8.1 SECTION 8 PIPING AND FLUID FLOW PRESSURE SURGE IN FLUID PIPING SYSTEMS 8.2 Pressure Surge in a Piping System From Rapid Valve Closure 8.2 Piping Pressure Surge with Different Material and Fluid 8.5 Pressure Surge in Piping System with Compound Pipeline 8.6 PIPE PROPERTIES, FLOW RATE, AND PRESSURE DROP 8.8 Quick Calculation of Flow Rate and Pressure Drop in Piping Systems 8.8 Fluid Head-Loss Approximations for All Types of Piping 8.10 Pipe-Wall Thickness and Schedule Number 8.11 Pipe-Wall Thickness Determination by Piping Code Formula 8.12 Determining the Pressure Loss in Steam Piping 8.15 Piping Warm-Up Condensate Load 8.18 Steam Trap Selection for Industrial Applications 8.20 Selecting Heat Insulation for High- Temperature Piping 8.27 Orifice Meter Selection for a Steam Pipe 8.29 Selection of a Pressure-Regulating Valve for Steam Service 8.30 Hydraulic Radius and Liquid Velocity in Water Pipes 8.33 Friction-Head Loss in Water Piping of Various Materials 8.33 Chart and Tabular Determination of Friction Head 8.36 Relative Carrying Capacity of Pipes 8.39 Pressure-Reducing Valve Selection for Water Piping 8.41 Sizing a Water Meter 8.42 Equivalent Length of a Complex Series Pipeline 8.43 Equivalent Length of a Parallel Piping System 8.44 Maximum Allowable Height for a Liquid Siphon 8.45 Water-Hammer Effects in Liquid Pipelines 8.47 Specific Gravity and Viscosity of Liquids 8.47 Pressure Loss in Piping Having Laminar Flow 8.48 Determining the Pressure Loss in Oil Pipes 8.49 Flow Rate and Pressure Loss in Compressed-Air and Gas Piping 8.56 Flow Rate and Pressure Loss in Gas Pipelines 8.57 Selecting Hangers for Pipes at Elevated Temperatures 8.58 Hanger Spacing and Pipe Slope for an Allowable Stress 8.66 Effect of Cold Spring on Pipe Anchor Forces and Stresses 8.67 Reacting Forces and Bending Stress in Single-Plane Pipe Bend 8.68 Reacting Forces and Bending Stress in a Two-Plane Pipe Bend 8.75 Reacting Forces and Bending Stress in a Three-Plane Pipe Bend 8.77 Anchor Force, Stress, and Deflection of Expansion Bends 8.79 Slip-Type Expansion Joint Selection and Application 8.80 Corrugated Expansion Joint Selection and Application 8.84 Design of Steam Transmission Piping 8.88 Steam Desuperheater Analysis 8.98 Steam Accumulator Selection and Sizing 8.100 Selecting Plastic Piping for Industrial Use 8.102 Analyzing Plastic Piping and Lining for Tanks, Pumps and Other Components for Specific Applications 8.104 Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS 8.2 PLANT AND FACILITIES ENGINEERING Friction Loss in Pipes Handling Solids in Suspension 8.111 Desuperheater Water Spray Quantity 8.112 Sizing Condensate Return Lines for Optimum Flow Conditions 8.114 Estimating Cost of Steam Leaks from Piping and Pressure Vessels 8.116 Quick Sizing of Restrictive Orifices in Piping 8.117 Steam Tracing a Vessel Bottom to Keep Its Contents Fluid 8.118 Designing Steam-Transmission Lines Without Steam Traps 8.119 Line Sizing for Flashing Steam Condensate 8.124 Determining the Friction Factor for Flow of Bingham Plastics 8.127 Time Needed to Empty a Storage Vessel with Dished Ends 8.130 Time Needed to Empty a Vessel Without Dished Ends 8.133 Time Needed to Drain a Storage Tank Through Attached Piping 8.134 Pressure Surge in Fluid Piping Systems PRESSURE SURGE IN A PIPING SYSTEM FROM RAPID VALVE CLOSURE Oil, with a specific weight of 52 lb /ft 3 (832 kg /m 3 ) and a bulk modulus of 250,000 lb/in 2 (1723 MPa), flows at the rate of 40 gal /min (2.5 L/s) through stainless steel pipe. The pipe is 40 ft (12.2 m) long, 1.5 in (38.1 mm) O.D., 1.402 in (35.6 mm) I.D., 0.049 in (1.24 mm) wall thickness, and has a modulus of elasticity, E,of 29 ϫ 10 6 lb/in 2 (199.8 kPa ϫ 10 6 ). Normal static pressure immediately upstream of the valve in the pipe is 500 lb/in 2 (abs) (3445 kPa). When the flow of the oil is reduced to zero in 0.015 s by closing a valve at the end of the pipe, what is: (a) the velocity of the pressure wave; (b) the period of the pressure wave; (c) the amplitude of the pressure wave; and (d ) the maximum static pressure at the valve? Calculation Procedure: 1. Find the velocity of the pressure wave when the valve is closed (a) Use the equation 68.094 a ϭ ͙ ␥ [(1/K) ϩ (D/Et)] where the symbols are as given in the notation below. Substituting, 68.094 a ϭ 46 ͙52 [(1/ 25 ϫ 10 ) ϩ (1.402 /29 ϫ 10 ϫ 0.049] ϭ 4228 ft/s (1288.7 m/s) An alternative solution uses Fig. 1. With a D/t ratio ϭ 1.402/ 0.049 ϭ 28.6 for Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. PIPING AND FLUID FLOW PIPING AND FLUID FLOW 8.3 5,000 4,000 3,000 2,000 1524 1219 914 610 Bulk modulus K = 250,000 psi Specific weight = 52 lb per cu ft ␥ S t a i n l e s s s t e e l p i p e , E = 2 9 x 1 0 6 p s i C o p p e r p i p e , E = 1 7 x 1 0 6 p s i A l u m i n u m p i p e , E = 1 0 . 7 x 1 0 6 p s i 0 14012010080604020 a, Velocity of pressure wave, ft per sec Velocity, m/sec D / t; I.D. of pipe / wall thickness 250,000 psi (1723 GPa) 300,000 psi (2.07 GPa) 52 lb/ft 3 (832 kg/m 3 ) 62.42 lb/ft 3 (998.7 kg/m 3 ) 29 ϫ 10 6 psi (199.8 GPa) 17 ϫ 10 6 psi (117.1 (GPa) 10.7 ϫ 10 6 psi (73.7 GPa) FIGURE 1 Velocity of pressure wave in oil column in pipe of different diameter-to- wall thickness ratios. (Product Engineering.) stainless steel pipe, the velocity, a, of the pressure wave is 4228 ft/s (1288.7 m/ s). 2. Compute the time for the pressure wave to make one round trip in the pipe b) The time for the pressure wave to make one round trip between the pipe ex- tremities, or one interval, is: 2L /a ϭ 2(40)/4228 ϭ 0.0189 s, and the period of the pressure wave is: 2(2L/a) ϭ 2(0.0189) ϭ 0.0378 s. 3. Calculate the pressure surge for rapid valve closure c) Since the time of 01015 s for valve closure is less than the internal time 2L/a equal to 0.0189 s, the pressure surge can be computed from: ⌬p ϭ ␥ aV/144g for rapid valve closure. Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. PIPING AND FLUID FLOW 8.4 PLANT AND FACILITIES ENGINEERING The velocity of flow, V ϭ [(40)(231)(4)]/[(60)( )(1.402 2 )(12)] using the stan- dard pipe flow relation, or V ϭ 8.3 ft/s (2.53 m/s). Then, the amplitude of the pressure wave, using the equation above is: 52 ϫ 4228 ϫ 8.3 2 ⌬p ϭϭ393.5 lb/ in (2711.2 kPa). 144 ϫ 32.2 4. Determine the resulting maximum static press in the pipe d) The resulting maximum static pressure in the line, p max ϭ p ϩ ⌬p ϭ 500 ϩ 393.5 ϭ 893.5 lb/in 2 (abs) (6156.2 kPa). Related Calculations. In an industrial hydraulic system, such as that used in machine tools, hydraulic lifts, steering mechanisms, etc., when the velocity of a flowing fluid is changed by opening or closing a valve, pressure surges result. The amplitude of the pressure surge is a function of the rate of change in the velocity of the mass of fluid. This procedure shows how to compute the amplitude of the pressure surge with rapid valve closure. The procedure is the work of Nils M. Sverdrup, Hydraulic Engineer, Aerojet- General Corporation, as reported in Product Engineering magazine. SI values were added by the handbook editor. Notation a ϭ velocity of pressure wave, ft/s (m/s) a E ϭ effective velocity of pressure wave, ft /s (m/ s) A ϭ cross-sectional area of pipe, in 2 (mm 2 ) A o ϭ area of throttling orifice before closure, in 2 (mm 2 ) c ϭ velocity of sound, ft/s (m/s) C D ϭ coefficient of discharge D ϭ inside diameter of pipe, in (mm) E ϭ modulus of elasticity of pipe material, lb /in 2 (kPa) F ϭ force, lb (kg) g ϭ gravitational acceleration, 32.2 ft/s 2 K ϭ bulk modulus of fluid medium, lb/in 2 (kPa) L ϭ length of pipe, ft (m) m ϭ mass, slugs N ϭ T/(2L/a) ϭ number of pressure wave intervals during time of valve clo- sure p ϭ normal static fluid pressure immediately upstream of valve when the fluid velocity is V , lb/in 2 (absolute) (kPa) ⌬p ϭ amplitude of pressure wave, lb/in 2 (kPa) p max ϭ maximum static pressure immediately upstream of valve, lb/in 2 (absolute) (kPa) p d ϭ static pressure immediately downstream of the valve, lb/in 2 (absolute) (kPa) Q ϭ volume rate of flow, ft 3 /s (m 3 /s) t ϭ wall thickness of pipe, in (mm) T ϭ time in which valve is closed, s v ϭ fluid volume, in 3 (mm 3 ) v A ϭ air volume, in 3 (mm 3 ) V ϭ normal velocity of fluid flow in pipe with valve wide open, ft /s (m /s) V E ϭ equivalent fluid velocity, ft /s (m/s) V n ϭ velocity of fluid flow during interval n, ft/s (m/s) Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. PIPING AND FLUID FLOW PIPING AND FLUID FLOW 8.5 W ϭ work, ft ⅐ lb (W) ␥ ϭ specific weight, lb /ft 3 (kg/m 3 ) n ϭ coefficient dependent upon the rate of change in orifice area and dis- charge coefficient ϭ period of oscillation of air cushion in a sealed chamber, s PIPING PRESSURE SURGE WITH DIFFERENT MATERIAL AND FLUID (a) What would be the pressure rise in the previous procedure if the pipe were aluminum instead of stainless steel? (b) What would be the pressure rise in the system in the previous procedure if the flow medium were water having a bulk modulus, K, of 300,000 lb/in 2 (2067 MPa) and a specific weight of 62.42 lb/ft 3 (998.7 kg/m 3 )? Calculation Procedure: 1. Find the velocity of the pressure wave in the pipe (a) From Fig. 2, for aluminum pipe having a D /t ratio of 28.6, the velocity of the pressure wave is 3655 ft /s (1114.0 m/s). Alternatively, the velocity could be com- puted as in step 1 in the previous procedure. 2. Compute the time for one interval of the pressure wave As before, in the previous procedure, 2 L /a ϭ 2 (40/3655) ϭ 0.02188 s. 3. Calculate the pressure rise in the pipe Since the time of 0.015 s for the valve closure is less than the interval time of 2 L/a equal to 0.02188, the pressure rise can be computed from: ⌬p ϭ ␥ aV/144g or, 52 ϫ 3655 ϫ 8.3 2 ⌬p ϭϭ340.2 lb/ in (2343.98 kPa) 144 ϫ 32.2 4. Find the maximum static pressure in the line Using the pressure-rise relation, p max ϭ 500 ϩ 340.2 ϭ 840.2 lb/in 2 (abs) (5788.97 kPa). 5. Determine the pressure rise for the different fluid (b) For water, use Fig. 2 for stainless steel pipe having a D/t ratio of 28.6 to find a ϭ 4147 ft/s (1264 m/s). Alternatively, the velocity could be calculated as in step 1 of the previous procedure. 6. Compute the time for one internal of the pressure wave Using 2 L/a ϭ 2 (40)/4147 ϭ 0.012929 s. Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. PIPING AND FLUID FLOW 8.6 PLANT AND FACILITIES ENGINEERING 5,000 4,000 3,000 2,000 1524 1219 914 610 a, Velocity of pressure wave, ft per sec Velocity, m/sec 0 14012010080604020 D / t; I.D. of pipe / wall thickness See Fig. 1 for SI values Bulk modulus K = 250,000 psi Specific weight = 52 lb per cu ft ␥ S t a i n l e s s s t e e l p i p e , E = 2 9 x 1 0 6 p s i C o p p e r p i p e , E = 1 7 x 1 0 6 p s i A l u m i n u m p i p e , E = 1 0 . 7 x 1 0 6 p s i FIGURE 2 Velocity of pressure wave in water column in pipe of different diameter-to- wall thickness ratios. (Product Engineering.) 7. Find the pressure rise and maximum static pressure in the line Since the time of 0.015 s for valve closure is less than the interval time 2 L /a equal to 0.01929 s, the pressure rise can be computed from ⌬p ϭ ␥ aV/144g for rapid valve closure. Therefore, the pressure rise when the flow medium is water is: 62.42 ϫ 4147 ϫ 8.3 2 ⌬p ϭϭ463.4 lb/ in (3192.8 kPa) 144 ϫ 32.2 The maximum static pressure, p max ϭ 500 ϩ 463.4 ϭ 963.4 lb/in 2 (abs) (6637.8 kPa). Related Calculations. This procedure is the work of Nils M. Sverdrup, as detailed in the previous procedure. PRESSURE SURGE IN PIPING SYSTEM WITH COMPOUND PIPELINE A compound pipeline consisting of several stainless-steel pipes of different diam- eters, Fig. 3, conveys 40 gal/min (2.5 L /s) of water. The length of each section of Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. PIPING AND FLUID FLOW PIPING AND FLUID FLOW 8.7 FIGURE 3 Compound pipeline consists of pipe sections having dif- ferent diameters. (Product Engineer- ing.) pipe is: L 1 ϭ 25 ft (7.6 m); L 2 ϭ 15 ft (4.6 m); L 3 ϭ 10 ft (3.0 m); pipe wall thickness in each section is 0.049 in (1.24 mm); inside diameter of each section of pipe is D 1 ϭ 1.402 in (35.6 mm); D 2 ϭ 1.152 in (29.3 mm); D 3 ϭ 0.902 in (22.9 mm). What is the equivalent fluid velocity and the effective velocity of the pressure wave on sudden valve closure? Calculation Procedure: 1. Determine fluid velocity and pressure-wave velocity in the first pipe D 1 /t 1 ratio of the first pipe ϭ 1.402/0.049 ϭ 28.6. Then, the fluid velocity in the pipe can be found from V 1 ϭ 0.4085(G n /(D n ) 2 , where the symbols are as shown below. Substituting, V 1 ϭ 0.4085(40)/(1.402) 2 ϭ 8.31 ft/s (2.53 m/s). Using these two computed values, enter Fig. 2 to find the velocity of the pressure wave in pipe 1 as 4147 ft/s (1264 m/s). 2. Find the fluid velocity and pressure-wave velocity in the second pipe The D 2 /t 2 ratio for the second pipe ϭ 1.152/0.049 ϭ 23.51. Using the same ve- locity equation as in step 1, above V 2 ϭ 0.4085(40)/(1.152) 2 ϭ 12.31 ft/s (3.75 m/s). Again, from Fig. 2, a 2 ϭ 4234 ft /s (1290.5 m/s). Thus, there is an 87-ft/s (26.5- m/s) velocity increase of the pressure wave between pipes 1 and 2. 3. Compute the fluid velocity and pressure-wave velocity in the third pipe Using a similar procedure to that in steps 1 and 2 above, V 3 ϭ 20.1 ft/ s (6.13 m / s); s 3 ϭ 4326 ft/s (1318.6 m/s). 4. Find the equivalent fluid velocity and effective pressure-wave velocity for the compound pipe Use the equation LV ϩ LV ϩ ⅐⅐⅐ ϩ L V 11 22 nn V ϭ E L ϩ L ϩ ⅐⅐⅐ ϩ L 12 n to find the equivalent fluid velocity in the compound pipe. Substituting, Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. PIPING AND FLUID FLOW 8.8 PLANT AND FACILITIES ENGINEERING 25 ϫ 8.3 ϩ 15 ϫ 12.3 ϩ 10 ϫ 20.1 V ϭ E 25 ϩ 15 ϩ 10 ϭ 11.9 ft/s (3.63 m/s) To find the effective velocity of the pressure wave, use the equation L ϩ L ϩ ⅐⅐⅐ L 12 n a ϭ g (L / a ) ϩ (L /a ) ϩ ⅐⅐⅐ ϩ (L /a ) 11 22 nn Substituting, 25 ϩ 15 ϩ 10 a ϭ g (25/4147) ϩ (15/4234) ϩ (10/4326) ϭ 4209 ft/s (1282.9 m/s) Thus, equivalent fluid velocity and effective velocity of the pressure wave in the compound pipe are both less than either velocity in the individual sections of the pipe. Related Calculations. Compound pipes find frequent application in industrial hydraulic systems. The procedure given here is useful in determining the velocities produced by sudden closure of a valve in the line. L 1 , L 2 , ,L n ϭ length of each section of pipe of constant diameter, ft (m) a 1 , a 2 , ,a n ϭ velocity of pressure wave in the respective pipe sections, ft/s (m/s) a g ϭ effective velocity of the pressure wave, ft /s V 1 , V 2 , V n ϭ velocity of fluid in the respective pipe sections, ft/s (m/s) V E ϭ equivalent fluid velocity, ft /s (m/s) G n ϭ rate of flow in respective section, U.S. gal/min (L/s) D n ϭ inside diameter of respective pipe, in (mm) The fluid velocity in an individual pipe is 2 V ϭ 0.4085G / D nnn This procedure is the work of Nils M. Sverdrup, as detailed earlier. Pipe Properties, Flow Rate, and Pressure Drop QUICK CALCULATION OF FLOW RATE AND PRESSURE DROP IN PIPING SYSTEMS A 3-in (76-mm) Schedule 40S pipe has a 300-gal/min (18.9-L/s) water flow rate with a pressure loss of 8 lb /in 2 (55.1 kPa)/100 ft (30.5 m). What would be the flow rate in a 4-in (102-mm) Schedule 40S pipe with the same pressure loss? What would be the pressure loss in a 4-in (102-mm) Schedule 40S pipe with the same Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. PIPING AND FLUID FLOW PIPING AND FLUID FLOW 8.9 flow rate, 300 gal /min (18.9 L/s)? Determine the flow rate and pressure loss for a 6-in (152-mm) Schedule 40S pipe with the same pressure and flow conditions. Calculation Procedure: 1. Determine the flow rate in the new pipe sizes Flow rate in a pipe with a fixed pressure drop is proportional to the ratio of (new pipe inside diameter/known pipe inside diameter) 2.4 . This ratio is defined as the flow factor, F. To use this ratio, the exact inside pipe diameters, known and new, must be used. Take the exact inside diameter from a table of pipe properties. Thus, with a 3-in (76-mm) and a 4-in (102-mm) Schedule 40S pipe conveying water at a pressure drop of 8 lb/in 2 (55.1 kPa)/100 ft (30.5 m), the flow factor F ϭ (4.026/3.068) 2.4 ϭ 1.91975. Then, the flow rate, FR, in the large 4-in (102- mm) pipe with the 8 lb/ in 2 (55.1 kPa) pressure drop /100 ft (30.5 m), will be, FR ϭ 1.91975 ϫ 300 ϭ 575.9 gal/min (36.3 L/s). For the 6-in (152-mm) pipe, the flow rate with the same pressure loss will be (6.065/3.068) 2.4 ϫ 300 ϭ 1539.8 gal/ min (97.2 L/s). 2. Compute the pressure drops in the new pipe sizes The pressure drop in a known pipe size can be extrapolated to a new pipe size by using a pressure factor, P, when the flow rate is held constant. For this condition, P ϭ (known inside diameter of the pipe/new inside diameter of the pipe) 4.8 . For the first situation given above, P ϭ (3.068 /4.026) 4.8 ϭ 0.27134. Then, the pressure drop, PD N , in the new 4-in (102-mm) Schedule 40S pipe with a 300-gal/ min (18.9-L /s) flow will be PD N ϩ P(PD K ), where PD K ϭ pressure drop in the known pipe size. Substituting, PD N ϭ 0.27134(8) ϭ 2.17 lb/in 2 /100 ft (14.9 kPa/ 30.5 m). For the 6-in (152-mm) pipe, using the same approach, PD N ϭ (3.068/6.065) 4.8 (8) ϭ 0.303 lb/ in 2 /100 ft (2.1 kPa/30.5 m). Related Calculations. The flow and pressure factors are valuable timesavers in piping system design because they permit quick determination of new flow rates or pressure drops with minimum time input. When working with a series of pipe- size possibilities of the same Schedule Number, the designer can compute values for F and P in advance and apply them quickly. Here is an example of such a calculation for Schedule 40S piping of several sizes: Nominal pipe size, new/known Flow factor, F Nominal pipe size, known/new Pressure factor, P 2/1 5.092 1/2 0.0386 3/2 2.58 2/3 0.150 4/3 1.919 3/4 0.271 6/4 2.674 4/6 0.1399 8/6 1.933 6/8 0.267 10/8 1.726 8/10 0.335 12/10 1.542 10/12 0.421 When computing such a listing, the actual inside diameter of the pipe, taken from a table of pipe properties, must be used when calculating F or P. The F and P values are useful when designing a variety of piping systems for chemical, petroleum, power, cogeneration, marine, buildings (office, commercial, Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. PIPING AND FLUID FLOW 8.10 PLANT AND FACILITIES ENGINEERING residential, industrial), and other plants. Both the F and P values can be used for pipes conveying oil, water, chemicals, and other liquids. The F and P values are not applicable to steam or gases. Note that the ratio of pipe diameters is valid for any units of measurement— inches, cm, mm—provided the same units are used consistently throughout the calculation. The results obtained using the F and P values usually agree closely with those obtained using exact flow or pressure-drop equations. Such accuracy is generally acceptable in everyday engineering calculations. While the pressure drop in piping conveying a liquid is inversely proportional to the fifth power of the pipe diameter ratio, turbulent flow alters this to the value of 4.8, according to W. L. Nelson, Technical Editor, The Oil and Gas Journal. FLUID HEAD-LOSS APPROXIMATIONS FOR ALL TYPES OF PIPING Using the four rules for approximating head loss in pipes conveying fluid under turbulent flow conditions with a Reynolds number greater than 2100, find: (a)A 4-in (101.6-mm) pipe discharges 100 gal/min (6.3 L /s); how much fluid would a 2-in (50.8-mm) pipe discharge under the same conditions? (b) A 4-in (101.6-mm) pipe has 240 gal/min (15.1 L/s) flowing through it. What would be the friction loss in a 3-in (76.2-mm) pipe conveying the same flow? (c) A flow of 10 gal/min (6.3 L /s) produces 50 ft (15.2 m) of friction in a pipe. How much friction will a flow of 200 gal/min (12.6 L/s) produce? (d) A 12-in (304.8-mm) diameter pipe has a friction loss of 200 ft (60.9 m) /1000 ft (304.8 M). What is the capacity of this pipe? Calculation Procedure: 1. Use the rule: At constant head, pipe capacity is proportional to d 2.5 (a) Applying the constant-head rule for both pipes: 4 2.5 ϭ 32.0; 2 2.5 ϭ 5.66. Then, the pipe capacity ϭ (flow rate, gal/min or L/s)(new pipe size 2.5 )/(previous pipe size 2.5 ) ϭ (100)(5.66)/ 32 ϭ 17.69 gal /min (1.11 L/s). Thus, using this rule you can approximate pipe capacity for a variety of con- ditions where the head is constant. This approximation is valid for metal, plastic, wood, concrete, and other piping materials. 2. Use the rule: At constant capacity, head is proportional to 1/d 5 (b) We have a 4-in (101.6-mm) pipe conveying 240 gal/min (15.1 L/s). If we reduce the pipe size to 3 in (76.2 mm) the friction will be greater because the flow area is smaller. The head loss ϭ (flow rate, gal/min or L/s)(larger pipe diameter to the fifth power)/(smaller pipe diameter to the fifth power). Or, head ϭ (240)(4 5 )/(3 5 ) ϭ 1011 ft/1000 ft of pipe (308.3 m /304.8 m of pipe). Again, using this rule you can quickly and easily find the friction in a different size pipe when the capacity or flow rate remains constant. With the easy availability of handheld calculators in the field and computers in the design office, the fifth power of the diameter is easily found. 3. Use the rule: At constant diameter, head is proportional to gal /min (L/s) 2 (c) We know that a flow of 100 gal /min (6.3 L /s) produces 50-ft (15.2-m) friction, h, in a pipe. The friction, with a new flow will be, h ϭ (friction, ft or m, at known Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. PIPING AND FLUID FLOW