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Mean Areal Rainfall Calculation

The document describes a storm event over a small urban watershed with 4 rainfall gages. It provides the rainfall recorded at each gage and the drainage area associated with each. It asks to calculate the mean areal rainfall for the storm using arithmetic averaging and the Thiessen method. Using arithmetic averaging, the mean rainfall is 2.965 inches. Using the Thiessen method, which weights the rainfall by the drainage area associated with each gage, the mean rainfall is calculated to be 3.093 inches.

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Rodilyn Basay
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427 views2 pages

Mean Areal Rainfall Calculation

The document describes a storm event over a small urban watershed with 4 rainfall gages. It provides the rainfall recorded at each gage and the drainage area associated with each. It asks to calculate the mean areal rainfall for the storm using arithmetic averaging and the Thiessen method. Using arithmetic averaging, the mean rainfall is 2.965 inches. Using the Thiessen method, which weights the rainfall by the drainage area associated with each gage, the mean rainfall is calculated to be 3.093 inches.

Uploaded by

Rodilyn Basay
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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BASAY, RODILYN G.

BSCE-3 CE 3107 MW 5:00-6:00 PM 09-21-2020

A small urban watershed has four rainfall gages as located in the figure. Total rainfall
recorded at each gage during a storm event is listed in Table 1. Compute the mean
areal rainfall for this storm using (a) arithmetic averaging and (b) the Thiessen method.

 
Table 1. Rainfall record at each gage.

Sol’n:

A.) Arithmetic Averaging

n
P 1+ P 2+, …+ Pn 1
Pave= = ∑ Pi
n n i

Given: B= 2.92 in.; C= 3.01 in.

Pave = 2.92 + 3.01 / 2 = 2.965 in.


B.) The Thiessen Method

B
C
A
a
D

Pi (in) Ai (mi2) Ai/Ar (Pi)(Ai/Ar)


3.26 2.6 0.186 0.66
2.92 3.4 0.244 0.712
3.01 6.36 0.456 1.37
3.05 1.6 0.115 0.351
∑ = 13.96 3.093 in.

Sample Computation:

AA= (0.04) (65) = 2.61 mi2

AA/AR= 2.6 mi2/13.96 mi2 = 0.186

(PA) (AA /AR) = (3.26) (0.186) = 0.66 in.

Therefore, 2.965 in. and 3.093 in. are approximately the same.

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