Civil Engineering - Water Resources Engineering - Discussion
Discussion Forum : Water Resources Engineering - Section 3 (Q.No. 22)
22.
The respective storm totals at three surrounding stations A, B and C are 110, 90 and 70 mm. If the normal annual precipitation amounts at stations X, A, B and C are respectively 1000, 1100, 1200 and 1250 mm, the estimated storm precipitation at X is
Discussion:
4 comments Page 1 of 1.
AKSHAY said:
7 years ago
There are 2 methods.
1)Arithmetic.
2)Normal Ratio.
1)Arithmetic:
Take 10% of the missing storm means for x.
(1000*(10/100))=100,
Add 1000+100=1100,
now subtract 10% from 1000,
1000-100=900,
The normal annual rainfall for B & C are beyond values 900 and 1100.
So second method should be used,
2)Normal ratio method:
x=1/3{ [1000*(110/1100)]+[1000*(90/1200)]+[1000*(70/1250)]},
x=77mm.
1)Arithmetic.
2)Normal Ratio.
1)Arithmetic:
Take 10% of the missing storm means for x.
(1000*(10/100))=100,
Add 1000+100=1100,
now subtract 10% from 1000,
1000-100=900,
The normal annual rainfall for B & C are beyond values 900 and 1100.
So second method should be used,
2)Normal ratio method:
x=1/3{ [1000*(110/1100)]+[1000*(90/1200)]+[1000*(70/1250)]},
x=77mm.
(3)
Ghost said:
8 years ago
Precipitation at X = Nx/m [P1/N1+P2/N2+....+Pm/Nm].
M = Number of stations. - 1(4-1=3).
Px = 1000/(4-1) [110/1100+90/1200+70/1250] = 77.
M = Number of stations. - 1(4-1=3).
Px = 1000/(4-1) [110/1100+90/1200+70/1250] = 77.
(1)
Vinay said:
10 years ago
Precipitation at X = Nx/m [P1/N1+P2/N2+....+Pm/Nm].
M = Number of stations minus 1.
Hence,
Px = 1000/(4-1) [110/1100+90/1200+70/18] = 77.
M = Number of stations minus 1.
Hence,
Px = 1000/(4-1) [110/1100+90/1200+70/18] = 77.
Ahdj said:
10 years ago
Any one please explain me how answer 77 mm right answer?
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