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?
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?
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F07182ae3-0332-46bc-bc8f-f103a9c03760%2Fa01b10d5-d704-4d36-b137-486f7b92d8f7%2Fwm680azm_processed.png&w=3840&q=75)
Transcribed Image Text: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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