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z ∂x ∂y � � ∂ ∂ ∂ � � ∂z ∂x ∂y � � � � F1 F2 F3 ∂2F2 ∂2F3 ∂x∂z ∂x∂y − = = − ∂2F3 ∂y∂x + ∂2F1 ∂y∂z + ∂2F2 ∂z∂x − ∂2F1 ∂z∂y = 0. This is (2). Example 18.8. The gravitational field � � F (x, y, z) = cx (x2 + y2 + z2)3/2 ˆı+ cy (x2 + y2 + z2)3/2 jˆ+ cz (x2 + y2 + z2)3/2 ˆ k, is a gr...
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2f = div(grad f ) = ∂2f ∂x + ∂2f ∂y + ∂2f . ∂z 3 A solution of the differential equation �2f = 0, is called a harmonic function. Example 18.11. The function f (x, y, z) = − c , (x2 + y2 + z2)1/2 is harmonic. 4 MIT OpenCourseWare http://ocw.mit.edu 18.022 Calculus of Several Variables Fall 2010 For...
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Figure removed for copyright reasons. See Fig. 2 in Bockstaller et al. "Size-selective Organization of Enthalpic Compatibilized Nanocrystals in Ternary Copolymer/Particle Mixtures." J. Amer. Chem. Soc. 125 (2003): 5276-5277. Figure removed for copyright reasons. Figure removed for copyright reasons. See Fig. 2 in Bocks...
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Perfect Conductivity Lecture 2 Terry P. Orlando Dept. of Electrical Engineering MIT September 13, 2005 Massachusetts Institute of Technology 6.763 2005 Lecture 2 Outline 1. Persistent Currents 2. Parts of a Physical Theory 3. Circuits and Time Constants 4. Distributive Systems and Time constants A.Quasista...
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2. Constitutive Relations Massachusetts Institute of Technology 6.763 2005 Lecture 2 4 3. Summary Relation 1 jωC jωL R Massachusetts Institute of Technology 6.763 2005 Lecture 2 Simpler Circuits and Time Constants L R LC RC Energy stored in inductor Resonant transfer of energy between L and C Energy s...
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5 Lecture 2 Quasistatic Limit Length scale of system Wavelength of E&M wave Speed of light Frequency (angular) If the dimensions of a structure are much less than the wavelength of an electromagnetic field interacting with it, the coupling between the associated electric and magnetic fields is weak and a qu...
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34, from Orlando, T., and K. Delin. Foundations of Applied Superconductivity. Reading, MA: Addison-Wesley, 1991. ISBN: 0201183234. Magnetic Diffusion Equation Massachusetts Institute of Technology 6.763 2005 Lecture 2 10
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Electronics A Joel Voldman Massachusetts Institute of Technology Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. JV: 2.372J/...
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� × = ⇒ 0 E E r ( , ) t = −∇ V ( , ) r t ( ) V b V a − ( ) = − b ∫ a E l d ⋅ ( ) V b V a − ( ) = V Eg = ⇒ = E V g g Q = ε A V g = V CV = A ε g A ε g C = Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit...
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dI dt V L = - - Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. JV: 2.372J/6.777J Spring 2007, Lecture 6E - 6 Source elemen...
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1 Cs V IZ = L = ILs V IZ = R = IR - - - - V Z V s ( ) = I s Z s ( ) ( ) I + I + I + + i Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD M...
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C V i Z − C C i = − = L i C 0 i R i Z R R = 0 Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. JV: 2.372J/6.777J Spring 2007,...
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CourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. JV: 2.372J/6.777J Spring 2007, Lecture 6E - 13 Nodal analysis > Element law approach becomes tedious for circuits with multiple loops > Nodal analysis is a KCL-based approach Cite as: Joel Voldman, course materials...
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− Z C i 1 + + = v V = 1 0 i 0 3 v 2 = 0 i 2 0 − Z R v 2 + − Z L Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. JV: 2.372J/6...
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determine 2. Replace elements with complex impedances 3. Assign node voltages & ground node 4. Write KCL at each node v1 i1 C i2 i3 v2 L V0 + - R LRs v 2 = V 0 LRs + R Ls 1 Cs 1 + Cs 2 LRCs 2 + LRCs Ls R + 5. Solve for node voltages v 2 = V 0 6. Use node voltages to find what you care about V C = v 1 − v 2 = V V − 0...
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Z 1 i i1 i2 i Z1 Z2 Z + V - + V - V i Z = 1 1 = i Z = Z Z 1 2 Z Z + 1 1 Z 2 1 Z 1 = Z Z 1 2 Z Z + 1 2 = Z 1 // Z 2 + 1 Z 2 Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute...
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in parallel Z3 and Z4 NOT in parallel Z3 and Z4 in parallel Z1 and Z3 NOT in series Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month...
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R L Z Z + R L + 1 ZZ C L 3. Collapse circuit in terms of series/parallel relations till circuit is trivial 4. Re-expand to find signal of = V 0 = V 0 interest Z Z R L + L Z Z R L ( Z + R RLs 1 ) Z Z Z C L L 1 Ls RLs + ( R Ls + 1 ) Cs i L = V 0 RCs 2 + RLCs Ls R + Cite as: Joel Voldman, course materials for 6.777J / ...
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anical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. JV: 2.372J/6.777J Spring 2007, Lecture 6E - 26 Superposition > For circuits with multiple v1 v2 sources, • Turn off all independent sources except one • Solve circuit • Repea...
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6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. JV: 2.372J/6.777J Spring 2007, Lecture 6E - 28 Conclusions > There are many ways to analyze equivalent circuits > Use t...
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18.336 spring 2009 lecture 15 02/13/08 Finite Difference Methods for the One-Way Wave Equation � ut = cux u(x, 0) = u0(x) Solution: u(x, t) = u0(x + ct) Information travels to the left with velocity c. Three Approximations: ⎧ c ⎨ n n +1 − Uj Uj Δx U n − Uj n jc −1 Δx ⎪⎪⎪⎪⎪⎪⎪⎪ ⎪⎪⎪⎪⎪⎪⎪⎪ U n ⎩ c j+1 − U...
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1 c2uxxΔt 2 2 Leading order error: 1 1 2 2 1 2 2 uttΔt − cuxxΔx = c uxxΔt − cuxxΔx = cuxxΔx(r − 1) 1 2 1 2 = 0 if r = 1 r = cΔt Δx First order if r = 1 Courant number 1 � Downwind: Analogous: first order Centered: ut + 1 uttΔt − cux − 1 cuxxxΔx2 + O(Δt2) + O(Δx4) 6 2 Δt → 2 Δx First order in time → Sec...
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− λ n +1 + (1 − λ)Uj 2 Uj Δt Uj U n j+1 − 2Uj n n + Uj −1 + θ (Δx)2 n + λ � n 2 Uj −1 = c where λ = 2 Δt (Δx)2 θ How much diffusion? Lax-Friedrichs: Eliminate Uj n by λ = 1 θ = ⇒ (Δx)2 2Δt U n+1 j = 1 + r 2 � �� � n +1 + Uj 1 − r 2 � �� ≥0(for |r|≤1) � n Uj −1 ≥0(for |r|≤1) Monotone scheme n ...
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+ O(Δx2) Accuracy: ut+ 1 uttΔt+ 1 utttΔt2−cux− 1 cuxxxΔx2 Δt c2uxx− Δt c2uxxxxΔx2 1 utttΔt2 − 6 = 6 ut − cux = 0 Δt 2 1 uttΔt − 2 2 Stability: λ = r2 c uxx = 0 − 2 24 G = r2 + r 2 e ikΔx + (1 − r 2) + r2 − r 2 e−ik(Δx) = (1 − r 2) + r 2 cos(kΔx) + ir sin(kΔx) Worst case: kΔx = π ⇒ G = 1 − 2r2 Stable if |r|...
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1 I-campus pro ject School-wide Program on Fluid Mechanics Modules on Waves in fuids T. R. Akylas & C. C. Mei CHAPTER SEVEN INTERNAL WAVES IN A STRATIFIED FLUID 1 Introduction. The atmosphere and o c e a n are continuously stratifed due to change in temperature, composition and pressure. These changes in the ocean and ...
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to satisfy the continuity equation 1 Dp @u @ v @w + + + 0 (2.1) p Dt @x @ y @ z and the momentum equations ...
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The motion that takes place is assumed to be isentropic and without change of phase, so that e and q are constant for a material element. Therefore Dp @ p De @ p Dq + : (2.6) 0 Dt @ e Dt @ q Dt In other words, p is constant for a material element because e and q are, and p depends only on e and q . Such a fuid is said...
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that the wave motion results from the perturbation of a state of equilibrium, which is the state of rest. So the distribution of density and pressure is the hydrostatic equilibrium distribution given by When the motion develops, the pressure and density changes to @ pp @ z ; g p p: (2.12) p p(z ) + p , p (2.13) 0 p p(...
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form ...
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@ z @ t ; ; If we substitute equations (2.21) and (2.22) into equation (2.20), we obtain 2 0 2 0 2 1 @ p @ p @ w ; pp @x @ y @ t@ z 2 2 + + : (2.24) 0 We can eliminate p from (2.23) by using equation (2.16) to obtain 0 2 2 0 @ w @ p @ pp pp + g w : (2.25) 2 @ t @ t@ z @ z ; Third, we apply the operator + to equation (2...
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2 2 2 2 2 @ @ w @ w 1 @ @w @ w @ w 2 2 2 2 2 2 @ t @x @ y pp @ z @ z @x @ y + + pp + N + , (2.27) 0 where we defne 2 g @ pp N (z ) , (2.28) ; p @ z which has the units of frequency (rad/sec) and is called the Brunt-Viaisialia frequency or buoyancy frequency. If we assume that w varies with z much more rapidly than pp...
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of the momentum equations. The Boussinesq approximation leads to equation (2.30) for the vertical velocity w . 3 The Buoyancy Frequency (Brunt-Vaaisaalaa frequency). Consider a calm stratifed fuid with a static density distribution pp(z ) which decreases with height z . If a fuid parcel is moved from the level z upward...
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Brunt V iasialia frequency. This elementary consideration shows that once a fuid is displaced from its equilibrium position, gravity and density gradient provide restoring force to enable oscillations. 4 Internal Gravity Waves in Unbounded Stratifed Fluid. Consider the case in which the buoyancy (Brunt-Viaisialia) freq...
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N . The dispersion relation for internal waves is of quite a diferent c haracter compared to that for surface waves. In particular, the frequency of surface waves depends only on the magnitude k of the wavenumb e r , whereas the frequency of internal waves is ;! j j independent of the magnitude of the wavenumber and de...
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p 0 0 p cos(kx + ly + mz ! t ): (4.41) 2 2 1/2 ; ; (k + l ) From equation (2.16) we have the perturbation density p given by 0 ...
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a fxed point. For instance, if the horizontal velocity components and perturbation pressure of a progressive w ave are measured, the horizontal component o f t h e wavenumb e r vector can be deduced from (4.44). A sketch showing the properties of a plane progressive internal wave in the vertical plane that contains the...
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to the vertical. The restoring force p e r unit displacement (cos <dp /dz ) is less than the case where < , so the frequency 0 0 of vibration is less. As < tends to / 2, the frequency of vibration tends to zero. The case < / 2 i s n o t a n i n ternal wave, but it represents an important form of motion that is often ...
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u u i d a s t i d o s t i d L B G B L o u o s i u o n y g a B n t u F l u i d o y a n c y a i n i n g B u o o y a n t F l u i s i n g B u o y y d a a n c y n c y G r o u p V e l o c i t y c it y e l o e V s a h P H i g h P r e s s u r e ( u , w , P = 0 ) L o w ( u , w , P P H i g h r e s s u r e = 0 ) P r e s s u r e ...
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. Small arrows indicate the perturbation velocities, which a r e always parallel to the lines of constant phase. Large heavy arrows indicate the direction of phase propagation and group velocity. ...
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e r vector. When the group velocity has an upward component, therefore, the phase velocity has a d o wnward component, and vice versa. The group velocity vector is N C sin <(sin < cos e , sin < sin e , cos <): (4.46) g Therefore, the magnitude of the group velocity is ( ) sin <, and its direction is at j k j k ; ! j j ...
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12 m θ 2 2 2 1/2 (k + l + m ) C g φ φ 2 (k 2 + l ) 1/2 m l k Figure 3: Wavenumb e r vector and group velocity vector. 13 wave packet. The phase velocity is given by the equation (4.45), where the wavenumb e r vector k makes an angle < with the horizontal direction (see fgure 1, but...
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-dimensional Fourier transform pair considered is 1 1 u^(k , m ) dx dz exp( ikx imz )u(x, z ) (4.49) ;1 ;1 f ; ; g Z Z and 1 1 1 u(x, z ) dk dm exp( ikx imz ) u^(k , m ) : (4.50) 2 4 ;1 ;1 f ; ; g Z Z The Fourier transform of the equation (2.30) is given by the equation 2 2 2 @ w^ N k + w^ , (4.51) 0 2 2 2 @ t k + m...
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^ : (4.54) 0 ^ 2 p N 0 p (k , m, t) iA(k , m ) ex p ( i! t) + iB (k , m ) exp(i! t) , (4.55) g ! (k , m ) f; g where the constants A and B are determined from the Fourier transform of the initial conditions for p , given by the equations 0 0 p (x, z , 0) f (x, z ), (4.56) 0 @ p @ t (x, z , 0) 0, (4.57) which implies t...
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2 2 ~ 2 2 2 ; ; 1 1 (k k ) 1 (m m~ ) 1 (k + k ) 1 (m + m~ ) 2 2 2 2 ~ ~ ^ f (k , m ) exp + exp : ; ; 2 2 2 2 2<T 2 < 2 T 2 < 2 T ; ; ; ; ( � ! � !) (4.62) ...
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p a c ket has the same modulation along the x and z directions (< T ), the two parts of the initial wave packet travel towards the middle of the second and fourth quadrants, as we see in the animation. For the wave packet in the second (fourth) quadrant the group velocity vector p o i n ts away from the origin towards ...
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wave packets for early times is more intense than what was observed in the previous example, as we can see in the animation. To see it, click here. The third animation has as initial condition a Gaussian wave packet with < /2, 1 T /20 and k m~ . This initial wave packet has a shape of an elongated 1 ellipse 2 ~ � in ...
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16 4.2 Saint Andrew's Cross. Here we discuss the wave pattern for internal waves produced by a localized source on a sinusoidal oscillation, like an oscillating cylinder for example, in a fuid with constant density gradient (the buoyancy frequency is constant). For sinusoidal internal waves, the wave energy fux I p u (...
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gravity surface waves. Instead, the crests and other surfaces of constant phase stretch radially outward from the source because wave energy travels with the group velocity, which is parallel to surfaces of constant phase. For a source of defnite frequency ! N (less than the buoyancy frequency), those surfaces are all ...
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is orthogonal to the group velocity, a n d that 17 C + C cos e , sin e , 0 : (4.66) g N ; ! ! ; f g k ; ! j j Then, given the direction of the group velocity, the orthogonality of the phase and group velocity plus the condition (4.66), the direction of the phase velocity i s specifed. If the group velocity has a positi...
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e h a vior. In this section we study free wave propagation in a continuously stratifed fuid in the presence of boundaries. Attention is restricted to the case in which the bottom is fat. The equilibrium state that is being perturbed is the one at rest, so density, and hence buoyancy frequency, is a function only of the...
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18 C+C g C g C C C g C + C g Figure 4: Phase and group velocities. C+C g C g C C C g C+C g 19 Figure 5: St Andrew's Cross in a stratifed fuid. In the top fgure ! /N 0:9 and in the left bottom fgure ! /N :7. 0 © sources unknown. All rights reserved. This content is excluded from our Creative ...
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tain a free-surface 0 boundary condition for w^(z ). We apply the operator to the equation (2.24), and then @ t @ we substitute equation (5.70) into the resulting equation. As a result, we obtain the equation 3 2 2 @ w @ w @ w 2 2 2 @ t @ z @x @ y g + at z (5.71) 0 Now, if we substitute equation (5.67) into the equati...
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efect of confning the wave energy to a region of fnite extent, so the ocean can b e considered as a that causes the energy to waveguide propagate horizontally. A useful piece of imaginary is to picture internal waves propagating obliquely through the ocean, refections at the upper and lower boundaries ensuing no loss o...
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by the equation w^ (z ) sin[m(z + H )] with m (k + l ), (5.76) ; 2 ! 2 2 2 2 2 (N ! ) which already satisfes the bottom boundary condition. If we substitute equation (5.76) into the free-surface boundary condition (5.72), we obtain the dispersion relation ...
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, so equation (5.72) reduces to w^ (z ) 0 at z : (5.78) 0 This b o u n d a r y condition gives a dispersion relation of the form sin(mH ) 0 (5.79) or 2 (k + l )N H 2 2 2 2 ! , n , 2, 3, : : : , (5.80) 1 2 2 2 2 2 n + ( k + l )H which is close to the result given by the dispersion relation given by the free-surface bo...
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modes. ...
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in the range 0 ! N , i . e., there : is a continuous spectrum of solutions. Superposition of such solutions can b e used to solve initial-value problems, and have the from of Fourier integrals. When N varies with z , there is another type of solution possible, namely, one that satisfes the condition at the ground yet d...
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umbers m and m . For this case, the spectrum is continuous 1 2 and ! can assume any value b e t ween 0 and N . This is not true for the case when 2 N ! N , when the frequency ! can assume only a fnite set of values in the range 2 1 N ! N . In this case, the solution of equation (5.73) for the frst layer is given by 2...
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2 2 2 2 2 (! N ) At t h e intersection z H b e t ween the two l a yers, the perturbation pressure p and the 0 vertical velocity w should be continuous. Alternatively, this condition can be expressed in terms of the ratio 0 p Z , (5.85) p w 0 which must b e the same on both sides of the boundary. It is convenient to ref...
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modes. The relative amplitude of the diferent modes depends on the initial state. 6 Energetics of Internal Waves. The energy equation for internal waves, under the assumption of small perturbations, incompressible and inviscid fuid and irrotational fow, can b e obtained by multiplying ...
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gp @ (p u) @ (p v ) @ (p w) 2 2 2 0 0 0 0 pp u + v + w + + + + : (6.87) 0 @ t 2 2 pNp @x @ y @ z 2 The term pp(u + v + w )/2 stands for the perturbation kinetic energy density. The ; ) 2 2 2 term (p u, p v , p w) stands for the perturbation energy density fux and the term 0 0 0 2 0 2 1 g � 2 2 ��N stands for the per...
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the energy due to the perturbation, which is equal to 1 1 2 2 p g + g (p p )h dxdy , (6.89) 2 1 2 2 2 ; Z Z � � ...
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.91) becomes 2 0 2 1 g p 2 pNp 2 Z Z Z dxdydz : (6.93) The connection with (6.87) is now clear. For periodic waves in a medium with uniform properties, the integral over each wavelength is the same, and so the mean over a large volume becomes equal to the mean over one wavelength in the limit as the volume tends to inf...
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the average over one wavelength, so it is convenient to consider the spatial mean for the fuxes as well. Thus the energy fux density vector F is defned by the equation ; ! 0 27 0 0 0 0 0 F ;! (u , v , w ) >, (6.95) p where (u , v , w ) is the perturbation velocity v ector. 0 0 0 7 Mountain Waves. Internal waves in the...
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velocity over single ridge in a fnite atmosphere or o c e a n (waveguide case). We assume that the atmosphere or ocean density stratifcation is such that the buoyancy frequency is constant. 7.1 Governing Equation. We consider an air or water moving at a constant velocity U (z ) over a periodic or localized topography g...
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write the horizontal velocity i n t h e form 1 D p p D t (7.100) 0 u(x, z ) U (z ) + u (x, z ), (7.101) 0 We substitute equation (7.101) in the equations (7.97) to (7.100). We assume the velocities u and w as small quantities, so we can linearize the resulting equations. 0 The linearized form of the continuity equatio...
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hydrostatic equilibrium distribution given by equation (2.12). When the motion devel- ops, the pressure and density are given, respectively, b y equations (2.13) and (2.14), and 0 0 p and p are the pressure and density perturbations of the \background state". Now the momentum equation assume the form 29 0 0 @ u @ u @ U...
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continuity equation (7.102) to obtain 2 0 2 @ u @ w + : (7.109) 0 @ t@x @ t@ z Second, we take the x derivative of the equation (7.106) to obtain 2 0 2 0 2 0 @ u @ u @w @U @ p pp + U + (7.110) @x@ t @x @x @ z @x ; 2 2 � � 0 Third, we eliminate the u variable from the equation (7.110) above. To do so, we use equation (...
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@ @ 30 2 2 2 2 0 2 0 0 0 @ w @ w @ w @ p @ p @ p @ p 2 pp + 2 U + U U g + U , (7.112) 2 2 @ t @x@ t @x @ t@ z @x@ z @ t @x ; ; ; � � � � and with equation (7.105) we can eliminate p from equation (7.112). The result is the 0 equation 2 2 2 2 0 2 0 @ w @ w @ w @ p @ p @ pp 2 pp + 2 U + U U gw , (7.113) 2 2 @ t @x@ t @x ...
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the vertical velocity w . We can simplify the equation above. We can write it in the form @ @ @ w 1 @ @w @ @ 1 @ @U @w @ w 2 2 2 2 + U + pp + U pp + N (z ) , 0 @ t @x @x pp @ z @ z @ t @x pp @ z @ z @x @x ; 2 2 � � � � (7.115) where N (z ) is the buoyancy frequency defned according to equation (2.28). If we assume t...
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t @x @x @ z @ t @x @ z @ z @x @x ; 2 2 2 � � � � (7.118) We can simplify this equation further by assuming that the velocity U is constant. In this case we end up with an equation of the form @ @ @ w @ w @ w 2 2 2 2 2 + U + + N (z ) , (7.119) 0 @ t @x @x @ z @x 2 2 2 � � Next, we discuss boundary conditions for the...
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) U on z (7.121) 0 @ ( @ x For an infnite atmosphere we need a radiation condition, which ensures that the energy fux is away from the ground. In other words, energy is radiated away from the ground by t h e i n ternal waves generated by the topography. For the case of a fnite atmosphere, we need a boundary condition ...
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z , t) A U k cos(kx ) at z (7.123) 0 0 In this example, we assume a constant b u o yancy frequency N for the entire atmosphere. Under such condition, we can assume a solution for the steady state regime of the form w(x, z , t) A cos(kx + mz ), (7.124) where k is the topography wavenumber, since the solution (7.124) ha...
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33 7.2.1 Case . > k U N There are two solutions for equation (7.126) (plus or minus sign of the square root of the right hand side of equation (7.126)), and to decide which o n e w e should use to represent energy being radiated away from the ground we are going to consider the energy fux in the vertical direction. Acc...
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away from the ground, we chose m as given by the equation 2 N 2 m k : (7.130) + 2 U ; r Since energy is b e i n g radiated away from the ground, there is drag exerted by the topography due to the generation of the internal waves. The magnitude of the drag force p e r unit area is equal to the rate T p e r unit area at...
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k equal / to the horizontal distance traveled by a fuid c particle in one buoyancy period. From equation (7.126) we have that Thus, the angle < b e t ween wave crests and the vertical changes according to the equation 0 N U 2 2 2 k + m k ) : ( c The angle < given by the equation above is illustrated in the fgure 7. 0...
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drag force exerted by the topography. To see this, we consider the expression for the pressure in this case, which is given by the equation 0 A p ip1p U exp( 1 z + ikx) (7.136) if; ; g k This is out of phase with the vertical velocity, i.e., is zero when w is a maximum or a minimum, and is a maximum or a minimum when w...
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speed U ), we will see the localized topography moving with speed U , and we will see waves with ; phase speed U matching the speed on the topography (steady disturbance downstream ; to the topography in the fxed reference frame). Here we consider the second reference frame, since in this reference frame we can easily ...
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related according to the equation 0 x x + U t: (7.139) The boundary condition on the ground, given by the equation (7.121), for this ref- erence frame assumes the form 0 0 @ ( w(x , z , t ) U (x + U t ) at z : (7.140) 0 @x To solve the boundary value problem given by equations (7.138) and (7.140) with the appropriate ...
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37 w^ (k , z , t ) ikU ( (k) exp(ikU t) (7.144) ^ Next, we consider a time dependence of the form w^ (k , z , t ) w^ (k , z ) exp( i! t) (7.145) 0 ; If we substitute equation (7.145) into the governing equation (7.143) we end up with the governing equation for w^ , which follows: 0 2 0 @ w^ 2 2 2 2 2 0 ! + ( N k ! k ) ...
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ersion relation, which follows kN ! (7.149) 2 2 p m + k From the dispersion relation above we can obtain the group velocity v ector. The group velocity is the speed with which energy is propagated by the internal waves, it is the gradient o f t h e wave frequency with respect to the wavenumber, then ...
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need m to b e negative and for k 0 we need m to b e positive. As a result, we chose the appropriate branch of m given by equation (7.148) as follows: For k 0 w e have that • 2 m(k , N /U ) , (7.152) ; ; j j : 2 1/2 U N 2 N U k if k 8 2 N N 1/2 h i i k if k > ; ) U U { : ; ; j j [ ; )J For k 0 we have that • m(k , N /U...
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39 2 m C U , (7.155) gx ; k ; ! 2 j j which implies a negative v alue for the horizontal component of the group velocity. From this equation we also realize that the horizontal component of the group velocity is in the same direction as the horizontal component of the phase speed, but smaller magnitude. This implies th...
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apply the inverse Fourier transform to the result- ing equation, we obtain 0 0 U N ^ 2 w(x , z , t ) ik ( (k) s i n ( k z + k U t + k x )dk N/U 2 ; 8 U ; 0 s Z { 1 : 2 N (7.158) + ik ( (k) e x p ( k z ) sin(+k U t + k x )dk ^ 2 0 N/U s ; ; U Z 9 } In terms of the fxed reference frame, the vertical velocity w is given ...
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( (x) (fxed reference fram e) is given by the equation A 0 ( (x) , (7.160) 1 + ( x/b) 2 and its Fourier transform is given by the equation ^ ( (k) b exp( A kb ): (7.161) 0 ;j j For this particular example, the vertical velocity is given by w(x, z ) U A b k exp( b k ) sin ( k z + kx )dk 0 N/U 2 N 2 ; ; j j ; U 8 0 s ...
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1 (x, z ) A b exp( b k ) sin ( k z + kx )dk 0 8 ; j j ; U 0 s N/U 2 N 2 Z { 1 : 2 2 N + exp( b k k z ) sin(+kx )dk N/U s ; j j ; ; U 9 Z } 41 (7.164) By inspecting equation (7.164), if the buoyancy frequency N is zero, the stream function 1 has a simple expression given by the equation 1 (x, z ) , (7.165) 1 + [ x/(b +...
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velocity w . For this governing equation we h a ve the boundary condition at the ground, given by equation (7.121), and we also consider a boundary condition at z h, which is the top of the fnite atmosphere. This boundary condition is given by the equation w(x, z ) 0 at z h: (7.166) In this case no energy is radiated i...
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x axis Figure 8: Stream lines for the case of zero value for the buoyancy frequency N . A 1 0 and b . 4 43 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 s i x a z -10 0 10 20 x axis Figure 9: Stream lines for the case of non-zero value for the buoyancy freque...
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(7.166). This implies that we should have ! kU . We assume a general solution for w^ of the form 0 ; 0 w^ B sinh(m(h z )), (7.168) ; which already satisfes the Fourier transform of equation (7.166). If we substitute equa- tion (7.167) into the ground boundary condition given by equation (7.144), we obtain ikU ( (k) ^ B...
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the integration contour of the inverse Fourier transform of w^(k , z , t ). By substituting equation (7.169) into equation (7.168), we obtain for w^ (k , z , t ) the expression 45 w^ (k , z , t ) ikU ( (k) exp(ikU t), (7.172) ; sinh(m(z h)) ^ sinh(mh) 0 and its inverse Fourier transform gives the expression for w(x , z...
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2 k l with l , 2, : : : (7.176) 1 ± ; U h ( ) ( r These are frst order poles which are usually pure imaginary numb e r s , but we can have real poles for non-zero buoyancy frequency values if > for some values of l (smaller U h N l� values of l). The real p o l e s are associated with waves generated by the presence o...
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half part of the complex k plane for (U t + x ) > 0, or a semi-circle 0 of infnite radius lying in the lower half part of the complex k plane for (U t + x ) 0. 0 We consider a deformation also of the real axis to take into account the real poles that we may have. We have to decide if we deform the real axis to pass abo...
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the case (U t + x ) > 0. Then, we deform the real axis 0 to pass below the real poles that we m i g h t h a ve. The integration contours are illustrated in fgure 10. The contribution of the integral along the part of the contour that lies at the infnite in the complex k plane is zero. Therefore, the only contribution c...
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case (U t + x ) > 0, (B) - Deformed 0 integration contour for the case (U t + x ) 0. 0 48 L 0 (ik(U t +x )) i sinh(m(k , N /U )(h z )) ^ 0 w(x , z , t ) Res ikU ( (k) e ; 2 2 ; 2 sinh(m(k , N /U )h) j k= (N/U ) ;(j� /h ) p j=1 X L � � i sinh(m(k , N /U )(h z )) ^ (...
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t ) iRes ikU ( (k) e ; 2 2 sinh(m(k , N /U )h) j k=;i (j� /h ) ;(N/U ) p j=L+1 X � � (7.181) Since, the minimum value of l is one, the critical speed for a given value of the buoyancy frequency N is U . For current values U > , there is no wave � � N h N h disturbance downstream of the localized topography. To illust...
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Case x > 0. We assume that the frst L poles are real, and the poles for l > L are • pure imaginary. We have that 49 L i sinh(m(k , N /U )(h z )) ^ 1 (x, z ) Res ( (k) exp(ikx) ; 2 2 ; 2 sinh(m(k , N /U )h) j k= (N/U ) ;(j� /h) p j=1 X L � � i sinh(m(k , N /U )(h z )) ^ + Res ( (k) exp(ikx) ; 2 2 ; 2 sinh(m(k , N /U ...
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(j� /h) ;(N/U ) p j=L+1 X � � (7.184) The equations (7.183) and (7.184) can b e written in a simple way in terms of the quantities e , 1 and a defned in the appendix A. j j j Case x > 0, where the frst L poles are assumed real numbers, and the other poles • are in the upper part of the complex k plane. L 1 1 (x, z ) /...
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1 exp(a x), (7.187) j j j ; j=L+1 X We chose for ( (x) the same topography we considered in the previous section. The stream lines for this fow are illustrated in fgure 11, 12 and 13. For stream speeds that approach the critical values from below, the group velocity l� N h C for the l-th lee wave approaches the stream ...
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j ia i (j /h ) (N /U ) (A.189) j 2 2 ± ± ; p ...
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i x a z 15 10 5 -10 -5 0 5 10 15 20 25 x axis 30 35 40 45 50 Figure 13: Stream functions for the \Witch of Agnasi" topography in a fnite atmosphere with U /N 1 and A /2. We expect to see a superposition of nine lee wave for this 1 0 value of N /U . ...
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