dirac Eq.
6.1. Dirac Equation Ref: M.Kaku, Quantum Field Theory, Oxford Univ Press (1993) η μν = η μν = diag(1, -1, -1, -1) p 0 = p 0 p = p i = -p i p μ p μ = p 0 p 0 + p i p i = E c 2 - p 2 = (m c) 2 H = c p 2
Διαβάστε περισσότεραThe kinetic and potential energies as T = 1 2. (m i η2 i k(η i+1 η i ) 2 ). (3) The Hooke s law F = Y ξ, (6) with a discrete analog
Lecture 12: Introduction to Analytical Mechanics of Continuous Systems Lagrangian Density for Continuous Systems The kinetic and potential energies as T = 1 2 i η2 i (1 and V = 1 2 i+1 η i 2, i (2 where
Διαβάστε περισσότεραThe challenges of non-stable predicates
The challenges of non-stable predicates Consider a non-stable predicate Φ encoding, say, a safety property. We want to determine whether Φ holds for our program. The challenges of non-stable predicates
Διαβάστε περισσότεραConcrete Mathematics Exercises from 30 September 2016
Concrete Mathematics Exercises from 30 September 2016 Silvio Capobianco Exercise 1.7 Let H(n) = J(n + 1) J(n). Equation (1.8) tells us that H(2n) = 2, and H(2n+1) = J(2n+2) J(2n+1) = (2J(n+1) 1) (2J(n)+1)
Διαβάστε περισσότεραProblem Set 3: Solutions
CMPSCI 69GG Applied Information Theory Fall 006 Problem Set 3: Solutions. [Cover and Thomas 7.] a Define the following notation, C I p xx; Y max X; Y C I p xx; Ỹ max I X; Ỹ We would like to show that C
Διαβάστε περισσότεραTechnical Information T-9100 SI. Suva. refrigerants. Thermodynamic Properties of. Suva Refrigerant [R-410A (50/50)]
d Suva refrigerants Technical Information T-9100SI Thermodynamic Properties of Suva 9100 Refrigerant [R-410A (50/50)] Thermodynamic Properties of Suva 9100 Refrigerant SI Units New tables of the thermodynamic
Διαβάστε περισσότεραAppendix to On the stability of a compressible axisymmetric rotating flow in a pipe. By Z. Rusak & J. H. Lee
Appendi to On the stability of a compressible aisymmetric rotating flow in a pipe By Z. Rusak & J. H. Lee Journal of Fluid Mechanics, vol. 5 4, pp. 5 4 This material has not been copy-edited or typeset
Διαβάστε περισσότεραCRASH COURSE IN PRECALCULUS
CRASH COURSE IN PRECALCULUS Shiah-Sen Wang The graphs are prepared by Chien-Lun Lai Based on : Precalculus: Mathematics for Calculus by J. Stuwart, L. Redin & S. Watson, 6th edition, 01, Brooks/Cole Chapter
Διαβάστε περισσότεραENGR 691/692 Section 66 (Fall 06): Machine Learning Assigned: August 30 Homework 1: Bayesian Decision Theory (solutions) Due: September 13
ENGR 69/69 Section 66 (Fall 06): Machine Learning Assigned: August 30 Homework : Bayesian Decision Theory (solutions) Due: Septemer 3 Prolem : ( pts) Let the conditional densities for a two-category one-dimensional
Διαβάστε περισσότεραSection 9.2 Polar Equations and Graphs
180 Section 9. Polar Equations and Graphs In this section, we will be graphing polar equations on a polar grid. In the first few examples, we will write the polar equation in rectangular form to help identify
Διαβάστε περισσότερα( ) 2 and compare to M.
Problems and Solutions for Section 4.2 4.9 through 4.33) 4.9 Calculate the square root of the matrix 3!0 M!0 8 Hint: Let M / 2 a!b ; calculate M / 2!b c ) 2 and compare to M. Solution: Given: 3!0 M!0 8
Διαβάστε περισσότεραDERIVATION OF MILES EQUATION FOR AN APPLIED FORCE Revision C
DERIVATION OF MILES EQUATION FOR AN APPLIED FORCE Revision C By Tom Irvine Email: tomirvine@aol.com August 6, 8 Introduction The obective is to derive a Miles equation which gives the overall response
Διαβάστε περισσότεραSpace-Time Symmetries
Chapter Space-Time Symmetries In classical fiel theory any continuous symmetry of the action generates a conserve current by Noether's proceure. If the Lagrangian is not invariant but only shifts by a
Διαβάστε περισσότεραThe Simply Typed Lambda Calculus
Type Inference Instead of writing type annotations, can we use an algorithm to infer what the type annotations should be? That depends on the type system. For simple type systems the answer is yes, and
Διαβάστε περισσότεραDuPont Suva 95 Refrigerant
Technical Information T-95 SI DuPont Suva refrigerants Thermodynamic Properties of DuPont Suva 95 Refrigerant (R-508B) The DuPont Oval Logo, The miracles of science, and Suva, are trademarks or registered
Διαβάστε περισσότερα5.4 The Poisson Distribution.
The worst thing you can do about a situation is nothing. Sr. O Shea Jackson 5.4 The Poisson Distribution. Description of the Poisson Distribution Discrete probability distribution. The random variable
Διαβάστε περισσότεραΗΜΥ 220: ΣΗΜΑΤΑ ΚΑΙ ΣΥΣΤΗΜΑΤΑ Ι Ακαδημαϊκό έτος Εαρινό Εξάμηνο Κατ οίκον εργασία αρ. 2
ΤΜΗΜΑ ΗΛΕΚΤΡΟΛΟΓΩΝ ΜΗΧΑΝΙΚΩΝ ΚΑΙ ΜΗΧΑΝΙΚΩΝ ΥΠΟΛΟΓΙΣΤΩΝ ΠΑΝΕΠΙΣΤΗΜΙΟ ΚΥΠΡΟΥ ΗΜΥ 220: ΣΗΜΑΤΑ ΚΑΙ ΣΥΣΤΗΜΑΤΑ Ι Ακαδημαϊκό έτος 2007-08 -- Εαρινό Εξάμηνο Κατ οίκον εργασία αρ. 2 Ημερομηνία Παραδόσεως: Παρασκευή
Διαβάστε περισσότεραΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 19/5/2007
Οδηγίες: Να απαντηθούν όλες οι ερωτήσεις. Αν κάπου κάνετε κάποιες υποθέσεις να αναφερθούν στη σχετική ερώτηση. Όλα τα αρχεία που αναφέρονται στα προβλήματα βρίσκονται στον ίδιο φάκελο με το εκτελέσιμο
Διαβάστε περισσότεραDuPont Suva 95 Refrigerant
Technical Information T-95 ENG DuPont Suva refrigerants Thermodynamic Properties of DuPont Suva 95 Refrigerant (R-508B) The DuPont Oval Logo, The miracles of science, and Suva, are trademarks or registered
Διαβάστε περισσότεραST5224: Advanced Statistical Theory II
ST5224: Advanced Statistical Theory II 2014/2015: Semester II Tutorial 7 1. Let X be a sample from a population P and consider testing hypotheses H 0 : P = P 0 versus H 1 : P = P 1, where P j is a known
Διαβάστε περισσότεραMath221: HW# 1 solutions
Math: HW# solutions Andy Royston October, 5 7.5.7, 3 rd Ed. We have a n = b n = a = fxdx = xdx =, x cos nxdx = x sin nx n sin nxdx n = cos nx n = n n, x sin nxdx = x cos nx n + cos nxdx n cos n = + sin
Διαβάστε περισσότεραInverse trigonometric functions & General Solution of Trigonometric Equations. ------------------ ----------------------------- -----------------
Inverse trigonometric functions & General Solution of Trigonometric Equations. 1. Sin ( ) = a) b) c) d) Ans b. Solution : Method 1. Ans a: 17 > 1 a) is rejected. w.k.t Sin ( sin ) = d is rejected. If sin
Διαβάστε περισσότεραSCHOOL OF MATHEMATICAL SCIENCES G11LMA Linear Mathematics Examination Solutions
SCHOOL OF MATHEMATICAL SCIENCES GLMA Linear Mathematics 00- Examination Solutions. (a) i. ( + 5i)( i) = (6 + 5) + (5 )i = + i. Real part is, imaginary part is. (b) ii. + 5i i ( + 5i)( + i) = ( i)( + i)
Διαβάστε περισσότεραSecond Order Partial Differential Equations
Chapter 7 Second Order Partial Differential Equations 7.1 Introduction A second order linear PDE in two independent variables (x, y Ω can be written as A(x, y u x + B(x, y u xy + C(x, y u u u + D(x, y
Διαβάστε περισσότεραCHAPTER (2) Electric Charges, Electric Charge Densities and Electric Field Intensity
CHAPTE () Electric Chrges, Electric Chrge Densities nd Electric Field Intensity Chrge Configurtion ) Point Chrge: The concept of the point chrge is used when the dimensions of n electric chrge distriution
Διαβάστε περισσότεραDuPont Suva. DuPont. Thermodynamic Properties of. Refrigerant (R-410A) Technical Information. refrigerants T-410A ENG
Technical Information T-410A ENG DuPont Suva refrigerants Thermodynamic Properties of DuPont Suva 410A Refrigerant (R-410A) The DuPont Oval Logo, The miracles of science, and Suva, are trademarks or registered
Διαβάστε περισσότεραExercises 10. Find a fundamental matrix of the given system of equations. Also find the fundamental matrix Φ(t) satisfying Φ(0) = I. 1.
Exercises 0 More exercises are available in Elementary Differential Equations. If you have a problem to solve any of them, feel free to come to office hour. Problem Find a fundamental matrix of the given
Διαβάστε περισσότεραw o = R 1 p. (1) R = p =. = 1
Πανεπιστήµιο Κρήτης - Τµήµα Επιστήµης Υπολογιστών ΗΥ-570: Στατιστική Επεξεργασία Σήµατος 205 ιδάσκων : Α. Μουχτάρης Τριτη Σειρά Ασκήσεων Λύσεις Ασκηση 3. 5.2 (a) From the Wiener-Hopf equation we have:
Διαβάστε περισσότεραthe total number of electrons passing through the lamp.
1. A 12 V 36 W lamp is lit to normal brightness using a 12 V car battery of negligible internal resistance. The lamp is switched on for one hour (3600 s). For the time of 1 hour, calculate (i) the energy
Διαβάστε περισσότεραThe ε-pseudospectrum of a Matrix
The ε-pseudospectrum of a Matrix Feb 16, 2015 () The ε-pseudospectrum of a Matrix Feb 16, 2015 1 / 18 1 Preliminaries 2 Definitions 3 Basic Properties 4 Computation of Pseudospectrum of 2 2 5 Problems
Διαβάστε περισσότεραForced Pendulum Numerical approach
Numerical approach UiO April 8, 2014 Physical problem and equation We have a pendulum of length l, with mass m. The pendulum is subject to gravitation as well as both a forcing and linear resistance force.
Διαβάστε περισσότεραECE Spring Prof. David R. Jackson ECE Dept. Notes 2
ECE 634 Spring 6 Prof. David R. Jackson ECE Dept. Notes Fields in a Source-Free Region Example: Radiation from an aperture y PEC E t x Aperture Assume the following choice of vector potentials: A F = =
Διαβάστε περισσότεραω ω ω ω ω ω+2 ω ω+2 + ω ω ω ω+2 + ω ω+1 ω ω+2 2 ω ω ω ω ω ω ω ω+1 ω ω2 ω ω2 + ω ω ω2 + ω ω ω ω2 + ω ω+1 ω ω2 + ω ω+1 + ω ω ω ω2 + ω
0 1 2 3 4 5 6 ω ω + 1 ω + 2 ω + 3 ω + 4 ω2 ω2 + 1 ω2 + 2 ω2 + 3 ω3 ω3 + 1 ω3 + 2 ω4 ω4 + 1 ω5 ω 2 ω 2 + 1 ω 2 + 2 ω 2 + ω ω 2 + ω + 1 ω 2 + ω2 ω 2 2 ω 2 2 + 1 ω 2 2 + ω ω 2 3 ω 3 ω 3 + 1 ω 3 + ω ω 3 +
Διαβάστε περισσότεραPARTIAL NOTES for 6.1 Trigonometric Identities
PARTIAL NOTES for 6.1 Trigonometric Identities tanθ = sinθ cosθ cotθ = cosθ sinθ BASIC IDENTITIES cscθ = 1 sinθ secθ = 1 cosθ cotθ = 1 tanθ PYTHAGOREAN IDENTITIES sin θ + cos θ =1 tan θ +1= sec θ 1 + cot
Διαβάστε περισσότεραAssalamu `alaikum wr. wb.
LUMP SUM Assalamu `alaikum wr. wb. LUMP SUM Wassalamu alaikum wr. wb. Assalamu `alaikum wr. wb. LUMP SUM Wassalamu alaikum wr. wb. LUMP SUM Lump sum lump sum lump sum. lump sum fixed price lump sum lump
Διαβάστε περισσότεραSecond Order RLC Filters
ECEN 60 Circuits/Electronics Spring 007-0-07 P. Mathys Second Order RLC Filters RLC Lowpass Filter A passive RLC lowpass filter (LPF) circuit is shown in the following schematic. R L C v O (t) Using phasor
Διαβάστε περισσότεραLinearized Lifting Surface Theory Thin-Wing Theory
13.021 Marine Hdrodnamics Lecture 23 Copright c 2001 MIT - Department of Ocean Engineering, All rights reserved. 13.021 - Marine Hdrodnamics Lecture 23 Linearized Lifting Surface Theor Thin-Wing Theor
Διαβάστε περισσότεραEulerian Simulation of Large Deformations
Eulerian Simulation of Large Deformations Shayan Hoshyari April, 2018 Some Applications 1 Biomechanical Engineering 2 / 11 Some Applications 1 Biomechanical Engineering 2 Muscle Animation 2 / 11 Some Applications
Διαβάστε περισσότερα5. Choice under Uncertainty
5. Choice under Uncertainty Daisuke Oyama Microeconomics I May 23, 2018 Formulations von Neumann-Morgenstern (1944/1947) X: Set of prizes Π: Set of probability distributions on X : Preference relation
Διαβάστε περισσότεραLecture 15 - Root System Axiomatics
Lecture 15 - Root System Axiomatics Nov 1, 01 In this lecture we examine root systems from an axiomatic point of view. 1 Reflections If v R n, then it determines a hyperplane, denoted P v, through the
Διαβάστε περισσότεραCHAPTER 48 APPLICATIONS OF MATRICES AND DETERMINANTS
CHAPTER 48 APPLICATIONS OF MATRICES AND DETERMINANTS EXERCISE 01 Page 545 1. Use matrices to solve: 3x + 4y x + 5y + 7 3x + 4y x + 5y 7 Hence, 3 4 x 0 5 y 7 The inverse of 3 4 5 is: 1 5 4 1 5 4 15 8 3
Διαβάστε περισσότεραBounding Nonsplitting Enumeration Degrees
Bounding Nonsplitting Enumeration Degrees Thomas F. Kent Andrea Sorbi Università degli Studi di Siena Italia July 18, 2007 Goal: Introduce a form of Σ 0 2-permitting for the enumeration degrees. Till now,
Διαβάστε περισσότεραJackson 2.25 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell
Jackson 2.25 Hoework Proble Solution Dr. Christopher S. Baird University of Massachusetts Lowell PROBLEM: Two conducting planes at zero potential eet along the z axis, aking an angle β between the, as
Διαβάστε περισσότεραStrain gauge and rosettes
Strain gauge and rosettes Introduction A strain gauge is a device which is used to measure strain (deformation) on an object subjected to forces. Strain can be measured using various types of devices classified
Διαβάστε περισσότεραTesting for Indeterminacy: An Application to U.S. Monetary Policy. Technical Appendix
Testing for Indeterminacy: An Application to U.S. Monetary Policy Technical Appendix Thomas A. Lubik Department of Economics Johns Hopkins University Frank Schorfheide Department of Economics University
Διαβάστε περισσότεραDurbin-Levinson recursive method
Durbin-Levinson recursive method A recursive method for computing ϕ n is useful because it avoids inverting large matrices; when new data are acquired, one can update predictions, instead of starting again
Διαβάστε περισσότεραOn a four-dimensional hyperbolic manifold with finite volume
BULETINUL ACADEMIEI DE ŞTIINŢE A REPUBLICII MOLDOVA. MATEMATICA Numbers 2(72) 3(73), 2013, Pages 80 89 ISSN 1024 7696 On a four-dimensional hyperbolic manifold with finite volume I.S.Gutsul Abstract. In
Διαβάστε περισσότεραFractional Colorings and Zykov Products of graphs
Fractional Colorings and Zykov Products of graphs Who? Nichole Schimanski When? July 27, 2011 Graphs A graph, G, consists of a vertex set, V (G), and an edge set, E(G). V (G) is any finite set E(G) is
Διαβάστε περισσότερα6.4 Superposition of Linear Plane Progressive Waves
.0 - Marine Hydrodynamics, Spring 005 Lecture.0 - Marine Hydrodynamics Lecture 6.4 Superposition of Linear Plane Progressive Waves. Oblique Plane Waves z v k k k z v k = ( k, k z ) θ (Looking up the y-ais
Διαβάστε περισσότερα