Based on the assumption that some annual flow data fit the normal probability distribution, the statisticalparameters (mean, standard deviation, and skewness coefficient) for the annual flow data were respectivelydetermined to be 750 m3/s, 110 m3/s, and 0.0. For this annual flow data, estimate the magnitude of flood flowscorresponding to recurrence intervals of 10-year and 100-year? 1 mile = 5280 ft.1 mile? = 640 acres1 acre = 43560 ft²1 inch = 2.54 cm1 ft. = 12 in1-hectometer =100 meter1 ha = 10,000 m²P -R-E -T-G=±ASP-R- ET =0Pe = P - E lossesNxPx = E1Ni() PiPx =E(Pi|Li?) /(1]Li?)E = C (eo – ea) [1+(W/10)]U=k, BB=E (t p/ 100)|f= f + (fo – f.) e-ktф%3D (Р— R)DΣ (Pi- Φ) Δt RQ =R= (P – 0.2S)²/ (P + 0.8S) S= (1000/CN) –10 Qpeak = CIA(I – 0) = AS/AtCo = (- KX + 0.5 At)/AWhere, A = K- KX + 0.5AtS= K [XI + (1-X) 0]C1 = (KX + 0.5 At)/AO2 = Co I2 + Cılı + C2 O1C2 = (K – KX – 0.5At)/A(I,+ I,) + [(2S,/At) – 0,] = [(2S,/At) + 0,]PE1UE2) -P(E1) +P E2 ) -PEl Ω Ε2)PP B/A)-P (Α^ B/ P (A)Risk = 1- [P, (E)"]=1-[1-÷j" Reliability = (1- Risk) = [1 - 1" P(Q2) = m/(n+1)n!P(x) = () (p)* (1-p)@=3)where, () =P(Q2)=1/T; Tr=1/P(Q2)х! x (п-х)!nE(X, - X)x=s- E(X – X)2(n-1)ΣΧCv = () x 100%C, = g1*(п - 1)(п - 2)sX -XZ=-,= Q + K,S and Q, = Q +K¸SQuX, = x+K„.sT0.5772+In In|Т-1K, =K+ /K2-abKUK- VK2-aba = [1-{z'/2(N-1)}; b = K² -2INKĻb = K -z/Naа

x = 750m3/s μ =110 m3/s T1 = 10 years T2 = 100 years

Answered: Based on the assumption that some…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Based on the assumption that some annual flow data fit the normal probability distribution, the statistical
parameters (mean, standard deviation, and skewness coefficient) for the annual flow data were respectively
determined to be 750 m3/s, 110 m3
/s, and 0.0. For this annual flow data, estimate the magnitude of flood flows
corresponding to recurrence intervals of 10-year and 100-year?

1 mile = 5280 ft.
1 mile? = 640 acres
1 acre = 43560 ft²
1 inch = 2.54 cm
1 ft. = 12 in
1-hectometer =100 meter
1 ha = 10,000 m²
P -R-E -T-G=±AS
P-R- ET =0
Pe = P - E losses
Nx
Px = E1
Ni
() Pi
Px =E(Pi|Li?) /(1]Li?)
E = C (eo – ea) [1+(W/10)]
U=k, B
B=E (t p/ 100)
|
f= f + (fo – f.) e-kt
ф%3D (Р— R)D
Σ (Pi- Φ) Δt R
Q =R= (P – 0.2S)²/ (P + 0.8S) S= (1000/CN) –10 Qpeak = CIA
(I – 0) = AS/At
Co = (- KX + 0.5 At)/A
Where, A = K- KX + 0.5At
S= K [XI + (1-X) 0]
C1 = (KX + 0.5 At)/A
O2 = Co I2 + Cılı + C2 O1
C2 = (K – KX – 0.5At)/A
(I,+ I,) + [(2S,/At) – 0,] = [(2S,/At) + 0,]
PE1UE2) -P(E1) +P E2 ) -PEl Ω Ε2)
PP B/A)-P (Α^ B/ P (A)
Risk = 1- [P, (E)"]=1-[1-÷j" Reliability = (1- Risk) = [1 - 1" P(Q2) = m/(n+1)
n!
P(x) = () (p)* (1-p)@=3)
where, () =
P(Q2)=1/T; Tr=1/P(Q2)
х! x (п-х)!
nE(X, - X)
x=
s- E(X – X)2
(n-1)
ΣΧ
Cv = () x 100%
C, = g1*
(п - 1)(п - 2)s
X -X
Z=-
,= Q + K,S and Q, = Q +K¸S
Qu
X, = x+K„.s
T
0.5772+In In|
Т-1
K, =
K+ /K2-ab
KU
K- VK2-ab
a = [1-{z'/2(N-1)}; b = K² -2IN
KĻ
b = K -z/N
a
а

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