Question

2. a) Try to find respectively the pressure head, elevation head at location A, B and C in arrangements as shown in Figure 1-a & -b under seepage (water exits at datum level). b) How much water could be expected from an arrangement Figure 1-b in 30 minutes if 0.5 litre of water was collected from Figure 1-a arrangement in one hour (assume the same type of soil in both arrangement). 0.4m |-0.5 mm|-0.5 -0.50 -0.5 B 1 1 m Exit Datum 0.75 m [0$m a. Soil sample only at central section in tube 0.4 m -0.5 -0.5 -0.5 -0.5 | Soll: A18 1 m Exit 0.75 m Datum 1 0.5 m b. soil sample extended in full in tube Figure 1 Water seepage through a sample of the same kind soil filled in a tube

Answer 1

solution Figurela 40 im fawomto sm datum Hevatron head at -0.15 0.05 m Elevation head act 0.25m Pressure head at A 09 to find 1. Calculate pressure head before soil Sample t in this care loss in head Apressive head at this point is equal to head total pressure 2. calculate precure head head al at given point soil upto given - Total pressure bead head locs due to head loss hydraulic gradient x length of soil Total pressure bead = 0.9m point pressure bead at before soil -09m pressure head at A - Total pressure head before soll sample head loce due to soil sample th 0.9-9xl vlon total head loce hydraulic gradient total solu length 0.9 709 P = 0.90.9%0.5 = burm

pressure head at Bire) Pe - Total pressure head head lose po before soil sample soil sample - ا ۳۷ - C - Q 0.9 0.98 1 figure 1-b Elevation head at 1-0-75-05-025m head at B= o-TS-01 -0-250 Elevation pressure head at A - Total pressure head before foil sample head lors due to Son samople uptoA -0.9 -9XL 16.505 = 0.9 -0.45*(0.570-5) (2 h = 0.45m -0.45 pressure head at B PR Toral pressure bead before' soil sample bead locs due to soil sample upto B 0.9 - TX L uptoB 0.9* 0:45*(0.570-5+0.5) -0.225 m pressure head figl-a Elevation head -А B 6.25m 0-25m 0.45m figi-b 0.25m 0.25 m o usm 0.2250

Goren for figure 05 latre ie on hout of water collected in Volume -0-5 litre tame 1 hour = 60 min discharge a= KEA Чехе ки tube constant Space soil is came and area came We ako Know That volume dischargea (KA)(0) time volume lime (KA) 0.9= 60 KA = 0.00925 Lit/min for figure 1-6 Amount of Water collected 30 min Volume time = 30 min (KAYU) = volume HE time (KA)0:45 - 30 0.00926x 0.45= 30 V = 0.125 lit