Question # 01

A Reservoir operator has to be release water from reservoir for being picked up at a distance of 50 Km for downstream users. The average width of the stream for the anticipator discharge of 40km, The mean daily class a tank evaporation for this season is 0.5cm.Estimate the mean daily evaporation from the stream in Acre-feet per day .If the pan co-efficient is 0.7. Assuming 15% loss due to seepage , find discharge at the head of canal if the required discharge at the tail is to be 50 m3/s.

SOLUTION :

Discharge at tail Qtail = 50m3/s Discharge at head Qhead = ?Surface area of stream = A = L x W

as L = 50 km= 50 x 1000= 50000 m W = 40 m A = 50000 x 40 = 2000000

Seepage Lossees = 15 %

Evaporation Losses :

PAN Evaporation E = 0.5 cmPAN Co-efficient Kp = 0.7 LAKE Evaporation El = ?

El = Kp.EEl = 0.7 x 0.5El = 0.35 = 3.5 x10-3m Or 0.0035m

So Evaporation losses from stream per day E stream = 0.0035 x 2000000

= 7000m3/day= _ 7000_

24 x 3600= 0.081m3/sec

As 1 Acer-ft = 43560ft1 meter = 3.282As = 7000 m 3

day= 7000 x (3.28)3

= 247013 ft3 / dayAs 1 Acer-ft = 43560 ft

So = 247013 43560

= 5.6 Acer-ft day

Question # 02

An Engineer-In-Charge of Reservoir operation has to release water from reservoir to provide irrigation supplies at a discharge of 40 km for downstream Users. The average width of the stream for the anticipates discharge is 25 km. The daily mean Class A pan evaporation for this season is 5mm per day.Estimate the daily evaporation losses from the stream in hector-meter-per day.

Solution: Stream Length = 40 km

= 40 x 1000 = 40000 mStream width (W) = 25m

Surface Area of stream = 40000 x 25 = 1000000m2

Assuming Kp = 0.7Pan Evaporation = E = 5mm/day

El = Kp E= 0.7 x 5 = 3.5mnm or 0.0035m

E Stream = 0.0035 x 1000000m3

= 3500m3

1 Hector = 10000m2

E Stream = 3500_ 10000 = 0.35 hector-meter

Question # 03A small catchment area 150 hectare received a rainfall of 10.5cm in 90 minutes, Due

to a storm draining the catchment was dry before the storm and experienced a runoff lasting for 10 hours with an average discharge Value of 2m3/sec. The stream was given dry after the runoff event.(a) What is the amount of water in acre-feet which was not available to runoff alone to combined effect of infiltration, evaporation and transpiration.(b) What is the ratio of total and direct runoff to precipitation.Solution :

Catchment Area “A” = 150 hectr as 1hectr = 10000m2

= 1500000m2

Precipitation “P” = 10.5cm = 0.105mTotal Runoff Volume = A x P

= 1500000 x .0.015= 1575000m3

Time Duration of Rain = 90 min = 90 x 60 = 5400 SecTotal Runoff Discharge Q Rain = 157500

5400= 29.17m3/Sec

Q Runoff = 2m3/secRunoff time = 10 hrs

= 10 x 3600 = 36000 secTotal Runoff Volume = 2 x 36000 = 72000m3

(a) For Water Lost :Water Lost = Q Rain – Q Runoff

= 157500 – 72000Water Lost = 85500 m3

as 1m3 = 35.315 ft3

1acre-f t = 43560 ft3

So,Water Lost = 85500 x 35.315

43560Water Lost = 69.3 Acer-Ft

(b) Ratio Between Total Runoff to Direct Runoff :Direct Runoff = 72000m3

Total Runoff = 157500m3

Ratio = 72000_ = 0.457 157500

Percentage = 72000 x100 157500= 45.7%

Question # 04For Data given Example 4.4 Find total infiltration during the storm period Using

Horton’s Equation assuming fo = 1.5cm/h and fc = 0.5cm/h.Solution :

Total Infiltration “F” = ?Rain = 50mmTime Duration = 2 HRS

Initial Infiltration Rate fo = 1.5cm/hConstant Infiltration Rate fc = 0.5cm/h.

ASTotal Infiltration F = [ fo – fc ] (1 – e-kt) + fet

K= [1.5 – 0.5] (1 –e -1x2) + 0.5 x 2

1= [1] (0.864) + 1 1= 1.864 cm

Question # 05A storm width 10cm precipitation produced a direct runoff as 5.8cm given the time

distribution of the storm in Table 4.3 Estimate the ɸ Index.Solution :

Table 4.3 Time Distribution of the Storm.

Hour(Time) 1 2 3 4 5 6 7 8

International Rainfall (cm) 0.4 0.9 1.5 2.3 1.8 1.6 1.0 0.5

Calculation for Φ Index

Rainfall (cm) “f” values (cm)

0.45 .50 0.6

0.4 - - -

0.9 0.45 0.4 0.3

1.5 1.05 1.0 0.9

2.3 1.85 1.8 2

1.8 1.35 1.3 1.2

1.6 1.15 1.10 1

1.0 0.55 0.5 0.4

0.5 0.05 -

Total 6.45 6.10 5.8

Question # 07For 3-Hours duration 225mm of total Rainfall we observed over 3200 km2 catchment

Area. The infiltration capacity curve for this area can be given by Horton’s Equation (Equation 4.6 & 4.7) in wich fo = 10mm/h and fc = 0.5mm/h. Evaporation and other losses during the storm period were observed to be 50mm.Find excess Rainfall over the Catchment (a) Estimate Direct Runoff Volume in m3 & Hectare –m from excess Rainfall(b) Total Run off in Hector-m

Solution :Time Duration “t” = 3HoursRainfall “P” =225mmArea “A” = 3200km2

Initial Infiltration Rate fo = 10mm/hConstant Infiltration Rate fc = 0.5mm/h

K = 1h-1

Total Infiltration F = [ fo – fc ] (1 – e-kt) + fet K

= [10 – 0.5] (1 –e -1x3) + 0.5 x 3 1= [9.5] (0.950) + 1.5 = 10.5 mm

Excess Rain Fall = Rain – F – Losses= 225 – 10.5 – 50= 164.5

Direct Runoff = Rainfall Excess x Area of Catchment= 164.5 x 3200 (1000)2

1000= 526400000m3

= 526400000 10000

= 52640 Hector – meter Total Runoff = [Rain – Losses other than infiltration ] x A

= [225-50] x 3200 (1000)2 m3

1000

= 0.175 x 3200 (1000)2

= 560000000 m3

As 1 Hectr – meter = 560000000 hector – meter 10000= 56000 hector – meter

Question # 08An infiltration Capacity Curve prepared for a Catchment indicates an initial capacity

of 2.5cm/h and attains a constant value of 0.5 cm/h after 10 hour of Rainfall. With the Horton’s constant K = 6 day-1

Solution :Time Duration “t” = 10 Hours

Initial Infiltration Rate fo = 2.5 mm/hConstant Infiltration Rate fc = 0.5 mm/h

K = 6 days-1 Or 6/24 = 0.25h-1

As Horton’s EquationTotal Infiltration F = [ fo – fc ] (1 – e-kt) + fet

K= [2.5 – 0.5] (1 –e -0.25x10) + 0.5 x 10 0.25= [8] (0.918) + 5 = 7.34 mm= 12.34 cm

Question # 09In a project related to rainfall – runoff stadiesw f – curve was plotted to established

an equation of the form of Horton’s Equation. If F = 8.50 sq Units on the graph with each sq represented 1 cm/h on the Vertical and 2 minute on the abscissa and fo = 4.5cm/h fc = 1.2cm/h Determine the Horton’s Equation and Calculate “f” for t = 10 minutesSolution :

Time Duration “t” = 10 MinInitial Infiltration Rate fo = 4.5 mm/hConstant Infiltration Rate fc = 1.2 mm/h

F = 8.50 Units = 8.50 (1cm x 2 h) h 60

As Horton’s EquationTotal Infiltration F = [ fo – fc ] (1 – e-kt) + fet Let K = 1

K= [4.5 – 0.5] (1 –e -1x.166) + 1.2 x 0.166 1= [3.3] (0.152) + 2= 5.3 mm

Question # 10In a storm, total rain fall is 2.29cm and the total infiltration loss is 0.88cm,

Calculate the rain fall excess. Neglect evaporation during the period.

Solution :Rainfall “P” = 2.29cmInfiltration “F” = 0.88 cm

Rainfall excess = ?

Rainfall excess = P – F

= 2.29-0.88 = 1.41cm

Question # 11Determine the runoff from a catchment of area 2.3k over wich 7.5cm of Rainfall

occurred during 1 day Storm.An infiltration of 0.6 cm/h and attained a constant Value of 0.15cm/h after 12 hours of Rainfall with Horton’s Constant Value of K=3k-1. A class A-pom installed in the catchment indicates a degree of 2.5 cm in water level on that day. All other losses were found to be negligible.Solution :

Catchment Area “A” = 2.3km2

Rainfall “P” = 7.5cmTime Duration “t” = 1 day 24hours

Infiltration Losses :Time Duration “t” = 12 Hours

Initial Infiltration Rate fo = 0.6 mm/hConstant Infiltration Rate fc = 0.15 mm/h

K = 3 h-1

As Horton’s Equation :Total Infiltration F = [ fo – fc ] (1 – e-kt) + fet Let K = 1

K= [0.6 – 0.15] (1 –e -3x12) + 0.15 x 24 1= [0.15] (1) + (3.6)= 3.75 mm

Evaporation Losses :E = 2.5cmKp = 0.7El = ?El = Kp x E

= 0.7 x 2.5 = 1.75cm

Rainfall Excess :

= P – F – El = 7.5 -3.75 – 1.75 = 2.00 cm

Direct Runoff : = Rainfall Excess x Area of Catchment= _2_ x 2.3 (1000)2

100= 46000 m3

= 46000 10000= 4.6 Hector – meter