FLOOD FREQUENCY ANALYSIS
1. Arrange the annual peak discharge highest to lowest (rank 1-n)
Normal Distribution Method ((x̄, σ, ∑)
Rank
Observed
Return Period, T
Discharge
1/p
p
p>
0.5
1-p
w
KT
X2
xT
x̄ + KTσx
2. Plotting position, p
(n = total , m = rank)
𝟏 𝟏/𝟐
𝐰 = [𝐥𝐧 ( 𝟐 )]
When p > 0.5, use (𝟏 − 𝐩) to solve for 𝐰
𝐩
3.
4.
KT = w − [
5.
X2
2.515517 + 0.802853w + 0.010328w2
1 + 1.432788w + 0.189269w2 + 0.001308w3
]
Log-Normal Distribution Method (x̄, σx, CS, CV ; y̅, σy, CS, CV)
Rank
Rank
x, Observed
y
Discharge
log(x)
Observed
Return
Period, T
1/p
p
Return Period, T
p>
0.5
1-p
w
KT
KT
yT
xT
y̅ + KTσy
10 yT
xT
X2
X2
Discharge
Gumbel distribution Method (x̄, σx, CS, CV)
x̄ + KTσx
1/p
Chi-squared
Log Pearson Type III (Gamma) Distribution (x̄, √σ𝟔x, CS, CV ; y̅, σy, CS,
𝐓 CV)
Rank
x,
Observed
Discharge
𝑧=w−[
y
log (x)
𝐊𝐓 = −
{𝟎. 𝟓𝟕𝟕𝟐 + 𝐥𝐧 [𝐥𝐧 (
)]}
𝛑
𝐓−𝟏
p
p>
w
z
k
KT
0.5
1-p
CSy/6
Return
Period, T
1/p
yT
xT
y̅ + KTσy
10 yT
X2
2.515517 + 0.802853w + 0.010328w 2
]
1 + 1.432788w + 0.189269w 2 + 0.001308w 3
When CSy = 0, 𝐊𝐓 = z
When CSy ≠ 0;
𝐊𝐓 = 𝐳 + (𝐳𝟐 − 𝟏)𝐤 +
𝟏
𝟑
𝟏
(𝐳𝟑 − 𝟔𝐳)𝐤𝟐 − (𝐳𝟐 − 𝟏)𝐤𝟑 + 𝐳𝐤𝟒 + 𝐤𝟓
𝟑
WATER DEMAND
Level II Water Service
No. of Faucets
Nf =
Total Domestic Consumption
Population
faucets
×
Person per HH
HH
= n faucets
Tdc = SL−II × Nf =
Unit Consumption (50 -60 lpcd, 4-6 HH)
Uc =
𝑛
𝑙𝑖𝑡𝑒𝑟
𝑐𝑎𝑝𝑖𝑡𝑎
𝑑𝑎𝑦
(lpcd) ×
persons
HH
=
Water Demand (Wd / ADD)
𝑛 𝑙𝑖𝑡𝑒𝑟
𝑑𝑎𝑦−𝐻𝐻
Wd=ADD= Tdc + NRW(%Wd)= Tdc / 1-%NRW
Supply per faucet
SL−II = 𝑈𝐶 ×
HH
faucets
𝑛 𝑙𝑖𝑡𝑒𝑟
𝑑𝑎𝑦
Source Capacity
=
𝑛
𝑙𝑖𝑡𝑒𝑟
𝑑𝑎𝑦
SC = 𝑊𝑑 ×
𝑓𝑎𝑢𝑐𝑒𝑡
𝑑𝑎𝑦
1 ℎ𝑟
× 36000 𝑠
operating hrs
=
𝑛 𝑙𝑖𝑡𝑒𝑟
𝑠𝑒𝑐𝑜𝑛𝑑
Level III Water Service
𝑚
3
𝑚
3
No. of Connections [lnstitutional (1.0 day), commercial(0.8 day), etc. ]
HH
Nc = (Population × Person ) + Other Connections = n connections
Unit Consumption (80-100 lcpd)
Uc =
𝑛
𝑙𝑖𝑡𝑒𝑟
𝑐𝑎𝑝𝑖𝑡𝑎
𝑑𝑎𝑦
(lpcd) ×
persons
HH
=
Total Domestic Consumption
T𝑑𝑐 = 𝑈𝐶 × 𝑛𝑜. 𝑜𝑓 𝐻𝐻 =
𝑛 𝑙𝑖𝑡𝑒𝑟
𝑑𝑎𝑦
Water Demand (Wd / ADD)
𝑛 𝑙𝑖𝑡𝑒𝑟
𝑑𝑎𝑦−𝐻𝐻
Wd = ADD = Tdc + NRW ; Tdc = ADD/ 1-NRW
Source Capacity
SC = 𝑊𝑑 ×
𝑑𝑎𝑦
1 ℎ𝑟
× 36000 𝑠
operating hrs
=
𝑛 𝑙𝑖𝑡𝑒𝑟
𝑠𝑒𝑐𝑜𝑛𝑑