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Unformatted text preview: 4.4. nan riti: rin I 10 11 9"?8313500 = 2.4?! 09;:
CruEEiﬁctiunll aria 0+ pipe:
5 = 111' x 0.05m": = 0.001% m!
3* = 0:13.000 11 0.001%: = 1.271. we
ﬁn I O 11', l ‘5 m
0(1 oil. 1 pi = p],
h;  2.: .1ng  2.5 mini" {see Eq 1151:
Frnm Eq. [111.41]1
“HI,  912‘, — It! + 1%} — ﬁg”: + In,
 0.3115 — IL} + {1.39:2} 4 2.5
 52.344 — 0.5124, .1er
Given: Parlor F' I 0.1 RH I 100 Jr:
41111,: PM as 100r2.4ﬁ5  40.03 am; 50: 52.344  0.1312, 40.00
2“ = 1152.344 — 40.DB]IQ.B1=125‘m 1.5. haul]. nibaw is in the 11: plan with thl Hater entering
11: the Fulﬁl“ : 011001100 :01 leaving in thin positive
: inaction. The Entries in 00011 0101.12 are found by 0011 010111; 341.1451 11:1 [452 3. :1 1n Exllple 45, t0 31w. {sime 1“ I 0}. 1014:111100th
"(BlightE? J'Ps Ph5h.:+r a 0.: 3 0,: 0.11 In I: dirt0:101:11 m {31: v0.5 ' E. 1iwhip"'4 50,:  Pb shlz * Fm: The quantities needed are: 2
m = 5 "1(0'1 x 993 = 41.03 kgfs
Ba = Bb = 1 (by assumption)
V = ﬁ = 6 mfs ﬂ = 0
a a,x a,z
Vb = Vb'z = 6 mfs Vbe = 0
_ 2
pa — pb = ?O,DOO him
_ _ 1 2 x 3
S —5 =5 —u:{0ul £430.00?35m
a h a,x
Sa,z = 0 Sb,x = 0 Substituting gives Fw x I ﬁ?.03 (0  6)  (?D,DOU x 0.00?85) + U  282.2 — 549.5 = —331.? N PH 2 = é?.03 (6 — 0) — D + (?0,000 x 0.00?35) = 282.2 + 5&9.5 = 331.? N Total force (see diagram):
F = 331.? ii = 1175.2 N acting in the direction indicated. Il?t.ﬂt
9317
»—'E3I.7
I:
LiI'.‘ 5.? Basis of solution: Pressure drop in bag—pass is the same as pressure drop in bypassed section.
Let subscript 1 refer to the bypass: subscript 4
refer to bypassed section. From Eq. [533}, since
2. = 2g, and fittings losses and contraction and
expansion losses are neglected. —l
Egupé...‘ ELL. i F!
P I‘D. 13 ”'9. 23. I: from which: —I BIL —
:7»: =93”?! Given 1:.1 = 22 ft. L; = 15ft; D1 = 0.1567 ft
DH a 0.6667 ft fﬂl’zz 1 (0.1657 1: 16] .I" (0.666? 1: 22]}frﬁﬂ  ousznifﬁ A second relation between F1 and F‘ is found from continuity:
mrFMI=nM where rs, total mass floraI rats in upstream section. Since :5: = pFS= F1== (FoDﬁ— F403) f DE Given 3} = 5 ft/s, F1 = {6  ?41{Dﬁ11{nﬁn
= {5 s in (0.666lzf0.16672} 96 —16F4 (Bl Assume values of El; find F¢; calculate the
corresponding Reynolds numbers; from Fig. (5.10}
find friction factors, using appropriate
roughness parameters. For steel pipe, k =
0.00015ft. For a 2inch pipe, assuming a
viscosity of 1 cP and a density of 62.3 lb/fta Re; 2 Divipr’p = 2 x 62.3F1f12 x 6.12 x 10“ = 1.55 x 104?1
k1/Dl= 0.00015 x 12f2 = 0.0009
For 8inch pipe, 3 x 52.3% R34 = — = 6.18 x 1o“?4
12 x 6.?2 x 10 1;qu = 0.00015 2: 12f0 = 0.00023
From Fig (5.10}, by trial, the following
quantities are found to satisfy Equation {A} and (B): 5 .8 Bygass ByEassed section F. = 2.02 ft/s F, = 5.87 ft/s
Rel = 31,310 Re4= 363,000 f1 : 0.0063 £,= 0.0041
flﬁlz = 0.0257 0.0182 0.5.2 = 0.0257 Mass flow rate through bypass: .. = 2.02 x 3600 x (n/q) * 0.16672 x 62.3
= 9,888 lblh
Total flow rate: so = 6 x 3600 x [ﬂ/4JX 0.66672 x 62.3 = 469,800 lb/h
Fraction of total flow through bypass: mufﬁn = 9,888 f469,800 = 0.021 or 2.1% V = 15ft/s D = 2ft
Assume n = 1 CP, p = 52.3 lb/ft3
Re = 2 x 15 x 62.3 = 2.73 x 106
6.72 x 10 “4
Let subscript 1 refer to the original pipe, 2 to
lined pipe.
For k/D 0.0003f2 = 0.00015, from Fig. 5.10:
f1 = 0.0032 For smooth pipe, f2= 0.0023. From Eq. (5.7), since D2: 1.9ft: 5.12 Equivalent diameter:
0,=4 x 15 x 20/(c2 x 151+:2 x 20})= 13.14 ft F = 50 ftfs AL = 250 ft [3
At 180°E‘, p = [102 czP (Appendix' 8] _ 17.14x50x0ﬂﬂt32
0.02.):6323010" {29 x 492:!{359 x 5401 =0.002 lbffts Re 2 3.95.110“ For such a large duct, kx’DwO. FIGm Fig. 5.10,
f = 0.0023 I From Table 51, Kf= 2 x 9.9 = 1.8. From Eq. (5.?2): f4nx00023x250 502
0.: ——————————+10
L 1704 ZxBZJT = 15.1 ftlbfflb
Pa  Ph= phi: 0.002 x 05.1 = 0.65 lbffinz
or 0.032 lbf/in2 01 = pFS = 0.062 x 50 x 15 x 20 = 030 lbfs 0 = 0:0 = 030 x 25.1;550 = 121 hp {04.0 kW) ' UtaEq. {432} between I and b
n‘tpl,
ﬁnGiu0 0.60 H,  200 + h;
Friction: Contraction at A. friction in pipe. expansion at B.
Pipe: D 3 4.025 {12 = 0.335 ft {Appendix 3}
Crosssect. area  0.0384 ft: {Appendix 3} 172' 400.117.4131 a 50 x 0.000: 11:11  10.1 it}:
KID  0.00D15fﬂ.336 = 0.0004
0.330 a 10.1 x 1.13 a: 52.3 Ra   3.1 x 105
1.2 x 0.12 x 10" f  0.0045 (Fig. 5.10}
K; for fittings: 2 gate valves 2 x 0.2 = 0.4
1 1:01:10 51 I 4 010003 4 x 0.0  3.6 4 tea: 4 x 1.§  7.2 Kg  11.2 Contraction lass: Kc I 0.! [Eq. (5.691]
Expansion 1055: K. = 1.0 (Sq. (5.57)] Pram Eq. (5.?2]. 4 x 0.0045 X 600 hfm {—_+0.4+1.0+11.2) x
0.0036 102 = 69.5 ft_lbf / lb
2 x 32.174 From Eq. (4. 65) , w = (AZ + hf)/n = (150 + 69.5)/0.60
= 366 ft—lbf/lb m==400x 8.33 x 1.18/60 = 65.5 ibis
P = 366 x 65.5f550 = 43.6 hp
Cost = (43.6 x 400M300 = $58.13,’day. 5.14.Mass flow: a: = 600 x 8.33 x 60 = 300,0001bfh
From Eq. (4.51), with LP= n x 6.6065 1’ 12 :
1.588 ft and p. = lcP: 41.3 00,000
Re = "~—
1 48816136142 kx’D = 0.00015/0_5 = 0.0003
From Fig. 5.10,. f = 0.0043 From Appeudix 3, V= 300,000!45,000= 6.6?ft/s = 312,300 6L = 2500 ft p = 62.3 lb/ft3 From Eq. (5T) r AP __ 4x0.0043x2500x6.672x62.3
: (6.065! 12)x2x32.174 = 3665 lbf/ftz or 25.4 lbffing ...
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 Fall '07
 Poehl

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