4.) GIVEN THE PHASE DIAGRAM SHOW. WHAT IS THE V₂ Va 3 AIR WATER SOLIDS W= 10% 25=165.416 f1³ 8moist a 127lb f3 VALUE OF POROSITY
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- Ans 12%Determined and fil-up the unknown data of the given table. Physical Properties Mix Design Specific Dry-rodded Materials Fineness Estimated Absolute gravity weight ft3 Absorption modulus Ib/yd3 Ib/yd3 (SSD) Ib/ft3 Cement 3.15 ? ? Sand 2.64 0.7 2.72 11/2 62.4 Ib/ft Gravel 2.67 0.5 100.5 Water ? Vol. of entrapped air w/C = ? GIVEN: Specification Requirements compressive strength, psi 4000 Slump, in. 3 4Show complete solutuon pls
- YALVE 3) Constont level constent level k 0.26 O-63 soyuB 0.48m According to the above figure given 760 me rater is collected in the container in 15 minutes.If kA is giver as kA= 2x 10m/s..calculate the permeability of isoil B (kR).Length of the Systen perpendicaelar to the Figure plane (thickness) is coefficent O.15 m. -0.2m13. At 15°C water temperature, vapor pressure of water is 1.70 kN/m“. The saturation vapor pressure at that temperature of water is nearly: [Given: specific weight of Mercury = 845.52 lb/ft') %3D (A)0.504 inches of Hg (B) 1.450 inches of Hg (C) 0.755 inches of Hg (D) None of the above 6ni noitsoot s in lo 000,21 001.0(A) EALO(8) (D)O 185 nowi odi 1obiano T ignsl ai 1ovo 1919mpib T91omsib s 2sd oqiq orlT ninismor odi not mm 021o 0.1-u 22ol tixo tot bas uit 0 oteT wofn stste-ybsote Inotont dondaioW-vOtsd diwMOITAKIMAKS MUTTGIM 1.) DETERMINE THE COEFFICIENT OF PERMEABILITY FROM THE FOLLOWING DATA (MM/S) LENGHT OF SAMPLE = 25cm ile browsit enitubon Monastale AREA OF CROSS SECTION OF THE SAMPLE = 30 cm² HEAD OF WATER = 40ст to rutemi DISCHARGE = 2.00 mL IN 110 seconds yan! apth inbom utne nit veroo atuato ed basebal 15 olsadneni give in to lua ant of oub nensum step mA'S a pronic, uns gesig te brs
- (1) 28 (2) (3) (4) 15" 2 12 6 8 fo= 3000 psi NWC for all questims. Mcr= ? (1) What is (2) 28" ftop ² foot if M= 45 4" 12 TH 12" Uncracked M = 30 ftop=? foot = ? 30" ft-k Mer=? 1 Uncra e ked M = 50 frap = ? flot = ? ft-k ft- Mer? Mer = ? Uncracked. +-k M = 10 ftop = ? foot = ?Mc = 3050g Ms = 800g Mss = 820g Mws = 3450g Calculate the values of: Mw = Ms 1. Bulk (O.D.) Specific Gravity = (Mc+Mss-Mws) Mss 2. Bulk (SSD) Specific Gravity =- (Mc+Mss-Mws) Ms 3. Apparent Specific Gravity =. (Mc+Ms-Mws) Mss-Ms 4. Percent Absorption = x 100% Mssoit-sample has the totah volume ep(470)cm³, to pal weight ef(790) gunr weight f Solid(730) gimr Efthe secifte gravity af salidis(2-7) Poyosity, ordratio, watess content, degreetsatavatien Find the fotlosing.
- Castor oil (SG = 0.96) uiny Problem 9: %3D A tank whose bottom is 18 inches x 18 inches is filled with 2 in. three liquid as shown: 12 inches of carbon (SG=1.59), 4 inches of water and 2 inches of (SG=0.96). Compute the total hydrostatic psi) acting at the bottom of the tank. tetrachloride water 4 in. castor oil Carbon tetrachloride 12 in. pressure (in (SG = 1.59)AA AAA EXAMPLE 1-1 a A Soll HAJ A of 10% and RESIDUAL WATER CONTENT 84.3 FT3 FLOW) INTO AN porosity of 30%, OF SOFT x Soft IS THE WATER LEVEL ARBA ▷ RISE? IP 2 6447Given that the absorption capacity of a sample of aggregates is 8.7% and the oven-dry mass of the sample is 1000 g. Determine the mass of water that this sample 'can absorb. O A. 43.50 O B. 8.70 O C. 870.00 O D. 0.87 O E. 87.00