Solution manual for chemical process equipment design by turton

Solution Manual for Chemical Process Equipment Design by Turton Full file at https:// Solutions Manual for Chemical Process Equipment Design (2017) by Richard Turton Joseph A Shaeiwitz Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc Full file at https:// Solution Manual for Chemical Process Equipment Design by Turton Full file at https:// Chapter 1  1   u3     gz  e f  Ws       u   dP first term: enthalpy term second term: kinetic energy, often written as (v)2)/2 third term: potential energy fourth term: frictional losses fifth term: shaft work mass flowrate constant, so, if larger pipe is and smaller pipe is   1 A1v1   A2 v2 , where A is cross sectional area for flow, if density constant, m then A1v1  A2 v2 , so if area goes down, velocity must go up in pipe Pressure head is a method for expressing pressure in terms of the equivalent height of a fluid, where the pressure head is the pressure at the bottom of a that height of that fluid From the mechanical energy balance, the value is obtained by dividing every term by g, such that all terms have units of length   Av , if the density is constant, since the cross-sectional area is constant, Since m the velocity must be constant 1-1 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc Full file at https:// Solution Manual for Chemical Process Equipment Design by Turton Full file at https://   Av  v m mass in = mass out, so if single pipe, mass flowrate must remain constant if density remains constant, volumetric flowrate remains constant if density and area remain constant, velocity must remain constant Re  inertial forces viscous forces inertial forces keep fluid flowing, viscous forces resist fluid flow 1-2 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc Full file at https:// Solution Manual for Chemical Process Equipment Design by Turton Full file at https:// all of the following have similar shape laminar flow flow in pipe f turbulent flow Re laminar flow CD turbulent flow flow past submerged object Re laminar flow flow in packed bed f turbulent flow Re 1-3 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc Full file at https:// Solution Manual for Chemical Process Equipment Design by Turton Full file at https:// pipe length: linear relationship, as pipe gets longer, friction increases proportionally velocity in pipe (or flowrate): in turbulent flow, friction increases as square of velocity or flowrate, linear in laminar flow pipe diameter: strong inverse relationship; in turbulent flow, friction goes with d-5; in laminar flow it is d-4 series: mass flowrate constant, pressure drops additive 10 parallel: mass flowrates additive, pressure drops equal 11 for compressible flow, density not constant as pressure changes, so must indicated integral dP   u   1     u    gz  e f  Ws  for incompressible flow, density constant, so first term is simplified P   u     gz  e f  Ws       u   12 frictional drag due to “skin friction,” which is contact with solid object/surface form drag is from energy loss due to flow around object (fluid changing direction takes energy) 13 in a packed bed, void fraction = volume of bed not occupied by solid (void space)/total volume of bed 1-4 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc Full file at https:// Solution Manual for Chemical Process Equipment Design by Turton Full file at https:// 14 total volume is volume of bed if empty solid volume is total volume of solids in bed void volume is volume of bed not occupied by solids total volume = solid volume + void volume 15 sphericity = surface area of sphere/surface area of particle, both having same volume 16 in a manometer, where the pressure drop is expressed as a positive number P  (  manometer fluid   fluid flowing ) gh , where h = is the difference in manometer fluid height So, for a large pressure drop, the height difference (and hence the height of the manometer required) decreases if the manometer fluid is dense, like mercury However, for very small pressure drops, accuracy is lost due to the small height difference in a mercury manometer, so a less dense manometer fluid is better 17 That is the flowrate at which the available net positive suction head equals the required net positive suction head For higher flowrates, the fluid will vaporize upon entering the pump, causing cavitation, which damages the pump However, it is physically possible to operate at higher flowrates 18 For a centrifugal pump, this is where the control valve is wide open, so there is minimal pressure drop across the control valve At this point, since the control valve is wide open, the flowrate is at its maximum possible value Therefore, it is physically impossible to operate at higher flowrates 1-5 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc Full file at https:// Solution Manual for Chemical Process Equipment Design by Turton Full file at https:// 19 The mechanical energy balance across a pump or compressor (neglecting any height difference between suction and discharge) reduces to dP Ws    Since vapor densities are 2-3 orders-of-magnitude lower than liquid densities, more shaft work is required, so the cost of power increases proportionally 20 Centrifugal compressors cannot achieve very large compression ratios (outlet pressure/inlet pressure), so staging is needed to get large compression ratios Positive displacement compressors can have larger compression ratios Energy in compression is minimized with isothermal compression, which is not possible, since compressing a gas causes the temperature to increase Isothermal operation could be approached with an infinite number of compression/intercooling stages with infinitesimal pressure and temperature increases, which is a nice limiting case, but impossible Staging compressors with intercooling is an attempt to approach the limiting case in a practical way The economics of a process determines the number of stages to use There is also the problem that if the temperature in a compressor stage increases too much, the seals will get damaged, which is a good reason the keep the compression ratio low 1-6 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc Full file at https:// Solution Manual for Chemical Process Equipment Design by Turton Full file at https:// 21 For fully developed turbulent flow, it is assume that the friction factor has reached its asymptotic value The proportionalities are Lv P  D a Since the pressure drop is proportional to flowrate squared, doubling the flowrate increases the pressure drop by a factor of four b Since the pressure drop is proportional to diameter to the negative fifth power, increasing the diameter by 25% changes the pressure drop by 1.25-5 = 0.33, so the pressure drop decreases by a factor of three Note that the friction factor may change slightly, since the roughness/pipe diameter value will change slightly This is ignored in all parts of this problem c v  D , so, for constant pressure drop, the flowrate increase significantly d it is exactly a proportional increase e v  L0.5 , so the flowrate decreases, but it is a one-half-power decrease, so the decrease in flowrate is less than the increase in length f subscript is for the original, long segment; subscript is for the shorter parallel segments; take the ratio of pressure drops in both cases, noting that the diameters are constant, that the length of pipe is half of pipe 1, and that the flowrates in each pipe are one-half of the original, so the pressure drop goes down by a factor of eight; note that minor losses due to the parallel piping are neglected P2 L2 v22 D15   0.5(0.5)  0.125 P1 L1 v1 D2 1-7 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc Full file at https:// Solution Manual for Chemical Process Equipment Design by Turton Full file at https:// 22 For laminar flow, the proportionalities are Lv P  D a Since the pressure drop is proportional to flowrate, doubling the flowrate doubles the pressure drop b Since the pressure drop is proportional to diameter to the negative fourth power, increasing the diameter by 25% changes the pressure drop by 1.25-4 = 0.41 c v  D , so, for constant pressure drop, the flowrate increase significantly, more than for turbulent flow d it is exactly a proportional increase e v  L1 , so the flowrate decreases in proportion to the increase in length f subscript is for the original, long segment; subscript is for the shorter parallel segments; take the ratio of pressure drops in both cases, noting that the diameters are constant, that the length of pipe is half of pipe 1, and that the flowrates in each pipe are one-half of the original, so the pressure drop goes down by a factor of four; note that minor losses due to the parallel piping are neglected P2 L2 v2 D14   0.5(0.5)  0.25 P1 L1 v1 D24 1-8 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc Full file at https:// Solution Manual for Chemical Process Equipment Design by Turton Full file at https:// 23 m  Av  v v  Av pipe – area from Table 1.1, density given m kg/s v1    0.00706 m /s  850 kg/m v1  v1 0.00706 m /s   3.26 m/s A1 21.65  10  m pipe – area from Table 1.1 m  v2  (850 kg/m )(0.0106 m /s)  9.01 kg/s v2  v2 0.0106 m /s   1.66 m/s A2 63.79  10 -4 m pipe – area from Table 1.1 v3  v3 A3  (4.032 m/s)(13.13  10 -4 m )  0.00529 m/s m  v3  (850 kg/m )(0.00529 m /s)  4.50 kg/s m  m  m  m m  10.51 kg/s pipe – area from Table 1.1 m 10.51 kg/s v4    0.0124 m /s  850 kg/m v4  v4 0.0124 m /s   2.59 m/s A4 47.69  10  m 1-9 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc Full file at https:// Solution Manual for Chemical Process Equipment Design by Turton Full file at https:// 32 P2 = 550 kPa P3 = 115 kPag P4 = 762.6 kPag a to get mass flowrate, MEB on pump P    pW s m 0 (762.6  115.6)(1000) N/m 0.75(6710 J/s)  0 m 900 kg/m m  7.0 kg/s b get area from Table 1.1 m 7.0 kg/s v2in    3.59 m/s A 900 kg/s(21,65  10 -4 m ) c MEB from 1-2 P    pW s v  gz  e f  0 m 550.000  P1 3.59  m / s 0.75(6710 J/s) m/s ( 25 m)    30 J/kg  50 J/kg 0 kg/s 900 kg/m P1  201.3 kPa 1-15 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc Full file at https:// Solution Manual for Chemical Process Equipment Design by Turton Full file at https:// d calculate velocity in 1-in pipe m 7.0 kg/s   0.946 m/s A 900 kg/s(82.19  10 -4 m ) across pump v1in  v 22  v12  m2 / s2 pressure term with answer to part c is 539 m2/s2, work term is 719 m2/s2, so KE term is small compared to them 1-16 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc Full file at https:// Solution Manual for Chemical Process Equipment Design by Turton Full file at https:// 33 P2 = 170.3 kPa P3 = 34.5 kPag P4 = 551.6 kPag a to get efficiency, MEB on pump P    pW s m 0  p (1000 J/s) (551.6  34.5)(1000) N/m  3 0 1000 kg/m 10 m/s(1000 kg/m )  p  0.52 b to get frictional loss, MEB from 1-2  pW s P  v  gz  e f  0  m 551,600  34,500 n/m 0.517(1000 J/s)  9.8 m/s (50 m)  e f 0 kg/s 1000 kg/m P1  490.1 J/kg 1-17 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc Full file at https:// Solution Manual for Chemical Process Equipment Design by Turton Full file at https:// 34  pW s  v  gz  e f  0  m P P1 = 25 psia a to get pump power, MEB across pump, so must determine pressure drop across pump from manometer   32.2 ft/sec  (12  10 ft)  2381.2 ft lb f / lb P  (13.6  0.88)(62.4 lb/ft )  32.2 ft lb/(lb sec ) f   3 ft / sec ft / sec  6.366 ft/sec  25.46 ft/sec v1  v2   (1 ft)  (0.5 ft)2 4  550 ft lb f   0.75W s  2 2 hp sec  2381.2 ft lb f / lb 25.46  6.366 ft sec    0 0.88(62.4 lb/ft ) ft /sec (62.4 lb/ft ) W  35.1 hp s 1-18 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc Full file at https:// Solution Manual for Chemical Process Equipment Design by Turton Full file at https:// b max height is z when pump is at 35.1 hp  550 ft lb f  75 ( 35 hp) 2 (14.7 - 25)lb f / ft 32 ft / sec  hp sec (144 in /ft )  z  3 0.88(62.4 lb/ft ) 32.2 ft lb/(lb f sec ) ft /sec (62.4 lb/ft )    0 z  79.8 ft z is difference in levels in tank, if destination tank is on ground, must add 10 ft for source tank and subtract ft for destination tank, so answer is 77.8 ft 1-19 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc Full file at https:// Solution Manual for Chemical Process Equipment Design by Turton Full file at https:// 35 a v2   m 6.5 kg/s   1.27 m/s A2 800 kg/m (63.79  10 -4 m ) b MEB from source tank level to pump inlet first need velocity in suction line m 6.5 kg/s v3    0.989 m/s A3 800 kg/m (82.19  10 -4 m ) P3  P1   v32  g ( z  z1 )  e f  P3  150,000 N/m (0.989 m/s)   9.81 m/s (5 m)  30 J/kg  800 kg/m P3  164.8 kPa c MEB on entire system P4  P1   pW s v 42   g ( z  z1 )   e f  0 m 200,000  150,000 N/m (1.27 m/s) 0.7(1500 J/s)   9.81 m/s (h m)  80 J/kg  0 6.5 kg/s 800 kg/m h  1.86 m so, height above ground = 6.86 m 1-20 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc Full file at https:// Solution Manual for Chemical Process Equipment Design by Turton Full file at https:// 36 a accumulation = in – out, nothing in, so dm   m dt b m  A2 v m  V  Atank h  A1 h d ( A1 h)   A2 v dt density constant since liquid dh  v A2 dt A1 dh  v A2 dt A1 c MEB on control volume tank level to point at pipe discharge v1 = in tank  pW s P  v  gz  e f  0  m v 22  v12  g ( z  z1 )  v 22  g (  h)  v  gh 1-21 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc Full file at https:// Solution Manual for Chemical Process Equipment Design by Turton Full file at https:// d integrate from initial height to any height from time zero to time t A define a  g A2 dh   ah 0.5 dt h t dh   a  0 dt h0 h at   h   h00.5   2  e at h = 2h00.5 t a 1-22 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc Full file at https:// Solution Manual for Chemical Process Equipment Design by Turton Full file at https:// 37 m  m  m  m P4  P3  P5  P4   v 42  v32   pW s   v 42  g ( z5  z )  e f  first equation suggests mixing second equation suggests pump with suction and discharge pipes having different diameters third equation suggest pressure loss due to friction and height change, with deceleration upon entering tank, since zero velocity at point 1-23 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc Full file at https:// Solution Manual for Chemical Process Equipment Design by Turton Full file at https:// 38 m  m  m  m P2  P1 v 22  g ( z  z1 )  0  g ( z  z1 )   pWs  P5  P4  P6  P5  v52  v 42    pW s  v52  g ( z6  z5 )  0 z  z1 first equation suggest mixing second equation suggests height change with acceleration, probably flow out of tank, since no pump, assume downward third equation suggests open-air tanks, since no KE term and no pressure terms fourth equation suggests flow across pump with different pipe diameters fifth equation suggest flow up to tank inequality suggests destination tank level higher than source tank level 1-24 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc Full file at https:// Solution Manual for Chemical Process Equipment Design by Turton Full file at https:// 39 P    pW s v  gz  e f  0 m P2  P1  v 22  v12 0  assume no friction, no height change, no pump (away from heart) denotes normal artery, denotes balloon region, where diameter is larger   Av , at so, if v2 < v1, since velocity goes down in larger diameter region (from m constant density and flowrate, of area goes up, velocity goes down), pressure must go up, because if velocity term is negative, pressure term must be positive, so they can add to zero pressure going up makes aneurysm worse, since it would keep expanding 1-25 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc Full file at https:// Solution Manual for Chemical Process Equipment Design by Turton Full file at https:// 40 fLv  pW s  v  gz   0 D m  since assumed horizontal and uniform diameter, M EB reduces to fLv  pW s  0 d m P fLv P   on pipe section, and pump power must equal D   m  v v v A 0.9    6.81    4 log 10     f  3.7 D  Re   P assume “generic” properties of water for density and viscosity  50 gal  ft    60 sec  7.48 gal  4.78 ft/sec v 0.02330 ft ft    50 gal    6.95 lb/sec m  (62.4 lb/ft )     60 sec  7.48 gal   2.067  ft (4.78 ft/sec)(62.4 lb/ft )  12   7.64  10 Re   4 6.72  10 lb/ft/sec   0.0018 in 0.9    6.81    4 log 10     f  3.7 D  Re   0.9  0.0018 in  6.81    4 log 10     f  3.7(2.067 in)  7.64  10   f  0.0056 1-26 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc Full file at https:// Solution Manual for Chemical Process Equipment Design by Turton Full file at https:// 0.8W s 2(0.0056)(5000 ft)(4.78 ft/sec)  0 (2.067 / 12 ft )(32.2 ft lb/lb f / sec ) 6.95 lb/sec W  2004 ft lb sec s f  hp sec W s  2004 ft lb f / sec  550 ft lb f    3.64 hp  41 fLv  pW s  v  gz   0  D m zero pressure drop since emerges at atmospheric and source tank is open no pump,so no work term P 2 fLv v  gz  0 D 0.9    6.81      4 log 10    f  3.7 D  dv   must solve two equations simultaneously location is tank level, location is discharge point, velocity in tank assumed zero v22 f (50 m)v22  9.81 m/s (10 m)  0 (0.0525m) 0.9  0.000045 m    6.81(0.001 kg/m/s)    4 log 10     f  3.7(0.0525 m)  (0.0525 m)v2 (1000 kg/m )   f  0.00788 v2  2.51 m/s v  2.51 m/s(21.65  10 -4 m )  0.0054 m / s 1-27 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc Full file at https:// Solution Manual for Chemical Process Equipment Design by Turton Full file at https:// 42 fLv  pW s  v  gz   0  D m tank to tank, so KE term zero fLeq v  pW s P   gz   0  D m P if use equivalent length method, or, if use velocity heads method fLv P v  pW s  gz    Ki  0  D m 0.9    6.81    4 log 10     f  3.7 D  Re   must look up 10 - in, schedule - 20 pipe informatio n     1500 gal   ft / 7.48 gal   5.83 ft/sec v  60 sec    10.25 ft         12 1500 gal ft / 7.48 gal (0.87)(62.4 lb/ft )  181.44 lb/sec m  60 sec (10.25 / 12 ft)(5.83 ft/sec)(0.87)(62.4 lb/ft ) Re   10057 40 cP(6.72  10 -4 lb/ft/sec/ cP)      0.0018  6.81    4 log 10     f  3.7(10.25)  10057   f  0.00776 equivalent length method  elbows  valves  tank entrance  tank exit 50(35)  15(9)  0.55(50)  1(50)(10.25 / 12)  1676.3 ft so Leq  70(5280)  1676.3  3.6713  10 ft 25(144)lb f / ft 32.2 ft/sec (100 ft) 2(0.00776)(3.713  10 ft)(5.83 ft/sec)   0.87(62.4 lb/ft ) 32.2 ft lb/lb f sec (10.25 / 12 ft)(32.2 ft lb/lb f sec ) 0.7W s (550 ft lb f /hp/sec)  0 181.44 lb/sec W s  3433 hp 1-28 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc Full file at https:// Solution Manual for Chemical Process Equipment Design by Turton Full file at https:// velocity heads method 25(144)lb f / ft 32.2 ft/sec (100 ft) 2(0.00776)(70)(5280 ft)(5.83 ft/sec)     0.87(62.4 lb/ft ) 32.2 ft lb/lb f sec (10.25 / 12 ft)(32.2 ft lb/lb f sec ) 0.7W s (550 ft lb f /hp/sec) (50(0.75)  15(0.17)  0.55  1) (5.83 ft/sec)  0 181.44 lb/sec 2(32.2 ft lb/lb f sec ) W  3438 hp s observe that both methods give virtually the same answer 1-29 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478 Copyright (c) 2017 by Pearson Education, Inc Full file at https://