Tiêu đề Eval 5 Nov 2024 Solutions.pdf ✅

SITUATION 3. ▣ 10. From the formula for shrinkage limit, SL = m1 − m2 m2 − (V1 − V2 )ρw m2 SL = 45 g − 31 g 31 g − (25 cm3 − 16 cm3 ) (1 g cm3 ) 31 g SL = 16.13% ▣ 11. Before solving for the specific gravity of soil solids, solve first for the shrinkage ratio: SR = m2 V2ρw SR = 31 g (16 cm3) (1 g cm3 ) SR = 1.9375 Therefore, to solve for the specific gravity of soil solids, Gs = 1 1 SR − SL Gs = 1 1 1.9375 − 0.1613 Gs = 2.82 SITUATION 4. Recall the formulas for equivalent hydraulic conductivity. For parallel flow (water seeps through all layers at the same time, the discharge is divided into all layers) k∥ = ∑ kihi ∑ hi For perpendicular flow (water seeps through the layers one after the other) k⊥ = ∑ hi ∑ hi ki