Lecture 6ENERGY LOSSES IN HYDRAULIC SYSTEMS Learning Objectives Upon completion of this chapter, the student should be able to: State the differences between laminar and turbulent flows. Define Reynolds number and state its importance. Explain the Darcy–Weisbach equation. Explain the various types of joints used in fluid power. Evaluate the head losses for laminar and turbulent flow. Explain the various types of losses in fittings and valves. Design a system considering all head losses in the system. 1.1 Introduction Liquids such as water or petrol flow much easily than other liquids such as oil. The resistance to flow is essentially a measure of the viscosity of a fluid. The greater the viscosity of a fluid, the less readily it flows and the more is the energy required to move it. This energy is lost because it is dissipated as heat. Energy losses occur in valves and fittings. Various types of fittings, such as bends, couplings, tees, elbows, filters, strainers, etc., are used in hydraulic systems. The nature of path through the valves and fittings determines the amount of energy losses. The more circuitous is the path, the greater are the losses. In many fluid power applications, energy losses due to flow in valves and fittings exceed those due to flow in pipes. Therefore,a proper selection of fitting is essential. In general, the smaller the size of pipe and fittings, the greater the losses. The resistance to flow of pipes, valves and fittings can be determined using empirical formulas that have been developed by experimentation. The energy equation and the continuity equation can be used to perform a complete analysis of a fluid power system. This includes calculating the pressure drops, flow rates and power losses for all components of the fluid power system. The purpose of this chapter is to study the detailed circuit analysis of energy losses in fluid power systems containing valves, fittings and other power transmission and energy conversion elements. 1.2 Laminar and Turbulent Flows When speaking of fluid flow, one refers to the flow of an ideal fluid. Such a fluid is presumed to have no viscosity. This is an idealized situation that does not exist. When referring to the flow of a real fluid, the effects of viscosity are introduced into the problem. This results in the development of shear stresses between neighboring fluid particles when they move at different velocities. In the case of an ideal fluid flowing in a straight conduit, all the particles move in parallel lines with equal velocity. In the flow of a real fluid, the velocity adjacent to the wall is zero; it increases rapidly within a short distance from the wall and produces a velocity profile such as shown in Figure1.1. u (a) (b) Figure 1.1 Typical velocity profile: (a) Ideal fluid.(b) Real fluid There are two types of flow in pipes: 1. Laminar flow:This is also known as streamline or viscous flow and is illustrated in Fig.1.2. In streamline flow, the fluid appears to move by sliding of laminations of infinitesimal thickness relative to adjacent layers; that is, the particles move in definite and observable paths or streamlines. The flow characteristic of a viscous fluid is one in which viscosity plays a significant part. U F, U Velocity profile Y y u Figure 1.2 Laminar flow 2.Turbulent flow: It is illustrated in Fig.1.3. It is characterized by a fluid flowing in random way. The movement of particles fluctuates up and down in a direction perpendicular as well as parallel to the mean flow direction. This mixing action generates turbulence due to the colliding fluid particles. This causes a considerable more resistance to flow and thus greater energy losses than those produced by laminar flow. A distinguishing characteristic of turbulence is its irregularity, there being no definite frequency, as in wave motion, and no observable pattern, as in the case of large eddies. Figure 1.3 Turbulent flow 3.1. Reynolds came to a significant conclusion that the nature of the flow depends on the dimensionless parameter. that is. gravity does not affect the flow pattern. If it is flowing in a 30-mm diameter pipe at a velocity of 6 m/s. If Reis less than 2000. the only forces involved are those of friction and inertia. . Reynolds number between 2000 and 4000 covers a critical zone between laminar and turbulent flow. If turbulent flow is allowed to exist.1 The kinematic viscosity of a hydraulic fluid is 0. Example 1. It is not possible to predict the type of flow that exists within a critical zone. in honor of Osborne Reynolds. we can write vD vD vD 6 0. higher fluid temperatures occur due to greater frictional energy losses. 2. the flow is laminar.0001 SinceRe is less than 2000. Therefore. and hence significant forces are inertial force and fluid friction due to viscosity. who presented this in a publication of his experimental work in 1882. It is also obvious that capillarity is of no practical importance. the flow is laminar. for a submarine submerged far enough so as not to produce waves on the surfaces. Considering the ratio of inertial forces to viscous forces. turbulent flow should be assumed. If Reis greater than 4000. He conducted a series of experiments to determine the conditions governing the transition from laminar flow to turbulent flow. what is the Reynolds number? Is the flow laminar or turbulent? Solution: From the definition of Reynolds number. Thus. the parameter obtained is called the Reynolds number. turbulent flow systems suffering from excessive fluid temperature can be helped by increasing the pipe diameter to establish laminar flow. vD Re Where v is the fluid velocity. The same is true for an airplane traveling at speed below that at which compressibility of air is appreciable. the flow is turbulent.D is the inside diameter of the pipe.0001 m2/s. 1. is the fluid density and is the absolute viscosity of the fluid.03 Re 1800 <COMP: v and nu both being used> / 0. if the Reynolds number lies in the critical zone. Also.3 Reynolds Number In the flow of a fluid through a completely filled conduit. Head losses ina long pipe in which the velocity distribution has become fully established or uniform along its length can be found by Darcy’s equation as 2 L v HL f D 2g wherefis the Darcy friction factor. the friction factor is computed analytically. If it is flowing in a 20-mm diameter commercial steel pipe. This pressure decrease is mainly due to the friction of the fluid against the pipe wall. the friction factor equals the constant 64 divided by the Reynolds number: 64 f Re Substituting this into Darcy’s equation gives the Hagen–Poiseuille equation: HL 64 L v 2 Re D 2 g Example 1. Friction is the main cause of energy losses in fluid power systems. It is a very complicated problem and only in special cases.02 400 / 0. D is the inside diameter of the pipe (m). Theactual dependence of f on Re has to be determined experimentally. 1.5 Frictional Losses in Laminar Flow Darcy’s equation can be used to find head losses in pipes experiencing laminar flow by noting that for laminar flow. then 64 64 0.0001 m2/s.16 Re 400 .2 The kinematic viscosity of a hydraulic fluid is 0.1. (b) The velocity is 10 m/s. then Re vD vD vD 2 0.4 Darcy–Weisbach Equation If a fluid flows through a length of pipe and pressure is measured at two stations along the pipe. find the friction factor in each case: (a) The velocity is 2 m/s. L is the length of pipe (m). The prediction of this friction loss is one of the important problems in fluid power. It should be apparent that friction factors determined do not apply near the entrance portion of a pipe where the flow changes fairly quickly from one cross-section to the next or to any other flow in which acceleration terms are not negligible.0001 The flow is laminar. one finds that the pressure decreases in the direction of flow. Now f (b) If the velocity is 10 m/s. v is the average velocity (m/s) and g is the acceleration of gravity (m/s2). Solution: a) If the velocity is 2 m/s. find the head loss due to friction in units of bars for a 100-m smooth pipe.3 The kinematic viscosity of a hydraulic fluid is 0. p H L 1000 kg/m3 0.92 MN/m 2 1.03 1800 / 0.5 1.6 Frictional Losses in Turbulent Flow Darcy’s equation can be used to find head losses in pipes experiencing turbulent flow.5 m Hence.81 217.2 bar 1. .02 2000 / 0. Now f 64 64 0.90 9.Re vD vD vD 10 0.90. The oil has a specific gravity of 0.0001 m2/s.030 2 9.0001 Theflow is laminar. the friction factor in turbulent flow is a function of Reynolds number and the relative roughness of the pipe.032 Re 2000 Example 1.92 MPa 19. If it is flowing in a 30-mm diameter pipe at a velocity of 6 m/s.81 m/s 2 217. However.0001 We can express the head loss in bar as Re HL 64 L v 2 Re D 2 g 64 100 62 1800 0. Solution: We have vD vD vD 6 0. 1Typical values of absolute roughness for various types of pipe Type of Pipe (mm) Glass or plastic Smooth Drawn tube 0.0015 Wrought iron 0.6.26 Riveted steel 1. Tests have shown that head losses in valves and fittings are proportional to the square of the velocity of the fluid: v2 HL K 2g WhereK is called the loss coefficient of valve or fittings.1 gives typical values of absolute roughness for various types of pipes. .Re lies in between 2000 and 4000because it is not possible to predict whether flow is laminar or turbulent in this region.12 Galvanized iron 0.2.046 Asphalted cast iron 0. the majority of the energy losses occur in valves and fittings in which there is a change in the cross-section of flow path and a change in the direction of the flow. the friction factor can be determined easily. Table 1. No curves are drawn in the critical zone.1.8 To determine the values of the friction factor for use in Darcy’s equation.15 Cast iron 0.K factors for commonly used valves are given in Table1.1 Effect of Pipe Roughness The relative roughness of pipe is defined as the ratio of inside surface roughness ( ) tothediameter: Relative roughness D Table 1. If we know the relative roughness and Reynolds number. the straight line curves give the relationship for laminar flow: 64 f Re 1.7 Frictional Losses in Valves and Fittings For many fluid power applications. we use the Moody diagram.046 Commercial steel 0.At the left end of the chart (Reynolds number less than 2000). 4 For the hydraulic system shown in the Fig.096 m. (b) The elevation difference between stations 1 and 2 is 6. (c) The pump flow rate is 0.4.8 Equivalent Length Technique We can find a length of pipe that for the same flow rate would produce the same head loss as a valve or fitting. Find the pressure available at the inlet to the hydraulic motor.88 m.9.75 4 0. (e) The kinematic viscosity of oil is 75 cS.984 kW). . The pressure at the oil top surface level in the hydraulic tank is atmospheric (0 Pa gauge). 1. (d) The specific gravity of oil is 0. (f) The pipe diameter is 19.04 1.42 0. the following data are given: (a) A pump adds 2.22 and 4.00158 m3/s.20 0. 1.90 0.2 1. Example 1. This length of pipe. which is called the equivalent length of a valve or fitting.8 0.5 0. can be found by equating head losses across the valve or fitting and the pipe: 2 v2 L v K f D 2g 2g Thisgives KD Le f where Le is the equivalent length of a valve or fitting.5 24 2.90 4.305. (g) Pipe lengths are as follows: 0.05 mm.984 kW to a fluid (pump hydraulic power = 2.2 K factors for commonly used valves Valve or Fitting Globe valve Wide open 1/2 open Gate valve Wide open 3/4 open 1/2 open 1/4 open Return bend Standard tee Standard elbow 45 elbow 90 elbow Ball check valve Union socket K Factor 10 12.Table 1. The friction is f 64 64 0. P H m 0.00158 5.096 m The velocity at point 2 is v2 Q (m3 /s) 0. Re vD vD vD 5.542 1. v1 0 and 1 0 as oil tank is vented to the atmosphere.01905 1400 / 75 106 So the flow is laminar.Coupling Hydraulic Motor Breather Load 2 Hydraulic pump Prime mover Elevation 1 C Figure 1.4 Solution: Kinematic viscosity = 75 cS = 75 × 10−6 m2/s We write the energy equation between stations 1 and 2: Z1 p1 v12 p v2 H p H m H L Z2 2 2 2g 2g Since there is no hydraulic motor between stations 1 and 2.0457 Re 1400 .54 0.57 m 2 g 2 9. Z 2 Z1 6.01905)2 A (m2 ) 4 The velocity head at point 2 is v22 5.54 m/s (0. Now.81 Also. 6 1.0457 6.5 The oil tank for a hydraulic system (shown in Fig.97 kPa gauge pressure.2 The pump head is given by HP P (W) γ (N/m3 ) Q (m3 /s) Now γ 8817 N/m3 and P 2984 W .2 181. we solve for the pressure at station 2: p2 181.01905 6. (b) The inlet to the pump is 3.88 0. (e) The kinematic viscosity of oil is 100 cS. .4057 6. So HP 2984 W 214.57 p2 ( Z1 Z 2 ) H p p1 H p 33.3 m 8817 (N/m ) 0.41 0.001896 m3/s.1m of oil Finally.3 33.57 25. (d) The specific gravity of oil is 0.048 m below the oil level.9.6 m 0.897 kPa.We can now find the head loss due to friction between stations 1 and 2: fLp v 2 HL Dp 2 g where KD Lp 4.5) has the following details: (a) The oil tank is air pressurized at 68.22 f std elbow 0.01905 Next use Bernoulli’s equation to solve for P2 / : HL and v12 p2 v22 Z1 H p H m H L Z2 2g 2g p1 v22 HL 2g 6.096 H p 0 25. (c) The pump flow rate is 0.79 m 0.00158 (m3 /s) 3 Now p2 H p 33.9 0.305 1.2 214. (f) Assume that the pressure drop across the strainer is 6. 1.18817 1600000 1600 kPa Example 1.79 1. B and C) A B Breather 38 mm (ID) 1 CC Pipe length = 6m 3m 2 Pump Figure 1.5 Solution: Givenp1 = 68.001896 v2 1. ps = 6. This problem can be solved by the application ofmodified Bernoulli’s (energy) equation: Z1 p1 v12 p v2 H p H m H L Z2 2 2 2g 2g Also. .1 mm. Assuming the oil tank area (cross-section) to be large.1mm.9. Dp = 38.Z1−Z2 = 3m. (h) The total length of the pipe is 6. given Z1 Z 2 3 m We can write H m 0. Find the pressure at station 2. Q = 0.097 m. The pressure head at station 1 is p1 68970 7. the motor head is zero. To solve forP2. We have to find P2. SG = 0. Lp= 6m. Three standard elbows (A.001896m3/s.03812 ) A (m ) 4 The velocity head at point 2 is .66 m/s 2 (0.97kPa. the velocity at point 1 is v1 0 (negligible velocity).(g) The pipe diameter is 38. let us compute and substitute different quantities into the energy equation.897kPa. as there being no motor between points1 and 2.82 m 8817 The velocity at point 2 is Q (m3 /s) 0. =100 cS = 100 × 106 m2/s. 44 0. We know that Power( P) p1 Q P 68.097 3 f std elbow 0.12 0. The head loss across strainer can be calculated as Head loss across strainer pstrainer 6900 0.101 Re 632 Point 2 is before the pump.101 Pressure drop across the strainer is 6.001896 (m3 /s) 0.097 3 7.97 (kN/m3 ) 0.0381 6.782 m 8817 Now use Bernoulli’s theorem Z1 p1 v12 p v2 H p H m H L Z2 2 2 2g 2g p2 p2 (Z1 Z 2 ) p1 HL v22 2g 3.29 So p2 7. We can now find the head loss due to friction between stations 1 and 2 fLp HL Dp v2 Head loss across the strainer 2g Three standard elbows are used: KD Lp 6.v22 1.81 Re vD vD vD 1. we solve for p2 : p2 7.3 kPa If point 2 is after the pump.29 8817 64300 Pa 64.13 kW So.048 7.82 3.141 = 7.9 0. pump head is .662 0.9 kPa. therefore.29m of oil Finally. H p 0.66 0.0381 632 / 100 106 So the flow is laminar.141 m The Reynolds number is 2 g 2 9. the pump head is given by P (W) Hp γ (N/m3 ) Q (m3 /s) Here γ 8817 N/m3 .The friction is f 64 64 0. 048 7.7 m 8817 (N/m3 ) 0.13 103 W 7.09 m/s 2 (0.90 kinematic viscosity ( 0. ID = 25mm The globe valve is 25mm in size and is wide open. Pipe 1 Pipe 2 Qoil =40 GPM 25 mm dia 8 m long Globe valve Figure 1. we solve for p2: p2 15 p2 15 8817 132255 Pa 132.6 Solution: The velocity can be calculated as Q (m3 /s) 0.7 0.82 3.7 For the system shown in Fig.0001 m2 /s) and Q =0.001896 (m3 /s) We also know that Z1 p1 p2 v12 p v2 H p H m H L Z2 2 2 2g 2g p2 (Z1 Z 2 ) p1 HL Hp v22 2g 3.0025 v 5.26 kPa Example 1. SG =0.141 15m of oil Finally. ID = 25mm Pipe 2: length =8m. 1.0252 ) A (m ) 4 We know that 25 mm dia 8 m long .0025 m3 /s Find p2 p1 in units of bars.Hp 0.44 7.6. the following new data are applicable: Pipe 1: length = 8m. 0001 So the flow is laminar.025 Lp 8 8 16 21 m 0. Pipe: total length = 15m and ID = 38mm Oil: SG =0. . 1.81) 55.0503 21 5.025 2 9.092 = 55.0503 Re 1272 Also.09 0.8 For the system shown in Fig. the following data are applicable: P1 7 bar. Now friction factor is 64 64 f 0.90 and kinematic viscosity ( 0.7.93 bar Example 1.Re vD vD vD 5. So HL Now 0. Q =0.9 9.81 p HL p H L (1000 0.002 m3 /s . head loss is HL f Lp Dp v2 2g Now KD 10 0.8 m of oil 0.0503 f std elbow K = 10 for globe valve (fully open).8 493000 N/m 2 So p2 p1 p 493000 N/m 2 493 kPa 4.025 1272 / 0.0001 m2 /s) Solve for p2 in units of bars. 762 7.0001 So the flow is laminar.6 19.038)2 A (m ) 4 vD vD vD 1.76 0.096 15 4 0.002 1.096 19.038 669 / 0.81 Now we can express the head loss due to friction in terms as pressure using HL .038 0.6 1.6 m Head loss due to friction is given by 0.Motor Pipe length = 3m 90˚ elbow Pump Q Breather 38 mm (ID) Pipe length = 2m 90˚ elbow Load force C Pipe length = 4 m Pipe length= 6m Figure 1. The friction is f 64 64 0.096 Re 669 Now we can calculate the equivalent length ( Le ) as KD KD Le Lpipe 2 f globe valve f 90 elbow 10 0.096 0.7 Solution: The velocity V is v We know that Re Q (m3 /s) 0.82 m of oil 0.76 m/s 2 (0.75 0.038 2 9.038 15 2 0. p HL p H L (1000 0.9 9.9 For the fluid power system shown in Fig.81) 7. determine the external load F that a hydraulic cylinder can sustain while moving in an extending direction. 1.102 m All elbows are at 90 with K factor = 0. The pump produces a pressure increase of 6.75.7. Motor Check valve k= 4 3 Cylinder Breather Pump External load 4 13 5 2 Reservoir 12 9 6 1 8 7 10 11 Directional controlvalveK =5 Figure 1.69 bar p2 6. Kinematic viscosity 0.69 bar From the pressure drop we can calculated p2 as p1 p2 p 0.31 bar Example 1.8.82 690000 N/m2 0. The following data are applicable.00253 m3/s.0000930 m2 /s Weight density of oil 7840 N/m3 Cylinder piston diameter 0. Take frictional pressure losses into account. 1. Pipe length and inside diameters are given in Fig.69 bar 7 p2 0.203 m Cylinder rod diameter 0.8 .9 MPa from the inlet port to the discharge port and a flow rate of 0. 83 0.01902 ) 2 m 4 v11 . 6 v7.00189 (m3 /s) 3.10 0.02542 ) 2 m 4 v4.0254 0.8 0.73 m/s (0.0254 12 6 1.3 3.52 0.00253 (m3 /s) 2.02542 ) 2 m 4 v9.12.0381 8 2 1.00253 (m3 /s) 8.00253 (0.92 m/s (0.0190 0.67 m/s (0.99 m/s (0.05 6.0190 0.52 18.00189 m3 /s 0.00189 (m3 /s) 6.22 m/s (0. (b) Determine the extending and retracting speeds of cylinder.10 Solution:We can use the following equations 13 fL v2 p HL K 2g 1 Dp Now v Q (m3 /s) A (m2 ) We know that Re vD vD / vD Also Qreturn 0.52 1.05 0.0190 .01902 ) 2 m 4 Diameter 0.0254 13 7 1.0190 0.610 0.203 Velocity calculation: v1.3 0.52 1.13 0.52 0.2032 0.610 0.0254 (a) Determine the heat generation rate.2 0.03812 ) 2 m 4 0.1022 ) 0.0381 9 3 0.0381 10 4 15.5. Length (m) 1.Pipe Number Length (m) Diameter Pipe Number 1 0.0254 11 5 3.00253 (m3 /s) 4.2.0190 0. 6) Re 7.000093 Re(9.92 0. 3) 2.81 2 64 19.8 m 469000 Pa 1362 0.000093 All flows are laminar.13) 6.10) 8.0254 1018 0.99 H L(4.12.019 1822 0.67 H L(11.9 m 203000 Pa 1822 0.6) 10.019 2 9.00189)} = 1740 + 2410 .73 0.5.79 m 14000 Pa 909 0.00253) (46200 1230000) (0. Now f 64 Re Now head loss can be calculated for each element: 2 64 3.05 2.2032 ) (0.2.2.Reynolds number calculation: Re(1.05 3.05 8.81 2 64 27.4 6.3) 1.019 1363 0.22 0.8 4.22 H L(1.0381 2 9.81 Nowexternal load F that a hydraulic cylinder can sustain while moving in an extending directioncan be calculated as F Pressure on piston side × Area Pressure on rod side × Area (0.81 2 64 3.89 m 46200 Pa 1018 0.1022 ) F [(6900000) (14000 469000 203000)] (46200 1230000) 4 4 F (201000) (30900) 170000 N (a) Heat generation rate (power loss in watts) = Pressure × Discharge Power loss = {(14000 469000 20300) (0.000093 3.10) 0.81 2 64 3.8 4.000093 Re(11.0254 2 9.92 H L(9.019 2 9.75 25.0381 909 0.99 0.0254 1362 0.5.5 157 m 1230000 Pa 1363 0.5 1.0254 2 9.000093 Re(4.12.67 0.75 5.8) 0.2032 0.13) 1.5 59.73 H L(7. 10162 ) 4 4 Knowing the annulus area and discharge we can calculate the return velocity of piston as Annulus area = Vextending Q (m3 /s) 0.20322 ) (0.00253 0.104 m/s (0.00253 0.2032 m Area of pistonA = (0.= 4150 W = 4.20322 ) A (m ) 4 Now Cylinder rod diameter = 0.10162 ) A (m2 ) 4 4 .0780 m/s 2 (0.00253 m3 / s Cylinder piston diameter = 0.1016 m Area of rod = (0.15 kW (b) Now to calculate forward and retracting speed of the cylinder Qpump 0.20322 ) (0.10162 ) 4 m2 (0.2032 2 ) 4 m2 Knowing the area of piston and discharge we can calculate the forward velocity of piston as Vextending Q (m3 /s) 0. 2. 6. 2. In the flow of a real fluid. What are the important conclusions of the Reynolds experiment? 8. Write an expression for pressure drop down a pipe in terms of friction factor. 5. gravity affects the flow pattern. In the flow of a fluid through a completely filled conduit.Objective-Type Questions Fill in the Blanks 1. no definite and no observable pattern. Name two causes of turbulence in fluid flow. In the case of an ideal fluid flowing in a straight conduit. Briefly explain the method to calculate the equivalent length of a valve or fitting. the velocity adjacent to the wall is maximum. Reynolds number is the ratio of viscous forces to inertial force. 3. A distinguishing characteristic of turbulence is its irregularity. all the particles move in parallel lines with equal ______. 5. State True or False 1. Why is it important to select properly the size of pipes. Define Reynolds number and list its range for laminar and turbulent flows. 4. 9. 5. Friction is the main cause of energy losses in fluid power systems. What is meant by the equivalent length of a valve or fitting? 10. 3. valves and fittings in hydraulic systems? . Review Questions 1. 7. energy losses due to flow in valves and fittings may ______ those due to flow in pipes. Define the relative roughness and K factor of a valve or fitting. The flow characteristic of a viscous fluid is one in which ______ plays a significant part. 3. Darcy’s equation can be used to find head losses in pipes experiencing ______ flow. List the main causes of turbulence in fluid flow. 4. In fluid power systems. 4. Energy loss in turbulent flow is ______ than laminar flow. 2. Differentiate between laminar flow and turbulent flow. Exceed 2.Turbulent 3.Answers Fill inthe Blanks 1.Less State True or False 1.False 3.Viscosity 5.True .Velocity 4.False 5.True 4.False 2.