Fluid MachineryLecture notes Csaba H˝os [email protected] Csaba Bazs´o Rox´ana Varga May 10, 2014 Contents 1 Some basic relationships of fluid mechanics and thermodynamics 3 1.1 Continuity equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.2 Bernoulli’s equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.3 Energy equation for compressible flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.4 Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.4.1 Specific heat capacities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.4.2 Some basic thermodynamic relationships . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.4.3 Input shaft work and useful work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.4.4 Specific work for hydraulic machines . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.4.5 Efficiency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.5 2 Incompressible turbomachinery 10 2.1 Euler’s turbine equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.2 Velocity triangles and performance curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2.2.1 Radial (centrifugal) machines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 2.2.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.2.3 Axial machines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 2.2.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.2.5 Real performance curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Losses and efficiencies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 2.3.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Dimensionless numbers and affinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.4.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 Forces on the impeller . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 2.5.1 Radial force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 2.5.2 Axial force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 2.3 2.4 2.5 2.6 1 Fluid Machinery 2.7 2 Cavitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 2.7.1 24 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Hydraulic Systems 26 3.1 Frictional head loss in pipes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 3.2 Head-discharge curves and operating point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 3.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 4 Control 4.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Positive displacement pumps 5.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Hydro- and wind power 6.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 30 34 34 36 36 we obtain Z Z ρvdA = A ρv⊥ dA. we have div(ρv) = 0. /∂t = 0) and one-dimensional.4) A Note that the surface is defined by its normal unit vector dA and one has to compute the scalar product vdA.5) . between which we have rigid walls. if we integrate (1. (1.Chapter 1 Some basic relationships of fluid mechanics and thermodynamics 1. e. Its general form is ∂ρ + div(ρv) = 0 ∂t (1.g. ρ2 respectively. (1.3) on a closed surface A. Let us denote the average perpendicular velocities and the densities at the inlet A1 and outlet A2 by v1 . pipes. matter can neither be created or destroyed. . Than. For example. (1. If the flow is steady (∂ . This is the principle of mass conservation and gives the continuity equation. in many engineering applications the density can be considered to be constant. compressors. Thus vdA = |v| |dA| cos α = v⊥ dA.1) where div(v) = ∇v = ∂vx /∂x + ∂vy /∂y + ∂vz /∂z. However. In many engineering applications. etc. for the steady-state case.2) Moreover.1 Continuity equation In the absence of nuclear reactions.3) The above forms are so-called differential forms of the continuity equation. One can resolve the velocity to a component parallel to and another perpendicular to the surface as v = v ⊥ + v k . pumps. one can derive the so-called integral forms. there is an inflow A1 and an outflow A2 . ρ1 and v2 . 3 (1. . leading to div(v) = 0. we have m ˙ = ρ1 v1 A1 = ρ2 v2 A2 = const. also. 2 2 (1. we have ρ ρ p1 + v12 + ρgh1 = p2 + v22 + ρgh2 . 1. i. we simply state that the energy equation for frictionless.e. regardless of the forces acting on it. because it continues to exemplify fluid properties no matter how fast it is stirred or mixed.. i.3 dv ρL | {zdt} . we have Q = m/ρ ˙ = v1 A1 = v2 A2 = const. stationary flow of a compressible ideal gas without heat transfer takes the following form v2 + cp T = constant.) The Bernoulli equation can be extended to include friction and unsteady effects: X ρ ρ ρ ζi vi2 + p1 + v12 + ρgh1 = p2 + v22 + ρgh2 + 2 2 | {z2 } friction 1. for which the energy content per unit volume is Energy per unit volume = mgh + 12 mv 2 + pV ρ = p + v 2 + ρgh = constant.2 Bernoulli’s equation In the case of steady frictionless flow. the stress versus strain rate curve is linear and passes through the origin. the energy of the fluid along a streamline remains constant. • the flow is ideal. • points 1 and 2 refer to two points on the same streamline and • the fluid is Newtonian. (1. The constant of proportionality is known as the viscosity: τ = µγ.). 2 (1.Fluid Machinery 4 The quantity m ˙ is called mass flow rate (kg/s) and it simply reflects to the fact that under steady-state conditions the amount of mass entering the machine per unti time has to leave it. water is Newtonian.8) Note that the above form can only applied if • the flow is incompressible. V 2 (1. this means the fluid continues to flow.e. ρ = const. ˙ (In common terms. i.7) Considering two points of the streamline (the flow is from 1 to 2).9) unsteady term Energy equation for compressible flow Without derivation.10) where cp [J/kgK] is the specific heat capacity taken at constant pressure and T [K] is the absolute (!) temperature. Mostly we deal with incompressible fluids. separation. (1.e. there are no losses (friction. For example. . If the density is constant. etc.6) where Q (m3 /s) is the volumetric flow rate. isentropic. There are some forms of expressing the change in enthalpy (v = 1/ρ): dh = d(u + pv) = δq + vdp = T ds + vdp. (1.4. δY + δq = d h + 2 | {z } (1.1 5 Thermodynamics Specific heat capacities Assume that a definite mass of gas m is heated from T1 to T2 at constant volume and thus its internal energy is raised from U1 to U2 .4. And e=h+ c2 + gz 2 is the energy. (1. both of them being processes. κ−1 (1. It expresses the sum of the internal energy u and the ability to do hydrodynamic work p p h=u+ . one has to know what kind of process takes place in the machine (adiabatic. isotherm.4 1.11) where u is the internal energy per unit mass and cV (J/(kgK)) is the specific heat capacity measured at constant volume. to compute the overall process (to integrate the above equation).13) Thus we see that it is useful to define a new quantity which includes both the change of the internal energy u and the pressure work p dv = p d (1/ρ). ρ (1. open system in differential form is c2 + gz . gives cp ∆T = ∆u + p∆V = cV ∆T + R∆T → cp = R + cV .16) Note that h = cp T and u = cV T . thus its volume changes and work is done on the fluid: mcp ∆T = ∆U + mp∆V. which is emphasised by the δ symbol. We have mcV ∆T = ∆U or cV ∆T = ∆u.17) . Now we do the same experiment but now at constant pressure. after rewriting for unit mass and combining with the previous equation for constant volume process. 1.15) e δY is the elementary shaft work. cV cp = R κ κ−1 and cV = R 1 . the integral is inexact). (1. The term enthalpy is often used in thermodynamics.Fluid Machinery 1. Some useful equations: R = cp − cV .12) which. δq is the elementary heat transferred towards the fluid. however. (1.14) Some basic thermodynamic relationships One possible form of the energy equation for a steady.2 κ= cp . etc.) and the results depends on it (thus. Note that the above equation describes an elementary process. 17) we obtain dh = δq + T dsirrev + vdp.3 Input shaft work and useful work The input shaft power is simply the work needed to change the enthalpy of the fluid: .15) turns into δYu(sef ul) 1.20) with which (1.19) c2 δY = vdp + d + gz + T dsirrev.18) with which.4. | {z } 2 {z } | losses (1. (1. using (1. T ds = (1.Fluid Machinery 6 The entropy 1 is for an elementary change in the equilibrium is δq + dsirrev . . c2 − c21 Pin = m∆e ˙ =m ˙ h2 − h1 + 2 + g(z2 − z1 ) . . between the suction and pressure side of a compressor). = mc ˙ p (T2 − T1 ) 2 z1 ≈z2 .c1 ≈c2 (1. We have 1 Entropy is the only quantity in the physical sciences that seems to imply a particular direction of progress. κ − 1 ρ1 p1 |κ −{z1 } |{z} | {z } RT1 T2 /T1 (1.24) Yisotherm = ρ1 1 p p1 The real processes are usually described by polytropic processes but formally we use the same equations as in the isentropic case. sometimes called an arrow of time. . Although the gas heats up during the compression but in the vessel it will cool back to the pressure of the surroundings. We have p/ρ = RT (ideal gas law) and T =const. we have p/ρ = RT (ideal gas law) and p/ρκ = const. A typical compression system consists of a compressor and a pressure vessel. In the case of an isentropic process. Hence. entropy measurement is thought of as a kind of clock.22) Note that the above equation gives Yisentr. which stores the compressed gas. thus Z p2 p1 2 1 dp = RT1 ln (1.g. (1. = κ−1 κ p1 κ p2 κ −1 = R (T2 − T1 ) .20) between points 1 and 2 (e.21) When computing the useful work. We still assume that z1 ≈ z2 and c1 ≈ c2 . (κ is the isentropic exponent).23) cp which is exactly the input specific work defined by (1. we loose the heat energy and the ’useful’ process is isotherm. with the slight change of using the polytropic exponent n instead of κ. thus Z Yisentr.. the second law of thermodynamics states that the entropy of an isolated system never decreases. we integrate the Yu part of (1. = 1 2 1/κ 1/κ p p1 p−1/κ dp = 1 ρ1 ρ Z 2 p −1/κ 1 κ p1 dp = κ − 1 ρ1 " p2 p1 κ−1 κ # −1 . from this perspective. In other words. As time progresses.21). thus Z 2 1 " n−1 # ..Fluid Machinery 7 p/ρ = RT (ideal gas law) and p/ρn =const. n p1 1 . . ρ . p2 n n = dp −1 = R (T2 − T1 ) . 2 − pt. [m] = g ρg 2g N (1.isentropic = cp (T2s − T1 ) + 1.30) for a polytropic process. When neglecting the losses.t .1 .26) Specific work for hydraulic machines In the case of pumps.28) In the case of compressors. For a given T2 compression final temperature. the fluid cannot be considered as incompressible. [P a] = ∆pt = Yu ρ = p2 − p1 + ρ 2 m3 (1.t − h1.4 1 ρ Z 2 1 dp = 1 p2 − p1 .31) . Note that the polytropic exponent n is typically a result of curve fit that allows the accurate computation of the outlet temperature.25) Polytropic processes are real. the specific work is: Yu.5 Problems Problem 1. we have ηpolytropic = 1.polytropic n − 1 ρ1 p1 n−1 (1. However.4. ρ (1. non-adiabatic processes.5. we have ηisentropic = T2s − T1 . 2 (1. we have Yincomp.4. instead of Y usually the head is used: H= Yu p2 − p1 c2 − c21 J = + 2 + z2 − z1 . = 1. n−1 κ (1.27) In the case of ventilators. the energy change due to the geodetic heigth difference between the suction and pressure side is neglegible (z2 ≈ z1 ) and usually the change of total pressure is used: J c22 − c21 = pt.5 c22 − c21 = h2s. the fluid can be considered as incompressible. if the fluid is incompressible. Finally.29) Efficiency The ratio of the useful power and the input power is efficiency. T2 − T1 (1.1 n n−1 R(T2 − T1 ) cp (T2 − T1 ) = n κ−1 . Find the isentropic exponent. Find the power needed to cover the losses.6 Along a natural gas pipeline compressor stations are installed L = 75 km distance far from each other.7K.6K.3 < p2 /p1 < 3) and compressors (3 < p2 /p1 ). The exponent describing the politropic compression is n = 1. (Solution: (ρ2 − ρ1 )/ρ1 = 20. (Solution: A1 = 0. Yisentropic = 111. • Find the pressure. Gas constant R = 288 J/kgK. (Solution: κ = 1.5 Gas is compressed from 1 bar to 5 bar.3 Gas is compressed from 1 bar absolute pressure to 3 bar relative pressure. On the pressure side of the compressor the pressure is pp = 80 bar.6%.016 m2 . The exponent describing the real state of change is n = 1. Ploss = 66. Assuming p1 = 1 bar inlet pressure. Assuming that the process along the pipeline is isotherm. The surrounding air is at rest with p0 = 1 bar and T0 = 290 K.018. ht ≈ h is a reasonable approximation.5. The density of atmospheric air is 1. .16 kg/m3 .3%. • Find the needed compressor power assuming that the compression is a politropic process and n = 1. Pisoth. the density is ρ = 85 kg/m3 . The ambient air temperature at the inlet T1 = 22◦ C while at the outlet T2 = 231◦ C. the specific heat at constant pressure is cp = 1005J/kgK.u = 446.Fluid Machinery 8 The turbomachines conveying air are classified usually as fans (p2 /p1 < 1. while the velocity of the gas is vp = 6. The velocity in the inlet section is c = 180 m/s. Find the exponent describing the politropic compression and the density of air at the inlet and the outlet.5. the specific input work and the isentropic efficiency.5.407 kg/m3 .44kg/m3 . The gas constant is 288J/kgK. Find the absolute temperature and density of the air at the outlet.45. ρ1 = 1.402. t1 = 20o C inlet temperature and isentropic process (κ = 1. the density.9%. find the inlet cross section area and the isotherm useful power of a compressor conveying m ˙ = 3 kg/s mass flow rate. blowers (1.) Problem 1.4 Ideal gas (gas constant R = 288 J/kgK) with 27o C and 1 bar pressure is compressed to 3 bar with compressor. t2 = 128.2 Assuming isentropic process of an ideal gas.177kg/m3 .4). the pressure loss is calculated as p2beg −p2end 2 L = pbeg λ D ρbeg 2 2 vbeg . (Solution: n = 1.48 kJ/kg. Tisentropic = 410.4 m/s. if the mass flow is 3 kg/s. Find the isentropic specific useful work.5 kW) Problem 1. ηisentropic = 83. • Find the mass flow through the pipeline.3).5.729 kJ/kg.5.8kW) Problem 1.) Problem 1. Find the isentropic outlet temperature. κ = 1. the isentropic efficiency and the isentropic useful specific work. cp = 1000J/kgK. find the the relative density change (ρ2 − ρ1 )/ρ1 at the fan-blower border and the t2 outlet temperature at the blower-compressor border.1o C) Problem 1. ρ2 = 3.54. Yinput = 188.289 kJ/kg.4. ρ = 2.5. and the velocity at the end of pipeline.5. (Solution: Treal = 432. ηisentropic = 77. Yisentropic = 146. The diameter of the pipe is D = 600 mm the friction loss coefficient is λ = 0. m ˙ = 153.42 MW.Fluid Machinery 9 • Find the ratio of the compressor power and the power that could be released by the complete combustion of the transported natural gas.58%) .8 kg/s. ρs = 12. Pcomp /Pcomb = 0. Pcomp = 38.54 bar.26 kg/m3 . The heating value of the natural gas is H = 43 MJ/kg. vs = 44. (Hint: Pcomb = mH) ˙ (Solution: ps = 11.4 m/s. pumps. this corresponds to p2 /p1 ≈ 1.Chapter 2 Incompressible turbomachinery We classify as turbomachines all those devices in which energy is transferred either to. water and wind turbines are essentially the same machines. In what follows. or from. These enthalpy changes are intimately linked with the pressure changes in the fluid. Assuming isentropic process and ideal gas. also gases are considered to be incompressible. fans. Essentially. Figure 2. a continuously flowing fluid by the dynamic action of one ore moving blase rows. we give two derivations of the equation. a rotating blade row.3. Up to 20% relative density change.1 Euler’s turbine equation Euler’s turbine equation (sometimes called Euler’s pump equation) plays a central role in turbomachinery as it connects the specific work Y and the geometry and velocities in the impeller. 2.1: Generalized turbomachine Derivation 1: Moment of momentum Let us compute the moment of the force that is applied at the inlet and outlet of the generalized turbomachine 10 . a rotor or an impeller changes the stagnation enthalpy of the fluid moving through it. Thus. which is the potential of a rotating frame fo refernce: U = −r2 ω 2 /2. (For compressible fluids.6) where U is the potential associated with the conservative force field. rothalpy is I = cp T + 2. ρ 2 (2.3) =m ˙ [|c2 ||u2 | − |c1 ||u1 |] = m ˙ (c2u u2 − c1u u1 ) (2.1 : F = d (mc) dt → M= d ˙ (r × c) (r × mc) = m dt (2. Let w stand for the relative velocity. • c velocity is considered constant in the sense that its length and angle are constant. (2.. |{z} ρ 2 2 ρ 2 2 ρ 2 (2. and c is the velocity of the fluid on the radius r.8) ρ 2 which is exactly Euler’s turbine equation. c for the absolute velocity and u = rω for the ’transport’ velocity.4) where ui = |ui |.2) =m ˙ [c2 (ω × r2 ) − c1 (ω × r1 )] = m ˙ [c2 u2 − c1 u1 ] (2.) . With this the power need of driving the machine is mY ˙ = P = ωM = (M out − M in ) ω = m ˙ [ω (r2 × c2 ) − ω (r1 × c1 )] (2. we see that the specific work is Y = c2u u2 − c1u u1 .7) cu u Thus we see that the above quantity is conserved in a rotating frame of reference. and ci = |c2u |cos(α) Comparing the beginning and the end of the equation.5) Derivation 2: Rotating frame and reference and rothalpy The Bernoulli equation in a rotating frame of reference reads p w2 + + U = const. Let us find now the change of energy inside the machine: p c2 Y =∆ + = ∆ (cu u) . and the outlet with radius r2 .1) where m is the mass flow. Thus ˙ (r2 × c2 ) − m ˙ (r1 × c1 ) M = M out − M in = m .2 Velocity triangles and performance curves IDE SZERINTEM KENE EGY HAROMSZOG From the Euler turbine equation we have: ∆pe = ρgH = ρ (c2u u2 − c1u u1 ) c2 2 − ucu . thus w2 = u2 + c2 − 2u c = u2 + c2 − 2u cu . We consider the following assumptions: • The inlet of the turbomachine is a circle with radius r1 . which we refer to as rothalpy (abbreviation of rotational enthalpy). (2. We have c = u + w.Fluid Machinery 11 shown in figure 2. which gives r2 ω2 p c2 + u2 − 2cu u2 p c2 p w2 + − = + − = + − c u = const. 2.Fluid Machinery 12 where H is the head of the pump. b2 is its flow-through width at the outlet. see Fig. that is when c2u is increased than Q decreases (c2m decreases). So our goal now is to find a relationship between the head and the flow rate of the pump. the angle between u and w is approximately the blade angle β. Obviously.11) where D2 is the impeller outer diameter. In the reality. Figure 2. b2 is its flow-through width at the outlet and c2m is the radial component of the outlet absolute velocity. Basic trigonometrical identities show that c2u = u2 − c2m / tan β2 . we get: c2u u2 − c1u u1 (2.2. 2.10 we have that c outlet absolute velocity is the connection between the head and the flow rate of the pump. Also one can notice. From 2. i. (2. the .10) where D2 is the impeller outer diameter. (2.2: Centrifugal pumps The velocity triangle describes the relationship between the absolute velocity c. ~ Moreover. The theoretical flow rate is Qth = c2m A2 Ψ = c2m D2 πb2 Ψ. Ψ < 1 is a constant called blockage factor that takes into account that the real flow through area is smaller due to the blockage of the blade width at the outlet. that if ∆pe increased.1 Radial (centrifugal) machines Let us consider a centrifugal pump and the velocity triangles at the impeller inlet and outlet. the circumferential velocity u and the relative velocity w. Known the velocity triangle’s components and the density of the fluid. It is usual to assume that the flow has no swirling (circumferential) component at the inlet (due to Helmholtz’s third theorem).9) H= g The flow rate is Q = c2m A2 = c2m D2 πb2 .e. we know that (a) the circumferential velocity is u = Dπn and that (b) the relative velocity is tangent to the blade.9 and 2. 2. we have ~c = ~u + w. And if ∆pe decreased (c2u decreased) than Q increases (c2m increases). 9m) Problem 2. outlet flow angle is not exactly β2 .3: Centrifugal impeller with outlet velocity components.2 Problems Problem 2. the outlet width is b2 = 20 mm.Fluid Machinery 13 Figure 2. β2 > 90o .e.12) Thus. The inlet is prerotation-free.8 . β2 < 90o . for backward curved blades. the theoretical performance curve Hth (Qth ) of a centrifugal machine is a straight line. we have 2 2 c2u u2 u2 u2 u2 w2u u2 c2m Hth = λ =λ − =λ − g g g g g g tan β2 2 u2 u2 − Qth . i. (Solution: Hth = 22.4) • decreasing as Q is increased.2. The diameter of the impeller is D = 240 mm.7 A radial impeller runs at n=1440/min revolution speed and conveys Q = 40 l/s of water. thus the head is decreased. for radial blades (β2 = 90o ) and • increasing (as Q is increased) for forward curved blades. which is (see Figure 2.e. i. which is taken into account with the help of the slip factor λ (sometimes denoted by σ in the literature).2. The blade angle at the outlet is β2 = 25 degrees. If there is no prerotation (i. 2. c1u = 0). Find the theoretical head and draw a qualitatively proper sketch of the velocity triangle at the outlet.e. • horizontal. =λ g g tan β2 D2 πb2 Ψ (2.2. The mean meridian velocity component of a radial impeller with D2 = 400 mm diameter at n = 1440rpm revolution speed is cm = 2. an important difference between axial and centrifugal pumps (fans) is that in the case of axial machines.6%) 2. we have c2m = c1 (due to continuity). if we wanted to obtain constant ∆pt along the radial coordinate. the the pressure rise changes along the radial coordinate of the blade: c2m ∆pt (r) = ρu(r) (c2u (r) − c1u (r))|c1u =0 = ρ (2rπn) 2rπn − . u1 = u2 because it is assumed that the flow moves along a constant radius. Figure 2. see Fig. the change of the circumferential velocity has to be compensated by varying β2 . Notice that in this case. that results in 10% drop of the meridian velocity component. where Do and Di stand for the outer and inner diameter of the lade. The flow-through area is Do2 − Di2 π/4. The angle between the relative and circumferential velocity components is β2 = 25 degrees.5: Axial pump (left) and axial fan (right) However. (2.5 m/s. respectively.Fluid Machinery 14 Figure 2. Assuming (again) prerotation-free inlet (c1u = 0).2. 2. With a geometrical change of the blade shape.3 Axial machines In the case of axial machines the flow leaves the impeller axially. .13) tan β2 Thus. this angle is increased to to 28 degrees.5. (Solution: (H25o − H28o )/H25o = 4. Find the relative head change.4: Effect of blade shapes β2 angle on the performance curve. The inlet is prerotation-free. 6: Axial impeller with outlet velocity components.ax = c2. thus it was necessary to introduce a twist along its length.00137 m2 = 0.25 kg/m3 . The density of the air is ρ = 1.03425 m • umean = u1 = u2 = Dmean πn = 4. Due to the careful design the hydraulic efficiency is ηh = 85% however the volumetric efficiency as consequence of leakage flow rate between the housing and the impeller is just ηvol = 75%.2.5 mm the revolution speed is n = 2740 rpm.4 Figure 2.7: World War I wooden propeller Problems Problem 2. some 100 years later. While both the blade element theory and the momentum theory had their supporters. • Aring = (Do2 −Di2 )π • Dmean = 4 Do +Di 2 = 0. (Source: Wikipedia) 2.9 The outer diameter of a CPU axial cooler ventilator is Do = 47 mm the inner diameter is Di = 21. Draw the velocity triangles at the inlet and the outlet at the mean diameter. Find the flow rate and the total pressure rise on the impeller.ax = u tan β1 = 1.913 ms • cax = c1.788 ms .2. the Wright brothers were able to combine both theories. The blade angle at the suction side is β1 = 20◦ while at the pressure side β2 = 40◦ . They found that a propeller is essentially the same as a wing and so were able to use data collected from their earlier wind tunnel experiments on wings. Their original propeller blades are only about 5% less efficient than the modern equivalent. The twisted airfoil (aerofoil) shape of modern aircraft propellers was pioneered by the Wright brothers. They also found that the relative angle of attack from the forward movement of the aircraft was different for all points along the length of the blade.Fluid Machinery 15 Figure 2. The specific work along the radius is constant.o = 8.7.i = 13.131 ms • ∆cu = u − q2u = 2.2 at the inner and outer diameter. β1.4 degrees) .2. The inlet is prerotation-free.7. while the outer one is Do = 400mm. At Q = 0.ideal = ρu∆cu = 17.i = 16.Fluid Machinery 16 3 • q = ηvol Aring cax = 0. Find the angles β1.36 m3 /s the hydraulic efficiency is 85%.10 The inner diameter of an axial impeller is Di = 250 mm.o = 9.00184 ms • w2u = cax tan β2 = 2. β2.782 ms • ∆ptotal.5 Pa Problem 2. the head is 6 m.1 Pa • ∆ptotal = ηh ∆ptotal. (Solution: β1.7 and β2.ideal = 14. The revolution number of the impeller is 1470rpm. we have separation on the suction side of the blade.Fluid Machinery 2. one has to subtract the above two losses from the theoretical head: H = Hth (Q) − K1 Q2 − K2 (Q − Qd )2 . one experiences separation on the two sides of the blade.2.8 for a constant circumferential velocity u as the flow rate and thus the inlet velocity c is varied. Figure 2. Note that at the design point and close to it. the friction losses are moderate and no separation occurs. For small flow rates. On the other hand. the relative velocity w also varies. However neither of these assumptions are true.5 17 Real performance curves Our analysis so far assumed that the flow inside the impeller is ideal (no losses) and that the streamlines are following the blade shape (thus. . For lower flow rates. the higher the friction losses will be.8. Thus we have Hseparation To obtain the real performance curve. There are significant friction losses inside the impeller. For higher flow rates. thus Hf0 riction ∝ Q2 . if the angle of attack deviates from the ideal one. both friction and separation losses increase. the narrower the flow passage is. the friction loss decreases while separation increases.8: Friction and separation losses in the impeller. larger flow rates the separation is on the pressure side of the blade. while for 0 ∝ (Q − Qd )2 . which is illustrated in 2. This is illustrated in Figure 2. Moreover. the volute also introduces friction losses. These losses are proportional to the velocity squared. blade angles are also the streamline angles). At the design flow rate Qd the angle of attack ideal. (if any). The remaining 0 power is the theoretical power of the impeller: Pth = Pi − Pdf = (1 − νdf )Pi . Pv0 The theoretical head Hth and flow rate Qth and is further decreased by the leakage flow rates (Ql(eakage) ) inside the pump (flow across the gaps between the impeller and the housing) and the internal frictional losses h0 (e. which is taken into account by the disc friction coefficient: Pdf = νdf Pi . We have than 0 These represent the friction loss in the bearings and the mechanical sealing losses Mechanical losses Pm 0 .Fluid Machinery 2.1 Problems Problem 2. Hydraulic and volumetric losses Ph0 .9).14) .9: Losses of the pump. Let the input mechanical power transmitted by the shaft be denoted by Pinput . We have Pth = Qth ρgHth = (Q + Ql ) ρg (H + h0 ) = QρgH + Ql ρgH + Qth ρgh0 | {z } | {z } | {z } Pv0 Pu = QρgH Q + Ql H + h0 Qth Hth = QρgH Q H Q H |{z} |{z} ηv−1 2.11 −1 ηh → Ph0 Pu = Pth ηh ηv (2. The remaining power is called internal power Pi = Pinput − Pm 0 Disc friction losses Pdf A significant shear force appears in the fluid entrapped between the housing and 0 the impeller. in the impeller and volute).3. Figure 2.3 18 Losses and efficiencies Let us analyse the losses that decrease the efficiency of a turbomachine (see Figure 2.g.3. 5kW. the flow rate is Q = 0.63 m Pth ρgHth → Hth = 45 + 4.9% • Pth = (1 − ν)Pi = 28. Ph0 ) → Pu(sef ul) 0 • Pi = Pinput − Pm = 30.9% ηhydr = 90.055m3 /s and the head is 0 H = 45m. including leakage flow rate and the theoretical head.2% .7 kW → ηm = 95.00395 m3 /s • ηoverall = ηv · ηh · (1 − ν) · ηm = 75. Solution: 0 0 The power flow chart is: Pinput (Pm ) → Pi (Pdf ) → Pth (Pv0 . The input power at this operating point is Pin = 32kW. Make a complete analysis of the losses.Fluid Machinery 19 The revolution number of a water pump is 1470 rpm.3kW.0589 m3 /s → → Qleakage = 0.065.63 = 49.6% → ηv = 93.63 m = 0. the disc friction coefficient is νt = 0. The hydraulic power loss is Ph0 = 2.7 kW • h0h = Ph0 ρgQ • Qth = = 4. the mechanical power loss is Pm = 1. 29 log 44 (2.32 . Let us start with the elimination of D2 . resulting in new dimensionless numbers. 2 where ψ is a dimensionless pressure rise.048Qopt − 0. This is called affinity law : H1 = H2 n1 n2 2 . 2. both ψ and ϕ contains two parameters. Similarly. D2 and u2 . out of which one can be eliminated.21) nq Note that σ depends only on the revolution number but takes different values along the performance curve. This diagram is called Cordier-diagram. The centre of the path can be assumed with . [m]) Specific speed defines the shape of the impeller. Moreover. Thus when actually computing it. turbomachines having good efficiency pass a narrow path.19) 2gH D22 π 2 n2 (2.Fluid Machinery 2. Q1 n1 = Q2 n2 P1 = P2 → n1 n2 3 (2.94 − 0. What we found is that H ∝ n2 and Q ∝ n allowing the transformation of the performance curve given at n1 to be computed to another revolution number n2 . [m3 /s] 3/4 (2. one takes the data of the best-efficiency point.10. low specific speed means low flow rate and high pressure rise (radial impeller) while high specific speed occurs when the flow rate is high and the pressure rise is low. the specific speed of a turbomachine is nq = n[rpm] 1/2 Qopt.22) (Hopt. we do not √ π √ include the constant term 4 2g3/4 .17) These dimensionless quantities are called pressure number ψ and flow number ϕ. we define dimensionless numbers as H = ηh Hth = 2ηh u2 c2u u22 := ψ 2 u2 2g 2g (2. ηmax = 0. in the case of fans ρ ∆pt = ψ u22 .15) or. Finally. we have Q = ηv Qth = ηv D2 πb2 c2m = ηv (2. see Fig. by definition.20) √ √ 3/2 2 Q D2 π 3/2 n3/2 ϕ1/2 π Q1/2 √ = = n 3/4 √ 4 3/2 3/4 3/4 3/4 ψ 2g D2 π n (2gH) | H{z } (2. ϕ= ψ= from which we have σ= Q D22 π 4 u2 H u22 2g = = 4Q D23 π 2 n (2.18) As we have seen. Based on experience the available maximum efficiency can be estimated in the knowledge of Qopt and nq as follows n 2 q −0.23) Representing δopt (σopt ).4 20 Dimensionless numbers and affinity Based on the previously obtained formulae for theoretical head.16) D2 π 4D2 πb2 c2m u2 D22 := ϕ 2 u2 2 4D2 u2 4 (2. Based on the specific speed. (2.34 .41 log(σ) 1.036. ϕ = 0. (Solution: nq = 30.10: Specific speed and shape of the impeller.4.02m3 /s.Fluid Machinery 21 Figure 2. D2 = 240mm.66 − 0. (2. nq =27. Find the head and the specific = 0. the head and flow rate at the best-efficiency point are 17m and 0. Find the diameter of the impeller if. the revolution number is 1440 rpm.06 m3 /s. (Solution: H=30. Find the specific speed. The efficiency is = 0.14 The head produced by a six stage pump type CR 8-60 is H[m] = 68 − 0.24) Experience moreover shows that for a given nq estimation can be given for the ideal value of ψ as follows ψ= 2.2Q2 .13 The revolution number of a pump is 1450 rpm.00731(Q − 9. Find the flow number ϕ. the pressure number at the best-efficiency point should be ψ = 1.1 1. The volumetric efficiency is estimated as ηv = 0. Find the specific speed.3m. The unit of the flow rate in the formulae is [m3/h].5)2 .3 kW. the flow rate is 0.4.1 300 270 + nq 9/4 . the mechanical loss is Pm disc friction power loss is Pdf speed and make a sketch of the impeller.12 The input mechanical power of a water pump is 25 kW.4.25) Problems Problem 2.85.4. Find the head and flow rate at 970rpm. the hydraulic efficiency is ηh = 0.03 m3 /s.92. H970rpm = 7. δ= 2. Q970rpm = 0. based on industrial experience.) Problem 2. find the type of the impeller. Determine the input power of the water delivering pump .3. the impeller is a thin radial one.9 kW. the 0 0 = 1.61m) Problem 2. 4.02855m3 /s. mixed impeller.Fluid Machinery 22 for zero delivery Q = 0 by extrapolation from calculated points in the range Q = 1. 01m3 /s− 0.) . 1.5.4.) Problem 2.11. Qopt = 0.5m3/h. Pin = 1334W . Calculate 5 points of the pump-characteristic for the rotor speed n2 = 2900/min in the flow rate range Q2 = 0.17 The performance curve of a pump at 1450 rpm is given by H = 100 − 30000 Q2 and the efficiency is given by η = −78000 Q2 + 4500 Q. if the revolution number is 3000 rpm. H1740rpm = 144 − 30000 Q2 .16 Find the specific speed of the pump given by 2.9. Find the performance curve at 1740 rpm. Give the equation of the characteristics H2 (Q2 ) for the rotor speed n2 ! (Solution: H2 (Q2 ) = 160 − 40000Q22 ) Problem 2. Find the head and flow rate of the best-efficiency point.4. (Solution: nq = 29. Make a sketch of the impeller. (Solution: nq = 92. and using L’Hopital’s rule.) Figure 2.11: Performance chart for Problem 2. According to laboratory tests the affinity law is valid in this range. hence the impeller is radial.15 The characteristic curve of a pump at n1 = 1450/min rotor speed is H1 = 40m−40000s2 /m5 Q2 . 01m3 /s intervals. 0.9%.4. (Hopt = 76m. etam ax = 64. Problem 2.16. 05m3 /s at 0. 26) Figure 2. (2. 2 The axial force becomes.2 Axial force The axial force results from two components: • Momentum force • Pressure distribution on the hub and shroud. The momentum force is Fm = mv ˙ = ρr12 πc21 . on the hub (back of the impeller) Z r2 ρ r2 − rs2 Fhub = 2rπp(r)dr = · · · = r22 − rs2 π p2 − ∆p02 − ωf2 2 .12: Pressure distribution on the hub. (2. The overall axial force is Fax = Fhub − Fshroud − Fm .5.29) and its direction is towards the suction side (the axial force tries to ’pull down’ the impeller from the shaft). .5 2.Fluid Machinery 2.27) (2. TODO: Extend explanations.1 23 Forces on the impeller Radial force TODO 2. e.g. The pressure distribution is p(r) = ρ 2 2 r ωf + K 2 p(R2 ) = p2 − ∆p02 → ρ p(r) = p2 − ∆p02 − ωf2 r22 − r2 .28) A similar result is obtained for the shroud (front of the impeller) with replacing rs by r1 .5. 2 2 rs (2. 19 Calculate the axial force acting on the supporting disc of a pump impeller of 280mm diameter if the pressure at the impeller exit is 2bar.6 + 13600 × 0. Calculate the required geodetic height of the reservoir to avoid cavitation! The pipeline losses are to be taken into account.7.6 + 13600 Q2 Solution: It’s easy to calculate that Q = m/ρ ˙ = 0.626[m] Bernoulli’s equation between a surface point in the tank and the suction side of the pump reads: pt 02 ps c2 + + Hs = + s + 0 + h0pipeline ρg ρg ρg ρg From the suction side of the pump to the impeller we have: ps c2 pvapour + s = + es + N P SH ρg 2g ρg .38[m] N P SH = 1.3bar and the revolution number is 1470rpm. d = 100[mm].7 2. There is no leakage flow through the gap between the rotor supporting disc and the casing.20 A pump delivers water from a low-pressure steam boiler as shown in the figure below.027472 = 78. The angular velocity of the circulating water is half of that of the rotor. The average angular velocity of the fluid is 85% of that of the impeller.027472 = 2. λ = 0.6.7. (Solution: F = 9. Find the formula of pressure distribution as a function of the radial coordinate! Draw the cross section of the impeller and the axial pressure force! 2.18 Find the axial force on the back of the impeller.6 24 Problems Problem 2. N P SH[m] = 1.02 and the sum of loss factors is ζ = 5 • pump: H[m] = 82 − 4800 Q2 . The hub diameter is 40mm.6. whose outer diameter is D2 = 300mm. density of the hot water: ρ = 983[kg/m3 ] • pipe: L = 10[m]. the shaft diameter is Ds = 50mm.5[m/s] H = 82 − 4800 × 0.Fluid Machinery 2. • mass flow rate: m ˙ = 27[kg/s]. The rotor speed is 1440/min.1 Cavitation Problems Problem 2.02747[m3 /s] cs = Q/A = 3. the outlet pressure is 2.36kN) Problem 2. Fluid Machinery 25 (Note that es = 0 as the configuration is horizontal.13: Installation of the apparatus. 50. the friction coefficient is λ = 0.107(t − 15) + 0. The equivalent pipe length on the suction side is 5m. The hydraulic loss of the suction side pipe can be calculated from h0s = 652[s2 /m5 ]Q2 while the vapour pressure pv = 1.7. ρg where h0pipe c2 = s 2g Hs + L λ +ζ . d thus. 80.481m.21 Calculate the required pipe diameter to avoid cavitation. σ = 0. Problem 2.004(t − 15)2 . where the pressure (above the water level) is p = 40 kPa.8 kPa. • N P SHa = pt −pv ρg − Hs − h0s c2 → h0s = pt −pv ρg − Hs − N P SHr 2 Le s Le 8Q • h0s = λ D = λD 5 2 s 2g s D gπ s • Ds = 0. (Solution: Hs = 3. Find the Thoma cavitation number.03355[−].154) .7. 90) Solution: The sketch of the installation is shown in Figure 2.03m.22 Find the required suction side height of the pump that conveys water from an open surface reservoir at Q = 180m3 /h flow rate the head is H = 30m the required net positive suction head N P SHr = 5.704 + 0.13 Figure 2. The vapour pressure at the water temperature is 2. −1 c2s λ c2s λL Hs = 1 + N P SH + +ζ = · · · = 6.42[m] 2g d 2g d Thoma’s cavitation coefficient is σ = N P SH/H = 0. we have pt − pvapour Hs = − + h0pipe + N P SH. the suction flange of the pump is 3 m below the water level.02. if the pump delivers Q = 30 dm3 /s water from a closed tank. 65. The temperature of the water is T = 23◦ the ambient pressure is p0 = 1023mbar.2 m. The required net positive suction head is N P SHr = 3.073m → Ds = 80 mm Problem 2. (The standard pipe diameter series is: DN 40.) Putting the above two equations together. The most common equation used to calculate major head losses is the Darcy Weisbach equation: h0f = λ L 8Q2 L v2 =λ . the energy loss manifests itself in head (pressure) loss.because of the continuity equation . fittings. instead of pressure p [P a]. and ”minor losses” associated with bends. Head loss is divided into two main categories. In real moving fluids.1 Frictional head loss in pipes p In hydraulic machinery.08 for commercial pipes. . we have λ = 64. this equation need iteration for computing the actual value of λ. • For laminar flow Re < 2300. we have different regimes based on the Reynolds number. in the range of 4000 < Re < 105 . the Blasiuss √ 4 formula is ususllay used: λ = 0. D 2g D D4 π2 (3. minor losses can easily exceed major losses. with a relatively large number of bends and fittings. as well as turbulence. but . However. Based on Nikuradses experiments. we have √1λ = 1.Chapter 3 Hydraulic Systems 3. usually the term head is used: H [m] = ρg . valves. etc.the flow rate is constant. ”major losses” associated with energy loss per length of pipe. 0.7D e √ Colebrook-White equation covers both the smooth and rough regime: √1λ = −2 log Re1. 2 26 (3. energy is dissipated due to friction.03 .2) .55. Karman-Prandtl equation may be used: √1λ = −2 log 3.316/ Re. . e The • For turbulent flow in rough pipes. √ • For turbulent flow in smooth pipes. the value of λ is uncertain and falls into the range of 0.1) where the friction coefficient λ (sometimes denoted by f ) depends on the Reynolds number (Re = vD/ν.7D λr For relatively short pipe systems. These losses are usually taken into account with the help of the loss factor ζ in the form of ρ h0 = ζ v 2 . • For transitional flow 2300 < Re < 4000. ν [m2 /s] = µ/ρ being the kinematic viscosity of the fluid) and the relative roughness e/D (e [m] being the roughness projections and D the inner diameter of the pipe).95 log(Re λ) − 0.88 + 3. Instead. Note that as the hydraulic power is P = ρgHQ. etc. the loss of the elbows.1: Simple pumping system. see Figure 3. minor losses (ζ) are usually estimated from tables using coefficients or a simpler and less accurate reduction of minor losses to equivalent length of pipe (giving the length of a straight pipe with the same head loss). Such relationships can be found in *[2]. The head Hs(ystem) needed to convey Q flow rate covers the pressure difference and the geodetic height difference between the starting and ending point and the losses of the flow: the friction of the pipe. 3. valves. valves.Fluid Machinery 27 In design. (3. P L • λ D stands for the frictional loss of the pipe • +1 represents the discharge loss at index 2.3) .2. that ends up in a reservoir. and the discharge loss. etc..2 Head-discharge curves and operating point Let us consider a single pipe with several elbows. Figure 3. etc. Hs(ystem) X 2 X L p2 − p1 v = + (z2 − z1 ) + ζ+ λ +1 = Hstat + BQ2 ρg D 2g where • Hstat = • B= P p2 − p1 + (z2 − z1 ) ρg ζ+ P L λD +1 8 D4 π2 g Note • Hstat is independent of the flow rate Q P • ζ denotes the loss of the elbows. fittings. *[1] or *[3]. 3.) is ζ = 0.Assuming 65% overall (pump+motor) efficiency. Calculate the flow velocity in the second pipe and the overall flow rate of the common pump feeding the two pipes. Lp = 12 + 20 + 8 = 40[m] 2 Hpipe = Hstat + KQ = Hstat + " Ls λ + ζA + ζB Ds c2s + 2g P 2# cp Lp λ = + ζF + ζC + ζD + 1 Dp 2g = 12[m] + 3.35.pressure = cApipe • The ’extra’ 1 in the pressure side (.25[s2 /m5 ] × Q2 [m3 /s]2 Problem 3.25 . ζB..36. • Hstat = 8 + 4 = 12[m]. the friction coefficient of the straight segments is λ = 0.Fluid Machinery 3.. (Solution: Without the bypass line: H = 22. ζF = 0. λ = 0. calculate the energy demand for 100 days and the cost of the operation if the energy tariff is 32HU F/kW h. Solution: • Static (geodetic) head + dynamic (friction) losses of the pipe: Hpipe = Hstat + Hf riction • Volume flow rate: Q = cs(uction) Apipe. P = 5.3.3 28 Problems Problem 3. The diameter of the pipes is D1 = 100mm and D2 = 70mm.24 The artificial fountain Beneath the St.26. Q = 0.D = 0. Assuming that the flow velocity in the first pipe is 1.23 Calculate the head loss of the pipe depicted in the figure below as a function of the volume flow rate! Parameters: ζA = 1. Ls = 7 + 6 = 14[m]. calculate the the required head.ζD + 1) represents the outflow losses. ζC = 0.4[m3 /s].0155.02 and the friction coefficient of the other segments (bends.3.6[m] and Q = 0. Gellert is fed by two pipelines of 30m length.5. etc.826m[].5m/s.) Problem 3.suction = cp(ressure) Apipe.5.78[kW ] and Cost = 443691HU F .01678[m3 /s]. The height distance between the pump and the fountain is 22m. Ds = Dp = D = 0. whose water level is 25[m] below the default level.7[m]) Problem 3.3.25. d = 160mm. (Solution: Hsystem = 11. λD = 0. Calculate the required pump head! (Solution: Hp.80m) .22.07m.Fluid Machinery 29 A pump delivers Q = 1200[dm3 /min] water from an open-surface well. ζbD = 0. Find the head of the system and the pump head! Further data are: D = 400mm. ζ2bd = 0.35.018. Hpump = 12. The pipe collecting the water of five equal pumps has a diameter D.021.6 and ζp = 14 (without the outflow losses). ζnrv = 0. The inner diameter of the pressure tube connecting the pump with the collecting pipe is d. The loss coefficients are ζs = 3. The pressure side ends 5[m] above the default level and the water flows into an open-surface swimming pool.5.26 The submergible pump shown in the picture below delivers Q = 30l/s water into the basin.req. The diameter of the pipe on the suction side is Ds = 120[mm] and Dp = 100[mm] on the pressure side. = 35. ζbd = 0. λd = 0. ζf ilter = 3 . Chapter 4 Control 4.27 A pump running at 1470[rpm] with Hpump = 45 − 2781Q2 head delivers water into a pipeline with Hpipe = 20 + 1125Q2 .5[rpm] ∗ Q 0.1 Problems Problem 4. while varying the revolutionary speed.2[m]. let’s denote this constant by a. Q∗ is given by the intersection of the affine parabola and the original pump characteristic: Hap (Q∗ ) = Hpump (Q∗ ).06148[m3 /s] with H ∗ = 34. also H/Q2 remains constant. the working point moves along the central parabola (see figure). Calculate the required revolution number for the reduced flow rate Q0 = 0.08[m3 /s] and H = 27.06148 and just for checking the calculation 30 . we know that H 0 = 20+1125·0. the parameter of the affine parabola is a = H 0 /Q02 = 9125. However. Thus.05 = 1470 × = 1195.052 = 22. • Affinity states that while varying the revolutionary speed.05[m3 /s]. H/n2 and Q/n remain constant. given by Hap = a Q2 .1. Now we can employ affinity between Q∗ and Q0 : n0 = n∗ Q0 0. Thus.5[m]. as Q0 is given and we also know that this point has to be located on the pipeline characteristic. which gives Q = 0. Solution: • The actual working point is given by the solution of Hpump = Hpipe . So.81[m]. which gives Q∗ = 0. desired flow rate: Q0 = 0.81[m]. pipeline: Hpipe = 20 + 1125Q2 .1.Fluid Machinery 31 0 H =H ∗ n0 n∗ 2 = 34. 14702 Problem 4.52 = 22.5 × 1195. Compare the resulting operations in terms of power loss! .28 Solve the previous control problem (pump: Hpump = 45 − 2781Q2 .05[m3 /s]) using a throttle at the pressure side of the pump and also with a bypass line. 1. and if it it opened. The static head of the second side pipe is 25[m].4−25 0.0193[m /s] • Thus.4 90000 = 3 0. so the flow rate of the pump is Hp (Qp ) = H1 (Q1 ) → → Qp = H1 (Q1 ) = q 70−36.008[m3 /s] 36. whose opening results in a flow rate of 480[l/min] in the main pipe.01132 = 89279 • The solution is H2 (Q2 ) = 25 + 89279Q2 . Problem 4. works together with two parallel pipes.32 .Fluid Machinery 32 Problem 4.1.1. Their characteristics are: HI = 45m − 24900s2 /m5 Q2 HII = 35m − 32200s2 /m5 Q2 H1 = 10m − 4730s2 /m5 Q2 H2 = 15m − 8000s2 /m5 Q2 Find the flow rates and heads if valve ”V ” is closed. Which arragement will deliver more liquid through the pipe Hp = 20m + 25000s2 /m5 Q2 ? Problem 4.4[m] → a= 36. Solution: • Head of the main pipe at the prescribed flow rate: Q1 = 480[l/min] = 0.31 Two pumps. Calculate the head-flow relationship H2 (Q) of the side pipe. H1 = 70m − 50000s2 /m5 Q2 and H2 = 80m − 50000s2 /m5 Q2 can be coupled parallel or in series. whose characteristic curve is given by Hpump = 70 − 90000[s2 /m5 ]Q2 .4[m] • The head is the same.29 A pump.1. Problem 4.0113[m3 /s] • The actual characteristic of the side pipe: H2 (Q2 ) = 25 + aQ22 = 36. the flow rate on the side pipe is Q2 = Qp − Q1 = 0.0193 − 0.30 Pumps I and II feed pipes 1 and 2 shown in the figure below.008 = 0. The main pipe is given by H1 = 30 + 100000[s2 /m5 ]Q2 . 275m3 /s and −0. [m3 /s].29m3 /s.159Q2 .1.14m3 /s.) Problem 4.15m3 /s. . Qtower = 0.03. is feeding an irrigation system consisting of parallel pipes. Here Q is positive if water flows down from the tower. How large is the hydraulic loss in the valves in the first and in the second case? The power consumption of the pump is Pinput = 9. • Draw the sketch of the irrigation system with 3 parallel pipes! • How much water is discharged if only one pipe is in operation? • How many parallel pipes can be fed if the overpressure before the sprinklers must be 2bar? Problem 4. its characteristics is HT = 40 − 55|Q|Q.33 The characteristics of a pump supplying a small village with water is Hp = 70 − 330Q2 .Fluid Machinery 33 Pump S.015m3 /s. A high water tower is attached to the delivery tube of the pump.4 + 6240Q − 50000Q2 . The sprinklers discharge 4m3 /h water at 2bar overpressure. How large is the specific energy consumption f in the two cases? The units in the formulae are: [m]. [H] = m. Find the flow rates of the pump. Hpump = 36m and 41m. Each pipe contains at its end a sprinkler. the friction coefficient is 0.6m3 /s and 0. village and tower both for day and night operation.33m3 /s and 0. The units are Q[m3/h] and H[m].1. their inner diameter is 25mm. Draw a sketch of the water system. Draw the pump-pipe-valve arrangements for both cases. [kW ]. Qvillage = 0.34 How much water is delivered by the pump Hp = 70 − 45000Q2 through the pipe system Hs = 20 + 20000Q2 ? The flow rate must be reduced to 0. with the characteristitc curve HS = 37 − 0. The village network is modeled by the curve Hcd = 25 + 30Q2 during the day while the night operation can be described by Hcn = 25 + 750Q2 . This can be done either by throttling control or by using a by-pass control. Find the head of the pump both for day and night! Use a millimeter paper to draw the charasteristics curves! (Solutions: Qpump = 0. The units in the formulae are [Q] = m3/s. The pipes are 20m long. their characteristics can be written as Hspr = Kspr Q2 . Solution: Qmean = 2 × Apiston × s × n = 2 × 0.abs.6−0. relative pressure) located 50[m] above the suction tank. − p0 + ρgH = ptank.37 The piston diameter of a hydraulic cylinder is 50mm.15 × 120 60 3 = 6. The volumetric efficiency at 35bar pressure difference is 92%. the stroke is 150[mm] and the driving motor runs at 120[rpm].35 Calculate the hydraulic power of the double-acting piston pump.e. Diameter of the piston is D = 120[mm].rel.72[kW ] Problem 5. which delivers water from an open-surface tank into a closed one with 500[kP a] gauge pressure (i. It may not be lowered faster than 64mm/s.122 π 4 × 0. Prepare a sketch of the gear pump showing the rotation direction of the shafts.1. intake and delivery ports! How large will be P . How large is the driving torque if the pump efficiency is 85%? Problem 5.1.93 − 0.78 × 10−3 [ ms ] ∆p = ptank.1. The hydraulic oil of 970kg/m3 density leaves the cylinder through 34 .Chapter 5 Positive displacement pumps 5. Q if the rotor speed is n = 1440/min? Problem 5. How large must be the flow rate Q of the gear pump rotating with n = 960/min speed if its volumetric efficiency is 92%? Find the geometric volume of the pump and the pressure rise produced by it! Find the power P and the torque M of the driving motor! The pump efficiency is 74%. The cylinder diameter is 50mm. + ρgH = 991[kP a] P = Q∆p = 6.1 Problems Problem 5. An 800kg load is lifted by the piston rod of 20mm diameter with 12m/min velocity. Find the volume flow rate and the geometric volume! The shaft speed is 80rev/min.36 The characteristic curve of a gear pump is Q[dm3 /min] = 11. The pump delivery curve is Q[liter/min] = 8. M .38 The piston diameter of a vertical hydraulic cylinder is supporting a mass of 700kg.0043δp[bar].1.0467p[bar]. the piston rod diameter is 28mm. Find the valve area at the maximal opening! . The discharge coefficient of this valve is = 0.Fluid Machinery 35 a throttle valve.7. The diameter of the rotor is D2 = 55 m.1.28 MW • P usef ul = ηturbine ηgenerator P input = 1. The measured average water depth is h = 2.and wind power 6. The calculated average velocity is v = 0.106 MW • Rh = Area P erimeter = Bh 2h+B = 2.Chapter 6 Hydro.1 Problems Problem 6.1. Problem 6.4 m/s.35 m Turbine type: Kaplan turbine. Find the calculated efficiency related to the theoretical maximum of the efficiency? Solution: • ∆v = v 1 − v 3 = 4 m/s 36 . The efficiency of the turbine is ηturbine = 90 % the efficiency of the generator is ηgenerator = 96 %.2 kg/m3 . the density of the air is ρair = 1.39 The cross-section of a plant water channel is given.5 m • P input = QρgH = 1. The measured average wind speed at the level of the rotor is v 1 = 12 m/s.40 The instantaneous efficiency of an existing wind turbine is to be calculated. The average speed of the air behind the rotor is v 3 = 8 m/s.5 m2 • Q = Av = 29 m3 /s • H = hupstream − hdownstream = 4.5 m.9 m. What is the value of the hydraulic radius? What type of turbine is suitable for this power plant? Solution: • A = hB = 72. The velocity of the water flow is measured at several locations of the crosssection using a cup-type anemometer. The input power and useful power of the power plant are to be calculated. the width of the channel is B = 25 m. The height difference between the upstream and downstream water depth at the dam is hupstream − hdownstream = 4. 140 MW 1 = 0.463 MW 2 ∆v ∆v 1 − 2v v1 = 1.Fluid Machinery • A2 = D22 π 4 37 = 2376 m2 • P input = ρair A2 v 1 v 21 2 • P usef ul = ρair A2 v 31 • η = CP = P usef ul P input = 2. .463 < 16 27 = 0.593.41 How large is the power of a wind turbine of 30m rotor diameter if the wind speed is 8m/s? The Betz limit of the power coefficient Cp is 0.593 Problem 6.1.