# SolidsNotes13 DESIGN EXAMPLE.pdf

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DESIGN EXAMPLE – DILUTE PHASE PNEUMATIC CONVEYING A plastics production plant wants to increase the capacity through an existing conveying system. The existing system has 6 inch ID pipes and is configured as shown in the diagram below. The High Density Polyethylene (HDPE) particles have an average size of 4 mm. The conveying gas is at 68oF. The existing blower can produce 1375 SCFM. The desired capacity increase is from 20,000 lbm/hr to 30,000 lbm/hr. Can the existing b
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1 DESIGN EXAMPLE – DILUTE PHASE PNEUMATIC CONVEYING A plastics production plant wants to increase the capacity through an existing conveying system. The existing system has 6 inch ID pipes and is configured as shown in the diagram below. The High Density Polyethylene (HDPE) particles have an average size of 4 mm. The conveying gas is at 68 o F. The existing blower can produce 1375 SCFM. The desired capacity increase is from 20,000 lbm/hr to 30,000 lbm/hr. Can the existing  blower and pipe system meet this increase in capacity? Assume the pressure drop across the cyclone is 5 inches of water. The pressure drop across the blower inlet pipe and silencers is 0.3 psi. The pipe bends have R/D = 6. Pipe roughness is k = 0.00015 ft. The particles have density  p  ρ   = 59 lbm/ft 3 . Terminal velocity of the particles is t  U   = 30.6 ft/s. a  b c d e f g HOPPER BLOWER CYCLONE 325 ft 50 ft (vertical rise) 100 ft   2 SOLUTION Push conveying systems are conveniently calculated from exit to inlet (the exit pressure boundary condition is known). Hence, on the figure the points at which the pipe line is segmented are labeled ‘a’ through ‘g’ from exit to inlet. a-b Cyclone P a  = 14.7 psia (exit atmospheric pressure) P  b -P a  = cyclone P ∆  = 5 inches water (given)   P  b  = P a  + cyclone P ∆  = ____________  b-c Horizontal Pipe L = 325 ft Use density and velocity values at point ‘b’   === sP psi ft  ft PP AQV  bbSTPSTPg 60714196301375 23 min..min /   _________ Gas friction factor, from Churchill’s equation ( ) 1616901212312 37530270714572182 =    +    =++    = Re.Reln.Re . /  /   B Dk  A B A f   k = 0.00015ft, Re = 363,037 f = 0.00419 Gas density at point ‘a’from ideal gas law 33 07530528297310714  ft lbm Rlbmollbm Rlbmol ft  psi psi RT PM  oog  / ... ===  ρ     psiinches psiwater  180682715 ..  =      P  b  = 0.18 + 14.7 = 14.88 psi 222 196304504  ft  ft  D A .).( ===  π π   Velocity at point ‘b’ s ft  ft V  g  / .min / ... 311569208814714196301375 ===  Viscosity is insensitive to pressure At 68 o F s ft lbmcP g 000011400170 ..  == µ   Gas density at point ‘b’ 3 /0762.0 7.1488.140753.0  ft lbmPP STPbgSTPg ===  ρ  ρ    037363000014050311507620  3 , /  / .. / . / . Re == s ft lbm ft s ft  ft lbm   ( ) cgg zhoriz gV  D L f P 24 2  ρ µ λ  +=∆   3 Solids mass to air mass ratio ___________minmin /  / .  /  === 6013750752030000 33 hr  ft  ft lbmhr lbmQw STPgSTPs  ρ µ     z λ   for particles with d   p  > 500 microns = _________ = _________  z λ   = ____________ Hence =∆  bc P  __________ At point ‘c’ P c  = P  b  + =∆  bc P  _________ === sP psi ft  ft PP AQV  ccSTPSTPg 60714196301375 23 min..min /   _________ 71407530 ..  cSTPcgSTPg PPP ==  ρ  ρ   = ________ µ   = 4.833 d   p  = 4mm = 0.01312 ft D = 0.5 ft ( ) 38265017432 3115 22 .. / . / . ==  ft s ft s ft Fr    ( ) 221801312.0/174.32 /6.30 22 ==  ft s ft s ft Fr   p    z λ   = 0.001564 ( )( )  psiin ft  xslbf  ft lbms ft  ft lbm  xP bc 7291144774322311507620 50325833400156400041904 22223 .. / . / . .... =⋅⋅+⋅=∆  At point ‘c’ P c  = 14.88 + 1.729 = 16.61 psi g V   = 101.3 ft/s g  ρ   = 0.0850 lbm/ft 3   1025086030 0820 .... .     =  −−  ps z d  DFr Fr  µ λ  gDV Fr   g 2 =  pt  p gd U Fr  2 =   4 c-d Pipe Bend cd  P ∆  = ___________ For 6 >  D R  B = ________ cd  P ∆  = ___________ At point ‘d’   P d   = ___________ === sP psi ft  ft PP AQV  d d STPSTPg 60714196301375 23 min..min /   _________ 71407530 ..  d STPd gSTPg PPP ==  ρ  ρ   = ________ d-e Vertical Pipe  z ∆  = 50 ft    f = _________ Fr   = _________  z λ   = ____________ = ________ = ________  p V   = _________ µ   = 4.833 from previous page (loading is constant) B = 0.5 for this R/D ( )  psiin ft  xP cd  28601441174322310308500 8334150 222 ....).(. =⋅+=∆  P d   = 16.61 + 0.286 = 16.89 psia g V   = 101.55 ft/s g  ρ   = 0.0850 lbm/ft 3   Note gg V   ρ   = constant Hence, Re does not change. Since  f = f(Re, k/D) then f does not change.  f = 0.00419 µ   = constant = 4.833  p Fr   = constant = 2218 ( ) ==  ft s ft s ft Fr  5017432 55101 22 . / . / . 641.05 = 0.001946 = 0.7425  p V   = 75.40 ft/s ( ) s ft  ft lbm ft  shr hr lbm  / . / .  /  / , 40755919630 3600100030 1 32 −= ε   =0.9905 ( ) cggbend  gV  BP 21 2  ρ µ  +=∆ ( ) ccgg zvert  gg zgV  D L f P  ∆++=∆  02 24  ρ  ρ µ λ  1.025.086.03.0 082.0     =  −−  p p z d  DFr Fr  µ λ  ( )  p ps pg V  Aw  ρ ε  ρ ε ερ  ρ  −=−+= 11 0 5030 12301  .. .  p pg p d V V   ρ  −= 5030 5901312012301  .. ).(. −= g p V V
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