OUTCOME 3 TUTORIAL 3 - THE FLOW OF REAL FLUIDS

Unit 4: Fluid Mechanics Unit code: T/60/445 QCF Level: 4 Credit value: 5 OUTCOME 3 TUTORIAL 3 - THE FLOW OF REAL FLUIDS 3 Be able to deterine the behavioural characteristics and paraeters o real luid low Head losses: head loss in pipes by Darcy s orula; Moody diagra; head loss due to sudden enlargeent and contraction o pipe diaeter; head loss at entrance to a pipe; head loss in valves; low between reservoirs due to gravity; hydraulic gradient; siphons; haerblow in pipes Reynolds nuber: inertia and viscous resistance orces; lainar and turbulent low; critical velocity Viscous drag: dynaic pressure; or drag; skin riction drag; drag coeicient Diensional analysis: checking validity o equations such as those or pressure at depth; thrust on iersed suraces and ipact o a jet; orecasting the or o possible equations such as those or Darcy s orula and critical velocity in pipes CONTENTS. PIPE FLOW.. Conservation o Mass.. Conservation o Energy Energy Fors.3. Bernoulli s Equation.4. Hydraulic Gradient.5. Lainar and Turbulent Flow.5.. Lainar Flow.5. Turbulent Flow.5.3. Lainar and Turbulent Boundary Layers.5.4. Critical Velocity Reynolds No..6. Derivation o Poiseuille s Equation or Lainar Flow.7. Friction Coeicient.7.. Dynaic Pressure.7.. Wall Shear Stress.7.3. Friction Coeicient or Lainar Flow.8. Darcy Forula.8.. Fluid Resistance.8.. Moody Diagra and Relative Surace Roughness.9. Minor Losses.9.. Coeicient O Contraction Cc.9.. Coeicient O Velocity C v.9.3. Exit ro a Pipe into a Tank..9.4. Entry to a Pipe ro a Tank.9.5. Sudden Enlargeent.9.6. Sudden Contraction.9.7. Bends and Fittings.0. Siphons.. Moentu and Pressure Forces. WATER HAMMER.. Bulk Modulus.. Speed o Sound in an Elastic Mediu.3. Pressure Surges Due to Gradual Closure.4. Pressure Surges Due to Sudden Closure.5. The Eect o Elasticity in Pipe.6 Daping Out Pressure Surges This is a very large outcoe requiring a lot o study tie. The tutorial ay contain ore aterial than needed by those who have already studied the appropriate pre-requisite aterial. Let's start by revising basics. D.J.Dunn www.reestudy.co.uk

. PIPE FLOW The solution o pipe low probles requires the applications o two principles, the law o conservation o ass (continuity equation) and the law o conservation o energy (Bernoulli s equation). CONSERVATION OF MASS When a luid lows at a constant rate in a pipe or duct, the ass low rate ust be the sae at all points along the length. Consider a liquid being puped into a tank as shown (ig.). The ass low rate at any section is = Au = density (kg/3) u = ean velocity (/s) A = Cross Sectional Area () Fig. For the syste shown the ass low rate at (), () and (3) ust be the sae so A u = A u = 3 A 3 u 3 In the case o liquids the density is equal and cancels so. CONSERVATION OF ENERGY ENERGY FORMS A u = A u = A 3 u 3 = Q FLOW ENERGY This is the energy a luid possesses by virtue o its pressure. The orula is F.E. = pq Joules p is the pressure (Pascals) Q is volue rate (3) D.J.Dunn www.reestudy.co.uk

POTENTIAL OR GRAVITATIONAL ENERGY This is the energy a luid possesses by virtue o its altitude relative to a datu level. The orula is P.E. = gz Joules is ass (kg) z is altitude () KINETIC ENERGY This is the energy a luid possesses by virtue o its velocity. The orula is K.E. = ½ u Joules u is ean velocity (/s) INTERNAL ENERGY This is the energy a luid possesses by virtue o its teperature. It is usually expressed relative to 0 o C. The orula is U = c c is the speciic heat capacity (J/kg o C) is the teperature in o C In the ollowing work, internal energy is not considered in the energy balance. SPECIFIC ENERGY Speciic energy is the energy per kg so the three energy ors as speciic energy are as ollows. F.E./ = pq/ = p/ Joules/kg P.E/. = gz Joules/kg K.E./ = ½ u Joules/kg ENERGY HEAD I the energy ters are divided by the weight g, the result is energy per Newton. Exaining the units closely we have J/N = N /N = etres. It is noral to reer to the energy in this or as the energy head. The three energy ters expressed this way are as ollows. F.E./g = p/g = h P.E./g = z K.E./g = u /g The low energy ter is called the pressure head and this ollows since earlier it was shown p/g = h. This is the height that the liquid would rise to in a vertical pipe connected to the syste. The potential energy ter is the actual altitude relative to a datu. The ter u /g is called the kinetic head and this is the pressure head that would result i the velocity is converted into pressure. D.J.Dunn www.reestudy.co.uk 3

.3 BERNOULLI S EQUATION Bernoulli s equation is based on the conservation o energy. I no energy is added to the syste as work or heat then the total energy o the luid is conserved. Reeber that internal (theral energy) has not been included. The total energy E T at () and () on the diagra (ig.) ust be equal so : u u ET pq gz pq gz Dividing by ass gives the speciic energy or E T p u p u gz gz Dividing by g gives the energy ters per unit weight E T p u p u z z g g g g g Since p/g = pressure head h then the total head is given by the ollowing. u u h T h z h z g g This is the head or o the equation in which each ter is an energy head in etres. z is the potential or gravitational head and u /g is the kinetic or velocity head. For liquids the density is the sae at both points so ultiplying by g gives the pressure or. The total pressure is as ollows. u u pt p gz p gz In real systes there is riction in the pipe and elsewhere. This produces heat that is absorbed by the liquid causing a rise in the internal energy and hence the teperature. In act the teperature rise will be very sall except in extree cases because it takes a lot o energy to raise the teperature. I the pipe is long, the energy ight be lost as heat transer to the surroundings. Since the equations did not include internal energy, the balance is lost and we need to add an extra ter to the right side o the equation to aintain the balance. This ter is either the head lost to riction h L or the pressure loss p L. u u h z h z h L g g The pressure or o the equation is as ollows. u u p gz p gz pl The total energy o the luid (excluding internal energy) is no longer constant. Note that i one o the points is a ree surace the pressure is norally atospheric but i gauge pressures are used, the pressure and pressure head becoes zero. Also, i the surace area is large (say a large tank), the velocity o the surace is sall and when squared becoes negligible so the kinetic energy ter is neglected (ade zero). D.J.Dunn www.reestudy.co.uk 4

WORKED EXAMPLE No. The diagra shows a pup delivering water through as pipe 30 bore to a tank. Find the pressure at point () when the low rate is.4 d 3 /s. The density o water is 000 kg/ 3. The loss o pressure due to riction is 50 kpa. SOLUTION Fig. Area o bore A = x 0.03 /4 = 706.8 x 0-6. Flow rate Q =.4 d 3 /s = 0.004 3 /s Mean velocity in pipe = Q/A =.98 /s Apply Bernoulli between point () and the surace o the tank. u u p gz p gz p L Make the low level the datu level and z = 0 and z = 5. The pressure on the surace is zero gauge pressure. P L = 50 000 Pa The velocity at () is.98 /s and at the surace it is zero. 000x.98 p 0 0 000x9.95 0 50000 p 93.9kPa gauge pressure WORKED EXAMPLE The diagra shows a tank that is drained by a horizontal pipe. Calculate the pressure head at point () when the valve is partly closed so that the low rate is reduced to 0 d 3 /s. The pressure loss is equal to head. Fig.3 D.J.Dunn www.reestudy.co.uk 5

SOLUTION Since point () is a ree surace, h = 0 and u is assued negligible. The datu level is point () so z = 5 and z = 0. Q = 0.0 3/s A = d /4 = x (0.05 )/4 =.963 x 0-3. u = Q/A = 0.0/.963 x 0-3 = 0.8 /s Bernoulli s equation in head or is as ollows. u u h z h z h L g g 0 5 0 h h 7.7 0.8 0 x 9.8 WORKED EXAMPLE 3 The diagra shows a horizontal nozzle discharging into the atosphere. The inlet has a bore area o 600 and the exit has a bore area o 00. Calculate the low rate when the inlet pressure is 400 Pa. Assue there is no energy loss. SOLUTION Fig. 4 Apply Bernoulli between () and () ρu ρu p ρgz p ρgz pl Using gauge pressure, p = 0 and being horizontal the potential ters cancel. The loss ter is zero so the equation sipliies to the ollowing. ρu ρu p Fro the continuity equation we have Q Q u 666.7Q -6 A 600 x 0 Q Q u 5 000 Q -6 A 00 x 0 Putting this into Bernoulli s equation we have the ollowing. 666.7Q 5000Q 400 000 x 000 x 400.389 x0 400. x0 Q 400. x0 Q 89.7 x 0-6 9 9 9 Q Q 36 x0 3.5 x 0 /s 9 9 Q or 89.7 c 3 /s D.J.Dunn www.reestudy.co.uk 6

.4 HYDRAULIC GRADIENT Consider a tank draining into another tank at a lower level as shown. There are sall vertical tubes at points along the length to indicate the pressure head (h). Relative to a datu, the total energy head is h T = h + z + u /g This is shown as line A. Fig.5 The hydraulic grade line is the line joining the ree suraces in the tubes and represents the su o h and z only. This is shown as line B and it is always below the line o h T by the velocity head u /g. Note that at exit ro the pipe, the velocity head is not recovered but lost as riction as the eerging jet collides with the static liquid. The ree surace o the tank does not rise. The only reason why the hydraulic grade line is not horizontal is because there is a rictional loss h. The actual gradient o the line at any point is the rate o change with length i = h /L SELF ASSESSMENT EXERCISE. A pipe 00 bore diaeter carries oil o density 900 kg/3 at a rate o 4 kg/s. The pipe reduces to 60 bore diaeter and rises 0 in altitude. The pressure at this point is atospheric (zero gauge). Assuing no rictional losses, deterine: i. The volue/s (4.44 d3/s) ii. The velocity at each section (0.566 /s and.57 /s) iii. The pressure at the lower end. (.06 MPa). A pipe 0 bore diaeter carries water with a head o 3. The pipe descends in altitude and reduces to 80 bore diaeter. The pressure head at this point is 3. The density is 000 kg/3. Assuing no losses, deterine i. The velocity in the sall pipe (7 /s) ii. The volue low rate. (35 d3/s) 3. A horizontal nozzle reduces ro 00 bore diaeter at inlet to 50 at exit. It carries liquid o density 000 kg/3 at a rate o 0.05 3/s. The pressure at the wide end is 500 kpa (gauge). Calculate the pressure at the narrow end neglecting riction. (96 kpa) 4. A pipe carries oil o density 800 kg/3. At a given point () the pipe has a bore area o 0.005 and the oil lows with a ean velocity o 4 /s with a gauge pressure o 800 kpa. Point () is urther along the pipe and there the bore area is 0.00 and the level is 50 above point (). Calculate the pressure at this point (). Neglect riction. (374 kpa) 5. A horizontal nozzle has an inlet velocity u and an outlet velocity u and discharges into the atosphere. Show that the velocity at exit is given by the ollowing orulae. u ={p/ + u } ½ and u ={gh + u } ½ D.J.Dunn www.reestudy.co.uk 7

.5 LAMINAR and TURBULENT FLOW The ollowing work only applies to Newtonian luids..5. LAMINAR FLOW A strea line is an iaginary line with no low noral to it, only along it. When the low is lainar, the strealines are parallel and or low between two parallel suraces we ay consider the low as ade up o parallel lainar layers. In a pipe these lainar layers are cylindrical and ay be called strea tubes. In lainar low, no ixing occurs between adjacent layers and it occurs at low average velocities..5. TURBULENT FLOW The shearing process causes energy loss and heating o the luid. This increases with ean velocity. When a certain critical velocity is exceeded, the strealines break up and ixing o the luid occurs. The diagra illustrates Reynolds coloured ribbon experient. Coloured dye is injected into a horizontal low. When the low is lainar the dye passes along without ixing with the water. When the speed o the low is increased turbulence sets in and the dye ixes with the surrounding water. One explanation o this transition is that it is necessary to change the pressure loss into other ors o energy such as angular kinetic energy as indicated by sall eddies in the low. Fig.6.5.3 LAMINAR AND TURBULENT BOUNDARY LAYERS A boundary layer is the layer in which the velocity grows ro zero at the wall (no slip surace) to 99% o the axiu and the thickness o the layer is denoted. When the low within the boundary layer becoes turbulent, the shape o the boundary layers waivers and when diagras are drawn o turbulent boundary layers, the ean shape is usually shown. Coparing a lainar and turbulent boundary layer reveals that the turbulent layer is thinner than the lainar layer. Fig.7 D.J.Dunn www.reestudy.co.uk 8

.5.4 CRITICAL VELOCITY - REYNOLDS NUMBER When a luid lows in a pipe at a voluetric low rate Q 3/s the average velocity is deined Q u A is the cross sectional area. A u D u D The Reynolds nuber is deined as R e I you check the units o Re you will see that there are none and that it is a diensionless nuber. You will learn ore about such nubers in a later tutorial. Reynolds discovered that it was possible to predict the velocity or low rate at which the transition ro lainar to turbulent low occurred or any Newtonian luid in any pipe. He also discovered that the critical velocity at which it changed back again was dierent. He ound that when the low was gradually increased, the change ro lainar to turbulent always occurred at a Reynolds nuber o 500 and when the low was gradually reduced it changed back again at a Reynolds nuber o 000. Norally, 000 is taken as the critical value. WORKED EXAMPLE 4 Oil o density 860 kg/ 3 has a kineatic viscosity o 40 cst. Calculate the critical velocity when it lows in a pipe 50 bore diaeter. SOLUTION R u e u D ν R eν D 000x40x0 0.05 6.6 /s D.J.Dunn www.reestudy.co.uk 9

.6 DERIVATION OF POISEUILLE'S EQUATION or LAMINAR FLOW Poiseuille did the original derivation shown below which relates pressure loss in a pipe to the velocity and viscosity or LAMINAR FLOW. His equation is the basis or easureent o viscosity hence his nae has been used or the unit o viscosity. Consider a pipe with lainar low in it. Consider a strea tube o length L at radius r and thickness dr. y is the distance ro the pipe wall. y R r Fig.8 dy dr du dy du dr The shear stress on the outside o the strea tube is. The orce (F s ) acting ro right to let is due to the shear stress and is ound by ultiplying by the surace area. Fs = x r L du du For a Newtonian luid,. Substituting or we get the ollowing. dy dr du F s - rl dr The pressure dierence between the let end and the right end o the section is p. The orce due to this (F p ) is p x circular area o radius r. F p = p x r du Equating orces we have - π r μ ΔL Δp π r dr Δp du rdr μ ΔL In order to obtain the velocity o the strealine at any radius r we ust integrate between the liits u = 0 when r = R and u = u when r = r. u 0 du u u Δp - μ ΔL Δp 4 μ ΔL Δp 4 μ L R r r R r rdr R D.J.Dunn www.reestudy.co.uk 0

This is the equation o a Parabola so i the equation is plotted to show the boundary layer, it is seen to extend ro zero at the edge to a axiu at the iddle. Fig.9 Δp R For axiu velocity put r = 0 and we get u 4 μ ΔL The average height o a parabola is hal the axiu value so the average velocity is Δp R u 8 μ ΔL Oten we wish to calculate the pressure drop in ters o diaeter D. Substitute R=D/ and rearrange. 3 μ ΔL u Δp D The volue low rate is average velocity x cross sectional area. π R Δp R Q 8 μ ΔL This is oten changed to give the pressure drop as a riction head. 4 π R Δp π D Δp 8 μ ΔL 8 μ ΔL 4 3 μ L u The riction head or a length L is ound ro h =p/g h ρ gd This is Poiseuille's equation that applies only to lainar low. D.J.Dunn www.reestudy.co.uk

WORKED EXAMPLE 5 A capillary tube is 30 long and bore. The head required to produce a low rate o 8 3/s is 30. The luid density is 800 kg/3. Calculate the dynaic and kineatic viscosity o the oil. SOLUTION Rearranging Poiseuille's equation we get h gd 3Lu d x A 0.785 4 4 Q 8 u 0.8 /s A 0.785 0.03 x 800 x 9.8x 0.00 0.04 N s/ or 4.cP 3 x 0.03 x 0.008 0.04-6 30. x 0 / s or 30.cSt 800 WORKED EXAMPLE No.6 Oil lows in a pipe 00 bore with a Reynolds nuber o 50. The dynaic viscosity is 0.08 Ns/. The density is 900 kg/3. Deterine the pressure drop per etre length, the average velocity and the radius at which it occurs. SOLUTION Re=u D/. Hence u = Re / D u = (50 x 0.08)/(900 x 0.) = 0.05 /s p = 3µL u /D p = 3 x 0.08 x x 0.05/0. p=.88 Pascals. u = {p/4lµ}(r - r) which is ade equal to the average velocity 0.05 /s 0.05 = (.88/4 x x 0.08)(0.05 - r) r = 0.035 or 35.3. D.J.Dunn www.reestudy.co.uk

SELF ASSESSMENT EXERCISE. Oil lows in a pipe 80 bore diaeter with a ean velocity o 0.4 /s. The density is 890 kg/3 and the viscosity is 0.075 Ns/. Show that the low is lainar and hence deduce the pressure loss per etre length. (50 Pa) Calculate the axiu velocity o water that can low in lainar or in a pipe 0 long and 60 bore. Deterine the pressure loss in this condition. The density is 000 kg/3 and the dynaic viscosity is 0.00 N s/. (0.0333 /s and 5.9 Pa) 3. Oil low in a pipe 00 bore diaeter with a Reynolds Nuber o 500. The density is 800 kg/3. The dynaic viscosity µ = 0.08 Ns/. Calculate the velocity o a strealine at a radius o 40. (0.36 /s) 4. a. When a viscous luid is subjected to an applied pressure it lows through a narrow horizontal passage as shown below. By considering the orces acting on the luid eleent and assuing steady ully developed lainar low, show that the velocity distribution is given by dp d u μ dx dy b. Using the above equation show that or low between two lat parallel horizontal suraces distance t apart the velocity at any point is given by the ollowing orula. u = (/)(dp/dx)(y - ty) c. Carry on the derivation and show that the volue low rate through a gap o height t and 3 dp t width B is given by Q B. dx dp d. Show that the ean velocity u through the gap is given by u t dx 5 The voluetric low rate o glycerine between two lat parallel horizontal suraces apart and 0 c wide is c3/s. Deterine the ollowing. i. the applied pressure gradient dp/dx. (40 kpa per etre) ii. the axiu velocity. (0.06 /s) For glycerine assue that =.0 Ns/ and the density is 60 kg/3. Fig.0 D.J.Dunn www.reestudy.co.uk 3

.7 FRICTION COEFFICIENT The riction coeicient is a convenient idea that can be used to calculate the pressure drop in a pipe. It is deined as ollows. Wall Shear Stress C Dynaic Pressure.7. DYNAMIC PRESSURE Consider a luid lowing with ean velocity u. I the kinetic energy o the luid is converted into low or luid energy, the pressure would increase. The pressure rise due to this conversion is called the dynaic pressure. KE = ½ u Flow Energy = p Q Q is the volue low rate and = /Q Equating ½ u = p Q p = u /Q = ½ u.7. WALL SHEAR STRESS o The wall shear stress is the shear stress in the layer o luid next to the wall o the pipe. Fig. du The shear stress in the layer next to the wall is τo μ dy The shear orce resisting low is F τ πld s The resulting pressure drop produces a orce o Equating orces gives D Δp τ o 4L o F p ΔpπD 4.7.3 FRICTION COEFFICIENT or LAMINAR FLOW wall C Wall Shear Stress Dynaic Pressure Dp 4Lu 3Lu Fro Poiseuille s equation p Hence D C D 4Lu 3Lu 6 D u D 6 R e D.J.Dunn www.reestudy.co.uk 4

.8 DARCY FORMULA This orula is ainly used or calculating the pressure loss in a pipe due to turbulent low but it can be used or lainar low also. Turbulent low in pipes occurs when the Reynolds Nuber exceeds 500 but this is not a clear point so 3000 is used to be sure. In order to calculate the rictional losses we use the concept o riction coeicient sybol C. This was deined as ollows. C Wall Shear Stress Dynaic Pressure Dp 4Lu Rearranging equation to ake p the subject 4C Lu p D This is oten expressed as a riction head h p 4C Lu h g gd This is the Darcy orula. In the case o lainar low, Darcy's and Poiseuille's equations ust give the sae result so equating the gives 4C Lu 3Lu gd gd C 6 u D 6 R This is the sae result as beore or lainar low..8. FLUID RESISTANCE e The above equations ay be expressed in ters o low rate Q by substituting u = Q/A h 4C Lu gd 4C LQ Substituting A =D /4 we get the ollowing. gda h 3C RQ R is the luid resistance or restriction. R 5 5 g LQ D 3 C L g π D I we want pressure loss instead o head loss the equations are as ollows. p 3C gh RQ R is the luid resistance or restriction. R 5 5 D LQ 3 ρ C L π D It should be noted that R contains the riction coeicient and this is a variable with velocity and surace roughness so R should be used with care. D.J.Dunn www.reestudy.co.uk 5

.8. MOODY DIAGRAM AND RELATIVE SURFACE ROUGHNESS In general the riction head is soe unction o u such that h = u n. Clearly or lainar low, n = but or turbulent low n is between and and its precise value depends upon the roughness o the pipe surace. Surace roughness prootes turbulence and the eect is shown in the ollowing work. Relative surace roughness is deined as = k/d where k is the ean surace roughness and D the bore diaeter. An Aerican Engineer called Moody conducted exhaustive experients and cae up with the Moody Chart. The chart is a plot o C vertically against R e horizontally or various values o. In order to use this chart you ust know two o the three co-ordinates in order to pick out the point on the chart and hence pick out the unknown third co-ordinate. For sooth pipes, (the botto curve on the diagra), various orulae have been derived such as those by Blasius and Lee. BLASIUS C = 0.079 R e 0.5 LEE C = 0.008 + 0.5 R e 0.35. The Moody diagra shows that the riction coeicient reduces with Reynolds nuber but at a certain point, it becoes constant. When this point is reached, the low is said to be ully developed turbulent low. This point occurs at lower Reynolds nubers or rough pipes. A orula that gives an approxiate answer or any surace roughness is that given by Haaland. 6.9 3.6 log 0 C R e 3. 7. Fig. CHART D.J.Dunn www.reestudy.co.uk 6

WORKED EXAMPLE 7 Deterine the riction coeicient or a pipe 00 bore with a ean surace roughness o 0.06 when a luid lows through it with a Reynolds nuber o 0 000. SOLUTION The ean surace roughness = k/d = 0.06/00 = 0.0006 Locate the line or = k/d = 0.0006. Trace the line until it eets the vertical line at Re = 0 000. Read o the value o C horizontally on the let. Answer C = 0.0067 Check using the orula ro Haaland. C C C C C 3.6 log 3.6 log 3.6 log.06 0.0067 0 0 0 6.9 R e 3.7 6.9 0000 6.9 0000. 0.0006 3.7 0.0006 3.7.. WORKED EXAMPLE 8 Oil lows in a pipe 80 bore with a ean velocity o 4 /s. The ean surace roughness is 0.0 and the length is 60. The dynaic viscosity is 0.005 N s/ and the density is 900 kg/ 3. Deterine the pressure loss. SOLUTION Re = ud/ = (900 x 4 x 0.08)/0.005 = 57600 = k/d = 0.0/80 = 0.0005 Fro the chart C = 0.005 h = 4C Lu/dg = (4 x 0.005 x 60 x 4 )/( x 9.8 x 0.08) =.7 p = gh = 900 x 9.8 x.7 =.3 kpa. D.J.Dunn www.reestudy.co.uk 7

SELF ASSESSMENT EXERCISE 3. A pipe is 5 k long and 80 bore diaeter. The ean surace roughness is 0.03. It caries oil o density 85 kg/3 at a rate o 0 kg/s. The dynaic viscosity is 0.05 N s/. Deterine the riction coeicient using the Moody Chart and calculate the riction head. (Ans. 3075.). Water lows in a pipe at 0.05 3/s. The pipe is 50 bore diaeter. The pressure drop is 3 40 Pa per etre length. The density is 000 kg/3 and the dynaic viscosity is 0.00 N s/. Deterine the ollowing. i. The wall shear stress (67.75 Pa) ii. The dynaic pressures (980 Pa). iii. The riction coeicient (0.00575) iv. The ean surace roughness (0.0875 ) 3. Explain briely what is eant by ully developed lainar low. The velocity u at any radius r in ully developed lainar low through a straight horizontal pipe o internal radius ro is given by u = (/4µ)(ro - r)dp/dx dp/dx is the pressure gradient in the direction o low and µ is the dynaic wall skin riction coeicient is deined as C = o /( u ). viscosity. The Show that C = 6/Re where Re = ud/µ an is the density, u is the ean velocity and o is the wall shear stress. 4. Oil with viscosity x 0- Ns/ and density 850 kg/3 is puped along a straight horizontal pipe with a low rate o 5 d3/s. The static pressure dierence between two tapping points 0 apart is 80 N/. Assuing lainar low deterine the ollowing. i. The pipe diaeter. ii. The Reynolds nuber. Coent on the validity o the assuption that the low is lainar. D.J.Dunn www.reestudy.co.uk 8

.9 MINOR LOSSES Minor losses occur in the ollowing circustances. i. Exit ro a pipe into a tank. ii. Entry to a pipe ro a tank. iii. Sudden enlargeent in a pipe. iv. Sudden contraction in a pipe. v. Bends in a pipe. vi. Any other source o restriction such as pipe ittings and valves. Fig.3 In general, inor losses are neglected when the pipe riction is large in coparison but or short pipe systes with bends, ittings and changes in section, the inor losses are the doinant actor. In general, the inor losses are expressed as a raction o the kinetic head or dynaic pressure in the saller pipe. Minor head loss = k u /g Minor pressure loss = ½ ku Values o k can be derived or standard cases but or ites like elbows and valves in a pipeline, it is deterined by experiental ethods. Minor losses can also be expressed in ters o luid resistance R as ollows. h L u Q 8Q k k k RQ Hence R 4 4 A D D 8k p L 8gQ 8kg k RQ hence R 4 4 D D Beore you go on to look at the derivations, you ust irst learn about the coeicients o contraction and velocity. D.J.Dunn www.reestudy.co.uk 9

.9. COEFFICIENT OF CONTRACTION Cc The luid approaches the entrance ro all directions and the radial velocity causes the jet to contract just inside the pipe. The jet then spreads out to ill the pipe. The point where the jet is sallest is called the VENA CONTRACTA. Fig.4 The coeicient o contraction C c is deined as C c = Aj/Ao Aj is the cross sectional area o the jet and Ao is the c.s.a. o the pipe. For a round pipe this becoes C c = dj /do..9. COEFFICIENT OF VELOCITY C v The coeicient o velocity is deined as C v = actual velocity/theoretical velocity In this instance it reers to the velocity at the vena-contracta but as you will see later on, it applies to other situations also..9.3 EXIT FROM A PIPE INTO A TANK. The liquid eerges ro the pipe and collides with stationary liquid causing it to swirl about beore inally coing to rest. All the kinetic energy is dissipated by riction. It ollows that all the kinetic head is lost so k =.0 Fig.5 D.J.Dunn www.reestudy.co.uk 0

.9.4 ENTRY TO A PIPE FROM A TANK The value o k varies ro 0.78 to 0.04 depending on the shape o the inlet. A good rounded inlet has a low value but the case shown is the worst..9.5 SUDDEN ENLARGEMENT Fig.6 This is siilar to a pipe discharging into a tank but this tie it does not collide with static luid but with slower oving luid in the large pipe. The resulting loss coeicient is given by the ollowing expression. d k d.9.6 SUDDEN CONTRACTION Fig.7 This is siilar to the entry to a pipe ro a tank. The best case gives k = 0 and the worse case is or a sharp corner which gives k = 0.5..9.7 BENDS AND FITTINGS Fig.8 The k value or bends depends upon the radius o the bend and the diaeter o the pipe. The k value or bends and the other cases is on various data sheets. For ittings, the anuacturer usually gives the k value. Oten instead o a k value, the loss is expressed as an equivalent length o straight pipe that is to be added to L in the Darcy orula. D.J.Dunn www.reestudy.co.uk

WORKED EXAMPLE 9 A tank o water epties by gravity through a horizontal pipe into another tank. There is a sudden enlargeent in the pipe as shown. At a certain tie, the dierence in levels is 3. Each pipe is long and has a riction coeicient C = 0.005. The inlet loss constant is K = 0.3. Calculate the volue low rate at this point. SOLUTION Fig.9 There are ive dierent sources o pressure loss in the syste and these ay be expressed in ters o the luid resistance as ollows. The head loss is ade up o ive dierent parts. It is usual to express each as a raction o the kinetic head as ollows. 3C L 3 x 0.005 x 6 5 Resistance pipe A R.038 x0 s 5 5 gd g x 0.0 Resistance in pipe B Loss at entry K=0.3 R R A 3C L 3 x 0.005 x 3 5 4.50x0 s 5 5 gd B g x 0.06 8K 8 x 0.3 5 3 58 s 4 4 g D g π x 0.0 A d A 0 Loss at sudden enlargeent. k 0. 79 d B 60 8K 8x0.79 R 4 407.7 s 4 4 gπ D gπ x 0.0 A 8K 8x Loss at exit K= R 5 6370 s 4 4 gπ D gπ x 0.06 Total losses. h h L L R Q (R B R R Q R R 3 R 4 Q R R 5 4 )Q Q R 5 5 5 Q.0x 0 6 Q D.J.Dunn www.reestudy.co.uk

BERNOULLI S EQUATION Apply Bernoulli between the ree suraces () and () u u h z h z h L g g On the ree surace the velocities are sall and about equal and the pressures are both atospheric so the equation reduces to the ollowing. z - z = h L = 3 3 =.0 x 0 6 Q Q =.74 x 0-6 Q =.65 x 0-3 3 /s D.J.Dunn www.reestudy.co.uk 3

.0 SIPHONS Liquid will siphon ro a tank to a lower level even i the pipe connecting the rises above the level o both tanks as shown in the diagra. Calculation will reveal that the pressure at point () is lower than atospheric pressure (a vacuu) and there is a liit to this pressure when the liquid starts to turn into vapour. For water about 8 etres is the practical liit that it can be sucked (8 water head o vacuu). Fig.0 WORKED EXAMPLE 0 A tank o water epties by gravity through a siphon. The dierence in levels is 3 and the highest point o the siphon is above the top surace level and the length o pipe ro inlet to the highest point is.5. The pipe has a bore o 5 and length 6. The riction coeicient or the pipe is 0.007.The inlet loss coeicient K is 0.7. Calculate the volue low rate and the pressure at the highest point in the pipe. SOLUTION There are three dierent sources o pressure loss in the syste and these ay be expressed in ters o the luid resistance as ollows. Pipe Resistance Entry Loss Resistance Exit loss Resistance R R R 3C L 3 x 0.007 x 6 gd π g x 0.05 π 8K 8 x 0.7 3 5.x 0 s 4 4 g D gπ x 0.05 8K 8x 3.57 x 0 s 4 4 g D g x 0.05 6 5.4 x 0 s 5 5 5 5 3 Total Resistance R T = R + R + R 3 =.458 x 0 6 s -5 Apply Bernoulli between the ree suraces () and (3) Flow rate u u3 h z h3 z3 h L 0 z 0 0 z3 0 h L z z3 h L 3 g g Q z z 3.458 x 0.434x0 3 3 6 R T 3 / s Bore Area A=D /4 = x 0.05 /4 = 490.87 x 0-6 Velocity in Pipe u = Q/A =.434 x 0-3 /490.87 x 0-6 =.9 /s D.J.Dunn www.reestudy.co.uk 4

Apply Bernoulli between the ree suraces () and () h h z u h g h L z.9 g u h g h L L h 0 0 0 h 0.435 h.9 g L h L.435 h L Calculate the losses between () and () Pipe riction Resistance is proportionally saller by the length ratio. R = (.5/6) x.4 x 0 6 = 0.593 x 0 6 Entry Resistance R = 5. x 0 3 as beore Total resistance R T = 608. x 0 3 Head loss h L = R T Q =.45 The pressure head at point () is hence h = -.435 -.45 = -3.68 head D.J.Dunn www.reestudy.co.uk 5

. MOMENTUM and PRESSURE FORCES Changes in velocities ean changes in oentu and Newton's second law tells us that this is accopanied by a orce such that Force = rate o change o oentu. Pressure changes in the luid ust also be considered as these also produce a orce. Translated into a or that helps us solve the orce produced on devices such as those considered here, we use the equation F = (pa) + u. When dealing with devices that produce a change in direction, such as pipe bends, this has to be considered ore careully and this is covered in a later tutorial. In the case o sudden changes in section, we ay apply the orula F = (p A + u )- (p A + u ) point is upstrea and point is downstrea. WORKED EXAMPLE A pipe carrying water experiences a sudden reduction in area as shown. The area at point () is 0.00 and at point () it is 0.00. The pressure at point () is 500 kpa and the velocity is 8 /s. The loss coeicient K is 0.4. The density o water is 000 kg/ 3. Calculate the ollowing. i. The ass low rate. ii. The pressure at point () iii. The orce acting on the section. SOLUTION Fig. u = u A /A = (8 x 0.00)/0.00 = 4 /s = A u = 000 x 0.00 x 4 = 8 kg/s. Q = A u = 0.00 x 4 = 0.008 3 /s Pressure loss at contraction = ½ ku = ½ x 000 x 0.4 x 4 = 300 Pa Apply Bernoulli between () and () u u p p p L 000 x 4 3 000 x 8 p 500 x 0 300 p 57. kpa F = (p A + u )- (p A + u ) F = [(57. x 0 3 x 0.00) + (8 x 4)] [500 x 0 3 x 0.00) + (8 x 8)] F = 054.4 +3 500 64 F = 5.4 N D.J.Dunn www.reestudy.co.uk 6

SELF ASSESSMENT EXERCISE 4. A pipe carries oil at a ean velocity o 6 /s. The pipe is 5 k long and.5 diaeter. The surace roughness is 0.8. The density is 890 kg/3 and the dynaic viscosity is 0.04 N s/. Deterine the riction coeicient ro the Moody chart and go on to calculate the riction head h. (Ans. C = 0.0045 h = 0. ). The diagra shows a tank draining into another lower tank through a pipe. Note the velocity and pressure is both zero on the surace on a large tank. Calculate the low rate using the data given on the diagra. (Ans. 7.6 d 3 /s) Fig. 3. Water lows through the sudden pipe expansion shown below at a low rate o 3 d3/s. Upstrea o the expansion the pipe diaeter is 5 and downstrea the diaeter is 40. There are pressure tappings at section (), about hal a diaeter upstrea, and at section (), about 5 diaeters downstrea. At section () the gauge pressure is 0.3 bar. Evaluate the ollowing. (i) The gauge pressure at section () (0.387 bar) (ii) The total orce exerted by the luid on the expansion. (-3 N) Fig.3 4. A doestic water supply consists o a large tank with a loss ree-inlet to a 0 diaeter pipe o length 0, that contains 9 right angles bends. The pipe discharges to atosphere 8.0 below the ree surace level o the water in the tank. Evaluate the low rate o water assuing that there is a loss o 0.75 velocity heads in each bend and that riction in the pipe is given by the Blasius equation C=0.079(Re)-0.5 (0.8 d3/s). The dynaic viscosity is 0.89 x 0-3 and the density is 997 kg/ 3. 5. A tank o water epties by gravity through a siphon into a lower tank. The dierence in levels is 6 and the highest point o the siphon is above the top surace level. The length o pipe ro the inlet to the highest point is 3. The pipe has a bore o 30 and length. The riction coeicient or the pipe is 0.006.The inlet loss coeicient K is 0.6. Calculate the volue low rate and the pressure at the highest point in the pipe. (Answers.378 d 3 /s and 4.3 ) D.J.Dunn www.reestudy.co.uk 7

. WATER HAMMER In this section, we will exaine the causes o water haer. The sudden acceleration or deceleration o luids in pipes is accopanied by corresponding changes in pressure that can be extreely large. In the extree, the pressure surge can split the pipe. The phenoenon is oten accopanied by load haer noises, hence the nae. First, we ust exaine the Bulk Modulus (K) and the derivation o the acoustic velocity in an elastic luid.. BULK MODULUS ( K) Bulk odulus is deined as ollows. Change in pressure K Voluetric strain Vp V Vp V V is volue and p is pressure. The ollowing work shows how this ay be changed to the or K = dp/d Fig.4 Consider a volue V that is copressed to volue V by a sall increase in pressure p. The reduction in volue is V. The initial density is and this increases by The ass o V is = V...() The initial ass o V is = V...() The inal ass o V is = ( + ) V...(3) The increase in ass is due to the ass o V being copressed into the volue V. Hence () = (3) - () V = ( + ) V - V = V + V - V V = V = (V - V) = V - V The product o two sall quantities ( V) is ininitesially sall so it ay be ignored. ρδv V δρ V δv K ρ δρ V δp δv substitutethis into the orula or K ρδp δρ δv V δρ ρ In the liit as V 0, we ay revert to calculus notation. Hence K = dp/d D.J.Dunn www.reestudy.co.uk 8

. SPEED OF SOUND IN AN ELASTIC MEDIUM Most students don t need to know the derivation o the orula or the speed o sound but or those who are interested, here it is. Consider a pipe o cross sectional area A ull o luid. Suppose a piston is pushed into the end with a velocity u /s. Due to the copressibility o the luid, urther along the pipe at distance L, the luid is still stationary. It has taken t seconds to achieve this position. The velocity o the interace is hence a = L/t /s. In the sae tie the piston has oved x etres so u = x/t. Fig. 5 The oving luid has been accelerated ro rest to velocity a. The inertia orce needed to do this is in the or o pressure so the oving luid is at a higher pressure than the static luid and the interace is hence a pressure wave travelling along the pipe at velocity a. The volue Ax has been copacted into the length L. The initial density o the luid is. The ass copacted into length L is d = Ax. substitute x = ut d = A u t...(4) The density o the copacted luid has increased by d so the ass in the length L has increased by d = A L d Substitute L = at Equate (4) and (5) d = A a t d...(5) A u t= A a t d u a....(6) d The orce to accelerate the luid ro rest to a /s is given by Newton's nd law F = ass x acceleration = A dp ass = AL acceleration = u/t A dp = A L u/t dp = L u/t dp Substitute L = at then dp = a u a...(7) u The velocity o the pressure wave a is by deinition the acoustic velocity. Multiplying (6) by (7) gives a. Hence a u dp d u / dp a...(8) d Previously it was shown that dp K d a K D.J.Dunn www.reestudy.co.uk 9

Students who have studied undaental therodynaics will understand the ollowing extension o the theory to gases. The ollowing section is not needed by those ollowing the basic odule. Two iportant gas constants are the adiabatic index and the characteristic gas constant R. For a gas, the pressure change is adiabatic and i dp is sall then the adiabatic law applies. pv = Constant Dividing through by we get V p p C dp p C d Dierentiating we get Fro (8) it ollows that Fro the gas law we have p a pv = RT p = RT/V = ρrt p a RT RT The velocity o a sound wave is that o a weak pressure wave. I the pressure change is large then dp/d is not a constant and the velocity would be that o a shock wave which is larger than the acoustic velocity. For air =.4 and R = 87 J/kg K, hence at 0 o C (93 K) the acoustic velocity in air is ound as ollows. a RT = (.4 x 87 x 93) ½ = 343 /s.3 PRESSURE SURGES DUE TO GRADUAL VALVE CLOSURE Consider a pipe line with a luid lowing at a steady velocity o u /s. A stop valve is gradually closed thus decelerating the luid uniorly ro u to zero in t seconds. Fig.6 Volue o luid = AL Mass o luid = AL Deceleration = u/t Inertia orce required F = ass x deceleration = AL u/t To provide this orce the pressure o the luid rises by p and the orce is A p. Equating orces we have A p = AL u/t p = L u/t D.J.Dunn www.reestudy.co.uk 30

.4 PRESSURE SURGES DUE TO SUDDEN VALVE CLOSURE I the valve is closed suddenly then as t is very sall the pressure rise is very large. In reality, a valve cannot close instantly but very rapid closure produces very large pressures. When this occurs, the copressibility o the luid and the elasticity o the pipe is an iportant actor in reducing the rise in pressure. First, we will consider the pipe as rigid. When the luid stops suddenly at the valve, the luid urther up the pipe is still oving and copacting into the static luid. An interace between oving and static luid (a shock wave) travels up the pipe at the acoustic velocity. This is given by the equation : a= (K/)½ K = Vdp/dV I we assue that the change in volue is directly proportional to the change in pressure then we ay change this to inite changes such that K = V p/v V= Vp/K The ean pressure rise is p/ The strain energy stored by the copression = p V/ The change in kinetic energy = ½ u Equating or energy conservation we get Mu / = p V/= (p)v(p)/k u = V(p) /K Ku /V = (p) (p) = (/V)K u (p) = K u p = u(k)½ Since Then a = K/ then K = a p = u(a )½ p = u a For a large inite change, this becoes p = a u WORKED EXAMPLE A pipe 500 long carries water at /s. Calculate the pressure rise produced when a) the valve is closed uniorly in 5 seconds. b) when it is shut suddenly. SOLUTION The density o water is 000 kg/3 and the bulk odulus is 4 GPa throughout. Unior closure. Sudden closure p = L u/t = 000 x 500 x /5 = 00 kpa a = (K/) ½ = (4 x 0 9 /000) ½ = 000 /s p = a u = 000 x x 000 = 4 MPa D.J.Dunn www.reestudy.co.uk 3

SELF ASSESSMENT EXERCISE 5 The density o water is 000 kg/3 and the bulk odulus is 4 GPa throughout.. A pipe 50 long carries water at.5 /s. Calculate the pressure rise produced when a) the valve is closed uniorly in 3 seconds. (5 kpa) b) when it is shut suddenly. (3 MPa). A pipe 000 long carries water at 0.8 /s. A valve is closed. Calculate the pressure rise when a) it is closed uniorly in 0 seconds. (60 kpa) b) it is suddenly closed. (.6 MPa).5 THE EFFECT OF ELASTICITY IN THE PIPE A pressure surge in an elastic pipe will cause the pipe to swell and soe o the energy will be absorbed by straining the pipe wall. This reduces the rise in pressure. The ore elastic the wall is, the less the pressure rise will be. Consider the case shown. Fig.7 Kinetic Energy lost by luid = ½ u The ass o luid is AL so substituting K.E. = ½ ALu Strain Energy o luid = p AL/K (ro last section) Now consider the strain energy o the pipe wall. The strain energy o an elastic aterial with a direct stress is given by Fig.8 S.E. = ( /E) x volue o aterial The pipe ay be regarded as a thin cylinder and suitable reerences will show that stress stretching it around the circuerence is given by the ollowing orula. = pd/t D.J.Dunn www.reestudy.co.uk 3

Δp D Volue o etal = DtL Hence S.E. x t Equating KE lost to the total S.E. gained yields ρalu ρu Δp DAL Δp te AL K Δp D Δp Δp D K Δp πdtl E Δp DAL te te K ete D te K The solution is usually given in ters o the eective bulk odulus K' which is deined as ollows. D K te K The pressure rise is then given by p = u[/k']½ The acoustic velocity in an elastic pipe becoes a' and is given as a' = (K'/)½ ρu Hence p = u a' WORKED EXAMPLE 3 A steel pipe carries water at /s. The pipe is 0.8 bore diaeter and has a wall 5 thick. Calculate the pressure rise produced when the low is suddenly interrupted. The density o water is 000 kg/3 and the bulk odulus is 4 GPa. The odulus o elasticity or steel E is 00 GPa. SOLUTION D K te Sudden closure K 0.8 0.005 x 00 x 0 9 4 x 0 a' = (K'/) ½ = (95.4 x 0 6 /000) ½ = 976 /s p = a' u = 976 x x 000 =.95 MPa 9 95.4 MPa D.J.Dunn www.reestudy.co.uk 33

DAMPING OUT PRESSURE SURGES Pressure surges or water haer occurs whenever there is a change in low rate. There are any causes or this besides the opening and closing o valves. Changes in pup speeds ay cause the sae eect. Piston pups in particular cause rapid acceleration and deceleration o the luid. In power hydraulics, changes in the velocity o the ra cause the sae eect. The proble occurs both on large scale plant such as hydroelectric pipelines and on sall plant such as power hydraulic systes. The principles behind reduction o the pressure surges are the sae or each, only the scale o the equipent is dierent. For exaple, on power hydraulic systes, accuulators are used. These are vessels illed with both liquid and gas. On piston pups, air vessels attached to the pipe are used. In both cases, a sudden rise in pressure produces copression o the gas that absorbs the strain energy and then releases it as the pressure passes. Fig.9 On hydroelectric schees or large puped systes, a surge tank is used. This is an elevated reservoir attached as close to the equipent needing protection as possible. When the valve is closed, the large quantity o water in the ain syste is diverted upwards into the surge tank. The pressure surge is converted into a raised level and hence potential energy. The level drops again as the surge passes and an oscillatory trend sets in with the water level rising and alling. A daping oriice in the pipe to the surge tank will help to dissipate the energy as riction and the oscillation dies away quickly. Fig.30 D.J.Dunn www.reestudy.co.uk 34

SELF ASSESSMENT EXERCISE 6 The density o water is 000 kg/3 and the bulk odulus is 4 GPa throughout. The odulus o elasticity or steel E is 00 GPa.. A steel pipeline ro a reservoir to a treatent works is bore diaeter and has a wall 0 thick. It carries water with a ean velocity o.5 /s. Calculate the pressure rise produced when the low is suddenly interrupted. (.73 MPa). On a hydroelectric schee, water ro a high lake is brought down a vertical tunnel to a depth o 600 and then connects to the turbine house by a horizontal high-pressure tunnel lined with concrete. The low rate is 5 3 /s and the tunnel is 4 diaeter. (i) (ii) Calculate the static pressure in the tunnel under noral operating conditions.(5.9 MPa) Explain the dangers to the high-pressure tunnel when the turbines are suddenly stopped. (iii) Assuing the tunnel wall is rigid, calculate the axiu pressure experienced in the high-pressure tunnel when low is suddenly stopped. (6.7 MPa) (iv) Explain the saety eatures that are used in such situations to protect the tunnel. D.J.Dunn www.reestudy.co.uk 35