NUMERICAL ANALYSIS of PROCESSES

NAP6Partial differential equations (PDE), classification to hyperbolic, parabolic and eliptic equationsHyperbolic PDE (oscillation of trusses, beams, water hammer)MOC-method of characteristics (compressible flow)Rudolf itn, stav procesn a zpracovatelsk techniky VUT FS 2010

NAP6PDE partial differential equationsL.Wagner

NAP6PDE partial differential equationsWhen a system is described by a greater number of independent variables (spatial coordinates and time) we have to deal with partial differential equations (PDEs). Most PDE are differential equations with maximum second derivatives of the dependent variable (except eg. biharmonic equation deformation of membranes with the fourth derivative):Hyperbolic equation (oscillations and waves, supersonic flow. Characterised by finite velocity of pressure waves)

Parabolic equation (evolution problems, e.g. time evolution of a temperature or concentration profile, but also for example evolution of a boundary layer from inlet.)

Eliptic equation (steady problems of distribution temperatures, deformations,)

Example: Poissons equationExample: Fourier eqauation of heat transferExample: vibration of an elastic beam, water hammer

NAP6PDE partial differential equationsTyp of PDE is determined by coefficients of second derivatives (in case of the second order PDE)where a,b,c,f are arbitrary functions of x,y, solution and its first derivatives. Coefficients a,b,c determine characteristics y(x), satisfying equation

b2-4ac>0 hyperbolic equation (two real roots, therefore two characteristics)b2-4ac=0 parabolic equation (one root, one characteristic)b2-4ac

NAP6PDE Hyperbolic PDEWherever the effect of inertia forces and compressibility is important, the problem is described by hyperbolic partial differential equations, characterized by real characteristics (whose properties are sometimes used in the solution of hyperbolic PDE method of characteristics).

Nj as a base functionNAP6PDE vibration of trussThe equation describing the vibration truss is obtained by adding the accelerating force to the balance of forcesWeighted residual method using approximation Per partes integrationStiffness matrix[[K]]Mass matrix [[M]]Excitation Forces [[b]]Ni as a weight

NAP6PDE vibration of trussFree oscillation of frames without excitationis described by ordinary differential equations of the second order. The equations have constant coefficients therefore analytical solution existsSubstituting this solution to previous equation results to systm of algebraic equations for amplitudesBecause the systm is homogeneous a nontrivial solution exists only for singular matrix of the system, therefore ifMatrices [[K]] and [[M]] have N rows (rank N). Determinant is in fact a polynomial of the N-th degree in variable 2 and the algebraic equation has N-roots, N-eigenfrequencies. These frequencies together with the vectors of amplitudes are solution of eigenproblemThe eigenvalue problem is solved in MATLAB by function [u,omega]=eig(m\k)

NAP6PDE vibration of trussLinear base functionsresults to stiffness matrixand mass martixMass matrix is often substituted by diagonalised matrix, corresponding to uniform distribution of mass to nodes of elementDiagonalised matrix is easily inverted.Sum of matrix elements is the mass of element

NAP6PDE vibration of truss MATLABExample: The system of steel trusses (only one element for test)function [kl,ml,ig]=kloc(ie,x,con,a)E=200e9;R=7000;ae=a(ie);i1=con(ie,1);i2=con(ie,2);ig=[i1 i2];le=abs(x(i1)-x(i2));kl=E*ae/le*[1 -1;-1 1];ml=R*ae*le/6*[2 1;1 2];x=[0 1];con=[1 2];a=[1e-4];nu=length(x);ne=length(a);n=nu;k=zeros(n,n);m=zeros(n,n);for e=1:ne[kl,ml,ig]=kloc(e,x,con,a);k(ig(1:2),ig(1:2))=k(ig(1:2),ig(1:2))+kl(1:2,1:2);m(ig(1:2),ig(1:2))=m(ig(1:2),ig(1:2))+ml(1:2,1:2);end[u,omega]=eig(m\k);

In this case, we can easily check the result(for diagonalised mass matrix is the lower eigenfrequency lower )

Stiffness matrix [[K]]Mass matrix [[M]]NAP6PDE vibration of beamsPerhaps more important is the case of transverse vibration of beams, e.g. the coil supported on several locations on the ends of baffles and stuck into the tube sheet. Use the results of the previous lecture, where continuous lateral load beam to replace the inertia forcesBeyond that is the same as the truss, using only previously defined matrix stiffness and mass derived for the beam element.And you need not worry with the right side, because the beam support and restraint in their heads are strong boundary conditions.

NAP6PDE Water hammerUnsteady flow of compressible fluids in pipe networks, the cardiovascular circulation in elastic blood vessels, or supersonic flow around bodies, are typical examples of hyperbolic equations. Special case is a hydraulic hammer phenomenon, which occurs after the sudden closure of the valve in the pipeline ... resulting sudden increase in pressure (shock wave) propagates at the speed of sound in the fluid and in the pipe wall. Do you know how to calculate the speed? Do you know what the maximum pressure may arise?M. Rohani, M.H. Afshar Simulation of transient flow caused by pump failure: Point-Implicit Method of CharacteristicsAnnals of Nuclear Energy, Volume 37, Issue 12, December 2010, Pages 1742-1750 Barbara Wagner

NAP6PDE water hammerFormulation of problem:Pipe with a variable cross-section A (t, x), compressible fluid velocity v(t, x). The relationship between density and pressure is characterized by a volume compressibility modulus K [Pa]Aim of solution is the same as in the previously analysed case of incompressible fluids: calculation of pressure and velocity profiles along pipelines. This time the pressure and velocity depend also upon time p(t,x), v(t,x). These quantities are described by two equations, by continuity equation and by momentum balance (Bernoullis equation). In a simplified formBernoullis equationSpeed of sound aContinuity equation, for explanation see next slideThis is how to express flexibility of tube (cross section as a function of pressure)

NAP6PDE water hammer1D continuity equation for the case, when the cross section A(t,x) of an elastic pipe is inflated by action of inner pressureIflow in-outaccumulationj(the same equation written in material derivative)This term is neglected in the previous equation (usual assumption, see Khamlichi, Wave motion 1995)Mass balance of element dxVariable density, cross-section and velocityDensity - pressure

NAP6PDE water hammerElimination of velocity (continuity equation derived with respect time, Bernoulli equation wiith respect x and subtracted). Resulting equation for pressure p(t,x) or velocity v(t,x)Speed of soundare the same as equation describing oscillating truss. A small technical problem appears because of possible mix of different boundary conditions (pressure-velocity), for example waveform of the flow at one end and a pressure waveform at a second end. In case of water hammer effect in pipes it is represented by an abrupt change of flow at one end (valve closing), and a constant pressure at the second end (container). The method of weighted residuals is therefore necessary to solve the system of two PDE.

NAP6PDEFor hyperbolic equations, however, a different method that uses a specific feature of hyperbolic equations that is the existence of two real characteristics is often used.It is called method of characteristic (MOC).ZbranskRecommended reading Wylie, Streeter: Fluid Transients. McGraw Hill, 1978

Problem that will be solved

Compressible fluid flow in a stiff pipe or the incompressible fluid in an elastic tube (which may be "inflated" by pressure increase) - resulting compression effect is exactly the same. There are two factors determining the flow: the inertia of the fluid and the pressure exerted by the compression. The objective is to determine the mean velocity (hence the flow rate q) and a pressure p, both these values (v, p) varies in time and along the length of the tube. We will consider various boundary conditions at the tube ends: either specified (and time-varying) flow, or specified (and time-varying) pressure.

Method of characteristic MOCNAP6

Let us return back to the continuity and Bernoulli equationand adding up these two equations (the first is multiplied by arbitrary constant )Select a curve in the plane t-x (parameter of the curve x () can be directly time t). The full differential pressure and velocity along the curve areIf the curve x(t=) satisfy the equations (simultaneously)Method of characteristic MOCNAP6It is assumed that the velocity of floww is much less than the speed of sound-The following final equations expressed in term of total differentials along the curve x(t=) will be obtained(this curve is called CHARACTERISTIC of differential equation)

For exa,pme.The equation has two rootsNAP6Method of characteristic MOCand the two characteristics with differential equations that should be integratedIntegration in the direction of characteristics results to system of two algebraic equations for unknown pressure and velocity pC, vCFriction losses expressed in terms of Fanning friction factor fThe only inaccuracy is neglection of variability of friction losses

NAP6Solution of these algebraic equationscan be immediately applied for calculation of velocities and pressures at a time level C, given values at points A,B at an old time level.Method of characteristic MOC

NAP6Method of characteristic MOCAn alternative (and perhaps illuminating) formulation is analogous procedure that we used in solving a system of differential equations of heat exchangers, see Whitham: Linear and nonlinear waves, Wiley, 1974. The system of two differential equations for pressure and velocity can be written in matrix formApplication of suitable transformation (eigenvalue problem) results to system of separated equations[[Q]] matrix of eigenvectors [[A]] is diagonal matrixFor our specific case the eigen problem (eigenvectors of [[A]] ) can be solved analyticallyfunction Z1 by integration of equationalong characteristic dx/dt=-aalong characteristic dx/dt=afunction Z2 by integration of equation

NAP6The following example illustrates solutions to the problem of compressible flow in the pipe, where the input is the specified pressure (e.g. constant p0) and the output waveform of the specified flow rate, e.g. rapidly decreasing flow, a situation corresponding to the rapid closing the valve. The question is primarily what pressure increase causes the valve closing (water hammer in pipes).MOC water hammer MATLAB

NAP6Let us return back to final algebraic equation of MOCthat describe velocities and pressures in inner nodes of mesh. Only one characteristic aims to the boundary nodes, therefore the second equation representing boundary condition is necessary (waveform of pressure at inlet and velocity at outlet, for example zero velocity after closing the valve)Function describing a continuous closing of valve function vrel=valve(t) if t

NAP6MOC water hammer MATLABl=1;d=0.01;rho=1000;f=0.1;a=1;p0=2e3;v0=(p0*2*d/(l*rho*f))^0.5n=101;h=l/(n-1);v(1:n)=v0;p(1)=p0;for i=2:n p(i)=p(i-1)-f*rho*v0^2*h/(2*d);enddt=h/a;tmax=3;itmax=tmax/dt;fhr=f*h/(2*a*d); for it=1:itmax t=it*dt; for i=2:n-1 pa=p(i-1);pb=p(i+1);va=v(i-1);vb=v(i+1); pc(i)=a/2*((pa+pb)/a+rho*(va-vb)+fhr*(vb*abs(vb)-va*abs(va))); vc(i)=0.5*((pa-pb)/(rho*a)+va+vb-fhr*(vb*abs(vb)+va*abs(va))); end pc(1)=p0; vb=v(2);pb=p(2); vc(1)=vb+(pc(1)-pb)/(a*rho)-fhr*vb*abs(vb); vc(n)=v0*valve(t); va=v(n-1);pa=p(n-1); pc(n)=pa-rho*a*(vc(n)-va)+f*h*rho/(2*d)*va*abs(va); vres(it,1:n)=vc(1:n); pres(it,1:n)=pc(1:n); p=pc;v=vc;endpmax=max(max(pres))/p0x=linspace(0,1,n);time=linspace(0,tmax,itmax);contourf(x,time,pres,30)Tube L=1m, D=0.01 m, speed of sound a=1 m/s, steady forward velocity v=0.6325 m/s.At t=1.1 backpressure returns to in letAt t=0.1 the valve at outlet is closing }gradually)

NAP6MOC water hammer MATLABtime (up to 6 s)Rate of closing the valve to maximum overpressure (not to big)vrel=exp(-10*(t-.1)); fast closing approx. 0.1 svrel=exp(-50*(t-.1)); very fast closingn=101v0 =0.6325 m/s a=1 m/s =1000 kg/m2pmax =2180 PaApproximate analysis of maximum pressure: kinetic energy of liquid moving at speed v is transformed into deformation energy after a collisionLvFor this case pmax=632 Pa, that is only about 30% of the actual value. This analysis is not very accurate because the cylinder is not compressed uniformlySpeed of soundvrel=exp(-0.1*(t-.1)); slow closing approx. 10sukovsky equation see next slide

NAP6MOC water hammer MATLABThe relationship between the speed of sound, velocity v and pressure of the liquid column p expresses centuries old water hammer equations (Young 1808, the modern form of the equation see ukovsky 1898 for the water hammer in the elastic tube)The formula is so well known that only seldom its derivation is presented: dv piston velocitya the speed of the front shock wavesConstant velocity up to the shock Prerequisite step down density and pressure in the shock wave zoneThe derivation is based on the mass and momentum balance of control volume that moves at a constant speed to the rightvp,Substituting momentum balance results to the water hammer equationfluid velocity relative to the interface, which moves at the speed of sound aMomentum flux is velocity times densitycontrol volume moving at speed of sound to the rightMass balanceMomentum balance

NAP6MOC water hammer MATLABThese black pages, you can safely skip. They are just an example of how to make a numerical model more realistic (calculating realistic friction coefficient) and also demonstrate that MATLAB is very slow (because it is only an interpret)The following program was rewritten to FORTRAN (MATLAB loops for i=2:n-1, are replaced by do i=2,n-1, and commands if re

NAP6MOC water hammer MATLABp0=1e3;l=1;d=0.01;rho=1000;mju=0.001;a=4;vlam=p0*d^2/(32*mju*l);re=vlam*d*rho/mjuif re

NAP6MOC water hammer MATLABVelocityPressureWater, a=4 m/s, L=1m, D=0.01 m, p0=1 kPa, v0=1.14 m/s, Re=31000Pressure at closing valveAccelerated pulsation is caused by a higher speed of sound (stiffer elastic tube), a = 4 m / s. Time of flight of pressure waves is reduced to 0.25 s.

NAP6MOC water hammer MATLABThe case with oscillating flowrate at inlet (displecement pump) and constant pressure at outlet (e.g. zero pressure)v(t) function vrel=pump(t) vrel=0.5*sin(3*t);

p=0vc(1)=pump(it*dt);vb=v(2);pb=p(2);pc(1)=pb+rho*a*(vc(1)-vb)+f*h*rho/(2*d)*vb*abs(vb);

va=v(n-1);pa=p(n-1); pc(n)=0;vc(n)=va-(pc(n)-pa)/(rho*a)-fhr*va*abs(va);BA

NAP6

I am not sure, the solution may be wrongMOC elastic and rigid pipe

NAP6MOC elastic and rigid pipev(t)pe(t)LLeElastic pipe. Speed of sound a is determined by elasticity of pipeRigid tube. Infinite speed of sound.Consider the case of an incompressible fluid flow elastic tube (e.g. blood vessel) which is connected to the tube (perhaps the same diameter), but perfectly rigid. This seemingly innocuous combination of boundary conditions poses a problem for numerics. The rigid tube and the incompressible fluid speed of sound is infinite, the characteristics are parallel to the axis of the tube and the corresponding time step is infinitesimaly shortabrupt change in impedance (resistance) causes reflection of waves that interfere with the waves that are moving towards the right end (in the direction of flow)

NAP6MOC elastic and rigid pipeIncorrect (?) boundary condition at end (this condition includes 1st derivative)ACeBernoulli rovnici, should be valid at outlet of elastic tubeSubtraction of Bernoulli equation for rigid pipe D

NAP6Program MATLABle=0.001;de=0.1;pe=0;l=1;d=0.01;rho=1000;f=0.1;a=1;p0=0;v0=0;n=401;h=l/(n-1);v(1:n)=v0;p(1)=p0;for i=2:n p(i)=p(i-1)-f*rho*v0^2*h/(2*d);enddt=h/a;tmax=3;itmax=tmax/dt;fhr=f*h/(2*a*d); for it=1:itmax t=it*dt; for i=2:n-1 pa=p(i-1);pb=p(i+1);va=v(i-1);vb=v(i+1); pc(i)=a/2*((pa+pb)/a+rho*(va-vb)+fhr*(vb*abs(vb)-va*abs(va))); vc(i)=0.5*((pa-pb)/(rho*a)+va+vb-fhr*(vb*abs(vb)+va*abs(va))); end vc(1)=pump(it*dt);vb=v(2);pb=p(2); pc(1)=pb+rho*a*(vc(1)-vb)+f*h*rho/(2*d)*vb*abs(vb); va=v(n-1);vb=v(n);pa=p(n-1); pc(n)=(pc(n-1)/h+pe/le-f*rho*vb*abs(vb)/(2*d*de)*(de-d))/(1/h+1/le); vc(n)=va+1/(rho*a)*(pa-pc(n)-f*h*rho/(2*d)*va*abs(va)); vres(it,1:n)=vc(1:n); pres(it,1:n)=pc(1:n); p=pc;v=vc;endpmax=max(max(pres))x=linspace(0,1,n);time=linspace(0,tmax,itmax);contourf(x,time,pres,30)MOC elastic and rigid pipe

NAP6MOC elastic and rigid pipePressure profiles in elastic tube (a=1 m/s, L=1 m, D=De=0.01 m)

NAP6MOC elastic and rigid pipeExperiments conducted in our laboratory did not operate with a harmonic flow, but only with the temporary (transient) increase. Also, the actual speed of sound is higher. This also increases problems with numerical solutions.

NAP6The elastic tube (aortic pulse wave propagation velocity = 8 m / s) and connected rigid tube (plexiglass). Puls flow at the input generates an electronically controlled SuperPump (Varimex). Output to stilling tank.MOC elastic and rigid pipeExample of experiments from our laboratory (pulse wave causes deformation of the aorta, and this is monitored by a pair of high-speed cameras, DIC = Digital Image Correlation). Finite differences Lax Wendroff and Muscle (Monotone Upstream Scheme for Conservation Laws) methods are used. The problem is that these methods do not work when a perfectly rigid tube is connected to an elastic tube (the rigid tube has to be modeled as a flexible tube of a high rigidity). Mathematicians say that in this hyperbolic equation can not be used a boundary condition with the derivative. If they are right, the following solution is wrong ...

NAP6MOC elastic and rigid pipeLinear change of flowrate at inlet (ramp function). Laminar and turbulent regime.le=0.1;de=0.015;pe=0;mju=0.001;l=.18;d=0.015;rho=1000;a=8;v0=pump(0);ve=v0*(d/de)^2;f=frict(v0,d,rho,mju);fe=frict(ve,de,rho,mju);re=v0*d*rho/mjudpe=0.5*fe*le/de*rho*ve^2;dp=0.5*f*l/d*rho*v0^2;n=401;h=l/(n-1);v(1:n)=v0;p(1)=pe+dp+dpe;for i=2:n p(i)=p(i-1)-f*rho*v0^2*h/(2*d);enddt=h/a;tmax=3;itmax=tmax/dt; for it=1:itmax t=it*dt; for i=2:n-1 pa=p(i-1);pb=p(i+1);va=v(i-1);vb=v(i+1); fa=frict(va,d,rho,mju);fb=frict(vb,d,rho,mju); ea=fa*h/(2*a*d)*va*abs(va); eb=fb*h/(2*a*d)*vb*abs(vb); pc(i)=a/2*((pa+pb)/a+rho*(va-vb)+eb-ea); vc(i)=0.5*((pa-pb)/(rho*a)+va+vb-eb-ea); end vc(1)=pump(it*dt);vb=v(2);pb=p(2); fb=frict(vb,d,rho,mju); pc(1)=pb+rho*a*(vc(1)-vb)+fb*h*rho/(2*d)*vb*abs(vb); va=v(n-1);vb=v(n);pa=p(n-1); fb=frict(vb,d,rho,mju); fa=frict(va,d,rho,mju); pc(n)=(pc(n-1)/h+pe/le-fb*rho*vb*abs(vb)/(2*d*de)*(de-d))/(1/h+1/le); vc(n)=va+1/(rho*a)*(pa-pc(n)-fa*h*rho/(2*d)*va*abs(va)); vres(it,1:n)=vc(1:n); pres(it,1:n)=pc(1:n); p=pc;v=vc;endfunction vrel=pump(t)if t

NAP6MOC elastic and rigid pipeLinear increase of flowrate at inlet (ramp). Laminar and turbulent regime.0.1810.180.1

NAP6MOC elastic and rigid pipeVelocity and pressure at the end of the elastic tube. Reduction of frequency seems logical to me, because increased inertial mass and unchanged elasticity (L=0.18, Le=0.01, pmax=2946 Pa) (L=0.18, Le=0.1, pmax=3843 Pa).0.180.10.180.01

NAP6The lecture was devoted to the classification of partial differential equations of second order, with special attention to the hyberbolic type. Remember at least what is the next slideWhat is important

NAP6What is importantType of equation is determined by coefficients at the highest (second) derivativesCharacteristics are real if b2-4ac>0 (hyperbolic equation)One characteristic if b2-4ac=0 (parabolic equation)Real characteristics do not exist if b2-4ac