# Lecture 3 Worked Example 2 ```Worked Example 2
(a) A gravity sewer line includes m manholes upstream of the outfall, Fig.1(a). It is
fully submerged under design conditions. The flow between manholes n and (n+1)
is Hn. A minimum freeboard f is required, (viz sewage level is to be below ground
level by at least an amount of f.) Let the ground level at manhole n be Yn. If YmHn &lt; f, find the minimum of diversion required to be made from manhole m.
Assume head loss is proportional to the square of flow rate in the sewer.
(b) It is proposed to lay a new sewer to relieve the overloading of an existing sewer line
as shown in Fig.1(b). The diameter of the new sewer is to be the same as that of the
existing sewer. The ground level and existing sewage level at the manholes are as
indicated in Table 1(b). A freeboard of 0.5m is required at all manholes. Find i) the
minimum diversion required from manhole No.3 to the new sewer, ii) the maximum
redevelopment potential of Site B in terms of the sewage discharge units after the
new sewer diversion is implemented.
(c) Discuss the political, social and legal factors in the planning of land use changes
and infrastructure improvement works of an urban renewal project.
(a)
Given assumption: ℎ𝑛 − 𝑘𝑛 − 𝑄𝑛2
(1)
2𝑞
𝑞2
𝑛
𝑛
∴ ℎ(𝑄𝑛 + 𝑞) = 𝑘𝑛 (𝑄𝑛 + 𝑞)2 = ℎ𝑛 (1 + 𝑄 + 𝑄 2 )
(2)
If additional discharge of 𝑞𝑚𝑚 is diverted from manhole m, where m ≥ n, the new
total flow between manholes n and (n-1) is 𝑄𝑚𝑛 = 𝑄𝑛 − 𝑞𝑚𝑛
From (2), ℎ𝑚𝑛 = ℎ(𝑄𝑚𝑛 ) = ℎ𝑛 (1 −
2𝑞𝑚𝑚
𝑄𝑛
+
𝑞𝑛𝑚 2
𝑄2
)
Where ℎ𝑚𝑛 is the head loss between MH’s n and (n-1) after the diversion of qmm
from MHm.
𝑚
∴ 𝐻𝑚𝑚 = 𝐻0 + ∑ ℎ𝑚𝑛
𝑛=1
𝑚
𝑚
𝑚
𝑛=1
2
𝑛=1
𝑛=1
ℎ𝑛
ℎ𝑛
= 𝐻0 + (∑ ℎ𝑛 ) − 2𝑞𝑚𝑛 (∑ ) + 𝑞𝑚𝑚 2 (∑ 2 )
𝑄𝑛
𝑄𝑛
= 𝐻𝑚 − 2𝐵𝑚 𝑞𝑚𝑚 + 𝐴𝑚 𝑞𝑚𝑚
ℎ
𝑛
Where 𝐴𝑚 = ∑𝑚
𝑛=1 𝑄 2
𝑛
ℎ
𝑛
𝐵𝑚 = ∑𝑚
𝑛=1 𝑄
𝑛
Let 𝑌𝑚 − 𝐻𝑚 = 𝑓 − 𝛿, where 𝛿 &gt; 0
Require a diversion at manhole m so as to reduce 𝛿 to zero.
𝐻𝑚𝑚 = 𝑌𝑚 − 𝑓
Sub. (6) into (3)
𝑌𝑚 − 𝑓 = 𝐻𝑚 − 2𝐵𝑚 𝑞𝑚𝑚 + 𝐴𝑚 𝑞𝑚𝑚 2
𝐴𝑚 𝑞𝑚𝑚 2 − 2𝐵𝑚 𝑞𝑚𝑚 + (𝐻𝑚 + 𝑓 − 𝑌𝑚 ) = 0
Or
𝐴𝑚 𝑞𝑚𝑚 2 − 2𝐵𝑚 𝑞𝑚𝑚 + 𝛿 = 0
1
𝑞𝑚𝑚 = 2𝐴 (2𝐵𝑚 &plusmn; √4𝐵𝑚 2 − 4𝐴𝑚 𝛿)
𝑚
(3)
(4)
(5)
(6)
(7)
The smallest +ve value of qmm is to be chosen as the diversion required to meet the
min. freeboard requirement.
1
∴ 𝑞𝑚𝑚 = 𝐴 (𝐵𝑚 − √𝐵𝑚 2 − 𝐴𝑚 𝛿)
(7A)
𝑚
(b)
(i) Calculate H2 for the existing sewer line 0-1-2-3.
1
𝑄3 = 𝑄2 (= 100)𝑔𝑖𝑣𝑒𝑛.
2
1
∴ ℎ3 = ℎ2
4
ℎ2 + ℎ3 = 10.725 − 10.25 = 0.475
0.475
∴ ℎ2 =
= 0.38
1
(1 + 4)
∴ 𝐻2 = 10.25 + 0.38 = 10.63
𝑌2 − 𝐻2 = 11 − 10.63 = 0.37 &lt; 0.5𝑚
∴ 𝐹𝑟𝑒𝑒𝑏𝑜𝑎𝑟𝑑 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑚𝑒𝑛𝑡 𝑖𝑠 𝑒𝑥𝑐𝑒𝑒𝑑𝑒𝑑 𝑎𝑡 𝑚ℎ2 𝑜𝑛𝑙𝑦.
𝛿 = 0.5 − 0.37 = 0.13𝑚
𝑇ℎ𝑒 𝑑𝑖𝑣𝑒𝑟𝑠𝑖𝑜𝑛 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑎𝑓𝑓𝑒𝑐𝑡 𝑚ℎ1
2
∴ 𝐴2 = ∑
𝑛=2
ℎ𝑛
𝑄𝑛
2
=
ℎ𝑛
𝑄2
2
=
0.38
= 0.095 &times; 10−4
2002
ℎ2 0.38
=
= 0.19 &times; 10−2
𝑄2 200
0.19 &times; 10−2 − √0.192 &times; 10−4 − 0.13 &times; 0.095 &times; 10−4
𝑞22 =
= 37.8
0.095 &times; 10−4
~ 𝑚𝑖𝑛. 𝑑𝑖𝑣𝑒𝑟𝑠𝑖𝑜𝑛 𝑓𝑟𝑜𝑚 𝑚ℎ2 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 ℎ𝑒𝑛𝑐𝑒 𝑎𝑙𝑠𝑜 𝑓𝑟𝑜𝑚 𝑚𝑎𝑛ℎ𝑜𝑙𝑒 2.
𝐵2 =
(ii)
For max. redev. potential of site B, all sewage from site A to be discharged to the new
sewer. ∴ 𝑀𝐻3 belongs to sewer line 0-1-4-3 only and is disconnected from sewer line
0-1-2.
1
After diversion, ℎ2 is reduced by .
2
ℎ12 =
0.38
= 0.19
2
W.C. at MH2 becomes
𝐻21 = 10.25 + 0.19 = 10.44
Freeboard at manholes 1 &amp; 2 are:
𝑓1 = 0.75(= 11.0 − 10.25)
𝑓2 = 0.50(= 11.0 − 10.44)
Hence manhole 2 is critical in line 0-1-2.
𝛿21 = 0.56 − 0.05 = 0.06𝑚
Corresponding space capacity is calculated from outfall to MH2.
When 𝛿 &lt; 0, the -ve sign in (7) still to be taken so that qmm(&lt; 0) is the spare capacity
When 𝛿 &gt; 0,
2
𝐴12 = ∑
𝑛=1
ℎ𝑛
𝑄𝑛
2
𝐵21
=∑
2
=
0.25 0.19
+
= 0.2525 &times; 10−4
2002 1002
ℎ𝑛 0.25 0.19
=
+
= 0.315 &times; 10−2
𝑄𝑛 200 100
𝑛=1
−2
− √0.3152 &times; 10−4 + 0.03 &times; 0.2525 &times; 10−4
0.2525 &times; 10−4
= −2.3 (𝑠𝑝𝑎𝑟𝑒 𝑐𝑎𝑝. 𝑎𝑡 𝑀𝐻2)
Clearly this additional discharge has little effect on WL at mh1, hence would not affect
sewer line 0-1-4-3 significantly
∴ Redev. cap. of site B becomes
100 + 2.3 = 102.3
1
𝑞22
=
0.315 &times; 10
```