Pipe Flow Calculation R. Shankar Subramanian epartment o Chemical and Biomolecular Engineering Clarkon Univerity We begin with ome reult that we hall ue when making riction lo calculation or teady, ully developed, incompreible, Newtonian low through a traight circular pipe. π Volumetric low rate Q= V where i the pipe diameter, and V i the average velocity. V ρ V Q m Reynold Number: Re = = = = where ρ i the denity o the µ ν π ν π µ luid, µ i it dynamic vicoity, and ν = µ / ρ i the kinematic vicoity. The preure drop P i related to the lo in the Engineering Bernoulli Equation, or equivalently, the rictional head lo h, through P= ρ lo = γ h Here, the peciic weight γ = ρ g, where g i the magnitude o the acceleration due to gravity. Power The power required to overcome riction i related to the preure drop through Power = PQ or we can relate it to the head lo due to pipe riction via Power = γ h Q Head o/preure rop The head lo h i related to the Fanning riction actor through h V = g = or alternatively we can write the preure drop a P ( ρ V ) Friction Factor 16 In laminar low, =. Re In turbulent low we can ue either the Colebrook or the Zigrang-Sylveter Equation, depending on the problem. Both give equivalent reult well within experimental uncertainty. In thee equation, ε i the average roughne o the interior urace o the pipe. A table o roughne 1
value recommended or commercial pipe given in a textbook on Fluid Mechanic by F.M. White i provided at the end o thee note. Colebrook Equation 1 ε / 1.6 =.0 log +.7 Re Zigrang-Sylveter Equation 1 ε / 5.0 ε / 1 =.0 log log +.7 Re.7 Re Non-Circular Conduit Not all low conduit are circular pipe. An example o a non-circular cro-ection in heat exchanger application i an annulu, which i the region between two circular pipe. Another i a rectangular duct, ued in HVAC (Heating, Ventilation, and Air-Conditioning) application. e common are duct o triangular or elliptical cro-ection, but they are ued on occaion. In all thee cae, when the low i turbulent, we ue the ame riction actor correlation that are ued or circular pipe, ubtituting an equivalent diameter or the pipe diameter. The equivalent diameter, which i et equal to our time the Hydraulic Radiu, R i deined a ollow. e h e Cro - Sectional Area = Rh = Wetted Perimeter In thi deinition, the term wetted perimeter i ued to deignate the perimeter o the croection that i in contact with the lowing luid. Thi applie to a liquid that occupie part o a conduit, a in ewer line carrying wate-water, or a creek or river. I a ga low through a conduit, the entire perimeter i wetted. Uing the above deinition, we arrive at the ollowing reult or the equivalent diameter or two common cro-ection. We aume that the entire perimeter i wetted. Rectangular uct b For the duct hown in the ketch, the cro-ectional area i ab, while the perimeter i ( a+ b) o that the equivalent diameter i written a ollow. a
e ab = = ( a+ b) 1 1 + a b I the low i laminar, a reult imilar to that or circular tube i available or the riction actor, which can be written a = C/ Re, where C i a contant that depend on the apect ratio a/ b, and the Reynold number i deined uing the equivalent diameter. A ew value o the contant C or elected value o the apect ratio are given in the Table below (Source: F.M. White, Fluid Mechanic, 7 th Edition). For other apect ratio, you can ue interpolation. a/ b C a/ b C 1.0 1. 6.0 19.70 1. 1.7 8.0 0.59.0 15.55.0 1.17.5 16.7 0.0.8.0 18..00 Annulu a b The cro-ectional area o the annulu hown i ( a b ) π ( a+ b). Thereore, the equivalent diameter i obtained a ( a b ) π e = = a b π ( a+ b) π, while the wetted perimeter i Again, or laminar low, we ind that = C/ Re, where C i a contant that depend on the apect ratio a/ b, and the Reynold number i deined uing the equivalent diameter. A with the rectangular cro-ection, a ew value contant C or elected value o the apect ratio are given in the Table that ollow (Source: F.M. White, Fluid Mechanic, 7 th Edition). For other apect ratio, you can ue interpolation.
a/ b C a/ b C 1.0.00,0. 1.5.98 0.0 1.57 1.67.90 0 0.0.5.68 00 18.67 5.0.09 16.00 Minor oe Minor loe i a term ued to decribe loe that occur in itting, expanion, contraction, and the like. Fitting commonly ued in the indutry include bend, tee, elbow, union, and o coure, valve ued to control low. Even though thee loe are called minor, they can be ubtantial compared to thoe or low through hort traight pipe egment. oe are V / g. Thereore, we can write commonly reported in velocity head. A velocity head i minor loe a h m V = K, where K i called the lo coeicient. g Typical value o K or ome common itting are given below. Uually, the value depend upon the nominal pipe diameter, the Reynold number, and the manner in which the valve i intalled (crewed or langed). Manuacturer data hould be ued wherever poible. Globe Valve (ully open): 5.5-1 Gate Valve (ully open): 0.0-0.80 Swing Check Valve (ully open):.0-5.1 Standard 5 o Elbow: 0. - 0. ong radiu 5 o Elbow: 0.1-0.1 Standard 90 o Elbow: 0.1 -.0 ong radiu 90 o Elbow: 0.07-1.0 Tee: 0.1 -. When olving homework problem, ue the value given in Table 1.1 in the textbook by Welty et al. Sudden Expanion and Sudden Contraction A udden expanion in a pipe i one o the ew cae where the loe can be obtained rom the baic balance. The expreion or K i given by K d = 1
Here, d and repreent the diameter o the maller and larger pipe, repectively. For a udden contraction, we can ue the ame reult i d / 0.76. For maller value o d / we can ue the empirical relation K = 0. 1 d /. In both cae, we hould multiply K by the velocity head in the pipe egment o diameter d. The loe would be maller i the expanion or contraction i gradual. When a pipe emptie into a reervoir, all the kinetic energy in the luid coming in i diipated, o that you can treat thi a a udden expanion with the ratio d / = 0, yielding K = 1. Typical Pipe Flow Problem In typical pipe low problem, we know the nature o the luid that will low through the pipe, and the temperature. Thereore, we can ind the relevant phyical propertie immediately. They are the denity ρ and the dynamic vicoity µ. Knowing thee propertie, we alo can calculate the kinematic vicoity ν = µ / ρ. The length o the pipe can be etimated rom proce equipment layout conideration. The nature o the luid to be pumped will dictate corroion contraint on the pipe material. Other conideration are cot and eae o procurement. Baed on thee, we can elect the material o the pipe to be ued, and once we do, the roughne ε can be peciied. Thi leave u with three unpeciied parameter, namely the head lo h or equivalently, the preure drop required to pump the luid p, the volumetric low rate Q (or equivalently the ma low rate), and the pipe diameter. Unle we plan to alo optimize the cot, two o thee mut be peciied, leaving only a ingle parameter to be calculated. Thu, pipe low problem that do not involve cot optimization will all into three broad categorie. 1. Given and Q, ind the head lo h. Given and h, ind the volumetric low rate Q. Given Q and h, ind the diameter Each o thee three type o problem i illutrated next with a numerical example. 5
Example 1 Find the head lo due to the low o 1,500 gpm o oil ( ν = 1.15 t / ) through 1,600 eet o 8" diameter cat iron pipe. I the denity o the oil ρ = 1.75 lug / t, what i the power to be upplied by a pump to the luid? Find the BHP o the pump i it eiciency i 0.85. Solution We have the ollowing inormation. ρ = 1.75 lug / t ν = 1.15 t / = 0.667 t Thereore, the cro-ectional area i A = π / = π 0.667 t / = 0.9 t 1 t / t Q = 1500 ( gpm) =. 8.8 ( gpm) Thereore, the average velocity through the pipe i V We can calculate the Reynold number. ( t ) ( t ) = Q. / A = 0.9 = 9.58 t ( t) ( t ) V 0.667 9.58 / Re = = = 5.55 ν ( t ) 1.15 / Thereore, the low i turbulent. For cat iron, ε = 8.5 t. Thereore, the relative roughne i ( t) ( t) ε 8.5 = = 1.7 0.667 Becaue we have the value o both the Reynold number and the relative roughne, it i eicient to ue the Zigrang-Sylveter equation or a once-through calculation o the turbulent low riction actor. 1 ε / 5.0 ε / 1 =.0 log log +.7 Re.7 Re 1.7 5.0 1.7 1 =.0 log log + 1.8 =.7 5.55.7 5.55 which yield = 0.0061 6
The head lo i obtained by uing ( t) ( t ) V 1, 600 9.58 / h = = 0.0061 = 8.7 t g 0.667 t. t / The ma low rate i m = ρ Q= 1.75. = 5.85 The power upplied to the luid i calculated rom lug t lug t lug t t lb Power to Fluid = m h g = 5.85 8.7 ( t). = 1.58 t lb We know that 1 HorePower = 550. Thereore, Power to Fluid = 8.7 hp The eiciency o the pump η = 0.85. Thereore, 8.7 ( hp) Power to Fluid Brake Hore Power = = =.7 hp η 0.85 Example Water at 15 C low through a 5 cm diameter riveted teel pipe o length 50 m and roughne ε =. mm. The head lo i known to be 7.0 m. Find the volumetric low rate o water in the pipe. Solution For water at 15 C, ρ = 999 kg / m 6 calculated a ν = µ / ρ = 1.16 m / µ = 1.16 Pa o that the kinematic vicoity can be The pipe diameter i given a = 0.5 m, o that the cro-ectional area i A= π / = π 0.5 m / =.91 m The length o the pipe i given a = 50 m We do not know the velocity o water in the pipe, but we can expre the Reynold number in term o the unknown velocity. ( m) V 0.5 V 5 Re = = =.16 V ν ( m ) 6 1.16 / where V mut be in m/. At thi point, we do not know whether the low i laminar or turbulent. Given the ize o the pipe and the head lo, it i reaonable to aume turbulent low and proceed. In the end, we need to check whether thi aumption i correct. 7
Now, we are given the head lo h. et u write the reult or h in term o the riction actor. V h = Subtitute the value o known entitie in thi equation. g 50 ( m) V 7.0 ( m) = 0.5 ( m) Thi can be rearranged to yield 9.81 ( m/ ) V m = 1.99 where V mut be in m/. Taking the quare root, we ind 0.11 = V We can ee that the product Re can be calculated, even though we do not know the velocity V. Re 5 0.11 =.16 V =.05 V Given ε =. mm, the relative roughne i ε. ( m) = = 1.8 0.5 m Thereore, the entire right ide in the Colebrook Equation or the riction actor i known. We can ue the Colebrook Equation to evaluate the riction actor in an once-through calculation. ε =.0 log + =.0 log 9.8.7 + = Re.7.05 1 / 1.6 1.8 1.6 Thereore, the riction actor i = 0.0 0.11 Uing =, we can evaluate the velocity a V 0.11 ( m/ ) 0.11 ( m/ ) V = = = 1.9 m/ o that the volumetric low rate i obtained a 0. Q = VA = 1.9 m /.91 m = 6.80 5 5 We mut check the Reynold number. Re =.16 V =.00. Thi i well over,000 o that we can conclude that the aumption o turbulent low i correct. m 8
Example etermine the ize o mooth 1-gage BWG copper tubing needed to convey gpm o a 5 proce liquid o kinematic vicoity ν =.0 t / over a ditance o 1 t at ground level uing a torage tank at an elevation o 0 t. You can aume minor loe rom itting in the line to account or 5 t o head. In thi problem, we are aked to calculate the diameter o the tube. We are given = 150 t 1 t / t and Q = gpm =.. Given that the torage tank i located at an 8.8( gpm) elevation o 0 t above ground, we can iner that the available head lo or riction in the low h = 0 5 t = 15 t. through the tube i The diameter appear in both the Reynold number and the reult or the head lo in term o the riction actor. et u begin with the head lo and write it in term o the volumetric low rate, which i known. ( Q π ) / V Q h = = = g g π g Subtituting known entitie in thi equation, we obtain 5. ( t / ) 1 t. t / 15 t = = 6.65 5 π where mut be in eet. 5 o that =.6 5 The Reynold number can be written a ( t ) Q. / 1.18 Re = = = 5 πν π.0 t / 9 where mut be in eet. We can make urther progre i we aume the type o low, o that we can ue a correlation or the riction actor. It i reaonable in proce ituation with thi low rate to aume turbulent low. So, we hall proceed with that aumption, to be veriied later when we can calculate the Reynold number. It doe not matter which correlation we ue, becaue we mut olve an implicit equation or the diameter in either cae. So, let u ue the Colebrook equation becaue it i impler. For a mooth tube, the roughne, ε = 0, o that we can et the relative roughne ε / = 0 in the Colebrook equation to obtain
1 1.6 =.0 log Re In thi equation, ubtitute or both the riction actor and the Reynold number in term o the diameter, to obtain 5.0 log 5/.0 log / 1 1.6.5 = = 5/ 7.5 ( 1.18 / ) ( 7.5 ) or = 190 log.5 5/ 5 / Solving thi equation, we obtain 7.91 0.99" = t = A table o tandard tubing dimenion or peciied nominal diameter and Birmingham Wire Gage (BWG) value can be ound in many place. The textbook by Welty et al. provide it a Appendix N. From the table, we ind that or 1-gage tubing with an outide diameter o 1", the 1 inide diameter i 0.8. The next higher outide diameter available i 1 inch, and or thi O, 1-gage tubing come with an inide diameter o 1.08. Thereore, we mut elect one o thee two tube. I we want to be ure to obtain the deired low rate, we mut chooe the value that i larger than 0.99. You may wonder why. Here i an approximate anwer. In turbulent low, the riction actor V a, where 0 a < 1. In laminar low, V 1. In b both cae, we can write V V where b > 0. Thereore, the head lo rom pipe low 1 1 1 b riction h = V V V g For a ixed volumetric low rate, a the diameter i increaed, V b decreae and 1/ alo decreae. Thereore, the head lo decreae or a given volumetric low rate a the diameter i increaed. Thi mean that with a ixed head lo available, we can comortably achieve the deired low rate uing a uitable valve. On the other hand, i we chooe a diameter that i maller than the calculated value, we would need a larger head available or driving the low than i available. Now, let u ue the actual inide diameter o the elected tube, evaluate the Reynold number o the low. = 1.08" = 9.0 t to
( t ) Q. / Re = = = 1.1 5 πν π.0 t / 9.0 t turbulent a aumed. Thereore, the low i The actual riction actor can be calculated rom the Zigrang-Sylveter equation. 1 ε / 5.0 ε / 1 =.0 log log +.7 Re.7 Re 5.0 1 =.0 log 0 log 0 + 11.8 = 1.1 1.1 yielding = 0.007 The actual head lo or the deired volumetric low rate will be ( t) ( t ) Q 1. / h = = 0.007 = 8.0 t 5 π g 5 π. t / 9.0 t which i le than available head o 15 t. Thereore, we mut peciy 1-gage, 1 1 inch tubing or thi application. 11
Roughne value or Commercial Pipe Thee roughne value are given in Table 6.1 rom a textbook by White (1). Becaue o the variation in roughne in thee material depending on the ource, the roughne value reported here have uncertaintie ranging rom ± 0 % or new wrought Iron to ± 70 % or riveted teel. A typical uncertainty in the roughne value can be aumed to be in the range ± 0 50 %. Material Condition t mm Steel Sheet metal, new 1.6 5 Stainle, new 6 7 Commercial, new 1.5.6 Riveted 1.0 Ruted 7.0 Iron Cat, new 8.5.6 1 Wrought, new 1.5.6 Galvanized, new 5 1.5 1 Aphalted, cat 1. 1 Bra rawn, new 6 7 Platic rawn tubing 6 5 1.5 Gla Smooth Smooth Concrete Smoothed 1. Rough 7.0 Rubber Smoothed 5. 1 Wood Stave 1.6 5 1 Reerence 1. F.M. White, Fluid Mechanic, 7 th Edition, McGraw-Hill, New York, 011. 1