CHAPTER I. Introduction (Part 7)

of 1 to the square of '784 equals 1 to '615 or "62 nearly. To find the actual velocity of efflux in feet per second through any circular orifice in a thin plate multiply the square root of the head of water in feet by 8, and the product by -615 or -62. To find the actual dis- charge in gallons per minute Box gives the rule—Multiply the square root of the head of water in feet by the square of the diameter of the orifice in inches, and multiply the product by the constant num- ber or co-efficient 10. When the plate through which water issues is thick we find an increased delivery of water. Compare this diagram with the vena contracta dia- gram of the thin plate. If a short tube be added, not less in length than three times the diameter of the ori- fice, a further gain is obtained, which may be calculated by substituting the number 13 for 10 in the preceding rules for thin plate orifices. ^"'- '-^«"j/tSl.^'' '"^"^ P'"" ^^v ,^? ^ — ^^ — '■] l—n^ ^^ f J ^ ...-_ __ .__ 60 DOMESTIC SANITARY DRAINAGE AND PLUMBING. The existence of the vacuum caused by the outflow of water in the tube can be shown by adding a short tube, as in the diagram (Fig. 11), and placing a cup full of water under the end. When the water is flowing under a good head from the cistern a column of water will rise in the added tube, as in the diagram. Through conical tubes or adjutt^es the efflux is further increased, and the full effect, or nearly the theoretical Fig. 10. — Short tube adjutage. delivery, can be obtained when the natural form of the vena contracta is given to the adjutage, and even a greater delivery than the theoretical amount may be gained by the use of the modified conical elongated form of adjutage, specially recommended from actual experiment by VenturL The effect of the shape of the orifice is not marked when the water has to pass through long pipes, but it is well to give the best shape to the entry in all cases, when possible. ELEMENTARY SCIENCE FOR PLUMBERS. 61 Torricelli gave us this formula two hundred years fi^o, sl2gR = V, easily remembered, and meaning — ^Multiply the head of water in feet, represented by H, by 322 (which we know is the constant dynamical measure of the force of gravity represented by g\ multiply this product by 2, and Fig. 11.— Vacuum proof experiment find the square root. The result gives the velocity in feet per second. Experunent has proved that only i^^ft^ths, or 62 of the efflux calculated from this theorem is actually dis- charged through a hole in a thin plate. If a cistern 16 feet deep have a hole opened in the bottom we apply the formula V2yH thus: 2x32-2x16-1,030; Vl,030 « 32 ft. nearly, theoretical velocity in feet per second 62 DOMESTIC SANITARY DRAINAGE AND PLUMBING. a result which corresponds to the velocity at the end of one second of any substance falling 16 feet freely in vacuo on the earth's surface. Having ascertained the theoretical velocity, to find the quantity of water discharged through any given orifice (fitted with a tube in shape of vena con- tracta), multiply the velocity in feet per second by the narrowest area of the orifice in square feet, and the result gives the quantity discharged in cubic feet per second. Multiply this by 63 for gallons per second. As before explained, this result will be much modified by the form of the delivery opening or adjutage. Venturi has experimented with a conical adjutage, which, he states, gives a real discharge 24 times greater than will an orifice in a thin plate, and 1'46 times greater than the theoretic discharge. This form has a length nine times greater than the diameter of smaller base, and an angle of divergence 5" 6'. Practical Rules for Hydraulic Engineers and Plumbers. 1. To find (V'), the theoretical velocity of water in feet per second issuing from an orifice in the side of any vessel with any given head of water pressure, multiply the square root of the head of water in feet (H) by the constant number 8 — V'= JH X 8. Thus, from any orifice with two feet head of pressure, the theoretical velocity (V') is — V'= J2 X 8 = 1-41421 X 8 = 11-3 feet per second. 2. To find (G), the theoretical discharge of water in gallons per minute issuing through any given circular orifice in the side of any vessel with any given head of water (H), multiply the square root of (H), the head of ELEMENTARY SCIENCE FOR PLUMBERS. 63 water in feet, by the square of (d), the diameter of the circular orifice in inches, and multiply the product by the constant number 16 — Thus, with two feet head of water and a circular orifice of three inches, the theoretical discharge (G*) is — G'= J2 X 32 X 16 = 1-41421 X 9 X 16 = 203 gallons per minute ; or, with nine feet head of water and a circular orifice of two inches, the theoretical discharge (G*) is — G' = ^/9 X 2- X 16 = 3 X 4 X 16 = 192 gallons per minute. These rules give the theoretical velocity and discharge due to force of gravity ; the actual velocities and discharges are found by modified rules. 3. To find (V), the actual velocity of water in feet per second issuing from an orifice in a thin plate, as in an iron cistern, reduce the theoretical velocity by multiplying it by •62— V = V'x-62, or >^x8x-62. Thus, from any orifice with two feet head of pressure, the actual velocity (V) is — V= V2x8x-62, or 141421 x 8 x 62 = 7014, or 7 feet per second. 4. To find (G), the actual discharge of water in gallons per minute issuing from any given circular orifice in a thin plate, as in an iron cistern, reduce the theoretical discharge by multiplying it by '62 — G = G'x-62, or ^xd2xl6x-62. Or, by a simpler formula used by Box, multiply the square root of (H), the head of water in feet, by the square of (d) 64 DOMESTIC SANITARY DRAINAGE AND PLUMBING. the diameter of the orifice in inches, and multiply the product by the constant number 10 — Box's formula : G= JExd^xlO. Thus, with two feet head of water and a circular orifice of three inches, the actual discharge (G) is — G = V5^x 3« X 10, or 1-41421 x 9 x 10 = 127 gallons per minute ; or, with nine feet head of water and a circular orifice of two inches, the actual discharge (G) is — G = V9 >< 2^ X 10, or 3 X 4 X 10 = 120 gallons per minute. 5. To find (H), the head of water necessary in order to discharge any given number of gallons per minute through any given circular orifice in a thin plate, as in an iron cistern, multiply the square of (d), the given diameter of the orifice in inches, by the constant number 10, and divide the product into (G), the given number of gallons per minute, and square the quotient — Box's formula: K = (—^--)\ \d^ X 10/ Thus, to pass 127 gallons of water through a 3-inch circular orifice, the required head (H) is — «=(3^ro) ' - iw^y "'■'''• ">' "«^^^y 2 '^'- 6. To find (rf), the diameter in inches of a circular orifice in a thin plate necessary in order to discharge any given numter of gallons of water per minute with any given head of water in feet, multiply the square root of the given head of water in feet by the constant number 10, and divide the product into the given number of gallons per minute, and find the square root of the quotient — Box's formula : d = (-^r- )* V^HxlO/ ELEMENTARY SCIENCE FOR PLUMBERS. 65 Thus, under two feet head of water, to pass 127 gallons per minute, the required diameter (d) is — rf=( — : r, or (z-rrr^ — t-^)* = 3 inches nearly. \J2xW \l-41421xlO/ ^ 7. To find (G), the adiuil discharge of water in gallons per minute issuing from any given circular orifice fitted with a short tube not less in length than thrice the diameter, multiply the square root of (H), the head of water in feet, by the square of (d), the diameter of the orifice in inches, and multiply the product by the constant number 13 — Box's formula : G = JS x rf* x 13. 8. To find (H), the head of water necessary in order to discharge any given number of gallons per minute through any given circidar orifice fitted with a short tube not less in length than thrice the diameter, multiply the square of (d), the given diameter of the orifice in inches, by the constant number 13, and divide the product into (G), the given number of gallons per minute, and square the quotient. /CI \2 Box's formula: H = (^4-g). 9. To find (rf), the diameter in inches of a circular orifice fitted with a short tube not less in length than thrice the diameter necessary in order to discharge any given number of gallons of water per minute with any given head of water in feet, multiply the square root of the given head of water in feet by the constant number 13, and divide the product into the given number of gallons per minute, and find the square root of the quotient — Box's formula : d = ( p. \ VH X 13/ Rules 7, 8, and 9 are identical with rules 4, 5, and 6, except as regards the constant number employed. F 66 DOMESTIC SANITARY DRAINAGE AND PLUMBING. In order to find the weight of water flowing from an orifice in the side of a cistern in a given time, multiply tlie velocity of efflux in seconds, as found by Torricelli's theorem, J2ffRf by the time, multiply the product by the weight of a cubic inch of water, 252458 grains, and multiply this product by the cross-section of the orifice in square inches. The final product gives the weight in grains, theoretically due to Torricelli's theorem ; but hydraulic engineers know that, owing to the form taken by the water in issuing from this kind of orifice, the real section area of the jet at its narrowest point is -rtftrths of the section area of the orifice, and that, therefore, only VW^ths of the quantity of the theo- retical efflux will be in fact discharged. 8 They substitute in their calculations the smallest cross-section area of the vena contracta for the cross-section area of the orifice, and thus secure a rule which holds good in practice, viz. multiply the theoretical result above by '62, for the actual result agreeing with actual experiment and practice. To sum up these matters, we find that the actual discharge from any orifice in a thin plate, as the side or bottom of an iron cistern, is only 62 per cent, of the theoretical dis- charge, viz. 62 gallons, instead of 100 gallons. If a short cylindrical tube, in length three times the diameter of the orifice, be added, the discharge increases to 82 per cent, of the theoretical dis- FiG. 12.— Various oriiices. charge, viz. 82 gallons, instead of 100 WMiMtm MriiV.'n/r/ ELEMENTARY SCIENCE FOR PLUMBERS. 67 gallons. If the added tube be made in the shape assumed by the water in vena contracta the discharge increases to 92 per cent, of the theoretical discharge, viz. 92 gallons actual for every 100 theoretical; and when Venturi's inverted nozzle is used the discharge exceeds the theoretical by 25 per cent., or 125 gallons actual for every 100 gallons theoretical. To find the time required to fill any tank when water is flowing in and out at the same time, deduct the number of cubic feet flowing out per minute from the number of cubic feet flowing in per minute, and divide the remainder into the contents of tank in cubic feet, and the quotient gives the number of minutes required for the tank to fill. Thus, if ten cubic feet are flowing in and three cubic feet are flowing out per minute of a tank 12 feet x 10 feet X 6 feet, containing 720 cubic feet — 720 10-3 = 103 minutes required to fill the tank. To find the time required to empty any tank when water is flowing in and out at the same time, deduct the number of cubic feet flowing in per minute from the number of cubic feet flowing out per minute, and divide the re- mainder into the contents of the tank in cubic feet, and the quotient gives the number of minutes required to empty the tank. Thus, if six cubic feet are flowing in and eight cubic feet are flowing out, the tank starting full, being 9 feet x 6 feet X 3 feet, containing 162 cubic feet — 1 £»n - — - = 81 minutes required to empty the tank. 8-6 The amount flowing out must be measured or calculated when the tank is half full, as the delivery will be greatest 68 DOMESTIC SANITARY DRAINAGE AND PLUMBING. when full, down to nil when empty, the amount flowing in being taken as constant. When the length of a water pipe is forty-eight times its diameter the discharge will be V\fctiJ^8 of the theoretical discharge, or very nearly equal to the actual flow from an orifice in a thin plate, every increase in length adding to the friction and reducing the discharge. To find the actual discharge of water in cubic feet per second from any tank through different orifices, with and without adjutages, take the head of water in feet from the surface to the centre of orifice, find the square root, multiply by the constant number 8 (or, for special accuracy, 8'025) ; the product will be the theoretical velocity of efflux per second. Multiply this by the constant number or co-efficient belonging to the particular form of orifice, as determined by experiments and given in the table below ; the product will be the actual velocity of efflux per second. Multiply this by the area of the aperture in square feet, and the product will be the actual quantity of water discharged in cubic feet per second. Co-efficients for Different Orifices. Vena contracta ..... ConTerging a(]jatage angle, 13}", measured across narrow end Diverging adjutage angle, 6, measured across narrow end Diverging adjutage angle, 5", measured across wide end Cylindrical adjutage, projecting outwards — Diameter | the length . Diameter } the length . Diameter } the length . Diameter i the length . Diameter iV the length Diameter %V the length Diameter ^^o the length Short cylindrical adjutage, projecting inwards . Orifice in a thin plate For sluices whose lower edge is level with bottom of reservoir, and walla in line with orifice . « . ... •97 •94 •92 •56 •81 •80 •78 •77 •74 •72 •68 •60 •62 •96 ELEMENTARY SCIENCE FOR PLUMBERS. 69 To find the contents of a cylindrical water cistern in gallons approximately, square the diameter in feet, multiply by 5, and the product by the depth in feet. Thus, a cylindrical tank 2 feet diameter by 3 feet deep — 2x2x5x3«60 gallons contents approximately. The ordinary rule is to multiply the diameter squared in feet by '7854, and the product by the depth in feet, and this again by 6'23 for gallons, which in the case given would be — 2 X 2 X 7854 X 3 X 623 « 58716504 gallons exactly. To find the pressure in pounds per square inch under any given head of water, multiply the head in feet by '433 ; or, by another rule, divide the head in feet by 7, and multiply the quotient by 3, and add y^th part where great accuracy is required. Thus, at the bottom of a pipe 130 feet high — 130 X -433 = 5629 lbs. pressure per square inch, or — 130 -^ 7 = 18'57 X 3 = 557 lbs. pressure approximately, adding yJiyth for correctness = 56257 lbs. exactly. To find the resistance against the plunger of a pump in motion it is usual to make allowance for friction of valves, etc., and to take half the height in feet of the rising main to equal the pressure in pounds per square inch on the plunger. Thus, in raising water 50 feet, the pressure is calculated as 25 lbs. per square inch on the plunger of the pump. Multiply this by the area of the plunger or of the cylinder in inches for the total pressure in pounds. A 3-inch pump has 7068 inches area ; therefore total pressure under 50 feet = 176J lbs. To find the head of water producing any given pressure, multiply the pressure in pounds per square inch by 231. 70 DOMESTIC SANITARY DRAINAGE AND PLUMBING. Thus, with 5629 lbs. per square inch pressure — 56-29 X 2-31 = 1300299 feet head. To find the weight of water in one yard length of any cylindrical pipe approximately, square the diameter in inches. Thus, in one yard of 6-inch pipe — 6« = 6x6 = 36 lbs. weight. To find the contents in one yard length of any cylindrical pipe approximately, square the diameter in inches and divide by 10. Thus, in one yard of 6-inch pipe — 6* = 6 X 6 = 36 -s- 10 = 3*6 gallons approximately. Plumbers constantly require to calculate the amount of water which will flow through long pipes of