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A Example fine sand with a permeability of 0.01 cm/sec and a porosity of 0.35 is put in a glass tube permeameter from point B to point C and contained with porous screens. (All dimensions are in cm). The tube is 10 cm in diameter. The permeameter is kept full by pouring in water at point A, and water runs out at point D. Determine the total head, pressure head, elevation heads, and pressure at points A, B, C, D and the midpoint of the soil column. Determine the gradient, discharge velocity, seepage velocity and flow rate.
A D 30 60 B 20 50 40 C
Solution: Point A is at a free surface 60 cm above the datum. Thus EH = 60, PH = 0, TH = 60 + 0 = 60 cm. The pressure is zero. The resistance to flow through the glass tube from A to B and C to D is negligible in comparison to the resistance to flow through the soil column from B to C. Thus it can be assumed that all energy loss and total head loss occurs through the soil column. At point B, the total head is 60 cm, the elevation head is 20 cm. Thus the pressure head is 60 - 20 = 40cm. The pressure p is p = (PH)(w) = (0.40m)(9.81 kN/m3) = 3.92 kN/m2 = 3.92 kPa. At point C, the total head is 50 cm and elevation the head is 40 cm. Hence the pressure head is 10 cm, and the pressure is (0.1)(9.81) = 0.98 kPa. At point D, the total head and elevation head are 50 cm, the pressure head is zero, and the pressure is zero. As the cross-section of the soil column is constant, total head is lost uniformly along the length and the total head would be 55 cm at the midpoint. Water in a piezometer tapped into the midpoint of the soil would rise to elevation 55 cm. The elevation head at the midpoint is 30 cm; hence the pressure head is 55 - 30 = 25 cm and the pressure is (0.25)(9.81) = 2.45 kPa.
THIS EXAMPLE PROBLEM ADAPTED FROM PROF. WOLFFS HANDOUT
The gradient is the rate of loss of total head with respect to distance along the flow path. Within the uniform soil column, water loses 10 cm of total head over 30 cm of travel; hence i = 10/30 = 0.333. The discharge velocity is v = ki = (0.01 cm/sec)(0.333) = 0.00333 cm/sec The seepage velocity is
v= ki ( 0. 01)( 0. 333) = = 0. 00951 cm / sec n 0. 35
The flow rate is:
Q = kiA = ( 0. 01)( 0. 333)
(10) 2 = 0. 262 cm3 / sec = 0. 262 106 m3 / sec 4
THIS EXAMPLE PROBLEM ADAPTED FROM PROF. WOLFFS HANDOUT

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