(1)if ζeis put for the expression in brackets.ω1/ω0=1.11.21.51.71.81.92.02.53.03.54.05.06.07.08.0d1/d0=1.051.101.221.301.341.381.411.581.731.872.002.242.452.652.83ζe=.01.04.25.49.64.811.002.254.006.259.0016.0025.0036.049.0Fig. 88.Fig. 89.Abrupt Contraction of Section.—When water passes from a larger to a smaller section, as in figs. 88, 89, a contraction is formed, and the contracted stream abruptly expands to fill the section of the pipe. Let ω be the section and v the velocity of the stream at bb. At aa the section will be ccω, and the velocity (ω/ccω) v = v/c1, where ccis the coefficient of contraction. Then the head lost isɧm= (v/cc− v)2/ 2g = (1/cc− 1)2v2/2g;and, if ccis taken 0.64,ɧm= 0.316 v2/2g.(2)The value of the coefficient of contraction for this case is, however, not well ascertained, and the result is somewhat modified by friction. For water entering a cylindrical, not bell-mouthed, pipe from a reservoir of indefinitely large size, experiment givesɧa= 0.505 v2/2g.(3)If there is a diaphragm at the mouth of the pipe as in fig. 89, let ω1be the area of this orifice. Then the area of the contracted stream is ccω1, and the head lost isɧc= {(ω/ccω1) − 1}2v2/2g= ζcv2/ 2g(4)if ζ, is put for {(ω/ccω1) − 1}2. Weisbach has found experimentally the following values of the coefficient, when the stream approaching the orifice was considerably larger than the orifice:—ω1/ω =0.10.20.30.40.50.60.70.80.91.0cc=.616.614.612.610.617.605.603.601.598.596ζc=231.750.9919.789.6125.2563.0771.8761.1690.7340.480Fig. 90.When a diaphragm was placed in a tube of uniform section (fig. 90) the following values were obtained, ω1being the area of the orifice and ω that of the pipe:—ω1/ω =0.10.20.30.40.50.60.70.80.91.0ce=.624.632.643.659.681.712.755.813.8921.00ξc=225.947.7730.837.8011.7531.796.797.290.060.000Elbows.—Weisbach considers the loss of head at elbows (fig. 91) to be due to a contraction formed by the stream. From experiments with a pipe 11⁄4in. diameter, he found the loss of headɧe= ζεv2/ 2g;(5)ζe= 0.9457 sin21⁄2φ + 2.047 sin41⁄2φ.φ =20°40°60°80°90°100°110°120°130°140°ζε=0.0460.1390.3640.7400.9841.2601.5561.8612.1582.431Hence at a right-angled elbow the whole head due to the velocity very nearly is lost.Fig. 91.Fig. 92.Bends.—Weisbach traces the loss of head at curved bends to a similar cause to that at elbows, but the coefficients for bends are not very satisfactorily ascertained. Weisbach obtained for the loss of head at a bend in a pipe of circular sectionɧb= ζbv2/ 2g;(6)ζb= 0.131 + 1.847 (d/2ρ)7/2,where d is the diameter of the pipe and ρ the radius of curvature of the bend. The resistance at bends is small and at present very ill determined.Valves, Cocks and Sluices.—These produce a contraction of the water-stream, similar to that for an abrupt diminution of section already discussed. The loss of head may be taken as before to beɧv= ζvv2/ 2g;(7)where v is the velocity in the pipe beyond the valve and ζva coefficient determined by experiment. The following are Weisbach’s results.Sluice in Pipe of Rectangular Section(fig. 92). Section at sluice = ω1in pipe = ω.ω1/ω =1.00.90.80.70.60.50.40.30.20.1ζv=0.00.09.39.952.084.028.1217.844.5193Sluice in Cylindrical Pipe(fig. 93).Ratio of height of openingto diameter of pipe1.07⁄83⁄45⁄81⁄23⁄81⁄41⁄5ω1/ω =1.000.948.856.740.609.466.315.159ζv=0.000.070.260.812.065.5217.097.8Fig. 93.Fig. 94.Cock in a Cylindrical Pipe(fig. 94). Angle through which cock is turned = θ.θ =5°10°15°20°25°30°35°Ratio ofcrosssections.926.850.772.692.613.535.458ζv=.05.29.751.563.105.479.68θ =40°45°50°55°60°65°82°Ratio ofcrosssections.385.315.250.190.137.0910ζv=17.331.252.6106206486∞Throttle Valve in a Cylindrical Pipe(fig. 95)θ =5°10°15°20°25°30°35°40°ζv=.24.52.901.542.513.916.2210.8θ =45°50°55°60°65°70°90°ζv=18.732.658.8118256751∞Fig. 95.§ 84.Practical Calculations on the Flow of Water in Pipes.—In the following explanations it will be assumed that the pipe is of so great a length that only the loss of head in friction against the surface of the pipe needs to be considered. In general it is one of the four quantities d, i, v or Q which requires to be determined. For since the loss of head h is given by the relation h = il, this need not be separately considered.There are then three equations (see eq. 4, § 72, and 9a, § 76) for the solution of such problems as arise:—ζ = α (1 + 1/12d);(1)where α = 0.005 for new and = 0.01 for incrusted pipes.ζv2/ 2g =1⁄4di.(2)Q =1⁄4πd2v.(3)Problem1. Given the diameter of the pipe and its virtual slope, to find the discharge and velocity of flow. Here d and i are given, and Q and v are required. Find ζ from (1); then v from (2); lastly Q from (3). This case presents no difficulty.By combining equations (1) and (2), v is obtained directly:—v = √ (gdi/2ζ) = √ (g/2α) √ [di / {1 + 1/12d}].(4)For new pipes√ (g/2α) = 56.72For incrusted pipes= 40.13For pipes not less than 1, or more than 4 ft. in diameter, the mean values of ζ areFor new pipes0.00526For incrusted pipes0.01052.Using these values we get the very simple expressions—v = 55.31 √ (di) for new pipes= 39.11 √ (di) for incrusted pipes.(4a)Within the limits stated, these are accurate enough for practical purposes, especially as the precise value of the coefficient ζ cannot be known for each special case.Problem2. Given the diameter of a pipe and the velocity of flow, to find the virtual slope and discharge. The discharge is given by (3); the proper value of ζ by (1); and the virtual slope by (2). This also presents no special difficulty.Problem3. Given the diameter of the pipe and the discharge, to find the virtual slope and velocity. Find v from (3); ζ from (1); lastly i from (2). If we combine (1) and (2) we geti = ζ (v2/2g) (4/d) = 2a {1 + 1/12d} v2/gd;(5)and, taking the mean values of ζ for pipes from 1 to 4 ft. diameter, given above, the approximate formulae arei = 0.0003268 v2/d for new pipes= 0.0006536 v2/d for incrusted pipes.(5a)Problem4. Given the virtual slope and the velocity, to find the diameter of the pipe and the discharge. The diameter is obtained from equations (2) and (1), which give the quadratic expressiond2− d (2αv2/gi) − αv2/6gi = 0.∴ d = αv2/gi + √ {(αv2/gi) (αv2/gi + 1/6)}.(6)For practical purposes, the approximate equationsd = 2αv2/gi + 1/12= 0.00031 v2/i + .083 for new pipes= 0.00062 v2/i + .083 for incrusted pipes(6a)are sufficiently accurate.Problem5. Given the virtual slope and the discharge, to find the diameter of the pipe and velocity of flow. This case, which often occurs in designing, is the one which is least easy of direct solution. From equations (2) and (3) we get—d5= 32ζQ2/ gπ2i.(7)If now the value of ζ in (1) is introduced, the equation becomes very cumbrous. Various approximate methods of meeting the difficulty may be used.(a) Taking the mean values of ζ given above for pipes of 1 to 4 ft. diameter we getd =5√ (32ζ/gπ2)5√ (Q2/i)= 0.22165√ (Q2/i) for new pipes= 0.25415√ (Q2/i) for incrusted pipes;(8)equations which are interesting as showing that when the value of ζ is doubled the diameter of pipe for a given discharge is only increased by 13%.(b) A second method is to obtain a rough value of d by assuming ζ = α. This value isd′ =5√ (32Q2/ gπ2i)5√ α = 0.63195√ (Q2/i)5√ α.Then a very approximate value of ζ isζ′ = α (1 + 1/12d′);and a revised value of d, not sensibly differing from the exact value, isd″ =5√ (32Q2/ gπ2i)5√ ζ′ = 0.63195√ (Q2/i)5√ ζ′.(c) Equation 7 may be put in the formd =5√ (32αQ2/ gπ2i)5√ (1 + 1/12d).(9)Expanding the term in brackets,5√ (1 + 1/12d) = 1 + 1/60d − 1/1800d2...Neglecting the terms after the second,d =5√ (32α / gπ2)5√ (Q2/i) · {1 + 1/60d}=5√ (32α / gπ2)5√ (Q2/i) + 0.01667;(9a)and5√ (32α / gπ2) = 0.219 for new pipes= 0.252 for incrusted pipes.Fig. 96.Fig. 97.§ 85.Arrangement of Water Mains for Towns’ Supply.—Town mains are usually supplied oy gravitation from a service reservoir, which in turn is supplied by gravitation from a storage reservoir or by pumping from a lower level. The service reservoir should contain three days’ supply or in important cases much more. Its elevation should be such that water is delivered at a pressure of at least about 100 ft. to the highest parts of the district. The greatest pressure in the mains is usually about 200 ft., the pressure for which ordinary pipes and fittings are designed. Hence if the district supplied has great variations of level it must be divided into zones of higher and lower pressure. Fig. 96 shows a district of two zones each with its service reservoir and a range of pressure in the lower district from 100 to 200 ft. The total supply required is in England about 25 gallons per head per day. But in many towns, and especially in America, the supply is considerably greater, but also in many cases a good deal of the supply is lost by leakage of the mains. The supply through the branch mains of a distributing system is calculated from the population supplied. But in determining the capacity of the mains the fluctuation of the demand must be allowed for. It is usual to take the maximum demand at twice the average demand. Hence if the average demand is 25 gallons per head per day, the mains should be calculated for 50 gallons per head per day.Fig. 98.§ 86.Determination of the Diameters of Different Parts of a Water Main.—When the plan of the arrangement of mains is determined upon, and the supply to each locality and the pressure required is ascertained, it remains to determine the diameters of the pipes. Let fig. 97 show an elevation of a main ABCD ..., R being the reservoir from which the supply is derived. Let NN be the datum line of the levelling operations, and Ha, Hb... the heights of the main above the datum line, Hrbeing the height of the water surface in the reservoir from the same datum. Set up next heights AA1, BB1, ... representing the minimum pressure height necessary for the adequate supply of each locality. Then A1B1C1D1... is a line which should form a lower limit to the line of virtual slope. Then if heights ɧa, ɧb, ɧc... are taken representing the actual losses of head in each length la, lb, lc... of the main, A0B0C0will be the line of virtual slope, and it will be obvious at what points such as D0and E0, the pressure is deficient, and a different choice of diameter of main is required. For any point z in the length of the main, we havePressure height = Hr− Hz− (ɧa+ ɧb+ ... ɧz).Where no other circumstance limits the loss of head to be assigned to a given length of main, a consideration of the safety of the main from fracture by hydraulic shock leads to a limitation of the velocity of flow. Generally the velocity in water mains lies between 11⁄2and 41⁄2ft. per second. Occasionally the velocity in pipes reaches 10 ft. per second, and in hydraulic machinery working under enormous pressures even 20 ft. per second. Usually the velocity diminishes along the main as the discharge diminishes, so as to reduce somewhat the total loss of head which is liable to render the pressure insufficient at the end of the main.J. T. Fanning gives the following velocities as suitable in pipes for towns’ supply:—Diameter in inches481218243036Velocity in feet per sec.2.53.03.54.55.36.27.0§ 87.Branched Pipe connecting Reservoirs at Different Levels.—Let A, B, C (fig. 98) be three reservoirs connected by the arrangement of pipes shown,—l1, d1, Q1, v1; l2, d2, Q2, v2; h3, d3, Q3, v3being the length, diameter, discharge and velocity in the three portions of the main pipe. Suppose the dimensions and positions of the pipes known and the discharges required.If a pressure column is introduced at X, the water will rise to a height XR, measuring the pressure at X, and aR, Rb, Rc will be the lines of virtual slope. If the free surface level at R is above b, the reservoir A supplies B and C, and if R is below b, A and B supply C. Consequently there are three cases:—I.R above b; Q1= Q2+ Q3.II.R level with b; Q1= Q3; Q2= 0III.R below b; Q1+ Q2= Q3.To determine which case has to be dealt with in the given conditions, suppose the pipe from X to B closed by a sluice. Then there is a simple main, and the height of free surface h′ at X can be determined. For this conditionha− h′ = ζ (v12/2g) (4l1/d1) = 32ζQ′2l1/ gπ2d15;h′ − hc= ζ (v32/2g) (4l3/d3) = 32ζQ′2l3/ gπ2d35;where Q′ is the common discharge of the two portions of the pipe. Hence(ha− h′) / (h′ − hc) = l1d35/ l3d15,from which h′ is easily obtained. If then h′ is greater than hb, opening the sluice between X and B will allow flow towards B, and the case in hand is case I. If h′ is less than hb, opening the sluice will allow flow from B, and the case is case III. If h′ = hb, the case is case II., and is already completely solved.The true value of h must lie between h′ and hb. Choose a new value of h, and recalculate Q1, Q2, Q3. Then ifQ1> Q2+ Q3in case I.,orQ1+ Q2> Q3in case III.,the value chosen for h is too small, and a new value must be chosen.IfQ1< Q2+ Q3in case I.,orQ1+ Q2< Q3in case III.,the value of h is too great.Since the limits between which h can vary are in practical cases not very distant, it is easy to approximate to values sufficiently accurate.§ 88.Water Hammer.—If in a pipe through which water is flowing a sluice is suddenly closed so as to arrest the forward movement of the water, there is a rise of pressure which in some cases is serious enough to burst the pipe. This action is termed water hammer or water ram. The fluctuation of pressure is an oscillating one and gradually dies out. Care is usually taken that sluices should only be closed gradually and then the effect is inappreciable. Very careful experiments on water hammer were made by N. J. Joukowsky at Moscow in 1898 (Stoss in Wasserleitungen, St Petersburg, 1900), and the results are generally confirmed by experiments made by E. B. Weston and R. C. Carpenter in America. Joukowsky used pipes, 2, 4 and 6 in. diameter, from 1000 to 2500 ft. in length. The sluice closed in 0.03 second, and the fluctuations of pressure were automatically registered. The maximum excess pressure due to water-hammer action was as follows:—Pipe 4-in. diameter.Pipe 6-in. diameter.Velocityft. per sec.Excess Pressure.℔ per sq. in.Velocityft. per sec.Excess Pressure.℔ per sq. in.0.5310.6432.91683.01734.12325.63699.25197.5426In some cases, in fixing the thickness of water mains, 100 ℔ per sq. in. excess pressure is allowed to cover the effect of water hammer. With the velocities usual in water mains, especially as no valves can be quite suddenly closed, this appears to be a reasonable allowance (see also Carpenter,Am. Soc. Mech. Eng., 1893).IX. FLOW OF COMPRESSIBLE FLUIDS IN PIPES§ 89.Flow of Air in Long Pipes.—When air flows through a long pipe, by far the greater part of the work expended is used in overcoming frictional resistances due to the surface of the pipe. The work expended in friction generates heat, which for the most part must be developed in and given back to the air. Some heat may be transmitted through the sides of the pipe to surrounding materials, but in experiments hitherto made the amount so conducted away appears to be very small, and if no heat is transmitted the air in the tube must remain sensibly at the same temperature during expansion. In other words, the expansion may be regarded as isothermal expansion, the heat generated by friction exactly neutralizing the cooling due to the work done. Experiments on the pneumatic tubes used for the transmission of messages, by R. S. Culley and R. Sabine (Proc. Inst. Civ. Eng.xliii.), show that the change of temperature of the air flowing along the tube is much less than it would be in adiabatic expansion.§ 90.Differential Equation of the Steady Motion of Air Flowing in a Long Pipe of Uniform Section.—When air expands at a constant absolute temperature τ, the relation between the pressure p in pounds per square foot and the density or weight per cubic foot G is given by the equationp/G = cτ,(1)where c = 53.15. Taking τ = 521, corresponding to a temperature of 60° Fahr.,cτ = 27690 foot-pounds.(2)Fig. 99.The equation of continuity, which expresses the condition that in steady motion the same weight of fluid, W, must pass through each cross section of the stream in the unit of time, isGΩu = W = constant,(3)where Ω is the section of the pipe and u the velocity of the air. Combining (1) and (3),Ωup/W = cτ = constant.(3a)Since the work done by gravity on the air during its flow through a pipe due to variations of its level is generally small compared with the work done by changes of pressure, the former may in many cases be neglected.Consider a short length dl of the pipe limited by sections A0, A1at a distance dl (fig. 99). Let p, u be the pressure and velocity at A0, p + dp and u + du those at A1. Further, suppose that in a very short time dt the mass of air between A0A1comes to A′0A′1so that A0A′0= udt and A1A′1= (u + du) dt1. Let Ω be the section, and m the hydraulic mean radius of the pipe, and W the weight of air flowing through the pipe per second.From the steadiness of the motion the weight of air between the sections A0A′0, and A1A′1is the same. That is,W dt = GΩu dt = GΩ (u + du) dt.By analogy with liquids the head lost in friction is, for the length dl (see § 72, eq. 3), ζ (u2/2g) (dl/m). Let H = u2/2g. Then the head lost is ζ(H/m)dl; and, since Wdt ℔ of air flow through the pipe in the time considered, the work expended in friction is −ζ (H/m)W dl dt. The change of kinetic energy in dt seconds is the difference of the kinetic energy of A0A′0and A1A′1, that is,(W/g) dt {(u + du)2− u2} / 2 = (W/g) u du dt = W dH dt.The work of expansion when Ωudt cub. ft. of air at a pressure p expand to Ω(u + du) dt cub. ft. is Ωp du dt. But from (3a) u = cτW/Ωp, and thereforedu / dp = −cτW / Ωp2.And the work done by expansion is −(cτW/p) dp dt.The work done by gravity on the mass between A0and A1is zero if the pipe is horizontal, and may in other cases be neglected without great error. The work of the pressures at the sections A0A1ispΩu dt − (p + dp) Ω (u + du) dt= −(p du + u dp) Ω dtBut from (3a)pu = constant,p du + u dp = 0,and the work of the pressures is zero. Adding together the quantities of work, and equating them to the change of kinetic energy,W dH dt = −(cτW/p) dp dt − ζ (H/m) W dl dtdH + (cτ/p) dp + ζ (H/m) dl = 0,dH/H + (cτ/Hp) dp + ζdl / m = 0(4)Butu = cτW / Ωp,andH = u2/2g = c2τ2W2/ 2gΩ2p2,∴ dH/H + (2gΩ2p / cτW2) dp + ζdl / m = 0.(4a)For tubes of uniform section m is constant; for steady motion W is constant; and for isothermal expansion τ is constant. Integrating,log H + gΩ2p2/ W2cτ + ζ l / m = constant;(5)forl = 0, let H = H0, and p = p0;and forl = l, let H = H1, and p = p1.log (H1/H0) + (gΩ2/ W2cτ) (p12− p02) + ζ l / m = 0.(5a)where p0is the greater pressure and p1the less, and the flow is from A0towards A1.By replacing W and H,log (p0/p1) + (gcτ / u02p02) (p12− p02+ ζ l/m = 0
(1)
if ζeis put for the expression in brackets.
Abrupt Contraction of Section.—When water passes from a larger to a smaller section, as in figs. 88, 89, a contraction is formed, and the contracted stream abruptly expands to fill the section of the pipe. Let ω be the section and v the velocity of the stream at bb. At aa the section will be ccω, and the velocity (ω/ccω) v = v/c1, where ccis the coefficient of contraction. Then the head lost is
ɧm= (v/cc− v)2/ 2g = (1/cc− 1)2v2/2g;
and, if ccis taken 0.64,
ɧm= 0.316 v2/2g.
(2)
The value of the coefficient of contraction for this case is, however, not well ascertained, and the result is somewhat modified by friction. For water entering a cylindrical, not bell-mouthed, pipe from a reservoir of indefinitely large size, experiment gives
ɧa= 0.505 v2/2g.
(3)
If there is a diaphragm at the mouth of the pipe as in fig. 89, let ω1be the area of this orifice. Then the area of the contracted stream is ccω1, and the head lost is
ɧc= {(ω/ccω1) − 1}2v2/2g= ζcv2/ 2g
ɧc= {(ω/ccω1) − 1}2v2/2g
= ζcv2/ 2g
(4)
if ζ, is put for {(ω/ccω1) − 1}2. Weisbach has found experimentally the following values of the coefficient, when the stream approaching the orifice was considerably larger than the orifice:—
When a diaphragm was placed in a tube of uniform section (fig. 90) the following values were obtained, ω1being the area of the orifice and ω that of the pipe:—
Elbows.—Weisbach considers the loss of head at elbows (fig. 91) to be due to a contraction formed by the stream. From experiments with a pipe 11⁄4in. diameter, he found the loss of head
ɧe= ζεv2/ 2g;
(5)
ζe= 0.9457 sin21⁄2φ + 2.047 sin41⁄2φ.
Hence at a right-angled elbow the whole head due to the velocity very nearly is lost.
Bends.—Weisbach traces the loss of head at curved bends to a similar cause to that at elbows, but the coefficients for bends are not very satisfactorily ascertained. Weisbach obtained for the loss of head at a bend in a pipe of circular section
ɧb= ζbv2/ 2g;
(6)
ζb= 0.131 + 1.847 (d/2ρ)7/2,
where d is the diameter of the pipe and ρ the radius of curvature of the bend. The resistance at bends is small and at present very ill determined.
Valves, Cocks and Sluices.—These produce a contraction of the water-stream, similar to that for an abrupt diminution of section already discussed. The loss of head may be taken as before to be
ɧv= ζvv2/ 2g;
(7)
where v is the velocity in the pipe beyond the valve and ζva coefficient determined by experiment. The following are Weisbach’s results.
Sluice in Pipe of Rectangular Section(fig. 92). Section at sluice = ω1in pipe = ω.
Sluice in Cylindrical Pipe(fig. 93).
Cock in a Cylindrical Pipe(fig. 94). Angle through which cock is turned = θ.
Throttle Valve in a Cylindrical Pipe(fig. 95)
§ 84.Practical Calculations on the Flow of Water in Pipes.—In the following explanations it will be assumed that the pipe is of so great a length that only the loss of head in friction against the surface of the pipe needs to be considered. In general it is one of the four quantities d, i, v or Q which requires to be determined. For since the loss of head h is given by the relation h = il, this need not be separately considered.
There are then three equations (see eq. 4, § 72, and 9a, § 76) for the solution of such problems as arise:—
ζ = α (1 + 1/12d);
(1)
where α = 0.005 for new and = 0.01 for incrusted pipes.
ζv2/ 2g =1⁄4di.
(2)
Q =1⁄4πd2v.
(3)
Problem1. Given the diameter of the pipe and its virtual slope, to find the discharge and velocity of flow. Here d and i are given, and Q and v are required. Find ζ from (1); then v from (2); lastly Q from (3). This case presents no difficulty.
By combining equations (1) and (2), v is obtained directly:—
v = √ (gdi/2ζ) = √ (g/2α) √ [di / {1 + 1/12d}].
(4)
For pipes not less than 1, or more than 4 ft. in diameter, the mean values of ζ are
Using these values we get the very simple expressions—
v = 55.31 √ (di) for new pipes= 39.11 √ (di) for incrusted pipes.
v = 55.31 √ (di) for new pipes
= 39.11 √ (di) for incrusted pipes.
(4a)
Within the limits stated, these are accurate enough for practical purposes, especially as the precise value of the coefficient ζ cannot be known for each special case.
Problem2. Given the diameter of a pipe and the velocity of flow, to find the virtual slope and discharge. The discharge is given by (3); the proper value of ζ by (1); and the virtual slope by (2). This also presents no special difficulty.
Problem3. Given the diameter of the pipe and the discharge, to find the virtual slope and velocity. Find v from (3); ζ from (1); lastly i from (2). If we combine (1) and (2) we get
i = ζ (v2/2g) (4/d) = 2a {1 + 1/12d} v2/gd;
(5)
and, taking the mean values of ζ for pipes from 1 to 4 ft. diameter, given above, the approximate formulae are
i = 0.0003268 v2/d for new pipes= 0.0006536 v2/d for incrusted pipes.
i = 0.0003268 v2/d for new pipes
= 0.0006536 v2/d for incrusted pipes.
(5a)
Problem4. Given the virtual slope and the velocity, to find the diameter of the pipe and the discharge. The diameter is obtained from equations (2) and (1), which give the quadratic expression
d2− d (2αv2/gi) − αv2/6gi = 0.
∴ d = αv2/gi + √ {(αv2/gi) (αv2/gi + 1/6)}.
(6)
For practical purposes, the approximate equations
d = 2αv2/gi + 1/12= 0.00031 v2/i + .083 for new pipes= 0.00062 v2/i + .083 for incrusted pipes
d = 2αv2/gi + 1/12
= 0.00031 v2/i + .083 for new pipes
= 0.00062 v2/i + .083 for incrusted pipes
(6a)
are sufficiently accurate.
Problem5. Given the virtual slope and the discharge, to find the diameter of the pipe and velocity of flow. This case, which often occurs in designing, is the one which is least easy of direct solution. From equations (2) and (3) we get—
d5= 32ζQ2/ gπ2i.
(7)
If now the value of ζ in (1) is introduced, the equation becomes very cumbrous. Various approximate methods of meeting the difficulty may be used.
(a) Taking the mean values of ζ given above for pipes of 1 to 4 ft. diameter we get
d =5√ (32ζ/gπ2)5√ (Q2/i)= 0.22165√ (Q2/i) for new pipes= 0.25415√ (Q2/i) for incrusted pipes;
d =5√ (32ζ/gπ2)5√ (Q2/i)
= 0.22165√ (Q2/i) for new pipes
= 0.25415√ (Q2/i) for incrusted pipes;
(8)
equations which are interesting as showing that when the value of ζ is doubled the diameter of pipe for a given discharge is only increased by 13%.
(b) A second method is to obtain a rough value of d by assuming ζ = α. This value is
d′ =5√ (32Q2/ gπ2i)5√ α = 0.63195√ (Q2/i)5√ α.
Then a very approximate value of ζ is
ζ′ = α (1 + 1/12d′);
and a revised value of d, not sensibly differing from the exact value, is
d″ =5√ (32Q2/ gπ2i)5√ ζ′ = 0.63195√ (Q2/i)5√ ζ′.
(c) Equation 7 may be put in the form
d =5√ (32αQ2/ gπ2i)5√ (1 + 1/12d).
(9)
Expanding the term in brackets,
5√ (1 + 1/12d) = 1 + 1/60d − 1/1800d2...
Neglecting the terms after the second,
d =5√ (32α / gπ2)5√ (Q2/i) · {1 + 1/60d}=5√ (32α / gπ2)5√ (Q2/i) + 0.01667;
d =5√ (32α / gπ2)5√ (Q2/i) · {1 + 1/60d}
=5√ (32α / gπ2)5√ (Q2/i) + 0.01667;
(9a)
and
5√ (32α / gπ2) = 0.219 for new pipes= 0.252 for incrusted pipes.
5√ (32α / gπ2) = 0.219 for new pipes
= 0.252 for incrusted pipes.
§ 85.Arrangement of Water Mains for Towns’ Supply.—Town mains are usually supplied oy gravitation from a service reservoir, which in turn is supplied by gravitation from a storage reservoir or by pumping from a lower level. The service reservoir should contain three days’ supply or in important cases much more. Its elevation should be such that water is delivered at a pressure of at least about 100 ft. to the highest parts of the district. The greatest pressure in the mains is usually about 200 ft., the pressure for which ordinary pipes and fittings are designed. Hence if the district supplied has great variations of level it must be divided into zones of higher and lower pressure. Fig. 96 shows a district of two zones each with its service reservoir and a range of pressure in the lower district from 100 to 200 ft. The total supply required is in England about 25 gallons per head per day. But in many towns, and especially in America, the supply is considerably greater, but also in many cases a good deal of the supply is lost by leakage of the mains. The supply through the branch mains of a distributing system is calculated from the population supplied. But in determining the capacity of the mains the fluctuation of the demand must be allowed for. It is usual to take the maximum demand at twice the average demand. Hence if the average demand is 25 gallons per head per day, the mains should be calculated for 50 gallons per head per day.
§ 86.Determination of the Diameters of Different Parts of a Water Main.—When the plan of the arrangement of mains is determined upon, and the supply to each locality and the pressure required is ascertained, it remains to determine the diameters of the pipes. Let fig. 97 show an elevation of a main ABCD ..., R being the reservoir from which the supply is derived. Let NN be the datum line of the levelling operations, and Ha, Hb... the heights of the main above the datum line, Hrbeing the height of the water surface in the reservoir from the same datum. Set up next heights AA1, BB1, ... representing the minimum pressure height necessary for the adequate supply of each locality. Then A1B1C1D1... is a line which should form a lower limit to the line of virtual slope. Then if heights ɧa, ɧb, ɧc... are taken representing the actual losses of head in each length la, lb, lc... of the main, A0B0C0will be the line of virtual slope, and it will be obvious at what points such as D0and E0, the pressure is deficient, and a different choice of diameter of main is required. For any point z in the length of the main, we have
Pressure height = Hr− Hz− (ɧa+ ɧb+ ... ɧz).
Where no other circumstance limits the loss of head to be assigned to a given length of main, a consideration of the safety of the main from fracture by hydraulic shock leads to a limitation of the velocity of flow. Generally the velocity in water mains lies between 11⁄2and 41⁄2ft. per second. Occasionally the velocity in pipes reaches 10 ft. per second, and in hydraulic machinery working under enormous pressures even 20 ft. per second. Usually the velocity diminishes along the main as the discharge diminishes, so as to reduce somewhat the total loss of head which is liable to render the pressure insufficient at the end of the main.
J. T. Fanning gives the following velocities as suitable in pipes for towns’ supply:—
§ 87.Branched Pipe connecting Reservoirs at Different Levels.—Let A, B, C (fig. 98) be three reservoirs connected by the arrangement of pipes shown,—l1, d1, Q1, v1; l2, d2, Q2, v2; h3, d3, Q3, v3being the length, diameter, discharge and velocity in the three portions of the main pipe. Suppose the dimensions and positions of the pipes known and the discharges required.
If a pressure column is introduced at X, the water will rise to a height XR, measuring the pressure at X, and aR, Rb, Rc will be the lines of virtual slope. If the free surface level at R is above b, the reservoir A supplies B and C, and if R is below b, A and B supply C. Consequently there are three cases:—
To determine which case has to be dealt with in the given conditions, suppose the pipe from X to B closed by a sluice. Then there is a simple main, and the height of free surface h′ at X can be determined. For this condition
ha− h′ = ζ (v12/2g) (4l1/d1) = 32ζQ′2l1/ gπ2d15;
h′ − hc= ζ (v32/2g) (4l3/d3) = 32ζQ′2l3/ gπ2d35;
where Q′ is the common discharge of the two portions of the pipe. Hence
(ha− h′) / (h′ − hc) = l1d35/ l3d15,
from which h′ is easily obtained. If then h′ is greater than hb, opening the sluice between X and B will allow flow towards B, and the case in hand is case I. If h′ is less than hb, opening the sluice will allow flow from B, and the case is case III. If h′ = hb, the case is case II., and is already completely solved.
The true value of h must lie between h′ and hb. Choose a new value of h, and recalculate Q1, Q2, Q3. Then if
Q1> Q2+ Q3in case I.,
or
Q1+ Q2> Q3in case III.,
the value chosen for h is too small, and a new value must be chosen.
If
Q1< Q2+ Q3in case I.,
or
Q1+ Q2< Q3in case III.,
the value of h is too great.
Since the limits between which h can vary are in practical cases not very distant, it is easy to approximate to values sufficiently accurate.
§ 88.Water Hammer.—If in a pipe through which water is flowing a sluice is suddenly closed so as to arrest the forward movement of the water, there is a rise of pressure which in some cases is serious enough to burst the pipe. This action is termed water hammer or water ram. The fluctuation of pressure is an oscillating one and gradually dies out. Care is usually taken that sluices should only be closed gradually and then the effect is inappreciable. Very careful experiments on water hammer were made by N. J. Joukowsky at Moscow in 1898 (Stoss in Wasserleitungen, St Petersburg, 1900), and the results are generally confirmed by experiments made by E. B. Weston and R. C. Carpenter in America. Joukowsky used pipes, 2, 4 and 6 in. diameter, from 1000 to 2500 ft. in length. The sluice closed in 0.03 second, and the fluctuations of pressure were automatically registered. The maximum excess pressure due to water-hammer action was as follows:—
In some cases, in fixing the thickness of water mains, 100 ℔ per sq. in. excess pressure is allowed to cover the effect of water hammer. With the velocities usual in water mains, especially as no valves can be quite suddenly closed, this appears to be a reasonable allowance (see also Carpenter,Am. Soc. Mech. Eng., 1893).
IX. FLOW OF COMPRESSIBLE FLUIDS IN PIPES
§ 89.Flow of Air in Long Pipes.—When air flows through a long pipe, by far the greater part of the work expended is used in overcoming frictional resistances due to the surface of the pipe. The work expended in friction generates heat, which for the most part must be developed in and given back to the air. Some heat may be transmitted through the sides of the pipe to surrounding materials, but in experiments hitherto made the amount so conducted away appears to be very small, and if no heat is transmitted the air in the tube must remain sensibly at the same temperature during expansion. In other words, the expansion may be regarded as isothermal expansion, the heat generated by friction exactly neutralizing the cooling due to the work done. Experiments on the pneumatic tubes used for the transmission of messages, by R. S. Culley and R. Sabine (Proc. Inst. Civ. Eng.xliii.), show that the change of temperature of the air flowing along the tube is much less than it would be in adiabatic expansion.
§ 90.Differential Equation of the Steady Motion of Air Flowing in a Long Pipe of Uniform Section.—When air expands at a constant absolute temperature τ, the relation between the pressure p in pounds per square foot and the density or weight per cubic foot G is given by the equation
p/G = cτ,
(1)
where c = 53.15. Taking τ = 521, corresponding to a temperature of 60° Fahr.,
cτ = 27690 foot-pounds.
(2)
The equation of continuity, which expresses the condition that in steady motion the same weight of fluid, W, must pass through each cross section of the stream in the unit of time, is
GΩu = W = constant,
(3)
where Ω is the section of the pipe and u the velocity of the air. Combining (1) and (3),
Ωup/W = cτ = constant.
(3a)
Since the work done by gravity on the air during its flow through a pipe due to variations of its level is generally small compared with the work done by changes of pressure, the former may in many cases be neglected.
Consider a short length dl of the pipe limited by sections A0, A1at a distance dl (fig. 99). Let p, u be the pressure and velocity at A0, p + dp and u + du those at A1. Further, suppose that in a very short time dt the mass of air between A0A1comes to A′0A′1so that A0A′0= udt and A1A′1= (u + du) dt1. Let Ω be the section, and m the hydraulic mean radius of the pipe, and W the weight of air flowing through the pipe per second.
From the steadiness of the motion the weight of air between the sections A0A′0, and A1A′1is the same. That is,
W dt = GΩu dt = GΩ (u + du) dt.
By analogy with liquids the head lost in friction is, for the length dl (see § 72, eq. 3), ζ (u2/2g) (dl/m). Let H = u2/2g. Then the head lost is ζ(H/m)dl; and, since Wdt ℔ of air flow through the pipe in the time considered, the work expended in friction is −ζ (H/m)W dl dt. The change of kinetic energy in dt seconds is the difference of the kinetic energy of A0A′0and A1A′1, that is,
(W/g) dt {(u + du)2− u2} / 2 = (W/g) u du dt = W dH dt.
The work of expansion when Ωudt cub. ft. of air at a pressure p expand to Ω(u + du) dt cub. ft. is Ωp du dt. But from (3a) u = cτW/Ωp, and therefore
du / dp = −cτW / Ωp2.
And the work done by expansion is −(cτW/p) dp dt.
The work done by gravity on the mass between A0and A1is zero if the pipe is horizontal, and may in other cases be neglected without great error. The work of the pressures at the sections A0A1is
pΩu dt − (p + dp) Ω (u + du) dt= −(p du + u dp) Ω dt
But from (3a)
pu = constant,p du + u dp = 0,
and the work of the pressures is zero. Adding together the quantities of work, and equating them to the change of kinetic energy,
W dH dt = −(cτW/p) dp dt − ζ (H/m) W dl dtdH + (cτ/p) dp + ζ (H/m) dl = 0,dH/H + (cτ/Hp) dp + ζdl / m = 0
(4)
But
u = cτW / Ωp,
and
H = u2/2g = c2τ2W2/ 2gΩ2p2,
∴ dH/H + (2gΩ2p / cτW2) dp + ζdl / m = 0.
(4a)
For tubes of uniform section m is constant; for steady motion W is constant; and for isothermal expansion τ is constant. Integrating,
log H + gΩ2p2/ W2cτ + ζ l / m = constant;
(5)
for
l = 0, let H = H0, and p = p0;
and for
l = l, let H = H1, and p = p1.
log (H1/H0) + (gΩ2/ W2cτ) (p12− p02) + ζ l / m = 0.
(5a)
where p0is the greater pressure and p1the less, and the flow is from A0towards A1.
By replacing W and H,
log (p0/p1) + (gcτ / u02p02) (p12− p02+ ζ l/m = 0