# EXAMPLES OF DESIGN OF DEWATERING AND. PRESSURE RELIEF SYSTEMS

 This appendix consists of figures D-l through D – 10, which follow.

 D-

 NOTES: 1. System symmetrical about 2. Bottom of excavation equals 100 X 750 ft. 3. Clay is impervious and not under artesian pressure. 4. Line source of seepage. PROBLEM: Determine spacing of 2-1/2-in. self-jetting wellpoints with 2-in.-diam riser pipes to lower GWL 4 ft below bottom of excavation, assuming vacuum (v) = 24 ft, head loss in collector (He) = 2 ft, and intake of pump is 2 ft above collector. Assume style D, Mesh E, wellpoints are to be used.

 I

 hr I hD(L)

 SOLUTION: Compute spacing of wellpoints so that available (net) vacuum in headers (20 ft) will lower water level at wellpoints below that required to produce the nec­essary hD. Assume two stages of wellpoints will be required and that each stage will be installed 2 ft above the groundwater table existing at the time of installation. Also, assume rw = 0.12 ft .

 Assume a = 10 ft, then Qw = aQp = 9.1 gpm.

 From a plan flow net it can be shown that the average flow for a finite line of wellpoints for this excavation will be about 35 percent greater than for an infinite line. Thus Qw = 1.35 (9.1 gpm) = 1.64 cfm.

 Calculate head at wellpoint, hw, from eq 1 (fig. 4-22).

 k2 = Qw

 4 = (25)2

 l** In!0 … 0.2" 2"(0.12)

 hw = 24.9 ft

 For Qw = 12.3 gpm and well screen length = 3 ft, the hydraulic head losses are as follows: He = 0.2 ft from fig. 46a, curve 7 H, = 0.9 ft from fig. 46b Hr + H, = 0.5 ft from fig. 4-26c which includes loss in swing. HW(U) = 1.6 ft. Thus Ь*(и) – H*(U) = 24.9 – 1.6 = 23.3 ft. Therefore, the required effective vacuum in the header = el 42 – 23.3 = 18.7 ft. Since this value is slightly less than the available 20 ft, a wellpoint spacing of 10 ft with header at el 42 and top of wellpoint at el 21 would be satisfactory. Lower stage. Install lower stage at el 27 and 62 ft from £ of the excavation, to lower the groundwater to el 16. Required hD(L) = 16 ft. Compute h0(L) at a partially penetrating slot from eq 4 (fig. 4-3).

 hD(L) = h0(L) [L« (H – h0) – 1] 16 = h^, [“* (40 – h„) ♦ 1] h0(L) = 15.2 ft Compute Qp to a partially penetrating slot from eq 3 (fig. 4-3). Qp = [o,73 – 0.27 І-(н2 – h2) = [°-73 * 0.27 (~P)] (402 – 15.22) = 0.175 cfm/ft = 1.3 gpm/ft Assume a = 15 ft, then Qw = aQp = 15 X 1.3 = 19.5 gpm for an infinite line of wellpoints. For finite line of wellpoints, increase Qw in this case by 35 percent. Qw = 1.35 X 26.3 gpm = 3.52 cfm. Calculate head at wellpoint, hW(L) , from еЧ 1 (fig- 4-22).

 Figure D-l. Open excavation; two-stage wellpoint system; gravity flow.

 PRQBLEM1 Design s system of 16-in. slotted screen wells, pumped by deep-well turbine pumps, for lowering the poundweter level 5 ft below the bottom of the excavation. Assume maximum allowable Qw = 1,200 gpm, wells located 5 ft from top of slope, well radius rw = 1 ft, and of gravel filter = 0.25 mm. SOLUTION: Estimate total flow required from eq 3(fif. 4-17) using, radius A« of an equivalent larfe-diameter well computed from eq 6 (fig. 4-14). A, = ± V770/2 x 370/2 = 340 ft

 Figure D-2. Open excavation; deep wells;gravity flow.

 PROBLEM: Given the flow net, the data in the figure, and the plan of wells as shown, compute the well flow required to reduce the head in the sand stratum to el 40 ft at point D, the corresponding head hw at the wells, Ьд, midway between wells, and hp at the center of the excavation Assume that wells fully penetrate the pervious stratum and that D = 40 ft, к = 500 X 10"* cm/sec – 0.1 fpm, and rw – 1.0 ft. SOLUTION: Flow to slot (or wells) from flow net, eq 5 (fig. 4-27)

 Section A-A

 Scale in Feet 100 0 200 400

 Ahn

 Thus

 H – ho = H – hw – AhD = 33.2 – 3.2 = 30.0 ft

 (Modified from "Foundation Engineering," G. A. Leonards, ed., 1962, McGraw-Hill Book Company. Used with permission of McGraw-Hill Book Company.)

 Figure D-3. Open excavation; artesian flow; pressure relief design by flow net.

 Wells should have about 15 ft of 12-in. well screen. From fig. 4-24′, estimate Hw = 0.9 ft : hw • Hw = 11.1 – 0.9 = 10.2 ft. Wellpoints. Use З-ft slotted wellpoints with filter, rw = 0.5 ft, and 2-in. riser pipes 21 ft long; set header pipe at el 25. Assume wellpoint pump vacuum equals 24 ft with 2-ft friction loss in header and pump suction set 2 ft above header pipe. Net vacuum in header pipe equals 20 ft. Install wellpoints 110 ft from I from eq 6 (fig. 4-14). Ag = 140 ft. Assume some head, h, at the line of wells; the flow to the combined system can be expressed as follows (eq 3 (fig. 4-13) and 2 (fig. 4-14). H2 – h2= *n R/Ад (flow to line of wells) h2 – h? = In (flow from line of wells to wellpoints, "k *4 R = A* for well, i. e., R = 230 ft) Equate h2 and solve for Qp(T) H2. 2,^.^Qp(T) ln ^ + о£Ш ln r_ •гг к Ag irk A,

 DEEP WELLS’^8

 PLAN

 SECTION

 In order to prevent excessive drawdown at the wells, with both wells and wellpoints operating, reduce Qw by 50 percent. Then Qw = 0.50(37.4) = 18.7 cfm.

 El 60 Initial

 П _ Qp(T1 1287 length 4(220)

 1.46 gpm/ft

 Assume a wellpoint spacing (a) of 8 ft. Thus the flow per wellpoint, Qw , is: Qw = 8(1.46) = 11.7 gpm = 1.56 cfm Compute head at wellpoint, hw, from eq 1 (fig. 4-22) (he = hp)

 SOLUTION: Deep wells. The deep-well system must be designed such that the ground – water level is lowered 2 ft below the elevation at which the header pipes for the wellpoint system will be set. Set header pipes for wellpoint system at el 25. Required drawdown: H – = 55 – 23 = 32 ft Locate fully penetrating wells in a circular array around the perimeter of the excavation. A, – 230 ft. Estimate radius of influence, R, from fig. 4-23 For к = 0.1 fom and final drawdown, H – hD = 55 – (10 – 2) = 47 ft, R = 3180 ft. u Calculate flow to well system from eq 3 (fig. 4-13) and 2 (fig. 4-14).

 hw = 7.7 ft For Qw = 11.7 gpm, the hydraulic head losses are as follows: H* = 0.1 ft, from fig. 4-25a, curve 5 Hs = 1.0 ft, from fig. 4-25b Hy + Hr = 0.4 ft, from fig. 4-25c Hw = 1.4 ft

 Thus hw – Hy, = 7.7 – 1.4 = 6.3 ft

 Therefore, the required effective vacuum at the header = el 25 – 6.3 =• 18.7 ft. Since this is less than the available 20 ft, a wellpoint spacing of 8 ft with the header at el 25 and the top of the wellpoint screen at el 4 would be satisfactory. Calculate drawdown at well from eq 3 (fig. 4-13) and 1 (fig. 4-14). Н2 " hw = ^ [ n ln R – ln nrw – (n – 1) ln Ae ] + ^Ш-ln (55)2 . h2 =1*7 [8 in 3170 – ln 8(1.0) – (8 – 1) In 230] + ^ ln hw= 11.7 ft From fig. 4-24, estimate H^, 0.7 ft: hw – = 11.7 – 0.7 = 11.0 ft. In order to provide adequate pump submergence, set deep-well pump at el 3. (Since the actual drawdown in a well may be greater than the computed drawdown, it is generally advisable to set the pump intake not less than 7 to 10 ft below the computed drawdown in the well.)

 nQw = 299 cfm = 2233 gpm Try eight wells with radius, rw = 1.0 ft (12-in. screen with 6-in. filter).

 Q = — = 2®= 37.4 cfm = 280 gpm Calculate drawdown at well from eq 3 (fig. 4-13) and 1 (fig. 4-14) и2 – = лг ln л c-T)1 = ^ [" In R – In nr« – A. i w e (55)2 – hi = [8 In 3180 – ln 8(1.0) – (8 – 1) ln 230] hw= П.1 ft

 U. S. Army Corps of Engineers Figure D-4. Open excavation; combined deep-well and wellpoint system; gravity flow.

 SOLUTION: Compute QD from eq 4-12 assuming hw =0. Since Qpmust equal QD, substitute this value of QD in eq 1 (fig. 4-22) and computeh0 for various values of a assuming hw = 0 . Using these values ofhQ and a, compute Qp from eq 3 (fig. 4-3). The required spacing a isthat which makes Qp from eq 3 (fig. 4-3) equal to QD computed fromeq 1 (fig. 4-22). From eq 4-12. qd = 0.20112 :ШШ = 0,118 cfm = 0.88 gpm Substituting this value of Qp in eq 1 (fig. 4-22) and assuming hw = 0 gives h2 _ 0,1 IS і _____ a___ no " ТГХ 0.001 Ш 2тт{0 5) Also, from fig. 4-23, L = 100 ft for H – hw = 15 ft. Substituting thisvalue and the other constants into eq 3 (fig. 4-3) results in the followingequation: Qp = [o.73 + ~?(‘55 : Vj (l5^ – hg) Compute h0 and Qp for various values of a from eq 1 (fig. 4-22) andeq 3 (fig. 4-3), respectively, which results in the values tabulated below.

 ho = hw +i! rln (where hD = hd) Note: To solve the equations above simultaneously, it is necessary to assumehw = 0 . NOTATIONS: Qd = vertical flow per drain Qp = seepage through stratum being drained per length a measured alongline of drains кр = vertical permeability of drainAd = sectional area of drain with radiusк = permeability of stratum being drainedh0 = head at equivalent slot simulating line of drainshw = head at sand draina = spacing of drains Other dimensions and symbols are as shown.

 a ho ____ 4. iL ft cfm gpm 5 4.17 0.045 0.34 10 6.58 0.086 0.65 15 7.65 0.117 0.88 20 8.33 0.143 1.07

 Figure D-5. Open excavation; pressure relief combined with sand drains; artesian and gravity flows.

 30 = ^ + <“ ‘ U ifrTST 0r *>. = 27.7 ftThe flow Qp per unit length of system as computed from eq 1 (fig. 4-4) is Qr, =——————- ————————– = 0.23 cfm =1.7 gpm per ft of trench r 200 + lo.4 Compute Ahw from eq 1 (fig. 4-20), hw from eq 2 (fig. 4-21), and Hw from fig. 4-25, andselect a so that 1^ – > 26 ft (M minus the vacuum at the top of the riser pipe).

 a Qw Ahw hw Wellpoint, ft hw Hw ft cfm ft ft He* Hr + Hv§ Hw ft 10 2.3 0.50 27.2 1.75 0.22 0.87 2.84 24.4 8 1.8 0.36 27.3 1.16 0.17 0.54 1.87 25.4 6 1.4 0.25 27.5 0.74 0.13 0.34 1.21 26.3

t From fig. 4-25b.

Ф From fig. 4-25a, assuming He same as that given by curve 7. § From fig. 4-25c, assuming C = 110.

 Thus a spacing of 6 ft would be required, since hw – Hw should not be less than 26 ft. Thetops of the wellpoint screens would be set slightly below the top of the sand stratum.

 (Modifiedfrom "Foundation Engineering," G. A. Leonards, ed., 1962, McGraw-HillBook Company. Used with permission of McGraw-Hill Book Company.)

 Figure D-6. Trench excavation; pressure relief by wellpoints; artesian flow.

 As 1,000 gpm is about the maximum that can be pumped in a normal 10-in.-deep well pump, 16 wells would be required. Try spacing shown on plan. Make computations for the 4 wells in one quadrant and multiply the results by 4. For 4 wells: drawdown at center of excavation H – hc is determined fromeq 1 and 2 (fig. 4-10) (Rj = R): Qwj = 15ff° = Ю00 gpm = 134 cfm

 Well R, ft 7j. ft in? ri 1 4,700 266 2.93 2 4,700 324 2.73 3 4,700 352 2.65 4 4,700 314 2.76 Z = 11.07

 2 ttKD

 H – hc

 For 16 wells:

 H – hc = 4(11.6) = 46.4 ft or hc = 140 – 46.4 = 93.6 ft Since the maximum allowable hc is 95 ft, the system shown in plan is adequate. The approximate head hw at a well is computed from eq 1 (fig. 4-20) using an average well spacing, a, of 2(51-0 + 610)/16 = 140 ft.

 SECTION

 PROBLEM: Determine the number of 10-in.-diam wells with 6-in. gravel filter required to lower the head in the sand stratum 5 ft below bottom of excavation, for wells located at the top of slope and pumped by deep-well turbine pump (assume rw = 1.0 ft). Use a fully penetrating system of wood-stave wells with 3/16-in. slots and a gravel filter with D}o size = 0.25 mm. Area of slot ~ 10 percent of circumferential area of well screen. Geologic and soil conditions indicate a circular source of seepage к = 1,350 x 10~4 cm/sec or 2,700 X 10‘4 fpm. SOLUTION: Determine equivalent radius A, of well system from eq 6 (fig. 4-14) with wells located 5 ft from crown of slope *. = № = * From fig. 4-23, R a 4,960 ft for к = 1,350 x 10~4 cm/sec and H – hw = 45 ft. Com­pute total required flow, Q, , from eq IV-22 for hw = 95 ft and rw = A* = 355 ft. = 2irkD(H-hJ = 2яС0.27ІІ75Ш«-_«1 = 2 д(ю c{m = 16 000 ^ ln(R/rw) In 4,960/355

 Qw. __a_ _ 134 2irkDl 2ттг„ 2тг(0.27)75

 h,

 93.6 – 3.3 = 90.3 ft

 Hydraulic head losses in wells are obtained from fig. 4-24, assuming intake of pump is about 85 ft above the bottom of the sand.

 H, = 0.60 ft (from fig. 4-24a, with Qw = 1,000 gpm/75 ft of screen = 13.3 gpm/ft Hs = 0.37 ft (from fig. 4-24b using a screen length of 0.5(75) = 37.5 ft) Hr = 0.07 ft (from fig. 4-24b, for 10 ft of riser pipe and C = 130) Hy – 0.26 ft (from fig. 4-24c) = 1.30 ft

 Thus the water surface in the wells would be about 90.3 – 1.3 = 89.0 ft above the bottom of sand. Set pump bowl about 85 ft above bottom of sand and provide with 10-ft suction pipe.

 (Modifiedfrom "Foundation Engineering," G. A. Leonards, ed., 1962, McGraw-Hill Book Company. Used with permission of McGraw-Hill Book Company.)

 Figure D – 7. Rectangular excavation; pressure relief by deep wells; artesian flow.

 Aquifer 3. For artesian flow, the head producing flow for the combined hydrostatic-vacuum system is H + V – he. Assuming the circular array of wells to be a continuous drainage slot, for which W/D = 50 percent and R/A = 133 , it can be seen from fig 4-7 that the head in the center of the circular drainage slot approaches the head in the slot as R/A increases. Therefore, the flow to the wells for this situation can be computed from eq 2 (fig. 4-Ю), in which (H – hw) = (H + V – he), G – 1, and rw = A. = 2irkD(H – V – h.) = 21Q. q2K«) 190 – 15 – 27) s g, = 5„ 4T In R/A In 4000/30 y

 Total flow to well system = 46.1 + 127.5 + 599.4 = 773 gpm.

 Use 12 wells located 30 ft from the center of the shaft, with a spacing (a) be­tween wells of 15.5 ft. Flow per well, Qw = 64, 4 gpm = 8 61 cfm

 ROCK

 PROBLEM. Design a deep well and vacuum system to dewater a 70-ft deep shaft to be sunk into stratified clays and sand below the groundwater table. Assume a ring of wells installed 15 ft out from perimeter of shaft with an equivalent radius of influence, A, – 30 ft. Wells to fully penetrate the sand strata penetrated by the shaft and pumps to have a capacity in excess of the flow to each well. Vacuum to be maintained in wells equals 15 ft. Assume radius of influence of vacuum (R) to be the same for seepage (i. e., vacuum varies from that at well от wells to zero at R). Maximum height of shaft exposed at any one time equals 30 ft. SOLUTION:

 Vacuum water flow, from eq 2 (fig 4-10) (rw = A; effective aquifer thickness, H + hw D —— 2—• and drawdown, H – hw, = V): 2(0 005)1,(^4^) (15) St-V-I = –ІГ-Ї77— = = 2 15 rf">

 Qa = Ap(D – hw) W

 Aquifer 1 Q. — 15 ft(30 ft – 2 ft) 2?S9 x 10-5 ІЬ-seq/ft» W = 39.7 cfm/well ^ 23.744 x to’7 Ib-tec/ft^ Total airflow = 12(39.7) = 47.6 cfm at the mean absolute pressure, p of 34 + (34 – 15) = 26 5 f, of water Compute the required vacuum pump capacity

 Total water flow, aquifer 1, Q-j-.j = Qt-H-1 * Qt-V-1 = 4.01 + 2.15 = 6.16 cfm = 46.1 gpm. Aquifer 2. Compute the flow of water assuming combined artesian-gravity flow condi­tions for "Hydrostatic" water flow. Compute the additional flow caused by vacuum in wells assuming an equivalent artesian flow condition under the net vacuum head existing in the gravity flow region. Assume hw = 2 ft.

 Hydrostatic water flow, from eq 1 (fig. 4-12): QT H 2 = О = 0 01.[2(25) (50) 125)‘-(2)г| = ”M’2 In R/A In 2000/30 Vacuum water flow; compute R from eq 3 (fig. 4-12) In R ^ (p2 – ln R * ZD(H – D) In A = (252 – 22) In 2000 – 2 (25) (50 – 25) In 30 = 4 2DH – D2 – h2, 2(25) (50) – (2S)2 – (2)2 then R = 121 ft. Estimate vacuum at artesian-gravity flow boundary by plotting the vacuum versus loa r

 Оа-vp = Qa £ = 476 = 371 cfm

 Qa = 15(25 2) 3^1 f2 359 x 10 5) 0.6 = 65.2 cfm/weU 2 V3.744 x 10’7/ Total airflow = 12(65.2) = 783 cfm at p = 26.5 ft of water Qa-vp = 783 (^)= 610 cfm Provide vacuum pumps with a total capacity of 610 cfm at 15 ft (of water) vacuum.

 U. S. Army Corps of Engineers Figure D-8. Shaft excavation; artesian and gravity flows through stratified foundation; deep-well vacuum system.

 PROBLEM: Design a deep-well system to dewater an excavation for a tunnel for the conditions shown, For a single-line source, use the method of image analysis. The layout, as shown, was determined from an approximate flow net sketched for pre­liminary design purposes.

 со Line Source

 PLAN

 SECTION A-A

 SOLUTION: Assume 12-in. fully penetrating wells with surrounding filter, rw = 1.0 ft. For an assumed Qw = 150 gpm, the drawdown at points A and В and at well 5 are computed from eq 2 and 3 (fig. 4-18): H2 – hp = 4- C" Qwi In ^ ™ j=l Г1

 Figure D-9. Tunnel dewatering; gravity flows; deep-well system.

PROBLEM: Determine sump and pump capacity to control surface water in an excava­tion, 4 acres in area, located in Little Rock, Ark., for a rainstorm frequency of 1 in 5 years and assuming c = 0.9. Assume all runoff to one sump in bottom of excava­tion.

SOLUTION:

VR = cRA

FROM FIGURE:

 Rainstorm, min R, in. Vr – (X 103 gal) 10 0.85 83 30 1.70 166 60 2.10 205

Assume sump pump capacity = 4000 gpm. From plot, required storage = 50,000 gal.

U. S. Array Corps of Engineers

Figure D-10. Sump and pump capacity for surface runoff to an excavation.

D-ll