Solution Assignment 10 TEP 4100

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1 8 9 Solion Assignment 10 TEP 4100 Solion The flow rate through a specified water pipe is given. The pressure drop, the head loss, and the pumping power requirements are to be determined. Assumptions 1 The flow is steady and incompressible. The entrance effects are negligible, and thus the flow is fully developed. The pipe involves no components such as bends, valves, and connectors. 4 The piping section involves no work devices such as pumps and turbines. Properties The density and dynamic viscosity of water are given to be = kg/m and = kg/ms, respectively. The roughness of stainless steel is 0.00 mm. Analysis First we calculate the average velocity and the Reynolds number to determine the flow regime: V V m / s V m / s A c / 4 (0.05 m) / 4 V (999.1 kg/m )(4.584 m/s)(0.05 m) Re kg/m s which is greater than Therefore, the flow is turbulent. The relative roughness of the pipe is 6 10 m 5 / m The friction factor can be determined from the Moody chart, b to avoid the reading error, we determine it from the Colebrook equation using an equation solver (or an iterative scheme), 5 1 / log.0 log f 5.7 Re f f f It gives f = Then the pressure drop, head loss, and the required power inp become P P h f P g W pump f V m 0.05 m (999.1 kg/m )(4.584 m/s) 5 1kN 1000 kg m/s 1kPa 1kN/m V g 0 m 0.05 m (4.584 m/s) 10. m (9.81 m/s ) V P (0.009 m 1kW / s)(100.4 kpa) 1kPa m kw /s kpa 100 kpa Therefore, useful power inp in the amount of kw is needed to overcome the frictional losses in the pipe. iscussion The friction factor could also be determined easily from the explicit Haaland relation. It would give f = , which is sufficiently close to Also, the friction factor corresponding to = 0 in this case is , which indicates that stainless steel pipes in this case can be assumed to be smooth with an error of abo %. Also, the power inp determined is the mechanical power that needs to be imparted to the fluid. The shaft power will be more than this due to pump inefficiency; the electrical power inp will be even more due to motor inefficiency. Water 9 /s = 0 m = 5 cm Page 1 of 6

2 8 41 Solion aminar flow through a square channel is considered. The change in the head loss is to be determined when the average velocity is doubled. Assumptions 1 The flow remains laminar at all times. The entrance effects are negligible, and thus the flow is fully developed. Analysis The friction factor for fully developed laminar flow in a square channel is 56.9 f where Re V Re V Then the head loss for laminar flow can be expressed as h 1 V 56.9 V 56.9 V, f 8. 46V g Re g V g g which shows that the head loss is proportional to the average velocity. Therefore, the head loss doubles when the average velocity is doubled. This can also be shown as h iscussion 8 h g g g,.46v 8.46(V ) 8.46V The conclusion above is also valid for laminar flow in channels of different cross sections.,1 Page of 6

3 8 4 Solion Turbulent flow through a smooth pipe is considered. The change in the head loss is to be determined when the average velocity is doubled. Assumptions 1 The flow remains turbulent at all times. The entrance effects are negligible, and thus the flow is fully developed. The inner surface of the pipe is smooth. Analysis The friction factor for the turbulent flow in smooth pipes is given as 0. f Re where V Re Then the head loss of the fluid for turbulent flow can be expressed as h V 0. V V V V f, Re g g g g which shows that the head loss is proportional to the th power of the average velocity. Therefore, the head loss increases by a factor of =.48 when the average velocity is doubled. This can also be shown as h, V g V g h,1 0. (V ) g.48h,1 0. For fully rough flow in a rough pipe, the friction factor is independent of the Reynolds number and thus the flow velocity. Therefore, the head loss increases by a factor of 4 in this case since V h, 1 f g and thus the head loss is proportional to the square of the average velocity when f,, and are constant. 0. V iscussion Most flows in practice are in the fully rough regime, and thus the head loss is generally assumed to be proportional to the square of the average velocity for all kinds of turbulent flow. Note that we use diameter here in place of hydraulic diameter h. For a square duct, it turns o that h =, so this is a valid approximation. 8 8 Solion Water is transported to a residential area through concrete pipes, and the idea of lining the interior surfaces of the pipe is being evaluated to reduce frictional losses. The percent increase or decrease in the pumping power requirements as a result of lining the concrete pipes is to be determined. Assumptions 1 The flow is steady and incompressible. The entrance effects are negligible, and thus the flow is fully developed. The pipe line involves no components such as bends, valves, and connectors, and thus no minor losses. 4 The flow is turbulent (to be verified). Properties The density and kinematic viscosity of water are given to be = 1000 kg/m and = m /s. The surface roughness is mm for concrete and 0.04 mm for the lining. Analysis CASE 1 (concrete pipe, = 0.70 m). The average velocity and the Reynolds number are: Page of 6

4 V V 1.5 m /s V.898 m/s A c / 4 (0.70 m) / 4 Water V (.898 m/s)(0.70 m) 6 = 70 cm Re m /s m / s Since Re > 4000, the flow is turbulent. The relative roughness of = 1500 m the pipe is 0.00 m / m The friction factor can be determined from the Moody chart, b to avoid the reading error, we determine it from the Colebrook equation using an equation solver (or an iterative scheme), 1 / log.0 log f 6.7 Re f f f It gives f = Then the head loss and the required useful power inp become V 1500 m (.898 m/s) h f (0.0904) m 48. m g 0.70 m (9.81 m/s ) 1kN 1kW W pump, u V gh (1.5 m /s)(1000 kg/m )(9.81m/s )(48.19 m) 709.1kW 709 kw 1000 kg m/s 1kN m/s CASE For the case of pipe with lining, = 0.66 m. Then the average velocity and the Reynolds number are: V V 1.5 m /s V 4.84 m/s A c / 4 (0.66 m) / 4 Water = 66 cm V (4.84 m/s)(0.66 m) m /s Re m / s Since Re > 4000, the flow is turbulent. The relative roughness of = 1500 m the pipe is m 5 / m The friction factor can be determined from the Moody chart, b to avoid the reading error, we determine it from the Colebrook equation using an equation solver (or an iterative scheme), 5 1 / log.0 log f 6.7 Re f f f which yields f = Then the head loss and the required useful power inp become h f V 1500 m (4.84 m/s) ( ) 6.16 m 6. m g 0.66 m (9.81 m/s ) W V gh pump, u (1.5 m /s)(1000 kg/m 1kN )(9.81m/s )(6.16 m) 1000 kg m/s 1kW 84.9 kw 85 kw 1kN m/s The required pumping power changes by (85 709)709 = 0.457; a decrease of 45.7%. iscussion Note that the pipe head losses and the required pumping power are reduced almost by almost half as result of lining the inner surface of the concrete pipes, despite the reduction in pipe diameter. This confirms the importance of having smooth surfaces in piping. Page 4 of 6

5 Oppg : a) Trykk i flaska = trykk i sykkelpumpa = F/Areal Kraft 4 100N pflaske 197kPa / 4 d (0.054m) overtrykk ( gage ). b) Bernoulli s likning langs en strømlinje fra vannoverflaten inne i flasken og gjennom dysa. Inkompressibel, stasjonær og friksjonsfri strømning, neglisjerer ev. høydeforskjell: pflaske pa v (pflaske p a ) v m / s V S : Hastighet i slangen finnes fra massebevarelse: S S S S 0.005m 1 Qv v v v v v m 4 1.9m / s v 0.01m S S Re m / s Reynoldstallet i slangen er større enn 00, altså er strømningen turbulent. c) Vi har to kontraksjoner (innsnevringer): Fra flasken til slangen: Fra slangen til dysa: S 99 KC flaske 100 KC S 4 I tapsleddene skal disse faktorene multipliseres med v / der hastigheten v er i det minste røret. Bernoulli med kontraksjonstap: pflaske pa v vs v 1 KC1 KC der vs 4 v 1 v 1 p flaske p a 1 K C1 16 K C (1) pflaske pa K K v 18.5m/s 1 C1 16 C Page 5 of 6

6 d) Friksjonstap: v f der 50cm og v v. S 1 S 4 S Vi har en ukjent hastighet v S i Reynoldstallet Re, og dermed er friksjonsfaktoren f også ukjent: v 1 flaske a p p f der log Re f 0.8 f S pflaske pa v 50 (pflaske p a ) f v 16 ( f ) f Vi løser oppgaven med en iterasjonsprosess: Start med å gjette et Reynoldstall, f.eks Re = Fra Colebrook (eller Moody-diagram) f = eretter: v f 18.07m / s eretter: v /4 S Re f eretter: v f 18.05m / s osv. Iterasjonsprosessen konvergerer svært hurtig. Uansett: friksjonstapet er minimalt i en så kort slange. Batterisyre: Større viskositet ν gir mindre Re gir større f gir mindre v ( m/s ) Bensin: Mindre viskositet gir større v ( m/s ), dog ikke større enn svaret i c). Iterasjonsprosess i Matlab: Re = 55000; % Starter med å gjette et Reynoldstall for iter=1:5 prandtl=@(f) 1/sqrt(f)-*log10(Re*sqrt(f))+0.8; f=fzero(prandtl,[1e-10 1]); V_=sqrt(480/(1+Kc1/16+Kc+f*/(16*))); Re=V_/16*/ny; fprintf('iterasjon %1.0f: f = %5.4f, V_ = %5.f, Re = %6.0f\n',... iter,f,v_,re) end Programmet gir følgende skrift: Iterasjon 1: f = 0.005, V_ = 18.07, Re = Iterasjon : f = 0.014, V_ = 18.05, Re = 451 Iterasjon : f = 0.014, V_ = 18.05, Re = 451 Iterasjon 4: f = 0.014, V_ = 18.05, Re = 451 Iterasjon 5: f = 0.014, V_ = 18.05, Re = 451 Iterasjonsprosessen konvergerer svært hurtig, og det spiller liten rolle hvilket Reynoldstall vi gjetter initiellt, 10 og går like fint. Page 6 of 6

7 OPPGAVE: Oljetank a) Friksjonsfri strømning kan benytte Bernoulli langs en strømlinje. Bernoulli fra væskeoverflaten i tanken til løpet av løpsrøret: G a 0 P P V gh Q0, der V Q0 PG Pa gh Trykket i A er lik det statiske: PA PG gh Trykket i B ved å bruke Bernoulli fra A til B: P P 1 V P B A b a b) Friksjon i røret. Minste rørlengde fås når rørlengden er lik innløpslengden: V Q0 6.7m V 4Q0 Krav: strømningen skal være laminær: m 00 c) Gasstrykket P G. Benytter Bernoulli med tap fra fri overflate til løpet. Vi har kun friksjonstap over rørlengden: G a V gh PG Pa gh 1 V Q0 P P 4Q der α = for fullt viklet laminær rørstrømning. I oppgave a) regnet vi friksjonsfritt, altså skal svaret i c) gi en større verdi for trykket P G. Sammenligning av de to rykkene viser at dette stemmer. Faktoren ½ i a)-svaret er nå blitt til [ ] Vi har to nye effekter her: α = betyr at vi har fått lagd et fullt viklet hastighetsprofil, og vi har friksjonstap over rørlengden. d) Trykket i B kan finnes som i a); Bernoulli(en tap) fra A til B: 1 PB PA V, her er 1 fordi hastigheten over tverrsnittet i B er konstant. Trykket i A er det samme som tidligere: lik det statiske trykket. PA PG gh, men nå er P G større. Videre: Bernoulli med tap fra B og : PB V Pa V V PB Pa V V V ( 1 1, )

8 p p A ½ρU : Trykkfall for å aksellerere opp oljen til U. p B ½ρU : Trykkfall for å gjøre om U til fullt vilket profil. p a A Trykktap fra friksjon over rørlengden B ca. e x