Solutions Manual for Chemical Process Equipment Design (2017) by Richard Turton Joseph A. Shaeiwitz Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. 1-1 Chapter 1 1. 0 2 1 3 2 1                     s f W e z g u 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 2. mass flowrate constant, so, if larger pipe is 1 and smaller pipe is 2 2 2 2 1 1 1 v A v A m      , where A is cross sectional area for flow, if density constant, then 2 2 1 1 v A v A  , 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. Since Av m    , if the density is constant, since the cross-sectional area is constant, the velocity must be constant. Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. 1-2 5. v Av 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. forces viscous forces inertial Re  inertial forces keep fluid flowing, viscous forces resist fluid flow Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. 1-3 7. all of the following have similar shape f Re laminar flow turbulent flow flow in pipe C D Re laminar flow turbulent flow flow past submerged object f Re laminar flow turbulent flow flow in packed bed Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. 1-4 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 0 2 1 3 2 1                     s f W e z g u u dP  for incompressible flow, density constant, so first term is simplified 0 2 1 3                     s f W e z g u u P  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 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. 1-5 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 h g P flowing fluid fluid manometer     ) (   , 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. Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. 1-6 19. The mechanical energy balance across a pump or compressor (neglecting any height difference between suction and discharge) reduces to    dP W s 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. Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. 1-7 21. For fully developed turbulent flow, it is assume that the friction factor has reached its asymptotic value. The proportionalities are 5 2 D v L P    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. 2 5 D v   , so, for constant pressure drop, the flowrate increase significantly d. it is exactly a proportional increase e. 5 . 0   L v  , 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 125 . 0 ) 5 . 0 ( 5 . 0 2 5 2 5 1 2 1 2 2 1 2 1 2      D D v v L L P P   Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. 1-8 22. For laminar flow, the proportionalities are 4 D v L P    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. 4 D v   , so, for constant pressure drop, the flowrate increase significantly, more than for turbulent flow d. it is exactly a proportional increase e. 1   L v  , 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 25 . 0 ) 5 . 0 ( 5 . 0 4 2 4 1 1 2 1 2 1 2      D D v v L L P P   Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. 1-9 23. Av v v Av m         pipe 1 – area from Table 1.1, density given m/s 26 . 3 m 10 65 . 21 /s m 00706 . 0 /s m 00706 . 0 kg/m 850 kg/s 6 2 4 3 1 1 1 3 3 1 1         A v v m v     pipe 2 – area from Table 1.1 m/s 66 . 1 m. 10 63.79 /s m 0106 . 0 kg/s 9.01 /s) m 0106 . 0 )( kg/m 850 ( 2 4 - 3 2 2 2 3 3 2 2        A v v v m     pipe 3 – area from Table 1.1 kg/s 4.50 /s) m 00529 . 0 )( kg/m 850 ( m/s 00529 . 0 ) m 10 m/s)(13.13 032 . 4 ( 3 3 3 3 2 -4 3 3 3        v m A v v     kg/s 51 . 10 4 4 3 2 1     m m m m m      pipe 4 – area from Table 1.1 m/s 59 . 2 m 10 69 . 47 /s m 0124 . 0 /s m 0124 . 0 kg/m 850 kg/s 51 . 10 2 4 3 4 4 4 3 3 4 4         A v v m v     Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. 1-10 24. Av m    assume  = 1000 kg/m 3 for all sections kg/s 3.94 m/s) 3 )( m 10 13 . 13 ( kg/m 1000 kg/s 2.79 m/s) 5 )( m 10 574 . 5 ( kg/m 1000 2 4 3 2 2 2 2 4 3 1 1 1           v A m v A m     a. kg/s 73 . 6 4 3 4 3 2 1      m m m m m m       b. m/s 818 . 0 ) m 10 19 . 82 ( kg/m 1000 kg/s 73 . 6 2 4 3 4 4 4      A m v   c. 2 4 3 3 3 3 m 10 85 . 30 m/s) 18 . 2 ( kg/m 1000 kg/s 73 . 6      A m A  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 Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. 1-11 25. 4 2 i i d A   , so A 4-in = 0.0873 ft 2 , A 3-in = 0.0491 ft 2 , A 2-in = 0.0218 ft 2 , A tank = 19.63 ft 2 a. in tank /sec ft 0.3927 ft/sec) (0.02 ft 63 . 19 3 2   v  b. at constant density, volumetric flowrates balance ft/sec 5 . 4 ft 0.0491 - ft/sec) 4 ( ft 0.0218 - ft/sec) 8 ( ft 0.0873 /sec ft 39 . 0 3 3 2 2 2 3        in in out in tank v v v v   26. 0 2 1 2         m W e z g v P s p f     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 gal/min 487 ) gal/ft 48 7 sec/min)( 60 ( lb/ft 62.4 lb/sec 7 . 67 lb/sec 7 . 67 0 sec)) (hp / lb ft hp)(550 0.8(20 - /lb lb ft 80 ft) 50 ( ) sec lb/(lb ft 32.2 ft/sec 2 . 32 3 3 f f 2 f 2      . v m m    Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. 1-12 27. 0 2 1 2         m W e z g v P t s f     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 M W 84 . 7 W 10 84 . 7 s m kg 10 84 . 7 0 ) kg/m /s)(1000 m 25 ( 80 . 0 ) m/s m(9.81 10 m) 50 ( m/s 81 . 9 2 0 m/s) 1 ( 6 3 2 6 3 3 2 2 2             s s W W   28. 0 2 1 2         m W e z g v P s p f      P = 0, since both tanks are open to the atmosphere velocities both zero since tank levels m 6 . 7 0 ) kg/m s)(1000 h/3600 /h(1 m 10 J/s) 0.75(100 - J/kg 3.5 m) 10 )( m/s 81 . 9 ( 3 3 2     z z Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. 1-13 29. 0 2 1 2         m W e z g v P t s f     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 hp 3723 0 lb/sec) 65000 ( 55 . 0 sec)) /(hp lb ft 550 ( ) /sec lb/lb ft 32.2 ( 2 sec / ft 0 13.26 ft) 60 ( /sec lb/lb ft 32.2 ft/sec 32.2 ft/sec 26 . 13 4 ft) (10 lb/ft 62.4 lb/sec 65000 f 2 f 2 2 2 2 f 2 2 3 2                 s s W W v    30. 0 2 1 2         s f W e z g v P  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 P 1 = 34,500 kPa, P 2 = 3.168 kPa   m 10 42 . 1 m/s) 67 . 262 ( 4 mL 10 m s/min 60 mL/min 250 m/s 67 . 262 0 2 /s m 10 305 . 5 kg/m 1000 N/m 34500 - 3168 m/s 10 305 . 5 4 m) 1 . 0 ( mL) 10 / s)(m /60 mL/min(min 250 4 2 2 2 2 2 6 3 2 2 2 2 2 4 2 2 3 2 4 2 6 3 1                                       d d v A v v v v    Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. 1-14 31. 0 2 1 2         m W e z g v P s p f      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/ft 3 )(ft 3 /7.48 gal) =8.3 lb/gal a. hp 37 . 1 0 lb/gal) sec)(8.3 in/60 gal/min)(m 30 ( sec)) (hp / lb ft hp)(550 0.8( - /lb lb ft 200 ft) 928 873 ( ) sec lb/(lb ft 32.2 ft/sec 2 . 32 f f 2 f 2     s s W W   b. without pump, need sufficient head from reservoir to overcome friction ft 728 928 - z ft 200 0 /lb lb ft 200 ft) ( ) sec lb/(lb ft 32.2 ft/sec 2 . 32 f 2 f 2         z z z Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478.