Tiêu đề 04 Seepage Solutions.pdf ✅

04 Seepage Solution 1. The results of a constant head permeability test for a fine sand and sample having a diameter of 150 mm and a length of 150 mm are as follows: Constant head difference = 40 cm Time of collection of water = 83 seconds Weight of water collected = 392 g Solving for Volume of Water: Vw = mw ρw = ( 392 1000 kg) 1000 kg m3 Vw = 0.000392 m3 [ (1000 mm) 3 (1m) 3 ] Vw = 392,000 mm3 Solving for hydraulic conductivity (k): k = VwL aht k = (392,000 mm3 )(150 mm) ( π 4 ) (150mm) 2(40 x 10 mm)(83 seconds) k = 0.10 mm s SITUATION (2-4): For a falling head permeability test, the following are given: Length of specimen = 380 mm Area of specimen = 6.5 cm2 Hydraulic conductivity of soil specimen = 0.175 cm/min 2. What should be the area (cm2 ) of the standpipe for the head to drop from 650 cm to 300 cm in 8 min? Solving for area of standpipe: k = aL At ln ( h1 h2 ) 0.175 cm min = (a) ( 380 10 cm) (6.5 cm2)(8 min) ln ( 650 cm 300 cm) a = 0.31 cm2
3. Compute the interstitial velocity under the test condition if the soil specimen has a void ratio of 0.50 in cm/sec? Solving for velocity: v = ki = k ( hL L ) v = (0.175 cm min ) ( 1 min 60 sec)( 650 cm − 300 cm 380 10 cm ) v = 0.0269 cm sec Solving for porosity (n): n = e 1 + e = 0.50 1 + 0.50 n = 0.3333 Solving for interstitial velocity: vs = v n = 0.0269 cm sec 0.3333 vs = 0.08 cm s 4. Compute the head difference at time equal at 6 mins. Solving for head difference at time equal at 6 mins: k = aL At ln ( h1 h2 ) 0.175 cm min = (0.31 cm2 ) ( 380 10 cm) (6.5 cm2)(6 min) ln ( 650 cm h2 ) h2 = 364.16 cm = 364 cm SITUATION (5-7): A confined aquifer is shown in the figure. This aquifer has a source of recharge located as shown. The hydraulic conductivity of the aquifer is 50 m/day with a porosity of 23%. The piezometric (head) surface in the two observation wells 1000 m apart are at elevation 75 m and 65 m, respectively from the common datum. The aquifer has an average thickness of 30 m and an average width of 3.5 km. 5. Determine the nearest value of the rate of flow of water through the aquifer, in cubic meters per day. Solving for rate of flow: Q = Aki = Ak ( hL L ) Q = (30 m)(3500 m) (50 m day) ( 75 m − 65 m 1000 m )

9. Determine the total flow in cubic meter per hour. Draw Figure: QT = QA + QB + QC For QA: Solving for equivalent flow (Vertical Flow in Stratified Soil): Htotal Keq = H1 K1 + H2 K2 + H3 K3 + H6 K6 0.8 m + 0.7 m + 1.5 m + 0.9 m Keq = 0.8 m ( 6.25 100 m hr) + 0.7 m ( 5.75 100 m hr) + 1.5 m ( 4.50 100 m hr) + 0.9 m ( 3.60 100 m hr) Keq = 0.0468 m hr QA = AAKeqi AA = H2 (1 m) = (0.30 m)(1 m) = 0.30 m2 QA = (0.30 m2 ) (0.0468 m hr) (0.4615) QA = 0.006479 m3 hr For QB: Solving for equivalent flow (Vertical Flow in Stratified Soil): Htotal Keq = H1 K1 + H2 K2 + H3 K4 + H6 K6 0.8 m + 0.7 m + 1.5 m + 0.9 m Keq = 0.8 m ( 6.25 100 m hr) + 0.7 m ( 5.75 100 m hr) + 1.5 m ( 6.25 100 m hr) + 0.9 m ( 3.60 100 m hr) Keq = 0.0527 m hr QB = ABKeqi AB = H3 (1 m) = (0.50 m)(1 m) = 0.50 m2