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Hydrologic processes evaporation infiltration
1. Hydrologic Processes INCI 4138 Introduction to Water Resources Engineering Evaporation and Infiltration Processes Dr. Jorge Rivera-Santos & Ra ú l E. Zapata-L ó pez INCI 4138 Hydrologic Processes
6. Evaporation Process INCI 4138 Hydrologic Processes Water molecules are always moving around. This is caused by solar radiation . When there is enough energy , water molecules are knocked away from the surface . Wind moves them out of the upper layer. Water molecule transferred to the atmosphere become water vapor.
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8. INCI 4138 Hydrologic Processes Evaporation Evaporation ( yr or monthly ) from water bodies are estimated by measuring daily evaporation from a pan and then corrected using a given factor based on local climatic conditions and measured daily rainfall (if present). K Lp = 0.7 K lp = 1.5 at z=4m
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17. Infiltration process: Equations INCI 4138 Hydrologic Processes Φ = Phi Index, SCS approach Horton Equation I ∑ (P i -Φ) ∆t = R
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22. Infiltration – Ponding Process INCI 4138 Hydrologic Processes f o ’ = 5.24 in./hr Example Horton-RZL The standard f curve can be given by the equation f = 1.2 + (9 - 1.2) e -4.56t where f in in in./hr and t is in hr . (a & b) Develop and plot the standard infiltration and the cumulative infiltration curve. Rainfall intensity is 1.5 in./hr for the first 40 minutes and then 6.0 in./hr thereafter. P 40 =1.5in./hr(40min)(hr/60min)= 1 in. f’ = 1.2 + (5.24 - 1.2) e -4.56t’ where t’ = t-40
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25. Infiltration: Φ - Index (SCS) INCI 4138 Hydrologic Processes P total =∑(i∆t) =∑(P i ∆t) R=∑R i =∑(Q i ∆t) ∑ (P i -Φ) ∆t =∑R i =R or solving for Φ Φ = (∑(P i ∆t) – R)/(m∆t) EXAMPLE 2.19 (Gupta) Φ-INDEX EVALUATION 5 4.5 RunoffExcess=2.52in Infiltration 4 P 3.5 R E 3 C I 2.5 P I 2 Φ value T A 1.5 I O 1 N 0.5 (in./hr) 0 0 0.5 1 1.5 2 2.5 3 3.5 TIME (hr) DATA: Area (acre) = 500 Direct Runoff (acre-ft) 105 ‘ =105/500*12 in. = 2.52 in. Assuming Φ <2.0 in./hr Time Time Time Precipitation Amount of Increment Intensity Precipitation (hr) (min) (hr) (in./hr) (in.) (in.) δ t P (P* δ t) (P- Φ)δ t 0 0 0.5 4.5 2.25 (4.5-Φ)*0.5 0.5 30 0.5 3.0 1.5 (3-Φ)*0.5 1 60 0.5 1.0 0.5 (0)*0.5 1.5 90 0.5 3.5 1.75 (3.5-Φ)*0.5 2 120 0.5 2.0 1 (2.0-Φ)*0.5 2.5 150 0.5 0.0 0 (0)*0.5 3 180 Sumation 14 7 (13-4Φ)0.5 Total Precipitation (in.) = 7 Runoff (in.) = 105/500/12 = 2.52 Infiltration = P*t-R (in.) = 4.48 Therefore: (13-4Φ)*0.5 = 2.52 Φ = (13 - 2.52*2) / 4 = 7.96/ 4 = 1.99 in./hr 1.99 OK.
26. Infiltration: Φ - Index (SCS) INCI 4138 Hydrologic Processes P total =∑(i∆t) =∑(P i ∆t) R=∑R i =∑(Q i ∆t) ∑ (P i -Φ) ∆t =∑R i =R or solving for Φ Φ = (∑(P i ∆t) – R)/(m∆t) EXAMPLE 2.19 (Gupta) Φ-INDEX EVALUATION 5 4.5 RunoffExcess=2.52in Infiltration 4 P 3.5 R E 3 C I 2.5 P I 2 Φ value T A 1.5 I O 1 N 0.5 (in./hr) 0 0 0.5 1 1.5 2 2.5 3 3.5 TIME (hr) DATA: Area (acre) = 500 Direct Runoff (acre-ft) 105 ‘ =105/500*12 in. = 2.52 in. Assuming Φ <2.0 in./hr Time Time Time Precipitation Amount of Increment Intensity Precipitation (hr) (min) (hr) (in./hr) (in.) (in.) δ t P (P* δ t) (P- Φ)δ t 0 0 0.5 4.5 2.25 (4.5-Φ)*0.5 0.5 30 0.5 3.0 1.5 (3-Φ)*0.5 1 60 0.5 1.0 0.5 (0)*0.5 1.5 90 0.5 3.5 1.75 (3.5-Φ)*0.5 2 120 0.5 2.0 1 (2.0-Φ)*0.5 2.5 150 0.5 0.0 0 (0)*0.5 3 180 Sumation 14 7 (13-4Φ)0.5 Total Precipitation (in.) = 7 Runoff (in.) = 105/500/12 = 2.52 Infiltration = P*t-R (in.) = 4.48 Therefore: (13-4Φ)*0.5 = 2.52 Φ = (13 - 2.52*2) / 4 = 7.96/ 4 = 1.99 in./hr 1.99 OK.
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