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Transcript
Math with fixed number of mantissa digits Example 6 3 digit mantissa numbers Objective is to multiply these two numbers together. 0.512 x10 1 0.106 x10 2 0.512 x10 1 0.106 x10 2 37020 0000 0542 0.054272 Round off to 3 mantissa digits 0.0543 x 10 3 0.543 x 10 The multiplication with no constraints 2 Note: 3 x 10 Computer does these calculations in the base 2 number system so it is actually done as outlined above but with 1’s and 0’s. Example shows the philosophy of how it is done. Alternate approach (when there isn’t enough digits for intermediate steps) Truncate least significant digit that is outside the “digit space” as soon as it shows up in the calculation error is created after every arithmetic step error is multiplied if previous value is put back into another multiplication. Bottom Line When computers calculate bad STUFF happens so be skeptical of your numerical method results! Lagrangian Control Volume The reference space moves and the stuff in that space moves because the space is moving. Other perspective has the stuff moving through the non-moving control volume. < <l linear velocity through control volume <g acceleration of gravity <a acceleration of control volume m Mass flow rate l = l (t) m <g = <F One set of symbol options to indicate a vector > < l = (l x , l y , l z ) lx l > = ( ly lz lx Gravitational force on the object ) Velocity in x direction n Example 7 I=1 resultant = force +y <a [ m(t) ] Fn < I=1 <g + m The actual force that is pulling the control volume up +x n Balance statement: <g <Fn +z Develop the mass transport model for material referenced to the following Lagrangian control volume <l resultant = force = [ m(t) ] < F1 + < F2 <a = [ m(t) ] l ( -0, -0, -[m(t)] g ) + ( +0, +0, m z ) = ( 0, 0, +[m(t)] a ) z z <a This vector equation contains three scalar equations. (one for each of the direction components of the vector) (i) For the x direction: ( 0) + (0) = 0 l (ii) For the y direction: ( 0) + (0) = 0 l (iii) For the x direction: (-[m(t)] gz ) + m z = ( 0, 0, +[m(t)] a z ) ( d( vz (t) ) = - g + z l m z [m(t)] ) dt Note: m(t) = m m(t) = m t=0 0 + mt + mt a = z -[m(t)] g + m z z [m(t)] l m z d v (t) = - g + z z dt [m(t)] Lagrangian Control Volume Example 7 +y <a <Fn I=1 [ m(t) ] <g + m The actual force that is pulling the control volume up +x n Balance statement: <g +z Develop the mass transport model for material referenced to the following Lagrangian control volume <l resultant = force = [ m(t) ] < F1 + < F2 <a = [ m(t) ] l ( -0, -0, -[m(t)] g ) + ( +0, +0, m z ) = ( 0, 0, +[m(t)] a ) z z <a This vector equation contains three scalar equations. (one for each of the direction components of the vector) (i) For the x direction: ( 0) + (0) = 0 l (ii) For the y direction: ( 0) + (0) = 0 l (iii) For the x direction: (-[m(t)] gz ) + m z = ( 0, 0, +[m(t)] a z ) ( d( vz (t) ) = - g + z d( vz (t) ) = (- gz l + m z [m(t)] l m z [m(t)] ) dt ) dt Note: m(t) = m m(t) = m t=0 0 + mt + mt a = z -[m(t)] gz+ m z [m(t)] l m z d v (t) = - g + z z dt [m(t)] scalar the magnitude of m mass flow rate lz ml z linear velocity acceleration [m(t)] l force l momentum m z m z Numerical methods for engineers includes units (i) The three “3 unit” systems ( lenght, mass, time ) cgs mks 1 gram of stuff 2 accelerating at 1 cm/s 1 gram of stuff 2 accelerating at 1 m/s Weighs 981 dynes mass system A mass of stuff accelerating at “standard” sea level gravitational value Weighs 1 Newton force system Weighs 9.8 kg force The amount stuff that has this weight has a mass of 1 kg force-s 2/ meter (ii) 2 “national” systems British Mass System 1 pound of stuff mks Object accelerates American Engineering System 1 slug of stuff at 1 ft/sec 2 Object weighs 32.2 poundals (iii) the “4 unit” system < F = ( 1 ) (mass)< a g 32.2 lb ft mass 1 lb s2 force Object weighs 32.2 pounds force ( force, length, mass, time ) Common in USA 1 lb mass 1 lb will make 1 lbmass force 2 accelerate at 32.2 ft/s Weighs 1 lb force Example using 4 unit system Pressure difference = (top to bottom) P = gc 1 acceleration (mass )(a)(ft ) ft 3 Z water density P = (62.4 lbmass 2 P = (2,020 x 10 ft s 2 water tower is in Tampa ft lb mass ft3 ) ( 32.2 2 s2 ) the force is lb mass 32.2 lb ft mass 1 lb s2 force <F conversion factor between lb force and lb mass = ( force < F s2 force (mass) = 32.2 lb ft mass 1 lb Note: < 10 ft Not typical pressure units but they are still pressure units. Since mass is in lb g = 2 Z = 100 ft )( 10 ft) 2 a P (2,020 x 10 lbmass ft s 2 1 lb )( 32.2 lb s2 ft ) force mass 1 g ) (mass) <a 2 = 62.4 x10 lb force ft2 The “4 unit” system entertains two density concepts. g < F = (mass) < a (force magnitude) ( 32.2 lb ft mass 1 lb s2 32.2 lb 32.2 ft ft mass ) = (mass) ( ) 2 1 lb s 1 s2 force force force density 62.4 lb force 3 1 ft Both look the same (have units of pounds per foot cubed) but each represents a different concept. mass density 62.4 lb mass 3 1 ft g l m az z m m < < F1 <F =( 1 ) g <a Math with fixed number of mantissa digits Example 6 3 digit mantissa numbers Objective is to 0.512 x10 1 multiply these two numbers together. The multiplication with no constraints 0.512 x10 1 0.106 x10 2 0.106 x10 2 37020 Round off to 3 mantissa digits 0000 3 2 0.0543 x 10 0.543 x 10 0542 3 0.054272 x 10 Note: computer does these calculations in the base 2 number system so it is not actually done as outlined above. Example shows the philosophy of how it is done. Alternate approach (when there isn’t enough digits for intermediate steps) Truncate least significant digit that is outside the “digit space” as soon as it shows up in the calculation error is created after error is multiplied if every arithmetic step previous value is put back into another multiplication. Bottom Line When computers calculate bad STUFF happens so be skeptical of your numerical method results!