Preview Extract

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 ๏ฆ ๏ผ u3 ๏พ ๏ถ
๏ซ
๏ฒ1 ๏ฒ 2 ๏๏ง๏ง๏จ ๏ผ u ๏พ ๏ท๏ท๏ธ ๏ซ g๏z ๏ซ e f ๏ญ Ws ๏ฝ 0
2
dP
first term: enthalpy term
second term: kinetic energy, often written as (๏๏จv)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
๏ฆ ๏ฝ ๏ฒ1 A1v1 ๏ฝ ๏ฒ 2 A2 v2 , where A is cross sectional area for flow, if density constant,
m
then
A1v1 ๏ฝ 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.
๏ฆ ๏ฝ ๏ฒAv , 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.
๏ฆ ๏ฝ ๏ฒAv ๏ฝ ๏ฒv๏ฆ
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 ๏ฝ
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 ๏ฆ ๏ผ u 3 ๏พ ๏ถ
๏ฒ1 ๏ฒ ๏ซ 2 ๏๏ง๏ง๏จ ๏ผ u ๏พ ๏ท๏ท๏ธ ๏ซ g๏z ๏ซ e f ๏ญ Ws ๏ฝ 0
for incompressible flow, density constant, so first term is simplified
๏P 1 ๏ฆ ๏ผ u 3 ๏พ ๏ถ
๏ท ๏ซ g๏z ๏ซ e f ๏ญ Ws ๏ฝ 0
๏ซ ๏๏ง
๏ฒ 2 ๏ง๏จ ๏ผ u ๏พ ๏ท๏ธ
12.
frictional drag due to โskin friction,โ 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 ๏ฝ ( ๏ฒ manometer fluid ๏ญ ๏ฒ 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 ๏ฝ ๏ฒ
๏ฒ
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๏ฆ 2
๏P ๏ต 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๏ฆ ๏ต D 2 , so, for constant pressure drop, the flowrate increase significantly
d. it is exactly a proportional increase
e. v๏ฆ ๏ต L๏ญ0.5 , 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๏ฆ22 D15
๏ฝ
๏ฝ 0.5(0.5) 2 ๏ฝ 0.125
2
5
๏P1 L1 v๏ฆ1 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๏ฆ
๏P ๏ต 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๏ฆ ๏ต 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๏ฆ ๏ต L๏ญ1 , 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๏ฆ2 D14
๏ฝ
๏ฝ 0.5(0.5) ๏ฝ 0.25
๏P1 L1 v๏ฆ1 D24
1-8
Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. Copyright (c) 2017 by Pearson Education, Inc.
23.
m๏ฆ ๏ฝ ๏ฒAv ๏ฝ ๏ฒv๏ฆ
v๏ฆ ๏ฝ Av
pipe 1 โ area from Table 1.1, density given
m๏ฆ
6 kg/s
v๏ฆ1 ๏ฝ 1 ๏ฝ
๏ฝ 0.00706 m 3 /s
3
๏ฒ 850 kg/m
v1 ๏ฝ
v๏ฆ1
0.00706 m 3 /s
๏ฝ
๏ฝ 3.26 m/s
A1 21.65 ๏ด 10 ๏ญ 4 m 2
pipe 2 โ area from Table 1.1
m๏ฆ 2 ๏ฝ ๏ฒv๏ฆ2 ๏ฝ (850 kg/m 3 )(0.0106 m 3 /s) ๏ฝ 9.01 kg/s
v2 ๏ฝ
v๏ฆ2
0.0106 m 3 /s
๏ฝ
๏ฝ 1.66 m/s
A2 63.79 ๏ด 10 -4 m. 2
pipe 3 โ area from Table 1.1
v๏ฆ3 ๏ฝ v3 A3 ๏ฝ (4.032 m/s)(13.13 ๏ด 10 -4 m 2 ) ๏ฝ 0.00529 m/s
m๏ฆ 3 ๏ฝ ๏ฒv๏ฆ3 ๏ฝ (850 kg/m 3 )(0.00529 m 3 /s) ๏ฝ 4.50 kg/s
m๏ฆ 1 ๏ซ m๏ฆ 2 ๏ฝ m๏ฆ 3 ๏ซ m๏ฆ 4
m๏ฆ 4 ๏ฝ 10.51 kg/s
pipe 4 โ area from Table 1.1
m๏ฆ
10.51 kg/s
v๏ฆ4 ๏ฝ 4 ๏ฝ
๏ฝ 0.0124 m 3 /s
3
๏ฒ 850 kg/m
v4 ๏ฝ
v๏ฆ4
0.0124 m 3 /s
๏ฝ
๏ฝ 2.59 m/s
A4 47.69 ๏ด 10 ๏ญ 4 m 2
1-9
Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. Copyright (c) 2017 by Pearson Education, Inc.
24.
๏ฆ ๏ฝ ๏ฒAv
m
assume ๏ฒ๏ = 1000 kg/m3 for all sections
m๏ฆ 1 ๏ฝ ๏ฒA1v1 ๏ฝ 1000 kg/m 3 (5.574 ๏ด 10 ๏ญ4 m 2 )(5 m/s) ๏ฝ 2.79 kg/s
m๏ฆ 2 ๏ฝ ๏ฒA2 v2 ๏ฝ 1000 kg/m 3 (13.13 ๏ด 10 ๏ญ4 m 2 )(3 m/s) ๏ฝ 3.94 kg/s
a.
m๏ฆ 1 ๏ซ m๏ฆ 2 ๏ฝ m๏ฆ 3 ๏ฝ m๏ฆ 4
m๏ฆ 3 ๏ฝ m๏ฆ 4 ๏ฝ 6.73 kg/s
b.
v4 ๏ฝ
m๏ฆ 4
6.73 kg/s
๏ฝ
๏ฝ 0.818 m/s
๏ฒA4 1000 kg/m 3 (82.19 ๏ด 10 ๏ญ4 m 2 )
c.
A3 ๏ฝ
m3
6.73 kg/s
๏ฝ
๏ฝ 30.85 ๏ด 10 ๏ญ4 m 2
3
๏ฒA3 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 ๏ฝ
๏ฐd 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๏ฆ ๏ฝ 19.63 ft 2 (0.02 ft/sec) ๏ฝ 0.3927 ft 3 /sec
b.
at constant density, volumetric flowrates balance
v๏ฆtank ๏ฝ ๏ฅ v๏ฆin ๏ญout
0.39 ft 3 /sec ๏ฝ 0.0873 ft 2 (8 ft/sec) – 0.0218 ft 2 (4 ft/sec) – 0.0491 ft 2 v3๏ญin
v3๏ญin ๏ฝ 4.5 ft/sec
26.
๏P
๏ฒ
๏ซ
๏จ pW๏ฆ s
1 2
๏v ๏ซ g๏z ๏ซ e f ๏ญ
๏ฝ0
2
m๏ฆ
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) ๏ซ 80 ft lb f /lb ๏ฝ0
2
m๏ฆ
32.2 ft lb/(lb f sec )
m๏ฆ ๏ฝ 67.7 lb/sec
v๏ฆ ๏ฝ
67.7 lb/sec
(60 sec/min)( 7.48 gal/ft 3 ) ๏ฝ 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๏ฆ
1
๏ซ ๏v 2 ๏ซ g๏z ๏ซ e f ๏ญ s ๏ฝ 0
๏ฒ 2
๏จ t m๏ฆ
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๏ฆ s
(1 m/s) 2 ๏ญ 0
๏ซ 9.81 m/s 2 (๏ญ50 m) ๏ซ 10 m(9.81 m/s 2 ) ๏ญ
๏ฝ0
2
0.80(25 m 3 /s)(1000 kg/m 3 )
2
kg m
W๏ฆ s ๏ฝ 7.84 ๏ด 10 6
๏ฝ 7.84 ๏ด 10 6 W ๏ฝ ๏ญ7.84 M W
3
s
28.
๏จ pW๏ฆ s
1
๏ซ ๏v 2 ๏ซ g๏z ๏ซ e f ๏ญ
๏ฝ0
๏ฒ 2
m๏ฆ
๏P
๏P = 0, since both tanks are open to the atmosphere
velocities both zero since tank levels
(9.81 m/s 2 )(10 ๏ญ z m) ๏ซ 3.5 J/kg –
0.75(100 J/s)
๏ฝ0
10 m /h(1 h/3600 s)(1000 kg/m 3 )
3
z ๏ฝ 7.6 m
1-12
Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. Copyright (c) 2017 by Pearson Education, Inc.
29.
๏P
W๏ฆ
1
๏ซ ๏v 2 ๏ซ g๏z ๏ซ e f ๏ญ s ๏ฝ 0
๏ฒ 2
๏จ t m๏ฆ
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 ๏ฝ
65000 lb/sec
๏ฝ 13.26 ft/sec
2
๏ฆ
๏ถ
๏ฐ
(10
ft)
๏ท๏ท
62.4 lb/ft 3 ๏ง๏ง
4
๏จ
๏ธ
32.2 ft/sec 2
13.26 2 ๏ญ 0 ft 2 / sec 2 W๏ฆ s (550 ft lb f /(hp sec))
(
๏ญ
60
ft)
๏ซ
๏ญ
๏ฝ0
0.55(65000 lb/sec)
32.2 ft lb/lb f /sec 2
2(32.2 ft lb/lb f /sec 2 )
W๏ฆ ๏ฝ 3723 hp
s
30.
๏P
1
๏ซ ๏v 2 ๏ซ g๏z ๏ซ e f ๏ญ Ws ๏ฝ 0
๏ฒ 2
This is actually a nozzle problem, so only pressure and kinetic energy terms remain.
must look up vapor pressure of water at 25ยฐC, 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 ๏ฝ
๏ฝ 5.305 ๏ด 10 ๏ญ 4 m/s
2
๏ฐ (0.1 m)
4
๏จ
3168 – 34500 N/m 2 v 22 ๏ญ 5.305 ๏ด 10 ๏ญ 4
๏ซ
2
1000 kg/m 3
๏ฉ m /s ๏ฝ 0
2
2
2
v 2 ๏ฝ 262.67 m/s
3
๏ฆ ๏ฐd 22 ๏ถ
๏ฆ 250 mL/min ๏ถ๏ฆ m ๏ถ
๏ฆv 2 ๏ฝ ๏ง
๏ง
๏ท
๏ท๏ง 6
๏ท ๏ฝ A2 v 2 ๏ฝ ๏ง๏ง 4 ๏ท๏ท(262.67 m/s)
๏จ 60 s/min ๏ธ๏จ 10 mL ๏ธ
๏จ
๏ธ
d 2 ๏ฝ 1.42 ๏ด 10 ๏ญ 4 m
1-13
Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. Copyright (c) 2017 by Pearson Education, Inc.
31.
๏จ pW๏ฆ s
1
๏ซ ๏v 2 ๏ซ g๏z ๏ซ e f ๏ญ
๏ฝ0
๏ฒ 2
m๏ฆ
๏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๏ฆ s hp)(550 ft lb f / (hp sec))
32.2 ft/sec 2
(
873
928
ft)
200
ft
lb
/lb
๏ญ
๏ซ
๏ฝ0
f
(30 gal/min)(m in/60 sec)(8.3 lb/gal)
32.2 ft lb/(lb f sec 2 )
W๏ฆ ๏ฝ 1.37 hp
s
b.
without pump, need sufficient head from reservoir to overcome friction
32.2 ft/sec 2
(๏z ft) ๏ซ 200 ft lb f /lb ๏ฝ 0
32.2 ft lb/(lb f sec 2 )
๏z ๏ฝ ๏ญ200 ft ๏ฝ z – 928
z ๏ฃ 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
๏ฒ
๏ญ
๏จ pW๏ฆ s
m๏ฆ
๏ฝ0
(762.6 ๏ญ 115.6)(1000) N/m 2 0.75(6710 J/s)
๏ญ
๏ฝ0
m๏ฆ
900 kg/m 3
m๏ฆ ๏ฝ 7.0 kg/s
b.
get area from Table 1.1
m๏ฆ
7.0 kg/s
v2๏ญin ๏ฝ
๏ฝ
๏ฝ 3.59 m/s
๏ฒA 900 kg/s(21,65 ๏ด 10 -4 m 2 )
c.
do MEB from 1-2
๏P
๏ฒ
๏ซ
๏จ pW๏ฆ s
1 2
๏v ๏ซ g๏z ๏ซ e f ๏ญ
๏ฝ0
2
m๏ฆ
550.000 ๏ญ P1
3.59 2 ๏ญ 0 m 2 / s 2
0.75(6710 J/s)
2
9
.
8
m/s
(
25
m)
๏ซ
๏ซ
๏ซ 30 J/kg ๏ซ 50 J/kg ๏ฝ0
3
2
7 kg/s
900 kg/m
P1 ๏ฝ 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๏ญin ๏ฝ
m๏ฆ
7.0 kg/s
๏ฝ
๏ฝ 0.946 m/s
๏ฒA 900 kg/s(82.19 ๏ด 10 -4 m 2 )
across pump
v 22 ๏ญ v12
๏ฝ 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
๏ฒ
๏ญ
๏จ pW๏ฆ s
m๏ฆ
๏ฝ0
๏จ p (1000 J/s)
(551.6 ๏ญ 34.5)(1000) N/m 2
๏ญ ๏ญ3
๏ฝ0
3
1000 kg/m
10 m/s(1000 kg/m 3 )
๏จ p ๏ฝ 0.52
b.
to get frictional loss, do MEB from 1-2
๏จ pW๏ฆ s
๏P 1 2
๏ซ ๏v ๏ซ g๏z ๏ซ e f ๏ญ
๏ฝ0
๏ฒ 2
m๏ฆ
551,600 ๏ญ 34,500 n/m 2
0.517(1000 J/s)
๏ซ 9.8 m/s 2 (50 m) ๏ซ e f ๏ฝ0
3
1 kg/s
1000 kg/m
P1 ๏ฝ 490.1 J/kg
1-17
Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. Copyright (c) 2017 by Pearson Education, Inc.
34.
๏จ pW๏ฆ s
1
๏ซ ๏v 2 ๏ซ g๏z ๏ซ e f ๏ญ
๏ฝ0
๏ฒ 2
m๏ฆ
๏P
P1 = 25 psia
a.
to get pump power, do MEB across pump, so must determine pressure drop across
pump from manometer
๏ฆ
๏ถ
32.2 ft/sec 2
๏ง
๏ท(12 ๏ญ 10 ft) ๏ฝ 2381.2 ft lb f / lb
๏P ๏ฝ (13.6 ๏ญ 0.88)(62.4 lb/ft )๏ง
2 ๏ท
32.2
ft
lb/(lb
sec
)
f
๏จ
๏ธ
3
3
5 ft / sec
5 ft / sec
๏ฝ 6.366 ft/sec
๏ฝ 25.46 ft/sec
v1 ๏ฝ
v2 ๏ฝ
2
๏ฐ (1 ft)
๏ฐ (0.5 ft)2
4
4
๏ฆ 550 ft lb f ๏ถ
๏ท
0.75W๏ฆ s ๏ง๏ง
2
2
2
2
hp sec ๏ท๏ธ
2381.2 ft lb f / lb 25.46 ๏ญ 6.366 ft sec
๏จ
๏ซ
๏ญ
๏ฝ0
2
0.88(62.4 lb/ft 3 )
5 ft 3 /sec (62.4 lb/ft 3 )
W๏ฆ ๏ฝ 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
๏ฆ 550 ft lb f ๏ถ
๏ง๏ง
๏ท๏ท
0
.
75
(
35
.
1
hp)
2
2
hp
sec
(14.7 – 25)lb f / ft 2
32
.
2
ft
/
sec
๏ธ ๏ฝ0
๏จ
(144 in 2 /ft 2 ) ๏ซ
๏z ๏ญ
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 ๏ฝ 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 ๏ฝ
๏ฆ
m
6.5 kg/s
๏ฝ
๏ฝ 1.27 m/s
3
๏ฒA2 800 kg/m (63.79 ๏ด 10 -4 m 2 )
b.
MEB from source tank level to pump inlet
first need velocity in suction line
m๏ฆ
6.5 kg/s
v3 ๏ฝ
๏ฝ
๏ฝ 0.989 m/s
3
๏ฒA3 800 kg/m (82.19 ๏ด 10 -4 m 2 )
P3 ๏ญ P1
๏ฒ
๏ซ
v32
๏ซ g ( z 3 ๏ญ z1 ) ๏ซ e f ๏ฝ 0
2
P3 ๏ญ 150,000 N/m 2 (0.989 m/s) 2
๏ซ
๏ซ 9.81 m/s 2 (๏ญ5 m) ๏ซ 30 J/kg ๏ฝ 0
3
2
800 kg/m
P3 ๏ฝ 164.8 kPa
c.
MEB on entire system
P4 ๏ญ P1
๏ฒ
๏จ pW๏ฆ s
v 42
๏ซ
๏ซ g ( z 4 ๏ญ z1 ) ๏ซ ๏ฅ e f ๏ญ
๏ฝ0
2
m๏ฆ
200,000 ๏ญ 150,000 N/m 2 (1.27 m/s) 2
0.7(1500 J/s)
๏ซ
๏ซ 9.81 m/s 2 (h m) ๏ซ 80 J/kg ๏ญ
๏ฝ0
3
2
6.5 kg/s
800 kg/m
h ๏ฝ 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 โ out, nothing in, so
dm
๏ญ ๏ฝ m๏ฆ 2
dt
b.
m๏ฆ ๏ฝ ๏ฒA2 v
m ๏ฝ ๏ฒV ๏ฝ ๏ฒAtank h ๏ฝ ๏ฒA1 h
d ( ๏ฒA1 h)
๏ฝ ๏ญ ๏ฒA2 v
dt
density constant since liquid
dh
๏ฝ ๏ญv 2 A2
dt
A1 dh
๏ฝ ๏ญv 2
A2 dt
A1
c.
MEB on control volume tank level to point 2 at pipe discharge
v1 = 0 in tank
๏จ pW๏ฆ s
๏P 1 2
๏ซ ๏v ๏ซ g๏z ๏ซ e f ๏ญ
๏ฝ0
๏ฒ 2
m๏ฆ
v 22 ๏ญ v12
๏ซ g ( z 2 ๏ญ z1 ) ๏ฝ 0
2
v 22
๏ซ g ( ๏ญ h) ๏ฝ 0
2
v 2 ๏ฝ 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 ๏ฝ 1 2 g
A2
dh
๏ฝ ๏ญ ah 0.5
dt
h
t
dh
๏ฝ
๏ญ
a
๏ฒ 0. 5
๏ฒ0 dt
h0 h
at ๏ถ
๏ฆ
h ๏ฝ ๏ง h00.5 ๏ญ ๏ท
2๏ธ
๏จ
2
e.
at h = 0
2h00.5
t๏ฝ
a
1-22
Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. Copyright (c) 2017 by Pearson Education, Inc.
37.
m๏ฆ 1 ๏ซ m๏ฆ 2 ๏ฝ m๏ฆ 3 ๏ฝ m๏ฆ 4
P4 ๏ญ P3
๏ฒ
P5 ๏ญ P4
๏ฒ
๏ซ
v 42 ๏ญ v32
๏ญ ๏จ pW s ๏ฝ 0
2
๏ญ
v 42
๏ซ g ( z5 ๏ญ z 4 ) ๏ซ e f ๏ฝ 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๏ฆ 3 ๏ซ m๏ฆ 2 ๏ฝ m๏ฆ 4 ๏ฝ m๏ฆ 5
P2 ๏ญ P1
v 22
๏ซ g ( z 2 ๏ญ z1 ) ๏ซ
๏ฝ0
๏ฒ
2
g ( z 6 ๏ญ z1 ) ๏ญ ๏จ pWs ๏ฝ 0
P5 ๏ญ P4
๏ฒ
P6 ๏ญ P5
๏ฒ
v52 ๏ญ v 42
๏ซ
๏ญ ๏จ pW s ๏ฝ 0
2
v52
๏ซ g ( z6 ๏ญ z5 ) ๏ญ
๏ฝ0
2
z 6 ๏พ 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
๏ฒ
๏ซ
๏จ pW๏ฆ s
1 2
๏v ๏ซ g๏z ๏ซ e f ๏ญ
๏ฝ0
2
m๏ฆ
P2 ๏ญ P1
๏ซ
v 22 ๏ญ v12
๏ฝ0
2
๏ฒ
assume no friction, no height change, no pump (away from heart)
1 denotes normal artery, 2 denotes balloon region, where diameter is larger
๏ฆ ๏ฝ ๏ฒAv , 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 ๏จ pW๏ฆ s
๏ซ ๏v 2 ๏ซ g๏z ๏ซ
๏ญ
๏ฝ0
D
m๏ฆ
๏ฒ 2
since assumed horizontal and uniform diameter, M EB reduces to
2 fLv 2 ๏จ pW๏ฆ s
๏ญ
๏ฝ0
d
m๏ฆ
๏P 2 fLv 2
๏P
๏ซ
๏ฝ 0 on pipe section, and pump power must equal
D
๏ฒ
๏ฒ
m๏ฆ ๏ฝ ๏ฒv๏ฆ
v๏ฆ
v๏ฝ
A
0.9
๏ฉ ๏ฅ
1
๏ฆ 6.81 ๏ถ ๏น
๏ฝ ๏ญ4 log 10 ๏ช
๏ซ๏ง
๏ท ๏บ
f
๏ช๏ซ 3.7 D ๏จ Re ๏ธ ๏บ๏ป
๏P
assume โgenericโ properties of water for density and viscosity
3
๏ฆ 50 gal ๏ถ ft
๏ง
๏ท
๏จ 60 sec ๏ธ 7.48 gal
๏ฝ 4.78 ft/sec
v๏ฝ
0.02330 ft 2
ft 3 ๏ถ
2 ๏ฆ ๏ฆ 50 gal ๏ถ
๏ท๏ท ๏ฝ 6.95 lb/sec
m๏ฆ ๏ฝ (62.4 lb/ft )๏ง๏ง ๏ง
๏ท
๏จ ๏จ 60 sec ๏ธ 7.48 gal ๏ธ
๏ฆ 2.067 ๏ถ
ft ๏ท(4.78 ft/sec)(62.4 lb/ft 3 )
๏ง
12
๏ธ
๏ฝ 7.64 ๏ด 10 4
Re ๏ฝ ๏จ
๏ญ4
6.72 ๏ด 10 lb/ft/sec
๏ฅ ๏ฝ 0.0018 in
0.9
๏ฉ ๏ฅ
1
๏ฆ 6.81 ๏ถ ๏น
๏ฝ ๏ญ4 log 10 ๏ช
๏ซ๏ง
๏ท ๏บ
f
๏ช๏ซ 3.7 D ๏จ Re ๏ธ ๏บ๏ป
0.9
๏ฉ 0.0018 in
1
๏ฆ 6.81 ๏ถ ๏น
๏ฝ ๏ญ4 log 10 ๏ช
๏ซ๏ง
๏บ
4 ๏ท
f
๏ซ๏ช 3.7(2.067 in) ๏จ 7.64 ๏ด 10 ๏ธ ๏ป๏บ
f ๏ฝ 0.0056
1-26
Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478. Copyright (c) 2017 by Pearson Education, Inc.
0.8W๏ฆ s
2(0.0056)(5000 ft)(4.78 ft/sec) 2
๏ญ
๏ฝ0
(2.067 / 12 ft )(32.2 ft lb/lb f / sec 2 ) 6.95 lb/sec
W๏ฆ ๏ฝ 2004 ft lb sec
s
f
๏ฆ hp sec ๏ถ
๏ท๏ท ๏ฝ 3.64 hp
W๏ฆ s ๏ฝ 2004 ft lb f / sec๏ง๏ง
๏จ 550 ft lb f ๏ธ
41.
1
2 fLv 2 ๏จ pW๏ฆ s
๏ซ ๏v 2 ๏ซ g๏z ๏ซ
๏ญ
๏ฝ0
๏ฒ 2
D
m๏ฆ
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 ๏ซ g๏z ๏ซ
๏ฝ0
2
D
0.9
๏ฉ ๏ฅ
๏ฆ 6.81๏ญ ๏ถ ๏น
1
๏ท๏ท ๏บ
๏ฝ ๏ญ4 log 10 ๏ช
๏ซ ๏ง๏ง
f
๏ช๏ซ 3.7 D ๏จ dv๏ฒ ๏ธ ๏บ๏ป
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
๏ซ 9.81 m/s 2 (๏ญ10 m) ๏ซ
๏ฝ0
2
(0.0525m)
0.9
๏ฉ 0.000045 m ๏ฆ
๏ถ ๏น
1
6.81(0.001 kg/m/s)
๏ท ๏บ
๏ฝ ๏ญ4 log 10 ๏ช
๏ซ ๏ง๏ง
3 ๏ท
f
๏ช๏ซ 3.7(0.0525 m) ๏จ (0.0525 m)v2 (1000 kg/m ) ๏ธ ๏บ๏ป
f ๏ฝ 0.00788
v2 ๏ฝ 2.51 m/s
v๏ฆ ๏ฝ 2.51 m/s(21.65 ๏ด 10 -4 m 2 ) ๏ฝ 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 ๏จ pW๏ฆ s
๏ซ ๏v 2 ๏ซ g๏z ๏ซ
๏ญ
๏ฝ0
๏ฒ 2
D
m๏ฆ
tank to tank, so KE term zero
2 fLeq v 2 ๏จ pW๏ฆ s
๏P
๏ซ ๏ซ g๏z ๏ซ
๏ญ
๏ฝ0
๏ฒ
D
m๏ฆ
๏P
if use equivalent length method, or, if use velocity heads method
2 fLv 2
๏P
v 2 ๏จ pW๏ฆ s
๏ซ g๏z ๏ซ
๏ซ ๏ฅ Ki
๏ญ
๏ฝ0
๏ฒ
2
D
m๏ฆ
0.9
๏ฉ ๏ฅ
1
๏ฆ 6.81 ๏ถ ๏น
๏ฝ ๏ญ4 log 10 ๏ช
๏ซ๏ง
๏ท ๏บ
f
๏ซ๏ช 3.7 D ๏จ Re ๏ธ ๏ป๏บ
must look up 10 – in, schedule – 20 pipe informatio n
๏ฆ
๏ถ
๏ง
๏ท
1500 gal 3
4
๏ง
๏ท
ft / 7.48 gal ๏ง
๏ฝ 5.83 ft/sec
v๏ฝ
2 ๏ท
60 sec
๏ง ๏ฐ ๏ฆ๏ง 10.25 ft ๏ถ๏ท ๏ท
๏ง
๏ท
๏ธ ๏ธ
๏จ ๏จ 12
1500 gal 3
ft / 7.48 gal (0.87)(62.4 lb/ft 3 ) ๏ฝ 181.44 lb/sec
m๏ฆ ๏ฝ
60 sec
(10.25 / 12 ft)(5.83 ft/sec)(0.87)(62.4 lb/ft 3 )
Re ๏ฝ
๏ฝ 10057
40 cP(6.72 ๏ด 10 -4 lb/ft/sec/ cP)
๏จ
๏ฉ
๏จ
๏ฉ
0 .9
๏ฉ 0.0018
1
๏ฆ 6.81 ๏ถ ๏น
๏ฝ ๏ญ4 log 10 ๏ช
๏ซ๏ง
๏ท ๏บ
f
๏ช๏ซ 3.7(10.25) ๏จ 10057 ๏ธ ๏บ๏ป
f ๏ฝ 0.00776
equivalent length method ๏ elbows ๏ซ valves ๏ซ tank entrance ๏ซ tank exit
๏50(35) ๏ซ 15(9) ๏ซ 0.55(50) ๏ซ 1(50)๏(10.25 / 12) ๏ฝ 1676.3 ft
so Leq ๏ฝ 70(5280) ๏ซ 1676.3 ๏ฝ 3.6713 ๏ด 10 5 ft
25(144)lb f / ft 2
32.2 ft/sec 2 (100 ft) 2(0.00776)(3.713 ๏ด 10 5 ft)(5.83 ft/sec) 2
๏ซ๏ซ
๏ซ
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๏ฆ s (550 ft lb f /hp/sec)
๏ญ
๏ฝ0
181.44 lb/sec
W๏ฆ s ๏ฝ 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
๏ซ
๏ซ
๏ซ
๏ซ
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๏ฆ s (550 ft lb f /hp/sec)
(50(0.75) ๏ซ 15(0.17) ๏ซ 0.55 ๏ซ 1)
(5.83 ft/sec) 2 ๏ญ
๏ฝ0
2
181.44 lb/sec
2(32.2 ft lb/lb f sec )
W๏ฆ ๏ฝ 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.

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