EDEXCEL NATIONAL CERTIFICATE/DIPLOMA PRINCIPLES AND APPLICATIONS of FLUID MECHANICS UNIT 3 NQF LEVEL 3 OUTCOME 3 - HYDRODYNAMICS TUTORIAL - PIPE FLOW CONTENT Be able to determine the parameters of pipeline flow and impact of a flid jet Pipeline flow: flid principles e.g. eqation of continity of mass, eqation of continity of volme for incompressible flow, Bernolli s eqation, D Arcy s eqation; systems and devices e.g. pipes of varying section and level (sch as an inclined tapering pipe), differential pressre flow measring devices (ventri meter, orifice meter, Pitot-static tbe) Flid jets: force exerted by a jet issing from a stationary nozzle that impinges normally on a stationary vane e.g. flat plate, hemispherical cp; reaction of jet nozzle D.J.Dnn freestdy.co.k
PIPE FLOW The soltion of pipe flow problems reqires the applications of two principles, the law of conservation of mass (continity eqation) and the law of conservation of energy (Bernolli s eqation) CONSERVATION OF MASS When a flid flows at a constant rate in a pipe or dct, the mass flow rate mst be the same at all points along the length. Consider a liqid being pmped into a tank as shown (fig.). The mass flow rate at any section is m = A m = density (kg/m3) m = mean velocity (m/s) A = Cross Sectional Area (m) Fig. For the system shown the mass flow rate at (), () and (3) mst be the same so A = A = 3 A 3 3 In the case of liqids the density is eqal and cancels so A = A = A 3 3 = Q D.J.Dnn www.freestdy.co.k
CONSERVATION OF ENERGY ENERGY FORMS FLOW ENERGY This is the energy a flid possesses by virte of its pressre. The formla is F.E. = pq Joles p is the pressre (Pascals) Q is volme rate (m3) POTENTIAL OR GRAVITATIONAL ENERGY This is the energy a flid possesses by virte of its altitde relative to a datm level. The formla is P.E. = mgz Joles m is mass (kg) z is altitde (m) KINETIC ENERGY This is the energy a flid possesses by virte of its velocity. The formla is K.E. = ½ m m Joles m is mean velocity (m/s) SPECIFIC ENERGY Specific energy is the energy per kg so the three energy forms as specific energy are as follows. F.E./m = pq/m = p/ Joles/kg P.E/m. = gz Joles/kg K.E./m = ½ Joles/kg ENERGY HEAD If the energy terms are divided by the weight mg, the reslt is energy per Newton. Examining the nits closely we have J/N = N m/n = metres. It is normal to refer to the energy in this form as the energy head. The three energy terms expressed this way are as follows. F.E./mg = p/g = h P.E./mg = z K.E./mg = /g The flow energy term is called the pressre head and this follows since earlier it was shown p/g = h. This is the height that the liqid wold rise to in a vertical pipe connected to the system. The potential energy term is the actal altitde relative to a datm. The term /g is called the kinetic head and this is the pressre head that wold reslt if the velocity is converted into pressre. D.J.Dnn www.freestdy.co.k 3
3 BERNOULLI S EQUATION Bernolli s eqation is based on the conservation of energy. If no energy is added to the system as work or heat then the total energy of the flid is conserved. Internal (thermal energy) is not been inclded so temperatre does not featre in this. The total energy E T at () and () on the diagram (fig.) mst be eqal so : ET pq mgz m pq mgz m Dividing by mass gives the specific energy form E T p p gz gz m Dividing by g gives the energy terms per nit weight E T p p z z mg g g g g Since p/g = pressre head h then the total head is given by the following. h T h z h z g g This is the head form of the eqation in which each term is an energy head in metres. z is the potential or gravitational head and /g is the kinetic or velocity head. For liqids the density is the same at both points so mltiplying by g gives the pressre form. The total pressre is as follows. pt p gz p gz In real systems there is friction in the pipe and elsewhere. This prodces heat that is absorbed by the liqid casing a rise in the internal energy and hence the temperatre. In fact the temperatre rise will be very small except in extreme cases becase it takes a lot of energy to raise the temperatre. If the pipe is long, the energy might be lost as heat transfer to the srrondings. Since the eqations did not inclde internal energy, the balance is lost and we need to add an extra term to the right side of the eqation to maintain the balance. This term is either the head lost to friction h L or the pressre loss p L. h z h z h L g g The pressre form of the eqation is as follows. p gz p gz pl The total energy of the flid (exclding internal energy) is no longer constant. Note that if one of the points is a free srface the pressre is normally atmospheric bt if gage pressres are sed, the pressre and pressre head becomes zero. Also, if the srface area is large (say a large tank), the velocity of the srface is small and when sqared becomes negligible so the kinetic energy term is neglected (made zero). D.J.Dnn www.freestdy.co.k 4
WORKED EXAMPLE No. The diagram shows a pmp delivering water throgh as pipe 30 mm bore to a tank. Find the pressre at point () when the flow rate is.4 dm 3 /s. The density of water is 000 kg/m 3. The loss of pressre de to friction is 50 kpa. Fig. Area of bore A = x 0.03 /4 = 706.8 x 0-6 m. Flow rate Q =.4 dm 3 /s = 0.004 m 3 /s Mean velocity in pipe = Q/A =.98 m/s Apply Bernolli between point () and the srface of the tank. ρ ρ p ρgz p ρgz pl Make the low level the datm level and z = 0 and z = 5. The pressre on the srface is zero gage pressre. P L = 50 000 Pa The velocity at () is.98 m/s and at the srface it is zero. 000x.98 p 0 0 000x9.95 0 50000 p 93.9kPa gage pressre WORKED EXAMPLE No. The diagram shows a tank that is drained by a horizontal pipe. Calclate the pressre head at point () when the valve is partly closed so that the flow rate is redced to 0 dm 3 /s. The pressre loss is eqal to m head. Fig.3 D.J.Dnn www.freestdy.co.k 5
Since point () is a free srface, h = 0 and is assmed negligible. The datm level is point () so z = 5 and z = 0. Q = 0.0 m3/s A = d /4 = x (0.05 )/4 =.963 x 0-3 m. = Q/A = 0.0/.963 x 0-3 = 0.8 m/s Bernolli s eqation in head form is as follows. h z h z hl g g 0 5 0 h 0.8 0 x 9.8 h 7.7m WORKED EXAMPLE No. 3 The diagram shows a horizontal nozzle discharging into the atmosphere. The inlet has a bore area of 600 mm and the exit has a bore area of 00 mm. Calclate the flow rate when the inlet pressre is 400 Pa. Assme there is no energy loss. Fig. 4 Apply Bernolli between () and () ρ ρ p ρgz p ρgz pl Using gage pressre, p = 0 and being horizontal the potential terms cancel. The loss term is zero so the eqation simplifies to the following. ρ ρ p From the continity eqation we have Q 4Q Q 4Q 3.537Q 3.83Q A π x 0.6 A π x 0. Ptting this into Bernolli s eqation we have the following. 400 000 x 400 6.55 x0 400 600.3 x0 Q 400 600.3x0 3.537Q 3.83Q 3 3 3 Q Q 506.6Q 666 x0 000 x 6 3 Q 0.058 m /s or 5.8 dm /s 3 D.J.Dnn www.freestdy.co.k 6
SELF ASSESSMENT EXERCISE No.. A pipe 00 mm bore diameter carries oil of density 900 kg/m3 at a rate of 4 kg/s. The pipe redces to 60 mm bore diameter and rises 0 m in altitde. The pressre at this point is atmospheric (zero gage). Assming no frictional losses, determine: i. The volme/s (4.44 dm3/s) ii. The velocity at each section (0.566 m/s and.57 m/s) iii. The pressre at the lower end. (.06 MPa). A pipe 0 mm bore diameter carries water with a head of 3 m. The pipe descends m in altitde and redces to 80 mm bore diameter. The pressre head at this point is 3 m. The density is 000 kg/m3. Assming no losses, determine i. The velocity in the small pipe (7 m/s) ii. The volme flow rate. (35 dm3/s) 3. A horizontal nozzle redces from 00 mm bore diameter at inlet to 50 mm at exit. It carries liqid of density 000 kg/m3 at a rate of 0.05 m3/s. The pressre at the wide end is 500 kpa (gage). Calclate the pressre at the narrow end neglecting friction. (96 kpa) 4. A pipe carries oil of density 800 kg/m3. At a given point () the pipe has a bore area of 0.005 m and the oil flows with a mean velocity of 4 m/s with a gage pressre of 800 kpa. Point () is frther along the pipe and there the bore area is 0.00 m and the level is 50 m above point (). Calclate the pressre at this point (). Neglect friction. (374 kpa) 5. A horizontal nozzle has an inlet velocity and an otlet velocity and discharges into the atmosphere. Show that the velocity at exit is given by the following formlae. ={p/ + } ½ and ={gh + } ½ D.J.Dnn www.freestdy.co.k 7
4 LAMINAR and TURBULENT FLOW The following work only applies to Newtonian flids. LAMINAR FLOW A stream line is an imaginary line with no flow normal to it, only along it. When the flow is laminar, the streamlines are parallel and for flow between two parallel srfaces we may consider the flow as made p of parallel laminar layers. In a pipe these laminar layers are cylindrical and may be called stream tbes. In laminar flow, no mixing occrs between adjacent layers and it occrs at low average velocities. TURBULENT FLOW The shearing process cases energy loss and heating of the flid. This increases with mean velocity. When a certain critical velocity is exceeded, the streamlines break p and mixing of the flid occrs. The diagram illstrates Reynolds colored ribbon experiment. Colored dye is injected into a horizontal flow. When the flow is laminar the dye passes along withot mixing with the water. When the speed of the flow is increased trblence sets in and the dye mixes with the srronding water. One explanation of this transition is that it is necessary to change the pressre loss into other forms of energy sch as anglar kinetic energy as indicated by small eddies in the flow. Fig. 5 CRITICAL VELOCITY - REYNOLDS NUMBER When a flid flows in a pipe at a volmetric flow rate Q m3/s the average velocity is defined Q m A is the cross sectional area. A md md The Reynolds nmber is defined as R e If yo check the nits of Re yo will see that there are none and that it is a dimensionless nmber. Reynolds discovered that it was possible to predict the velocity or flow rate at which the transition from laminar to trblent flow occrred for any Newtonian flid in any pipe. He also discovered that the critical velocity at which it changed back again was different. He fond that when the flow was gradally increased, the change from laminar to trblent always occrred at a Reynolds nmber of 500 and when the flow was gradally redced it changed back again at a Reynolds nmber of 000. Normally, 000 is taken as the critical vale. D.J.Dnn www.freestdy.co.k 8
WORKED EXAMPLE No. 4 Oil of density 860 kg/m 3 has a kinematic viscosity of 40 cst. Calclate the critical velocity when it flows in a pipe 50 mm bore diameter. R e m md ν R eν D 000x40x0 0.05 6.6 m/s SELF ASSESSMENT EXERCISE No.. Oil flows in a pipe 80 mm bore diameter with a mean velocity of 0.4 m/s. The density is 890 kg/m3 and the viscosity is 0.075 Ns/m. Is the flow trblent or laminar?. Oil flow in a pipe 00 mm bore diameter with a Reynolds Nmber of 500. The density is 800 kg/m3. The dynamic viscosity µ = 0.08 Ns/m. Calclate the mean velocity. 5 PIPE FRICTION Becase of the viscosity of flids flowing in pipes, the flid sticks to the inside wall and hence shearing takes place between the static layers and the moving layers. Pressre is needed to overcome the shear force and as a reslt this pressre is lost as heat. We call this FLUID FRICTION. This is the tem p L and h L in Bernolli's eqation. One of the simplest ways of dealing with this is to se the concept of FRICTION COEFFICIENT. In order to deine this we mst first define the DYNAMIC PRESSURE and THE WALL SHEAR STRESS. DYNAMIC PRESSURE Consider a flid flowing with mean velocity m. If the kinetic energy of the flid is converted into flow energy, the pressre wold increase. The pressre rise de to this conversion is called the dynamic pressre. KE = ½ m m Flow Energy = p Q Q is the volme flow rate and = m/q Eqating ½ m m = p Q p = m /Q = ½ m D.J.Dnn www.freestdy.co.k 9
WALL SHEAR STRESS o The wall shear stress is the shear stress in the layer of flid next to the wall of the pipe. Fig. 6 d The shear stress in the layer next to the wall is τo μ dy The shear force resisting flow is F τ πld s The reslting pressre drop prodces a force of Eqating forces gives D Δp τ o 4L FRICTION COEFFICIENT o F p ΔpπD 4 The friction coefficient is a convenient idea that can be sed to calclate the pressre drop in a pipe. It is defined as follows. Wall Shear Stress Wall Shear Stress D Δp C f Dynamic Pressre Dynamic Pressre 4LρL m DARCY FORMULA This is the important part for this nit. The formla is mainly sed for calclating the pressre loss in a pipe de to trblent flow bt it can be sed for laminar flow. Trblent flow in pipes occrs when the Reynolds Nmber exceeds 500 bt this is not a clear point so 3000 is sed to be sre. We rearrange the formla for Cf as follows. DΔD 4Cf L m Cf p 4LρL D m p 4Cf L m This is often expressed as a friction head hf h f g gd This is the Darcy formla. In the case of laminar flow, it can be shown that: 6μ 6 Cf ρ D R m e wall D.J.Dnn www.freestdy.co.k 0
WORKED EXAMPLE No. 5 Oil flows in a pipe 80 mm bore and is 60 m long. The density is 900 kg/m 3 and the mean velocity is 4 m/s. Given C f = 0.005, calclate pressre loss. The dynamic viscosity is 0.005 N s/m. Determine the Reynolds nmber. h f = 4C f L/dg = (4 x 0.005 x 60 x 4 )/( x 9.8 x 0.08) =.7 m p = gh f = 900 x 9.8 x.7 =.3 kpa. R e = d/ = (900 x 4 x 0.08)/0.005 = 57600 WORKED EXAMPLE No. 6 The diagram shows a large tank draining into another lower tank throgh a pipe. The tanks are open to atmosphere. The levels may be regarded as constant. Given the friction coefficient C f is 0.0058, calclate the flow rate sing the data given on the diagram. Fig. 7 Apply Bernolli between A and B becase the pressre is the same at both (0 gage) and the velocity at both points is zero. A B ha za hb zb hl g g 0 90 0 0 0 0 h L h L = 90 m Assme that the entire head loss is de to friction we can se the D'Arcy formla. 4Cf L hf gd g d hf x 9.8 x 0. x 90.685 4 Cf L 4 x 0.0058 x 600 = 3.56 m/s Q = A = 3.56 x π x 0. /4 = 0.08 m 3 /s or 8 dm 3 /s D.J.Dnn www.freestdy.co.k
SELF ASSESSMENT EXERCISE No. 3. A pipe is 5 km long and 80 mm bore diameter. It caries oil of density 85 kg/m3 at a rate of 0 kg/s. The dynamic viscosity is 0.05 N s/m. Given the friction coefficient C f is 0.0083 calclate the friction head. (Ans. 3075 m.). Oil with viscosity 0.0 Ns/m and density 850 kg/m3 is pmped along a straight horizontal pipe 50 mm diameter and 0 m long with a flow rate of 5 dm3/s. Show that the flow is laminar and determine the following. i. The mean velocity (0.83 m/s) ii. The Reynolds nmber. ( 804) iii. The friction coefficient. (0.00887) iv. The pressre loss. (80 Pa) 3. A pipe carries oil at a mean velocity of 6 m/s. The pipe is 5 km long and.5 m diameter. The density is 890 kg/m3. Given the friction coefficient C f = 0.0045 determine the friction head hf. (Ans. hf = 0. m) 4. The diagram shows a reservoir draining into another throgh a pipe. The friction coefficient C f is 0.0. Calclate the flow rate assming that the srface levels are constant and the srfaces are both at atmospheric pressre. (Ans. 76 dm 3 /s) Fig. 8 D.J.Dnn www.freestdy.co.k
6. DIFFERENTIAL PRESSURE DEVICES Differential pressre devices prodce differential pressre as a reslt of changes in flid velocity. They have many ses bt mainly they are sed for flow measrement. Here are some pictres of a complete flow meter based on an orifice. Fig.9 GENERAL RELATIONSHIP Many devices make se of the transition of flow energy into kinetic energy. Consider a flow of liqid which is constrained to flow from one sectional area into a smaller sectional area as shown below. Fig. 0 The velocity in the smaller bore is given by the continity eqation as = A/A Let A/A = r = r If we apply Bernolli (head form) between () and () and ignoring energy losses we have h z h z g g For a horizontal system z = z so h h g g g h h r g h h r Vol/s Q A In terms of pressre rather than head we get, by sbstitting p = gh To find the mass flow remember m = A = Q A g h h r Δp Q A ρ r Becase we did not allow for energy loss, we introdce a coefficient of discharge Cd to correct the answer reslting in Δp Q CdA ρ r The vale of Cd depends pon many factors and is not constant over a wide range of flows. For a given device, if we regard Cd as constant then the eqation may be redced to: Q = K(p) 0.5 where K is the meter constant. D.J.Dnn www.freestdy.co.k 3
ORIFICE METERS These are simple differential devices sed to measre the flow of flids. A simple orifice forces the flid to contract into a jet and hence the pressre in the region of the jet is redced to give the pressre difference. Figre Typically the pressre is measred at points (A) and (B) bt there are variations. When a flid flows throgh an orifice it experiences frictional energy loss and a contraction in the diameter of the jet, both of which affect the vale of Cd. Also the shape of the orifice lip is important and this may be sharp or flat. Withot developing this we can apply the formla between the inlet and the jet and se Δp Q CdA where r is the ratio A /A o and A o is ρ r the area of the orifice. The flow formla is sally based on the orifice area and it is often possible to assme that is a negligible vale. If we apply Bernolli between () and () we have as before h h and neglecting g g Δp h h gh h gh h or g ρ Q = A o Introdcing C d to allow for friction and other factors not discssed here we have: Q C d A o g h h or Q C d A o Δp ρ WORKED EXAMPLE No. 7 An orifice 0 mm diameter is fixed in a horizontal pipe of diameter 5 mm. The differential pressre is 750 kpa. It may be assmed that the discharge coefficient is 0.64. The density of water is 998 kg/m3. Calclate the flow rate in m 3 /s. Use both formla to see if ignoring the approach velocity makes a difference. A o = πd /4 = π x 0.0 /4 = 0.34 x 0-3 m A = πd /4 = π x 0.05 /4 = 0.49 x 0-3 m r = A / A o =.563 D.J.Dnn www.freestdy.co.k 4
Ignoring the approach velocity Q C d A o Δp ρ Otherwise Δp Q CdA ρ r 0.64 x 0.34 x0 3 0.64 x 0.49 x0 3 x 750000 998 x 750000 998.563 0.007795 m /s 0.0045 m /s 3 3 Ignoring the approach velocity prodces significant error becase the diameters are similar. VENTURI METERS The Ventri Meter is designed to taper down to the throat gradally and then taper ot again. This prodces a differential pressre between () and () with very little friction so C d is closer to. The otlet (diffser) is designed to expand the flow gradally so that the kinetic energy at the throat is reconverted into pressre with little friction. The overall pressre loss is mch better than for an orifice meter. Fig. Δp The flow rate is given by the formla Q CdA K ρ r p for a given meter and flid. Cd is abot 0.97 for a good meter. Ventri meters are generally more expensive than orifice meters. WORKED EXAMPLE No. 8 A Ventri meter has a meter constant of K = 0.008 m 4 N -0.5 s -. Calclate the flow rate when p = 80 Pa Q = K(p)0.5 = 0.008 m 4 N -0.5 s - (80)0.5 = 0.073 (m 4 N -0.5 s - )(N 0.5 m - ) or m 3 /s D.J.Dnn www.freestdy.co.k 5
WORKED EXAMPLE No. 9 A Ventri meter is sed to meter the flow of oil of density 90 kg/m 3. The meter inlet bore is 80 mm and the throat bore is 70 mm. The coefficient of discharge C d = 0.97. Calclate the flow rate of oil when p = 580 Pa A = π x 0.08 /4 = 5.07 x 0-3 m A = π x 0.07 /4 = 3.848 x 0-3 m r = A /A =.306 Δp Q CdA ρ r 0.97 x 5.07 x0 x 580 90.306-3 3 6.55 x0 3 m /s PITOT TUBE Pitot tbes are basically designed to measre the velocity of a flid at a given point. The figre shows the arrangement for measring the velocity inside a pipe. The flid is rammed down the tbe and all the kinetic energy is converted into pressre p. The ambient pressre p is trapped from the wall of the pipe. The pressre rise enables the velocity to be calclated. Kinetic energy lost = m / Fig. 3 Flow energy gained = pv In the case of liqids, the volme is nchanged when it is rammed into the tbe so the change in flow energy may be written as V P. Eqating energy we have m Q Δp ΔpQ m Δp ρ Δp ρ WORKED EXAMPLE No. 0 A pitot tbe is pointed into a stream of air and a differential pressre of 380 Pa is measred. What is the velocity of the air. Assme the density changes are negligible and the density of the air is. kg/m 3. Δp ρ x 380. 5.7 m/s D.J.Dnn www.freestdy.co.k 6
SELF ASSESSMENT EXERCISE No. 4. An Orifice meter has a meter constant of K = 0.004 m 4 N -0.5 s -. Calclate the flow rate when a differential pressre of 00 Pa is obtained. (Answer 0.0566 m 3 /s). A Ventri meter is sed to meter the flow of water of density 000 kg/m 3. The meter inlet bore is 50 mm and the throat bore is 40 mm. The coefficient of discharge C d = 0.97. Calclate the flow rate of oil when p = 970 Pa (Answer. x 0-3 m 3 /s) 3. A Ventri meter is 50 mm bore diameter at inlet and 0 mm bore diameter at the throat. Oil of density 900 kg/m3 flows throgh it and a differential pressre head of 80 mm is prodced. Given Cd = 0.9, determine the flow rate in kg/s. (Ans. 0.085 kg/s) 4. A Ventri meter is 60 mm bore diameter at inlet and 0 mm bore diameter at the throat. Water of density 000 kg/m3 flows throgh it and a differential pressre head of 50 mm is prodced. Given Cd = 0.95, determine the flow rate in dm3/s. (Ans. 0.55dm3/s) 5. Calclate the differential pressre expected from a Ventri meter when the flow rate is dm3/s of water. The area ratio is 4 and Cd is 0.94. The inlet c.s.a. is 900 mm. (Ans. 4.96 kpa) 4. Calclate the mass flow rate of water throgh a Ventri meter when the differential pressre is 980 Pa given Cd = 0.93, the area ratio is 5 and the inlet c.s.a. is 000 mm. (Ans. 0.66 kg/s) 5. Calclate the flow rate throgh an orifice meter with an area ratio of 4 given Cd is 0.6, the pipe area is 900 mm and the d.p. is 586 Pa. (Ans. 0.56 kpa) 6. A pitot tbe is sed to measre the velocity of water in a stream. The differential pressre is 45 Pa and the density is 000 kg/m 3. What is the velocity? (Ans. 0.7 m/s) D.J.Dnn www.freestdy.co.k 7