Chemical Process Equipment Design Solution Manual

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. Chapter 1 1. 1   u3    1  2   u    gz  e f  Ws  0 2 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 2. mass flowrate constant, so, if larger pipe is 1 and smaller pipe is 2   1 A1v1   2 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 2 3. 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. 4.   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. 5.   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 6. 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. 7. 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. 8. 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 9. 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 do indicated integral 2 dP 1   u 3   1   2   u    gz  e f  Ws  0 for incompressible flow, density constant, so first term is simplified P 1   u 3     gz  e f  Ws  0    2   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. 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. 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. 21. For fully developed turbulent flow, it is assume that the friction factor has reached its asymptotic value. The proportionalities are Lv 2 P  5 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. 5 c. v  D 2 , 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 1 is for the original, long segment; subscript 2 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 2 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) 2  0.125 2 5 P1 L1 v1 D2 1-7 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. Copyright (c) 2017 by Pearson Education, Inc. 22. For laminar flow, the proportionalities are Lv P  4 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 4 , 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 1 is for the original, long segment; subscript 2 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 2 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. 23. m  Av  v v  Av pipe 1 – area from Table 1.1, density given m 6 kg/s v1  1   0.00706 m 3 /s 3  850 kg/m v1  v1 0.00706 m 3 /s   3.26 m/s A1 21.65  10  4 m 2 pipe 2 – area from Table 1.1 m 2  v2  (850 kg/m 3 )(0.0106 m 3 /s)  9.01 kg/s v2  v2 0.0106 m 3 /s   1.66 m/s A2 63.79  10 -4 m. 2 pipe 3 – area from Table 1.1 v3  v3 A3  (4.032 m/s)(13.13  10 -4 m 2 )  0.00529 m/s m 3  v3  (850 kg/m 3 )(0.00529 m 3 /s)  4.50 kg/s m 1  m 2  m 3  m 4 m 4  10.51 kg/s pipe 4 – area from Table 1.1 m 10.51 kg/s v4  4   0.0124 m 3 /s 3  850 kg/m v4  v4 0.0124 m 3 /s   2.59 m/s A4 47.69  10  4 m 2 1-9 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. Copyright (c) 2017 by Pearson Education, Inc. 24.   Av m assume = 1000 kg/m3 for all sections m 1  A1v1  1000 kg/m 3 (5.574  10 4 m 2 )(5 m/s)  2.79 kg/s m 2  A2 v2  1000 kg/m 3 (13.13  10 4 m 2 )(3 m/s)  3.94 kg/s a. m 1  m 2  m 3  m 4 m 3  m 4  6.73 kg/s b. v4  m 4 6.73 kg/s   0.818 m/s A4 1000 kg/m 3 (82.19  10 4 m 2 ) c. A3  m3 6.73 kg/s   30.85  10 4 m 2 3 A3 1000 kg/m (2.18 m/s) area is closest to 2.5-in, schedule-40 pipe, and that pipe has a slightly larger xs area, so it is a good choice 1-10 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. Copyright (c) 2017 by Pearson Education, Inc. 25. Ai  d i2 4 , so A4-in = 0.0873 ft2, A3-in = 0.0491 ft2, A2-in = 0.0218 ft2, Atank = 19.63 ft2 a. in tank v  19.63 ft 2 (0.02 ft/sec)  0.3927 ft 3 /sec b. at constant density, volumetric flowrates balance vtank   vin out 0.39 ft 3 /sec  0.0873 ft 2 (8 ft/sec) – 0.0218 ft 2 (4 ft/sec) – 0.0491 ft 2 v3in v3in  4.5 ft/sec 26. P    pW s 1 2 v  gz  e f  0 2 m assume inlet and outlet at same pressure since no information provided uniform pipe diameter, so kinetic energy term zero pipe length not needed, since frictional loss given 0.8(20 hp)(550 ft lb f / (hp sec)) 32.2 ft/sec 2 (50 ft)  80 ft lb f /lb 0 2 m 32.2 ft lb/(lb f sec ) m  67.7 lb/sec v  67.7 lb/sec (60 sec/min)( 7.48 gal/ft 3 )  487 gal/min 3 62.4 lb/ft 1-11 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. Copyright (c) 2017 by Pearson Education, Inc. 27. P W 1  v 2  gz  e f  s  0  2  t m inlet and discharge both at atmospheric pressure, so pressure term is zero inlet velocity is zero, since reservoir is like tank, so level is assumed constant multiply 20 m of head by g to get frictional loss in correct units W s (1 m/s) 2  0  9.81 m/s 2 (50 m)  10 m(9.81 m/s 2 )  0 2 0.80(25 m 3 /s)(1000 kg/m 3 ) 2 kg m W s  7.84  10 6  7.84  10 6 W  7.84 M W 3 s 28.  pW s 1  v 2  gz  e f  0  2 m P P = 0, since both tanks are open to the atmosphere velocities both zero since tank levels (9.81 m/s 2 )(10  z m)  3.5 J/kg – 0.75(100 J/s) 0 10 m /h(1 h/3600 s)(1000 kg/m 3 ) 3 z  7.6 m 1-12 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. Copyright (c) 2017 by Pearson Education, Inc. 29. P W 1  v 2  gz  e f  s  0  2  t m pressure term is zero since reservoir and discharge are both at atmospheric pressure frictional loss assumed zero, since nothing stated location 1 is reservoir, location 2 is discharge v2  65000 lb/sec  13.26 ft/sec 2    (10 ft)  62.4 lb/ft 3  4   32.2 ft/sec 2 13.26 2  0 ft 2 / sec 2 W s (550 ft lb f /(hp sec)) (  60 ft)   0 0.55(65000 lb/sec) 32.2 ft lb/lb f /sec 2 2(32.2 ft lb/lb f /sec 2 ) W  3723 hp s 30. P 1  v 2  gz  e f  Ws  0  2 This is actually a nozzle problem, so only pressure and kinetic energy terms remain. must look up vapor pressure of water at 25°C, which is 3.168 kPa location 1 is before the orifice, so P1 = 34,500 kPa, P2 = 3.168 kPa 250 mL/min(min /60 s)(m 3 / 10 6 mL) v1   5.305  10  4 m/s 2  (0.1 m) 4  3168 – 34500 N/m 2 v 22  5.305  10  4  2 1000 kg/m 3  m /s  0 2 2 2 v 2  262.67 m/s 3  d 22   250 mL/min  m  v 2      6   A2 v 2   4 (262.67 m/s)  60 s/min  10 mL    d 2  1.42  10  4 m 1-13 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. Copyright (c) 2017 by Pearson Education, Inc. 31.  pW s 1  v 2  gz  e f  0  2 m P = 0, since both tanks are open to the atmosphere velocities both zero since tank levels density of water in lb/gal = (62.4 lb/ft3)(ft3/7.48 gal) =8.3 lb/gal P a. 0.8(W s hp)(550 ft lb f / (hp sec)) 32.2 ft/sec 2 ( 873 928 ft) 200 ft lb /lb   0 f (30 gal/min)(m in/60 sec)(8.3 lb/gal) 32.2 ft lb/(lb f sec 2 ) W  1.37 hp s b. without pump, need sufficient head from reservoir to overcome friction 32.2 ft/sec 2 (z ft)  200 ft lb f /lb  0 32.2 ft lb/(lb f sec 2 ) z  200 ft  z – 928 z  728 ft 1-14 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. Copyright (c) 2017 by Pearson Education, Inc. 32. P2 = 550 kPa P3 = 115 kPag P4 = 762.6 kPag a. to get mass flowrate, do MEB on pump P    pW s m 0 (762.6  115.6)(1000) N/m 2 0.75(6710 J/s)  0 m 900 kg/m 3 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 2 ) c. do MEB from 1-2 P    pW s 1 2 v  gz  e f  0 2 m 550.000  P1 3.59 2  0 m 2 / s 2 0.75(6710 J/s) 2 9 . 8 m/s ( 25 m)    30 J/kg  50 J/kg 0 3 2 7 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. d. calculate velocity in 1-in pipe v1in  m 7.0 kg/s   0.946 m/s A 900 kg/s(82.19  10 -4 m 2 ) across pump v 22  v12  6 m2 / s2 2 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. 33. P2 = 170.3 kPa P3 = 34.5 kPag P4 = 551.6 kPag a. to get efficiency, do MEB on pump P    pW s m 0  p (1000 J/s) (551.6  34.5)(1000) N/m 2  3 0 3 1000 kg/m 10 m/s(1000 kg/m 3 )  p  0.52 b. to get frictional loss, do MEB from 1-2  pW s P 1 2  v  gz  e f  0  2 m 551,600  34,500 n/m 2 0.517(1000 J/s)  9.8 m/s 2 (50 m)  e f 0 3 1 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. 34.  pW s 1  v 2  gz  e f  0  2 m P P1 = 25 psia a. to get pump power, do MEB across pump, so must determine pressure drop across pump from manometer   32.2 ft/sec 2  (12  10 ft)  2381.2 ft lb f / lb P  (13.6  0.88)(62.4 lb/ft ) 2  32.2 ft lb/(lb sec ) f   3 3 5 ft / sec 5 ft / sec  6.366 ft/sec  25.46 ft/sec v1  v2  2  (1 ft)  (0.5 ft)2 4 4  550 ft lb f   0.75W s  2 2 2 2 hp sec  2381.2 ft lb f / lb 25.46  6.366 ft sec    0 2 0.88(62.4 lb/ft 3 ) 5 ft 3 /sec (62.4 lb/ft 3 ) W  35.1 hp 3 s 1-18 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. Copyright (c) 2017 by Pearson Education, Inc. b. max height is z when pump is at 35.1 hp  550 ft lb f    0 . 75 ( 35 . 1 hp) 2 2 hp sec (14.7 – 25)lb f / ft 2 32 . 2 ft / sec  0  (144 in 2 /ft 2 )  z  3 2 3 3 0.88(62.4 lb/ft ) 32.2 ft lb/(lb f sec ) 5 ft /sec (62.4 lb/ft ) 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 2 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. 35. a. v2   m 6.5 kg/s   1.27 m/s 3 A2 800 kg/m (63.79  10 -4 m 2 ) 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 3 A3 800 kg/m (82.19  10 -4 m 2 ) P3  P1   v32  g ( z 3  z1 )  e f  0 2 P3  150,000 N/m 2 (0.989 m/s) 2   9.81 m/s 2 (5 m)  30 J/kg  0 3 2 800 kg/m P3  164.8 kPa c. MEB on entire system P4  P1   pW s v 42   g ( z 4  z1 )   e f  0 2 m 200,000  150,000 N/m 2 (1.27 m/s) 2 0.7(1500 J/s)   9.81 m/s 2 (h m)  80 J/kg  0 3 2 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. 36. a. accumulation = in – out, nothing in, so dm   m 2 dt b. m  A2 v m  V  Atank h  A1 h d ( A1 h)   A2 v dt density constant since liquid dh  v 2 A2 dt A1 dh  v 2 A2 dt A1 c. MEB on control volume tank level to point 2 at pipe discharge v1 = 0 in tank  pW s P 1 2  v  gz  e f  0  2 m v 22  v12  g ( z 2  z1 )  0 2 v 22  g (  h)  0 2 v 2  2 gh 1-21 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. Copyright (c) 2017 by Pearson Education, Inc. d. integrate from initial height to any height from time zero to time t A define a  1 2 g A2 dh   ah 0.5 dt h t dh   a  0. 5 0 dt h0 h at   h   h00.5   2  2 e. at h = 0 2h00.5 t a 1-22 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. Copyright (c) 2017 by Pearson Education, Inc. 37. m 1  m 2  m 3  m 4 P4  P3  P5  P4   v 42  v32   pW s  0 2  v 42  g ( z5  z 4 )  e f  0 2 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 5 1-23 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. Copyright (c) 2017 by Pearson Education, Inc. 38. m 3  m 2  m 4  m 5 P2  P1 v 22  g ( z 2  z1 )  0  2 g ( z 6  z1 )   pWs  0 P5  P4  P6  P5  v52  v 42    pW s  0 2 v52  g ( z6  z5 )  0 2 z 6  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. 39. P    pW s 1 2 v  gz  e f  0 2 m P2  P1  v 22  v12 0 2  assume no friction, no height change, no pump (away from heart) 1 denotes normal artery, 2 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. 40. 1 2 fLv 2  pW s  v 2  gz   0 D m  2 since assumed horizontal and uniform diameter, M EB reduces to 2 fLv 2  pW s  0 d m P 2 fLv 2 P   0 on pipe section, and pump power must equal D   m  v v v A 0.9   1  6.81    4 log 10     f  3.7 D  Re   P assume “generic” properties of water for density and viscosity 3  50 gal  ft    60 sec  7.48 gal  4.78 ft/sec v 0.02330 ft 2 ft 3  2   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 3 )  12   7.64  10 4 Re   4 6.72  10 lb/ft/sec   0.0018 in 0.9   1  6.81    4 log 10     f  3.7 D  Re   0.9  0.0018 in 1  6.81    4 log 10    4  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. 0.8W s 2(0.0056)(5000 ft)(4.78 ft/sec) 2  0 (2.067 / 12 ft )(32.2 ft lb/lb f / sec 2 ) 6.95 lb/sec W  2004 ft lb sec s f  hp sec    3.64 hp W s  2004 ft lb f / sec  550 ft lb f  41. 1 2 fLv 2  pW s  v 2  gz   0  2 D m zero pressure drop since emerges at atmospheric and source tank is open no pump,so no work term P 1 2 2 fLv 2 v  gz  0 2 D 0.9    6.81   1    4 log 10    f  3.7 D  dv   must solve two equations simultaneously location 1 is tank level, location 2 is discharge point, velocity in tank assumed zero v22 2 f (50 m)v22  9.81 m/s 2 (10 m)  0 2 (0.0525m) 0.9  0.000045 m    1 6.81(0.001 kg/m/s)    4 log 10    3  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 2 )  0.0054 m 3 / s 1-27 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. Copyright (c) 2017 by Pearson Education, Inc. 42. 1 2 fLv 2  pW s  v 2  gz   0  2 D m tank to tank, so KE term zero 2 fLeq v 2  pW s P   gz   0  D m P if use equivalent length method, or, if use velocity heads method 2 fLv 2 P v 2  pW s  gz    Ki  0  2 D m 0.9   1  6.81    4 log 10     f  3.7 D  Re   must look up 10 – in, schedule – 20 pipe informatio n     1500 gal 3 4   ft / 7.48 gal   5.83 ft/sec v 2  60 sec    10.25 ft         12 1500 gal 3 ft / 7.48 gal (0.87)(62.4 lb/ft 3 )  181.44 lb/sec m  60 sec (10.25 / 12 ft)(5.83 ft/sec)(0.87)(62.4 lb/ft 3 ) Re   10057 40 cP(6.72  10 -4 lb/ft/sec/ cP)     0 .9  0.0018 1  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 5 ft 25(144)lb f / ft 2 32.2 ft/sec 2 (100 ft) 2(0.00776)(3.713  10 5 ft)(5.83 ft/sec) 2   0.87(62.4 lb/ft 3 ) 32.2 ft lb/lb f sec 2 (10.25 / 12 ft)(32.2 ft lb/lb f sec 2 ) 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. velocity heads method 25(144)lb f / ft 2 32.2 ft/sec 2 (100 ft) 2(0.00776)(70)(5280 ft)(5.83 ft/sec) 2     0.87(62.4 lb/ft 3 ) 32.2 ft lb/lb f sec 2 (10.25 / 12 ft)(32.2 ft lb/lb f sec 2 ) 0.7W s (550 ft lb f /hp/sec) (50(0.75)  15(0.17)  0.55  1) (5.83 ft/sec) 2  0 2 181.44 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.