Precipitation Data Analysis

CE322 Basic Hydrology
Jorge A. Ramirez

Precipitation Data Analysis - Interception and Depression Storage - Examples

The coordinates of four precipitation gauging stations are A = (3,4), B = (9,4), C = (3,12), and D = (9,12). The observed precipitation amounts at these gauges are P_A = 15 mm, P_B = 23 mm, P_C = 10 mm, P_D = 19 mm, respectively. These stations are located in a rectangular basin whose boundaries are defined by the following coordinates (0,0), (14,0), (14,16), (0,16). Compute the mean areal precipitation over this basin using the Thiessen polygons method and the arithmetic average method.

(1)

(2)

Using Thiessen polygons, the areas of influence are A_A = 48; A_B = 64; A_C = 48; A_D = 64. The basin area is A_basin = 224.Thus, the mean areal precipitation is obtained as:

MAP = (48/224)15 + (64/224)23 + (48/224)10 + (64/224)19 = 17.36 mm

Arithmetic Average method: MAP = (15 + 23 + 10 + 19)/4 = 16.75 mm

2. The precipitation amounts for the months of June, July, and August are missing from the record for one gauging station in a basin. This station belongs to a network of four in that basin. For those three months, the other three stations recorded the following:

	Station
	1	2	3
June	55	65	75
July	47	50	45
August	45	40	55

Estimate the missing precipitation values if the long-term annual average precipitation at the four stations is:

	Station
	1	2	3	4
June	60	65	70	67
July	50	55	65	60
August	45	47	60	55

Use the Normal Ratio Method,

(3)

June: P_x = ((67/60)55+(67/65)65+(67/70)75)/3 = 66.73
July: P_x = ((60/50)47+(60/55)50+(60/65)45)/3 = 50.83
August: P_x = ((55/45)45+(55/47)40+(55/60)55)/3 = 50.74

3. Assume that a rainfall event of intensity 1.25 cm/h falls over a uniformly forested watershed of area 20 km². If the only water losses during this event are those due to interception, compute the volume of water that leaves the basin as storm runoff for a 1-hour and a 2-hour rainfall. Interception volume as a function of precipitation volume P is given by,

(4)

Assume that K is 1.5, S is 0.2 cm and that the evaporation rate is zero.

Compute total precipitation volume:

P = (1.25 cm/h) (1 h) = 1.25 cm

P = (1.25 cm/h) (2 h) = 2.5 cm

Use equation 4 to obtain:

L_i = (0.2 cm) (1 - exp(-1.25/0.2)) = 0.1996 cm

L_i = (0.2 cm) (1 - exp(-2.5/0.2)) = 0.2 cm

Assuming that there is no change in basin storage, then the output of the basin is equal to:

Volume of Output = (P - L_i)*A_basin = ((1.25 cm - 0.1996 cm)/100 cm/m ) (20 10⁶ m²) = 210,080.0 m³

Volume of Output = (P - L_i)*A_basin = ((2.50 cm - 0.2 cm)/100 cm/m ) (20 10⁶ m²) = 460,000.0 m³

4. Repeat Problem 3 but assuming that in addition to interception there are also losses due to depression storage. The depression storage equation is:

(5)

Assume that S_d is 0.2 cm. What is the runoff ratio for these conditions? How much runoff would have been produced halfway into the storm?

P_e = P - L_i = 1.0504 cm. Thus, V = (0.2 cm) (1 - exp(-1.0504/0.2)) = 0.1989 cm

P_e = P - L_i = 2.3 cm. Thus, V = (0.2 cm) (1 - exp(-2.3/0.2)) = 0.2 cm

Again, assuming that there is no change in basin storage, then the output of the basin is equal to:

Volume of Output = (P_e - V)*A_basin = ((1.0504 cm - 0.1989 cm)/100 cm/m ) (20 10⁶ m²) = 170,300.0 m³

Volume of Output = (P_e - V)*A_basin = ((2.30 cm - 0.2 cm)/100 cm/m ) (20 10⁶ m²) = 420,000.0 m³