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Unformatted text preview: CIVL6049 Unban Development Management bv Engineering Amoach Worked Example_s
Exp.l  Answer to Misc. Prob. No. 5 of Part B of Exercise (21) 10 marks Given assumption: )1:n — kn — an (l) h (Q, + a) = kn (Q + q)2 261 92
:11 1— _ 2
r1(+QH+QHZ () If additional discharge of qmn is diverted from manhole m where m 2n , the new total flow between manholes n and (n. — 1) is Q,“ IQ” — qmm Where m 2 n From (2), h m" = h(Q m”) 2
qum + qnm Q n Q2 )
is the head loss betweenMH‘ s n and (n — 1) after the diversion of :12" (1 where h m H qmm from MHm. m 'I'Hmm=H0+thn
n=l
—H Him—2 (ihm Ni ’1")
— 0 ":1 n qmn "=1Qn qmm i.’=1Qn2
=Hm —ZBm gm +Am gm2 (3)
m h
where Am = Z "2 (4)
Q"
m h
= n 5
B Q” < ) Let Ym—Hm=f~§ where (3)0
Require a diversion at manhole m so as to reduce 5 to zero. ..H =Y —f (6) Sub. (6) into (3) Ym _f:Hm_2Bm qmm+Am qmm2
Am qmm2_2‘Bm qmm +(Hm+f—Ym):0 or Am gm2 — 23m qm +5 :0 qm =2; (23,” :\/4Bm2 —4A,,, a) (7) The smallest +ve value of qmm is to be chosen as the diversion required to meet the min. freeboard requirement. .‘.qmm=A—1— (3m —1/Bm2 — Am 5) (7A) (b) (i) Calculate H2 for the existing sewer line 0123. (5 marks)
Q3 251—Q2 (:100) given. 1 [13 =th 112 + I13 = 10.725 — 10.25 = .475 I22 = '475 =.38 1
(1+1) H2 =10.25+.38=10.63
1'2 «H2=11—10.63=.37(0.5m Freeboard requirement is exceeded at m h 20n1y.
5 =0.5—.37=.13m. * The diversion does not affect m h] 32=£=J—3§=.19x10"2
200 .19x10—2 — .192 x10‘4 —.13x.095x10*4 q”— .095 x 104 =37.8 ~ min. diversion from m 142 required Hence also from manhole 2. (b)(z:i) For max. redev. potential of site B,
(5 marks) all sewage from site A to be discharged to the new sewer. MH 3
belongs to sewer line 0—143 only and is disconnected from sewer line 0 12.
After diversion, I22 is reduced by2i
" h21=%3§=0.19 .
. 5  . q’
(ea—=17 .99 ‘?
WC. at MH2 becomes 1  (60
H2 =10.25+.19 = 10.44 toe—a
hmb Lia9
Z. I O Freeboard at manholes l & 2 are: fl =O.75(= 11.0—10.25)
f2 = 0.50 (:1 1.0 —10.44)
Hence manhole 2 is critical in line 012.
621 = (.56 — .5) = — .06m‘
Corresponding space capacity is calculated from outfall to MH 2. When 6(0, the we sign in (7) still to be taken so that qmm (( 0 ) is the spare capacity instead of the required. diversion when 5 )0. h .25 .19
+ M Q} =2002 1002 = .2525 x104 Bl_§hn_£i+;2
2 "=1Qn 200 100 =.315x10‘2 1_ .315><10‘2 —\/.3152 ><10_4 +.06><.2525><10*4 «122 ,2525 x 10‘4 _.315—.338
.2525 x102 = 2.3 (spare cap. at MHZ).
Clearly this additional discharge has little effect on WL at mhl, hence
would not affect sewer line 0143 signiﬁcantly ‘. Redev. cap. of site B becomes 100 + 2.3 = 102.3 after sewer diversion. CIVL6049
Worked Examples Exp 2 (Course Exercise Part B. Misc. Problem No.6) 1. A new town is being planned at the outskirt of an existing city to cater for future
population growth. The city consists of two zones and rail extension to the new town
is planned as the principal transport mode, Fig. 2(a). Assuming work trips can be
taken as an appropriate measure of the total daily trips and that all zones are of similar
development, the relevant results of a recent transport survey are summarized in Table
2(a). The generalized cost of Raw! upon rail extension are shown in Table 2(b). (a) Assuming the gravity model with iexp(3x generalized c050} as the
distribution function, find the next better approximate value of [3 starting with
an initial approximate value of 0.04. (10%) (1)) Estimate the minimum number of potential employments which should be
provided in the new town so that at least 33% of all trips originating from the
new town will be internal trips. (10%) (c) Brieﬂy outline Hong Kong’s strategy for population buildup in the ﬁrst and
second generation new towns. Discuss possible adaptation of the strategy for
future new towns in the light of current changes in economic structure. (13%)
Rail Extension
Zone 1 Zone 2 _ Zone 3
Existing Preposed
Districts New town
Fig. 2(a)
Table 21a!
(1) Population (x 104)
Zone 1 : 3O ZoneZ : 60 (2) Generalized costs tij(min.) (3) Trip matrix tij (daily persontrip) Table 219) Generalized cost upon rail extension:
1131: ﬁg; = 1132 = t23 = 30
t33 = 15 1(a) P_op zone 1: 30x104
zone 2: 60x104 Generalized cost tij (min) As all zones are of similax dev., the given gravity model takes the following form: Ta = aiGiAje 43m given: where ai = 1/ ZAje‘mij
j T 111(2Ajew]
Let FE = = _J'__._
aiGiAj GiAJ. Using ( 1),  m1 = etu
2 u .
Let S[ 2 ZAjCﬂnj , $2 = ZAjeﬂl“
Fl .1 T.. .
Using (2) & (3), = 3tij (for all i, j) ' J Given [3 '.= 0.04, I.‘ Sl * 156.O4X13,44+255 X email»sz = 8.76 + 9.72
= 18.48 82 I: 156v.04x24.l+ 25.5 e—mxmu = 5.72 + 12.3
= 18.02 _ Tijsi _ .573 .335
" .384 .481 (1) (2) (3) (4) (5) Plot Fij values against correéponding tij values on semilog graph paper, where tij are ti_ 13.44 24.1
’ 24.1 13.24. From Eq. 3, the slope of the graph gives: [3 = 0.033 (b) For planning the new town, assume the same gravity model _ Ajewmj I ‘
TU — G5 ZA'eﬂﬂhij (13] = 1 t0 3)
j J
Rail extension
(Ex CltY) (Proposed new town)
.038x40
T33 = Ga f3: —.038Iij
2 A13
1: HOSE l5
T31 Ase x . 38 0 —. _
GJ Ale 0 x4 038x30+AJc ‘038x15 where A1 = 15, A2 = 25.5, A3 = unknown k is to be 33%
G3 A,x.566 . .33 =
3.28 +8.16+.566A3 = (323+ 8.16)x.33 '. A3
.566x(1 — .33) = 9.95 Min. employment potential of 9.95 x 104 to be provided in zone 3. P:S/1EJ I. Exp 3 (Course Exercise Part B. Misc. Problem No.7) 5 Exisﬁng population and maximum capacity are
(6, 6.5) it 10ti for Zone I, and
i p (1, 1.6)x106forZoneZ. Intra and inter zonal trawl times are .
tn = 30, tlz = tn =40, ‘ _ €22 = 20 mimics
Capacity of existing link = I x 10‘ person  trips/day in each direction .  
It would be very costly to increase the capacity of the link. Each zone is of balanced
development. It is planned to expand the system to accommodate an additional
pepulation of 0.5 x 105. The land development cost (including local infrastructure (:1) Show that the total travel time is a concave mnctton of it, where x is the additional
population to be allocated to Zone I. . (13%) [Hint IA curve y(x) is concave if it is monotone45nd its 'slope decreagés witn‘it.) ()3) Hence determine the optimum allocation to Zone. 1 which minimizos the total cost
of land development and capitalized mspon costs. (10%) (c) Explain the basis of the algorithm for optimum land use allocation in the LUTO
model. ‘ . (10%) 2(a) Zone I Zone 2
605"” = 0.223
t“ = t2; = min
eﬂ.5x40 __,, t]; = 40 min
= t2. 505*“ = 0.368
Zone 1 2
Ex. Pop. 6 1
Max. cap. 6.5 1.6
(x106) As balanced land uses are provided in each zone, the given singly constrained gravity
model becomes: _ ~0.5n"
Tij — aiPine J u
"U
H
f—\
:U
m
.é
LA as
+ a
.36 3

O
E‘
\_/ Existing Condition: P. = 6, 132 = 1,(i.e. 106 units) 1x0.135 6x 0.223+lx0.l35
0.135 x1339441135
= 0.55 T12(ex) = 6 X =6 .6x0.135 T210”) = 1 x ————_._ 6x O.l35+1x 0.368
0.81 0.81 + 0.368
= 0.69 link loading max (T12, T21) 0.69 ll II This is signiﬁcantly less than the link capacity of 1 (x 10" persontrip/day) Planning Calculations x = additional pop. allocated to zone 1,
P1 = 6 +x 0 Sx S 0.5
P2 = 1+(0.5x)=l.5x Rae—0.511] T“ = (6”) (6+x x0223 (6+x)xo.223 +(1.5 —x)x0.135
* 1.338 +0.223x ‘
— (6 + x) Tij I.541+0.88x
(1.5'—x)x 0.135
T = ~——.______
‘2 (6 + ’0 (6+x)x 0.223 +(1.5—~x)x 0.135 0.2025 —0.135x
== (6 + x)
. 1.541+ 0.881: ‘ [Cont’d] (6+x)x 0.135
(6+ x)x 0.135 + (1.5—x)><0.363
0.81+0.135x 1.5— —————
( “1362—02339 E
II (LS—x) (1.5 — x) x 0.3687
(6 + x)>< 0.135 + (1.5 ——x)x0.368
0552—0368:: = 1.5— ————.—————
( 70(1362—0233x) T22 = (1.5 x) 30 x (1.338 + 0.223x) + 40 x (0.2025 — 0.13510] [ 1541+ 0.0883:
(6 + I“: 48.24 + 1.291: ] 1.541 + 0.0881: 3 31.3346 + x)(1+ 0.0267x) _ 1+0.057lx tle11+ttzT12 x(6+x) tnTllﬁnTn = (1.5_x)[40x(0.81+0.1133522)+g0.552—0.368x)]
.  . x 43.44 —1.96x
=3 1.5 _ __—__.._
( x)(1.362 — 02331:) 1—0.045x
= 31.9 .5 —————————
x0 J‘)(1—»0.1171x) The above expressions can be rewritten as follows: ¢1= I11Tu+f12T12
31.3x(6+x)[1+(LW 1+0.0571x} — 121T21+122T22 31.9x(1.5—x)[1+ 3'
1 (0.126)x
10.171x 1 The non—linear part of 4:1 &¢2 aredue to the 2“d term. in the square brackets. These are respectively proportional to:
¢11 =—x(5+x) 0r_x(1+ 0.16715)
1+ 0.5713: 1+0.057x __ x(1.5 —x) 1— 0.667;:
1—0.171x ¢£ orx(1—0.171x) The terms in the parentheses in gill and respectively increases with x and decreases with x. Hence both 9511 and ¢é are concave‘as shown. b. As the travel cost is a concave function of x, and values will be at
the extremities of the interval (0, 0.5). As the land development cost is proportional to x, the
concavity is not affected. Therefore the allocation to zone 1 should be eithertC'O or 0.5. (1) Scenario 1 Let x = 0; P1 = 6, P2 = 1.5 (Le. all allocation to zone 2) Tu = 6x0.2025
1.541 = 0.79
T2] 3 1.362 = 0.89 max(t12, T21) = 0.89 (unit: 106 pt/d). Link cap. need not be increased. Total travel cost $200x(¢1+¢2)
200x(31.3x6+31.9x1.5x1)x10‘ 47.13(x109$) ll II II Land development cost = 05x1 06x$10,000 = 5(x109s)
Tom capitalized COSt : 5213(X109S) for scenario 1 10 (2) Scenario 2 Let I = 0.5, P1=6.5, P2=1 ‘
i.e. assume all allocation to zone 1 Ta = 65 x (0.2025 —O.135 x 0.5)
l.54+0.088x0.5 0.135 1.585
0.55 6.5): ll T21: Ix(0.31+0.13sxo.5)
1.362 — 0.233 x 0.5
= 0.8775 1.2455
— 0.70 max.(T12,T21)=0.70
Existing link cap. is sufﬁcient Capitalized travel cost = $200x(¢1+¢2) ~= 200x[31.3x 6.5x (1 00364" 0'5
(unit:106$) = 200[3l.3x6.5x0.985+3‘1.9x1.069] = 200x234.5 = 46.91(x109S) 0.126x0.5
————— +3129 1+——————
1+0.0571x0.5) x( 1—0.171x0.5)] Land development cost = 0.5x10‘xss,000
= 4(x109s) Total capitalized cost for scenario 2 I
= 46.9 1 +4 = 50.91(x109$) Comparing the 2 scenarios, full allocation should be made to zone 1 & nothing to zone 2
i.e. x=0.5 Answer to include o a large no. of zones is equivalent to successive splitting of 2 zones. 0 Land use optimization algorithms in LUTO model is based on comparison of total costs
between 2 zones at a time. o Concavity of travel cost leads to “all or zero” allocation to eachizone. Hence iterative
process using linerized objective function (transport cost) will converge. ...
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This note was uploaded on 03/26/2012 for the course CIVIL 6049 taught by Professor Can'tremember during the Spring '12 term at HKU.
 Spring '12
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