CHAPTER 4 : FLUID DYNAMICS FLUID MECHANICS 46 h (1) (2) 4.4.3.2 Venturi Meter Venturi meter is a device used for measuring the rate of flow of a non-viscous, incompressible fluid in non-rotational and steady-stream lined flow. Although venturi meters can be applied to the measurement of gas, they are most commonly used for liquids. I II III 210 5 0 -7 0 • • fluid x fluid y Venturi meter is divided into parts : I Converging Cone – This section is also called the entry, which the diameter is equal to the exit. Converging cone angle is 210 . II Throat – A small portion of the venturi meter in which the throat is in the same length with a diameter of the pipe. The pipe diameter is uniform. III Diverging cone – It also known as the exit. Diameter of the cone enlarge from the throat. Corner of this section is between 5 0 -7 0 . The angles of the conical pipes are established to limit the energy losses due to flow separation. This mechanism is designed to reduce energy loss in the U-tube flow at the inlet and connected to the throat to get the pressure difference between points (1) and point (2). The arrangement and mode of action of a Venturi Meter. ▪ The venture meter consists of a short converging conical tube leading to a cylindrical portion called “throat” which is followed by a diverging section. ▪ The entrance and exit diameter is the same as that of the pipeline into which it is inserted. The angle of the convergent cone is usually 21o , the length of throat is equal to the throat diameter, and the angle of the divergent cone is 5o to 7o to ensure the minimum loss of energy. ▪ Pressure tapping are taken at the entrance and at the throat, either from the single holes or from a number of holes around the circumference connecting to an annular chamber or Piezometer ring, and the pressure difference is measured by a suitable gauge. ▪ For continuity of flow velocity v1 at the entry (section 1) will be less than the velocity v2 at the throat (section 2) since a1v1 = a2v2 and a1 is greater than a2. The kinetic energy in the throat will be greater than at the entrance and since by Bernoulli’s theorem the total energy at the two sections is the same, the pressure energy at the throat will be less than that at the entrance. The pressure difference thus created is dependent on the rate of flow through the meter.
CHAPTER 4 : FLUID DYNAMICS FLUID MECHANICS 47 h Hg H2O (1) (2) h1 Z A B Horizontal Venturi Meter • • From Bernoulli Equation : => 2g 2 2 V ρg 2 P 2 Z 2g 2 1 V ρg 1 P 1 Z + + = + + ; (Z1 = Z2) 2g 2 1 V 2 2 V ρg 2 P 1 P − = − _____ equation 1 substitute equation 2 into equation 1 => ρg 2 P 1 P − = 2g 2 1 V 2 A 1 V 1 A − 2 ( ) ( ) 1 2 2 A 1 A ρg 2 P 1 P 2g 1 V − − = ( ) ( ) 1 2 2 A 1 A ρg 2 P 1 P 2g 1 A Q − − = => 1 A Q = 1 2 m 2gH − QT = A1 1 2 m 2gH − => QAct = Cd x QT QAct = Cd x A1 1 2 m 2gH − where : -> ρg 2 P 1 P − = H -> 2 A 1 A = m (area ratio) From Continuity Equation : => Q1 = Q2 A1V1 = A2V2 2 A 1 V1 A 2 V = ______ equation 2
CHAPTER 4 : FLUID DYNAMICS FLUID MECHANICS 48 in the uniformity of pressure, H PA = PB PA = P1 + water h1 PB = P2 + water (h1 – h) + Hg h P1 + water h1 = P2 + water (h1 – h) + Hg h P1 - P2 = air h1 – water h + Hg h - water h1 = h (Hg - water) because = g , then => = − − water water ωwater w ater ω ω Hg ω h ω 2 P 1 P where : H = pressure head in terms of height of fluid = − 1 ω ωHg H h water h = difference of mercury levels in the U-tube Incline Venturi Meter ( ) c Q c A m gh c A m g Z Z P P g Act d T Act d Hg water Act d = = − − = − − + − 1 2 1 2 1 2 1 2 1 2 1 1 2 Q Q Q water Z1 Z2 h mercury
CHAPTER 4 : FLUID DYNAMICS FLUID MECHANICS 49 4.4.3.3 Orifice Meter The Venturi Meter described as a reliable flow measuring device. Furthermore, it causes little pressure loss. For these reasons, it is widely used, particularly for large-volume liquid and gas flows. However, this meter is relatively complex to construct and hence expensive especially for small pipelines. The cost of the Venturi Meter seems prohibitive, so simpler device such as Orifice Meter is used. The Orifice Meter consists of a flat orifice plate with a circular hole drilled in it. There is a pressure tap upstream from the orifice plate and another just downstream. There are three recognized methods of placing the taps and the coefficient of the meter will depend upon the position of the taps. The principle of the orifice meter is identical with that of the venturi meter. The reduction at the cross section of the flowing stream in passing through the orifice increases the velocity head at the expense of the pressure head, and a manometer measures the reduction in pressure between the taps. Bernoulli's equation provides a basis for correlating the increase in velocity head with the decrease in pressure head. From figure, the orifice meter is attached to the manometer. There are Section 1 (entrance of the orifice) and Section 2 (exit of the orifice also known as vena contracta). Types of orifice 1. Sharp-edged orifice, Cd = 0.62 2. Rounded orifice, Cd = 0.97 3. Borda Orifice (running free), Cd = 0.50 4. Borda Orifice (running full), Cd = 0.75
CHAPTER 4 : FLUID DYNAMICS FLUID MECHANICS 50 Coefficient of Velocity, Cv Base on the diagram, x = horizontal falls at point B from vena contracta, A at orifice y = vertical falls at point B above the orifice level h = head of liquid above the orifice t = time for particle to travel from vena contracta, A to point B Cv = coefficient of velocity = theory v actual v v C => actual velocity, vact = 2y 2 gx ; theory velocity, vt = 2gh Cv = t v act v = 4yh 2 x coefficient of contraction, Cc = area of jet at vena contracta area of orifice coefficient of flow rate, Cd = Cv X Cc => QAct = Cd x QT QAct = Cd x A1 1 2 m 2gH −
CHAPTER 4 : FLUID DYNAMICS FLUID MECHANICS 51 4.4.3.4 Pitot Tube - The Pitot Tube is a device used to measure the local velocity along a streamline. The Pitot Tube has two tubes: one is a static tube (b), and another is an impact tube (a). - The opening of the impact tube is perpendicular to the flow direction. The opening of the static tube is parallel to the direction of flow. - The two legs are connected to the legs of a manometer or an equivalent device for measuring small pressure differences. The static tube measures the static pressure, since there is no velocity component perpendicular to its opening. - The impact tube measures both the static pressure and impact pressure (due to kinetic energy). - In terms of heads, the impact tube measures the static pressure head plus the velocity head. Actual Velocity, V ▪ From figure, if the velocity of the stream at A is v, a particle moving from A to the mouth of the tube B will be brought to rest so that v0 at B is zero. By Bernoulli’s Theorem : Total Energy at A = Total Energy at B or g p v g p v 2 2 2 2 2 2 1 1 + = + ————(1) ▪ Now p d = and the increased pressure at B will cause the liquid in the vertical limb of the pitot tube to rise to a height, h above the free surface so that 0 p h + d = . ▪ Thus, the equation (1) h or v gh p p g v 2 0 2 2 = = − = ▪ Although theoretically v ( gh) T = 2 , pitot tubes may require calibration. The actual velocity is then given by v C ( gh) ACTUAL = 2 where C is the coefficient of the instrument. Pitot Tube a b A B h H
CHAPTER 4 : FLUID DYNAMICS FLUID MECHANICS 52 EXCERSICE 4.2 : 1. Water flows through a pipe 36 m from the sea level as shown in figure. Pressure in the pipe is 410 kN/m2 and the velocity is 4.8 m/s. Calculate total energy of every weight of unit water above the sea level. [H = 78.96 m] Answer : 2. A bent pipe labeled MN measures 5 m and 3 m respectively above the datum line. The diameter M and N are both 20 cm and 5 cm. The water pressure is 5 kg/cm2 . If the velocity at M is 1m/s, determine the pressure at N in N/m2 . [PN = 382620 N/m2 ] Answer : 36 m m N M 5 m 3 m
CHAPTER 4 : FLUID DYNAMICS FLUID MECHANICS 53 3. A pipe with 15 m length is installed in first floor of a hostel as shows in figure below. The pipe is tilted by 200 from the ground floor accommodation. Diameter of the pipe is 30 cm and the velocity is 2m/s. Calculate the total energy of the pipe if water pressure recorded is 5kN/m2 . Answer : 4. A venture tube tapers from 300 mm in diameter at the entrance to 100 mm in diameter at the throat; the discharge coefficient is 0.98. A differential mercury U-tube gauge is connected between pressures tapping at the entrance at throat. If the meter is used to measure the flow of water and the water fills the leads to the U-tube and is in contact with the mercury, calculate the discharge when the difference of level in the U-tube is 55 mm. [QActual = 0.0285 m3 /s] Answer : 200 15 m
CHAPTER 4 : FLUID DYNAMICS FLUID MECHANICS 54 5. A horizontal venturi meter measures the flow of oil of specific gravity 0.9 in a 75 mm diameter pipeline. If the difference of pressure between the full bore and the throat tapping is 34.5 kN/m2 and the area ratio, m is 4, calculate the rate of flow, assuming a coefficient of discharge is 0.97. [QActual = 0.0106 m3 /s] Answer : 6. An incline venturi meter measures the flow of oil of specific gravity 0.82 and has an entrance of 125 mm diameter and throat of 50 mm diameter. There are pressure gauges at the entrance and at the throat, which is 300 mm above the entrance. If the coefficient for the meter is 0.97 and pressure difference is 27.5 kN/m2 , calculate the actual discharge in m3 /s. [QActual = 0.0153 m3 /s] Answer :
CHAPTER 4 : FLUID DYNAMICS FLUID MECHANICS 55 7. A venturi meter measures the flow of water in a 75 mm diameter pipe. The difference between the throat and the entrance of the meter is measured by the U-tube containing mercury, which is being in contact with the water. What should be the diameter of the throat of the meter in order that the difference in the level of mercury is 250 mm when the quantity of water flowing in the pipe is 620 dm3 /min? Assume coefficient of discharge is 0.97. [diameter = 40.7 mm] Answer : 8. A meter orifice has a 100 mm diameter rectangular hole in the pipe. Diameter of the pipe is 250 mm. Coefficient of discharge, Cd = 0.65 and specific gravity of oil in the pipe is 0.9. The pressure difference that is measured by the manometer is 750 mm. Calculate the flow rate of the oil through the pipe. [QActual = 0.074 m3 /s] Answer :
CHAPTER 4 : FLUID DYNAMICS FLUID MECHANICS 56 9. An orifice meter consists of a 100 mm diameter in a 250 mm diameter pipe as shown in figure, and has a coefficient discharge of 0.65. The pipe conveys oil of specific gravity 0.9. The pressure difference between the two sides of the orifice plate is measured by a mercury manometer, which leads to the gauge being filled with oil. If the difference in mercury levels in the gauge is 760 mm, calculate the flowrate of oil in the pipeline. [Qactual = 0.0762 m3 /s] Answer : Pipe Area, A1 P1 P2 V1 V2 X C C Orifice area A2
CHAPTER 4 : FLUID DYNAMICS FLUID MECHANICS 57 10. A tank 1.8 m high, standing on the ground, is kept full of water. There is an orifice in its vertical site at depth, h m below the surface. Find the value of h in order the jet may strike the ground at a maximum distance from the tank. [h = 0.9 m] Answer : 11. A Pitot Tube is used to measure air velocity in a pipe attached to a mercury manometer. Head difference of that manometer is 6 mm water. The weight density of air is 1.25 kg/m3 . Calculate the air velocity if coefficient of the pitot tube, C = 0.94. [v = 9.12 m/s] Answer :
CHAPTER 4 : FLUID DYNAMICS FLUID MECHANICS 58 12. A pitot-static tube placed in the centre of a 200 pipe line conveying water has one orifice pointing upstream and the other perpendicular to it. If the pressure difference between the two orifices is 38 mm of water when the discharge through the pipe is 22 dm3 /s, calculate the meter coefficient. Take the mean velocity in the pipe to be 0.83 of the central velocity. [c = 0.977] Answer : 13. A sharp-edged orifice, of 50 mm diameter, in the vertical side of a large tank, discharges under a head of 4.8 m. If Cc = 0.62 and Cv = 0.98, determine; (a) the diameter of the jet, (b) the velocity of the jet at the vena contracta, (c) the discharge in dm3 /s. Answer :
CHAPTER 5 : ENERGY LOSS IN PIPELINES FLUID MECHANICS 59 CHAPTER 5 : ENERGY LOSS IN PIPELINES OBJECTIVES · know, understand and apply Bernoulli’s equation to pipeline systems · sketch the velocity profile in circular pipe system · explain and calculate energy loss in pipeline system · calculate and apply energy loss equation from reservoir · solve problem related to the pipeline system INTRODUCTION A pipe is defined as a closed conduit of circular section through which the fluid flows, filling the complete cross-section. The fluid in the pipe has no free surface. It will be at a pressure which may vary along the pipe. Losses of energy in a pipeline cannot be ignored. When the shock losses and friction loss have been determined, they are inserted in Bernoulli’s equation in the usual way.
CHAPTER 5 : ENERGY LOSS IN PIPELINES FLUID MECHANICS 60 5.1 THE ROUND PIPE SYSTEM 1 2 h g v g p z g v g p z + + = + + + 2 2 2 2 2 1 2 1 1 1 2 2 5.1.1 VELOCITY PROFILE IN THE CIRCULAR PIPE Velocity profile in circular pipe system Lamina flow mean velocity smooth pipe wall rough pipe Not all fluid’s particles travel at the same velocity within a pipe. The shape of the velocity curve (the velocity profile across any given section of the pipe) depends upon whether the flow is laminar. If the flow in a pipe is laminar, the velocity distribution at a cross section will be parabolic in shape with the maximum velocity at the centre being about twice the average velocity in the pipe. The velocity of the fluid in contact with the pipe wall is essentially zero and increases the further away from the wall. 5.2 HEAD LOSS 5.2.1 TYPES OF HEAD LOSS Losses of energy in pipe line are due to : a) shock loss at sudden enlargement b) shock loss at sudden contraction c) frictional resistance to flow d) loss at entry (inlet) e) loss at rounded and sharp exit (outlet)
CHAPTER 5 : ENERGY LOSS IN PIPELINES FLUID MECHANICS 61 a. (i) Head Loss caused by Sudden Enlargement (hL) P1 P2 v1 v2 => h v v g L = ( − ) 1 2 2 2 a1 a2 (ii) Head Loss caused by Sudden Contraction (hC) P1 P2 v1 v2 => 2g 2 2 v 2 1 C C 1 C h = − a1 a2 ~ Cc = coefficient of contraction b. Head Loss caused by Friction (hF) L => hf = 2g 2 v d 4fL ~ f = coefficient of friction d v hf = d2g 2 4fLv Darcy Equation P1 P2
CHAPTER 5 : ENERGY LOSS IN PIPELINES FLUID MECHANICS 62 c. Head Loss caused at the Sharp Inlet (hL) if the exit/outlet is into reservoir 2g 2 1 v L h = 2 1 d. Head Loss caused at the Sharp Outlet (hC) if the entry/inlet is from the reservoir = 2g 2 2 v 2 1 C h 1 2
CHAPTER 5 : ENERGY LOSS IN PIPELINES FLUID MECHANICS 63 EXCERSICE 5.1 : 1. A pipe carrying 1800 l/min of water increases suddenly from 10 cm to 15 cm diameter. Find : (a) the head loss due to the sudden enlargement [0.2294 m of water] (b) the difference in pressure in kN/m2 in the two pipes [3602.56 N/m2 ] Answer : 2. A pipe carrying 0.06 m3 /s suddenly contracts from 200 mm to 150 mm diameter. Assuming that the vena contracta is formed in the smaller pipe, calculate the coefficient of contraction if the pressure head at a point upstream of the contraction is 0.655 m greater than at a point just downstream of the vena contracta. [Cc = 0.615] Answer :
CHAPTER 5 : ENERGY LOSS IN PIPELINES FLUID MECHANICS 64 3. Determine the loss of head due to friction in a pipe 14 m long and 2 m diameter which carries 1.5 m/s oil. Take into consideration f = 0.05. [hF =0.16 m of oil] Answer : 4. Water flows vertically downwards through a 150 mm diameter pipe with a velocity of 2.4 m/s. The pipe suddenly enlarges to 300 mm in diameter. Find the loss of head. If the flow is reversed, find the loss of head, assuming the coefficient of contraction now being 0.62. [hL = 0.165 m, hC = 0.01 m] Answer :
CHAPTER 5 : ENERGY LOSS IN PIPELINES FLUID MECHANICS 65 5. Water from a large reservoir is discharged to atmosphere through a 100 mm diameter pipe 450 m long. The entry from the reservoir to the pipe is sharp and the outlet is 12 m below the surface level in the reservoir. Taking f = 0.01 in the Darcy formula, calculate the discharge. [Q = 8.96x10-3 m3 /s] Answer : 6. Water is discharged from a reservoir into the atmosphere through a pipe 39 m long. There is a sharp entrance to the pipe and the diameter is 50 mm for the first 15 m from the entrance. The pipe then enlarges suddenly to 75 mm in diameter for the remainder of its length. Taking into consideration the loss of head at entry and at the enlargement, calculate the difference of level between the surface of the reservoir and the pipe exit which will maintain a flow of 2.8 dm3 /s. Take f as 0.0048 for the 50 mm pipe and 0.0058 for the 75 mm pipe. [0.853 m of water] Answer :
CHAPTER 5 : ENERGY LOSS IN PIPELINES FLUID MECHANICS 66 7. Two reservoirs are connected by a pipeline which is 150 mm in diameter for the first 6 m and 225 mm in diameter for the remaining 15 m. The entrance and exit are sharp and the change of section is sudden. The water surface in the upper reservoir is 6 m above that in the lower. Tabulate the losses of head which occur and calculate the rate of flow in m3 /s. Friction coefficient f is 0.01 for both pipes. [0.185 m3 /s] Answer : 8. Water is discharged from a reservoir into the atmosphere through a pipe 80 m long. There is a sharp entrance to the pipe and the diameter is 250 mm for the first 50 m. The outlet is 35 m below the surface level in the reservoir. The pipe then enlarges suddenly to 450 mm in diameter for the remainder of its length. Take f = 0.004 for both pipes. Calculate the discharge. [Q = 0.623 m3 /s] Answer :
CHAPTER 5 : ENERGY LOSS IN PIPELINES FLUID MECHANICS 67 9. Two reservoirs have a difference in level of 9 m and are connected by a pipe line, which is 38 mm in diameter for the first 13 m and 23 mm for the remaining 6 m. Take f = 0.01 for both pipes and CC = 0.66. Calculate the discharge. [Q = 0.00345 m3 /s] Answer : 10. A pipe carrying 0.056 m3 /s suddenly changes diameter from; (a) 200 mm to 150 mm [0.19 m, 0.540 m] (b) 300 mm to 150 mm [0.19 m, 0.673 m] (c) 450 mm to 150 mm [0.19 m, 0.699 m] Find the loss of head and the pressure difference across the contraction in each case, given CC = 0.62. Answer :