Chapter 1Some basic relationships of ﬂuidmechanics and thermodynamics1.1 Continuity equationIn the absence of nuclear reactions, matter can neither be created or destroyed. This is the principle of massconservation and gives the continuity equation. Its general form is ∂ρ + div(ρv) = 0 (1.1) ∂twhere div(v) = v = ∂Fx /∂x + ∂Fy /∂y + ∂Fz /∂z. If the ﬂow is steady (∂ . . . /∂t = 0) and one-dimensional,we have div(ρv) = 0. (1.2)Moreover, in many engineering applications the density can be considered to be constant, leading to div(v) = 0. (1.3)The above forms are so-called diﬀerential forms of the continuity equation. However, one can derive theso-called integral forms. For example, for the steady-state case, if we integrate (1.3) on a closed surface A,we obtain ρvdA = ρv⊥ dA. (1.4) A ANote that the surface is deﬁned by its normal unit vector dA and one has to compute the scalar productvdA. One can resolve the velocity to a component parallel to and another perpendicular to the surface asv = v ⊥ + v . Thus vdA = |v| |dA| cos α = v⊥ dA.In many engineering applications, there is an inﬂow A1 and an outﬂow A2 , between which we have rigid walls,e.g. pumps, compressors, pipes, etc. Let us denote the average perpendicular velocities and the densities atthe inlet A1 and outlet A2 by v1 , ρ1 and v2 , ρ2 respectively. Than, we have m = ρ1 v1 A1 = ρ2 v2 A2 = const. ˙ (1.5) 3
Hydraulic Machines 4The 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, also. If the density isconstant, we have Q = m/ρ = v1 A1 = v2 A2 = const., ˙ (1.6)where Q (m3 /s) is the volumetric ﬂow rate.1.2 Bernoulli’s equationIn the case os steady frictionless ﬂow, the energy of the ﬂuid along a streamline remains constant. Mostlywe deal with incompressible ﬂuids, for which the energy content per unit volume is 1 mgh + 2 mv 2 + pV ρ Energy per unit volume = = p + v 2 + ρgh = constant. (1.7) V 2Considering two points of the streamline (the ﬂow is from 1 to 2), we have ρ 2 ρ 2 p1 + v1 + ρgh1 = p2 + v2 + ρgh2 . (1.8) 2 2Note that the above form can only applied if • the ﬂow is incompressible, i.e. ρ = const, • the ﬂow is ideal, i.e. there are no losses (friction, separation, etc.), • points 1 and 2 refer to two points on the same streamline and • the ﬂuid is Newtonian, i.e. the stress versus strain rate curve is linear and passes through the origin. The constant of proportionality is known as the viscosity: τ = µγ. (In common terms, this means the ˙ ﬂuid continues to ﬂow, regardless of the forces acting on it. For example, water is Newtonian, because it continues to exemplify ﬂuid properties no matter how fast it is stirred or mixed.)The Bernoulli equation can be extended to include friction and unsteady eﬀects: ρ 2 ρ 2 ρ 2 dv p1 + v1 + ρgh1 = p2 + v2 + ρgh2 + ζ i vi + ρL . (1.9) 2 2 2 dt friction unsteady term1.3 Thermodynamics1.3.1 Speciﬁc heat capacitiesAssume that a deﬁnite mass of gas m is heated from T1 to T2 at constant volume and thus its internal energyis raised from U1 to U2 . We have mcV ∆T = ∆U or cV ∆T = ∆u, (1.10)where u is the internal energy per unit mass and cV (J/(kgK)) is the speciﬁc heat capacity measured atconstant volume.Now we do the same experiment but now at constant pressure, thus its volume changes and work was doneon the ﬂuid: mcp ∆T = ∆U + mp∆V, (1.11)
Hydraulic Machines 5which, after rewriting for unit mass and combining with the previous equation for constant volume process,gives cp ∆T = ∆u + p∆V = cp ∆T + R∆T → cp = R + cV . (1.12)Thus we see that it is useful to deﬁne a new quantity which includes both the change of the internal energyu and the pressure work p dv = p d (1/ρ). Some useful equations: cp κ 1 R = cp − cV , κ= , cp = R and cV = R . (1.13) cV κ−1 κ−11.3.2 Some basic thermodynamic relationshipsOne possible form of the energy equation for a steady, open system in diﬀerential form is c2 δY + δq = d h + + gz , (1.14) 2 eδY is the elementary shaft work, δq is the elementary heat transferred towards the ﬂuid, both of thembeing processes, which is emphasised by the δ symbol. Note that the above equation describes an elemntaryprocess, however, to compute the overall process (to integrate the above equation), one has to know whatkind of process takes place in the machine (adiabatic, isentropic, isotherm, etc.) ad the results depends onit (thus, the integral is inexact).The term enthalpy is often used in thermodynamics. It expresses the sum of the internal energy u and theability to do hydrodynamic work p p h=u+ . (1.15) ρNote that h = cp T and u = cV T . There are some forms of expressing the change in enthalpy (v = 1/ρ): dh = d(u + pv) = δq + vdp = T ds + vdp. (1.16)The entropy 1 is for an elementary change in the equilibrium is δq ds = + dsirrev , (1.17) Twith which, using (1.16) we obtain dh = δq + T dsirrev + vdp, (1.18)with which (1.14) turns into c2 δY = vdp + d + gz + T dsirrev. (1.19) 2 losses δYu(sef ul) 1 Entropy is the only quantity in the physical sciences that seems to imply a particular direction of progress, sometimes calledan arrow of time. As time progresses, the second law of thermodynamics states that the entropy of an isolated system neverdecreases. Hence, from this perspective, entropy measurement is thought of as a kind of clock.
Hydraulic Machines 61.3.3 Input shaft work and useful workThe input shaft power is simply the work needed to change the enthalpy of the ﬂuid: c2 − c2 2 1 Pin = m∆e = m h2 − h1 + ˙ ˙ + g(z2 − z1 ) = mcp (T2 − T1 ) ˙ (1.20) 2 z1 ≈z2 ,v1 ≈v2When computing the useful work, we integrate the Yu part of (1.19) between points 1 and 2 (e.g. betweenthe suction and pressure side of a compressor). We still assume that z1 ≈ z2 and c1 ≈ c2 .In the case of an isentropic process, we have p/ρ = RT (ideal gas law) and p/ρκ =const., thus κ−1 1/κ 2 1/κ 2 p −1/κ p −1/κ κ p1 p2 κ Yisentr. = 1 p dp = 1 p dp = −1 . (1.21) ρ1 1 ρ 1 κ − 1 ρ1 p1Note that the above equation gives κ−1 κ p1 p 2 κ κ Yisentr. = −1 = R (T2 − T1 ) , (1.22) κ − 1 ρ1 p1 κ−1 RT1 T2 /T1 cpwhich is exactly the input speciﬁc work deﬁned by (1.20).A typical compression system consists of a compressor and a pressure vessel, which stores the compressedgas. Although the gas heats up during the compression but in the vessel it will cool back to the pressure ofthe surroundings. In other words, we loose the heat energy and the ’useful’ process is isotherm. We havep/ρ = RT (ideal gas law) and T =const., thus 2 p1 1 p2 Yisotherm = dp = RT1 ln (1.23) ρ1 1 p p1The real processes are usually described by polytropic processes but formally we use the same equationsas in the isentropic case, with the slight change of using the polytropic exponent n instead of κ. We havep/ρ = RT (ideal gas law) and p/ρn =const., thus n−1 2 1 n p1 p2 n n dp = −1 = R (T2 − T1 ) . (1.24) 1 ρ polytropic n − 1 ρ1 p1 n−1Polytropic processes are real, non-adiabatic processes. Note that the polytropic exponent n is typically aresult of curve ﬁt that allows the accurate computation of the outlet temperature.Finally, if the ﬂuid is incompressible, we have 2 2 1 1 p2 − p1 Yincomp. = 1 dp = dp = . (1.25) ρ 1 ρ 1 ρ1.3.4 Speciﬁc work for hydraulic machinesIn the case of pumps, the ﬂuid can be considered as incompressible. However, instead of Y usually the headis used: Yu p2 − p1 c2 − c2 J H= = + 2 1 + z2 − z1 . [m] = (1.26) g ρg 2g N
Hydraulic Machines 7In the case of ventillators, the energy change due to the geodetic heigth diﬀerence between the suction andpressure side is neglegible (z2 ≈ z1 ) and usually the change of total pressure is used: c2 − c2 2 1 J ∆pt = Yu ρ = p2 − p1 + ρ = pt,2 − pt,1 . [P a] = (1.27) 2 m3In the case of compressors, the ﬂuid cannot be considered as incompressible. When neglecting the losses,the speciﬁc work is: c2 − c2 2 1 Yu,isentropic = cp (T2s − T1 ) + = h2s,t − h1,t . (1.28) 21.3.5 EﬃciencyThe ratio of the useful power and the input power is eﬃciency. For a given T2 compression ﬁnal temperature,we have T2s − T1 ηisentropic = , (1.29) T2 − T1for a polytropic process, we have n n−1 R(T2 − T1 ) n κ−1 ηpolytropic = = . (1.30) cp (T2 − T1 ) n−1 κ1.4 ProblemsProblem 1.4.1The turbomachines conveying air are classiﬁed usually as fans (p2 /p1 < 1.3), blowers (1.3 < p2 /p1 < 3)and compressors (3 < p2 /p1 ). Assuming p1 = 1 bar inlet pressure, t1 = 20o C inlet temperature andisentropic process, ﬁnd the the relative density change (ρ2 − ρ1 )/ρ1 at the fan-blower border and the t2outlet temperature at the blower-compressor border. (Solution: (ρ2 − ρ1 )/ρ1 = 20.6%, t2 = 128.1o C)Problem 1.4.2Assuming isentropic process of an ideal gas, ﬁnd the inlet cross section area of a compressor conveyingm = 3 kg/s mass ﬂow rate. The velocity in the inlet section is c = 180 m/s. The surrounding air is at rest˙with p0 = 0 bar and T0 = 290 K.Problem 1.4.3Gas is compressed from 1 bar absolute pressure to 4 bar relative pressure. The gas constant is 288J/kgK, thespeciﬁc heat at constant pressure is cp = 1000J/kgK. The exponent describing the politropic compression isn = 1.5. Find the isentropic exponent. Find the isentropic speciﬁc useful work, the speciﬁc input work andthe isentropic eﬃciency. The density of atmospheric air is 1.16 kg/m3 . ht ≈ h is a reasonable approximation.(Solution: κ = 1.4, Yisentropic = 176.3 kJ/kg, Yinput = 212.5 kJ/kg, ηisentropic = 83%.)
Hydraulic Machines 8Problem 1.4.4Ideal gas (gas constant R = 288 J/kgK) with 27o C and 1 bar pressure is compressed to 3 bar with compressor.The exponent describing the real state of change is n = 1.5. Find the absolute temperature and density of theair at the outlet. Find the isentropic outlet temperature, the isentropic eﬃciency and the isentropic usefulspeciﬁc work. Find the power needed to cover the losses, if the mass ﬂow is 3 kg/s. (Solution: Treal = 432.7K,ρ = 2.407 kg/m3 , Tisentropic = 410.6K, ηisentropic = 83.3%, Yisentropic = 111.48 kJ/kg, Ploss = 66.8kW)Problem 1.4.5Gas is compressed from 1 bar to 5 bar. The ambient air temperature at the inlet T1 = 22◦ C while at theoutlet T2 = 231◦ C. Gas constant R = 288 J/kgK. Find the exponent describing the politropic compressionand the density of air at the inlet and the outlet. (Solution: n = 1.45, ρ1 = 1.177kg/m3 , ρ2 = 3.57kg/m3 .)
Chapter 2Incompressible turbomachineryWe classify as turbomachines all those devices in which energy is transferred either to, or from, a continuouslyﬂowing ﬂuid by the dynamic action of one ore moving blase rows. Essentially, a rotating blade row, a rotoror an impeller changes the stagnation enthalpy of the ﬂuid moving through it. These enthalpy changes areinitimately linked with the pressure changes in the ﬂuid.Up to 20% relative density change, the also gases are considered to be incompressible. Assuming isentropicprocess and ideal gas, this corresponds to p2 /p1 ≈ 1.3. Thus, pumps, fans, water and wind turbines areessentially the same amchines.2.1 Euler’s turbine equationEuler’s turbine equation (sometimes called Euler’s pump equation) plays a central role in turbomachinry asit connects the speciﬁc work Y and the geometry and velocities in the impeller. In what follows, we give twoderivations of the equation. Figure 2.1: Generalized turbomachineDerivation 1: Moment of momentumLet us compute the moment of the force that is applied at the inlet and outlet: 9
Hydraulic Machines 10 d d F = (mc) → M= (r × mc) = m (r × c) ˙ (2.1) dt dtThe power need of driving the machine is P = mY = (M out − M in ) ω = m∆ [ω (r × c)] = m∆ [c (ω × r)] = m∆ (c u) ˙ ˙ ˙ ˙ = m∆ (|c||u| cos α) = m (c2u u2 − c1u u1 ) ˙ ˙ (2.2)Comparing the beginning and the end of the equation, we see that the speciﬁc work is Y = c2u u2 − c1u u1 . (2.3)Derivation 2: Rotating frame and refernce and rothalpyThe Bernoulli equation in a rotating frame of refernce reads p w2 + + U = const., (2.4) ρ 2where U is the potential associated with the conservative force ﬁeld, which is the potential of a rotatingframe fo refernce: U = −r2 ω 2 /2. Let w stand for the relative velocity, c for the absolute velocity and u = rωfor the ’transport’ velocity. We have c = u + w, thus w2 = u2 + c2 − 2u c = u2 + c2 − 2u cu , which gives p w2 r2 ω2 p c2 + u2 − 2cu u2 p c2 + − = + − = + − c u = const. (2.5) ρ 2 2 ρ 2 2 ρ 2 cu uThus we see that the above quantity is conserved in a rotating frame of reference, which we refer to asrothalpy (abbriviation of rotational enthalpy). Let us ﬁnd now the change of energy inside the machine: p c2 Y =∆ + = ∆ (cu u) , (2.6) ρ 2 c2which is exactly Euler’s turbine equation. (For compressible ﬂuids, rothalpy is I = cp T + 2 − ucu .)2.2 Velocity triangles and performance curves2.2.1 Radial (centrifugal) machinesLet us consider a centrifugal pump and the velocity triangles at the impeller inlet and outlet, see Fig. 2.2.The theoretical ﬂow rate is Qth = c2m A2 Ψ = c2m D2 πb2 Ψ, (2.7)where D2 is the impeller outer diameter, b2 is its ﬂow-through width at the outlet and c2m is the radialcomponent of the outlet absolute velocity. Ψ < 1 is a constant that takes into account that the real ﬂowthrough area is smaller due to the blockage of the blade width at the outlet.
Hydraulic Machines 11 Figure 2.2: Centrifugal pumpsThe velocity triangle describes the relationship be-tween the absolute velocity c, the circumferentialvelocity u and the relative velocity w. Obviously,we have c = u + w. Moreover, we know that (a) thecircumferential velocity is u = Dπn and that (b)the relative velocity is tangent to the blade, i.e. theangle between u and w is approximately the bladeangle β.Basic trigonometrical identities show that c2u =u2 − c2m / tan β2 . It is usual to assume that theﬂow has no swirling (circumferential) component atthe inlet (due to Helmholtz’s third theorem). Inthe reality, the outlet ﬂow angle is not exactly β2 ,thus the head is decreased, which is taken into ac-count with the help of the slip factor λ (sometimesdenoted by σ in the literature). Figure 2.3: Centrifugal impeller with outlet velocityIf there is no prerotation (i.e. c1u = 0), we have components. c2u u2 u22 u2 w2u u2 2 u2 c2m Hth = λ =λ − =λ − g g g g g g tan β2 u22 u2 =λ − Qth . (2.8) g g tan β2 D2 πb2 ΨThus, the theoretical performance curve Hth (Qth ) of a centrifugal machine is a straight line, which is (seeFigure 2.4) • decreasing as Q is increased, for backward curved blades, i.e. β2 < 90o , • horizontal, for radial blades (β2 = 90o ) and • increasing (as Q is increased) for forward curved blades, i.e. β2 > 90o .
Hydraulic Machines 12 Figure 2.4: Eﬀect of blade shapes β2 angle on the performance curve.2.2.2 Axial machinesIn the case of axial machines the ﬂow leaves the impeller axially, see Fig. 2.5. The ﬂow-through area is 2 2 Do − Di π/4, where Do and Di stand for the outer and inner diameter of the lade, respectively. Noticethat in this case, u1 = u2 because it is assumed that the ﬂow moves along a constant radius. Assuming(again) prerotation-free inlet (c1u = 0), we have c2m = c1 (due to continuity). Figure 2.5: Axial pump (left) and fan (right)However, an important diﬀerence between axial and centrifugal pumps (fans) is that in the case of axialmachines, 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 − . (2.9) tan β2Thus, if we wanted to obtain constant ∆pt along the radial coordinate, the change of the circumferentialvelocity has to be compensated by varying β2 .
Hydraulic Machines 13 Figure 2.6: Axial impeller with outlet velocity components.The twisted airfoil (aerofoil) shape of modern air-craft propellers was pioneered by the Wright broth-ers. While both the blade element theory andthe momentum theory had their supporters, theWright brothers were able to combine both theo-ries. They found that a propeller is essentially thesame as a wing and so were able to use data col-lected from their earlier wind tunnel experimentson wings. They also found that the relative angleof attack from the forward movement of the aircraftwas diﬀerent for all points along the length of theblade, thus it was necessary to introduce a twistalong its length. Their original propeller blades areonly about 5% less eﬃcient than the modern equiv- Figure 2.7: World War I wooden propelleralent, some 100 years later. (Source: Wikipedia)2.2.3 Real performance curvesOur analysis so far assumed that the ﬂow inside the impeller is ideal (no losses) and that the streamlinesare following the blade shape (thus, blade angles are alos the streamline angles). However neither of theseassumptions are true.There are signiﬁcant friction losses inside the impeller, the narrower the ﬂow passage is, the higher thefriction losses will be. Moreover, the volute also introduces friction losses. These losses are proportional tothe velocity squared, thus Hf riction ∝ Q2 .On the other hand, if the angle of attack deviates from the ideal one, one experiences separation on the twosides of the blade. This is illustrated in TODO Figure BCS : for a constant circumferential velocity u as theﬂow rate and thus the inlet velocity c is varied, the relative velocity w also varies. At the design ﬂow rate Qdthe angle of attack ideal. For small ﬂow rates, we have separation on the suction side of the blade, while forlarger ﬂow rates the seperation is on the pressure side of the blade. Thus we have Hseparation ∝ (Q − Qd )2 .To obtain the real performance curve, one has to substract the above two losses from the theoretical head:H = Hth (Q) − K1 Q2 − K2 (Q − Qd )2 , which is illustrated in 2.8. Note that at the design point and close toit, the friction losses are moderate and no separation occurs. For lower ﬂow rates, the friction loss decreases
Hydraulic Machines 14while separation increases. For higher ﬂow rates, both friction and separation losses increase. Figure 2.8: Friction and separation losses in the impeller.2.2.4 ProblemsProblem 2.2.6An impeller runs at n=1440/min revolution speed and conveys Q = 40 l/s of water. The diameter of theimpeller is D = 240 mm, the outlet width is b2 = 20 mm. The blade angle at the outlet is β2 = 25 degrees.The inlet is prerotation-free. Find the theoretical head and draw a qualitatively proper sketch of the velocitytriangle at the outlet. (Solution: Hth = 22.9m)Problem 2.2.7The inner diameter of an axial impeller is Di = 250 mm, while the outer one is Do = 400mm. The revolutionnumber of the impeller is 1470rpm. The inlet is prerotation-free. At Q = 0.36 m3 /s the hydraulic eﬃciencyis 85%, the head is 6 m. The speciﬁc work along the radius is constant. Find the angles β1,2 at the innerand outer diameter. (Solution: β1,i = 13.7, β2,i = 16.7, β1,o = 8.7 and β2,o = 9.4 degrees)Problem 2.2.8The mean meridian velocity component of a radial impeller with D2 = 400 mm diameter at n = 1440rpmrevolution speed is cm = 2.5 m/s. The angle between the relative and circumferential velocity componentsis β2 = 25 degrees. With a geometrical change of the blade shape, this angle is increased to to 28 degrees,that results in 10% drop of the meridian velocity component. The inlet is prerotation-free. Find the relativehead change. (Solution: (H25o − H28o )/H25o = 4.6%)2.3 Losses and eﬃcienciesLet us analyse the losses that decrease the eﬃciency of a turbomachine. Let the input mechanical powertransmitted by the shaft be denoted by Pinput . We have thanMechanical losses P m These represent the friction loss in the bearings and the mechanical sealing losses (if any). The remaining power is called internal power Pi = Pinput − P m.
Hydraulic Machines 15Disc friction losses Pdf A signiﬁcant shear force appears in the ﬂuid entrapped between the housing and the impeller, which is taken into account by the disc friction coeﬃcient: Pdf = νdf Pi . The remaining power is the theoretical power of the impeller: Pth = Pi − Pdf = (1 − νdf )Pi .Hydraulic and volumetric losses Ph , Pv The theoretical head Hth and ﬂow rate Qth and is further decreased by the leakage ﬂow rates (Qleakage ) inside the pump (ﬂow across the gaps between the impeller and the housing) and the internal frictional losses h (e.g. in the impeller and volute). We have Pth = Qth ρgHth = (Q + Ql ) ρg (H + h ) = QρgH + Ql ρgH + Qth ρgh Pu Pv P h Q + Ql H + h Qth Hth = QρgH = QρgH → Pu = Pth ηh ηv (2.10) Q H Q H −1 −1 ηv ηh2.3.1 ProblemsProblem 2.3.9The input mechanical power of a water pump is 25 kW, the revolution number is 1440 rpm, the ﬂow rate is0.06 m3 /s. The volumetric eﬃciency is estimated as ηv = 0.92, the hydraulic eﬃciency is ηh = 0.85, the discfriction power loss is Pdf = 0.9 kW, the mechanical loss is Pm = 1.3 kW. Find the head and the specidicspeed and make a sketch of the impeller. (Solution: H=30.3m, nq =27.3, the impeller is a thin radial one.)Problem 2.3.10The revolution number of a water pump is 1470 rpm, the ﬂow rate is Q = 0.055m3 /s and the head is H = 45m.The hydraulic power loss is Ph = 2.5kW, the mechanical power loss is Pm = 1.3kW, the disc friction coeﬃcientis νt = 0.065. The input power at this operating point is Pin = 32kW. Make a complete analysis of the losses,including leakage ﬂow rate and the theoretical head. (Solution: Pinternal = 30.7kW, Pusef ul = 24.3kW,ηv = 93.2%, ηm = 95.9%, ηh = 93.2%, ηpump = 75.9%, Ql = 0.004m3 /s, Qth = 0.059m3 /s, Hth = 49.63m,h = 4.63m.)2.4 Dimensionless numbers and aﬃnityBased on the previously obtained formulae for theoretical head, we deﬁne dimensionless numbers as c2u u2 2 u2 H = ηh Hth = 2ηh := ψ 2 (2.11) u2 2g 2gor, in the case of fans ρ ∆pt = ψ u2 , (2.12) 2 2where ψ is a dimensionless pressure rise. Similarly, we have 4D2 πb2 c2m D2 π Q = ηv Qth = ηv D2 πb2 c2m = ηv 2 u2 D2 := ϕ 2 u2 2 (2.13) 4D2 u2 4These dimensionless quantities are called pressure number ψ and ﬂow number ϕ. What we found is thatH ∝ n2 and Q ∝ n allowing the transformation of the performance curve given at n1 to be computed toanother revolution number n2 . This is called aﬃnitiy law : 2 3 H1 n1 Q1 n1 P1 n1 = , = → = (2.14) H2 n2 Q2 n2 P2 n2
Hydraulic Machines 16As we have seen, both ψ and ϕ contains two parameters, D2 and u2 , out of which one can be eliminated,resulting in new dimensionless numbers. Let us start with the elimination of D2 . Q 4Q ϕ= 2 = 3 (2.15) D2 π D2 π 2 n 4 u2 H 2gH ψ= u2 = 2 (2.16) 2 D2 π 2 n2 2gfrom which we have √ √ 3/2 ϕ1/2 2 Q D2 π 3/2 n3/2 π Q1/2 σ = 3/4 = 3/2 √ 3/4 = √ 4 n 3/4 (2.17) ψ D2 π n (2gH) 2g 3/4 H nqNote that σ depends only on the revolution number but takes diﬀerent values along the performance curve.Thus when actually computing it, one takes the data of the best-eﬃciency point. Moreover, we do not √include the constant term √2gπ . Finally, by deﬁnition, the speciﬁc speed of a turbomachine is 4 3/4 1/2 Qopt. [m3 /s] nq = n[rpm] 3/4 (2.18) (Hopt. [m])Speciﬁc speed deﬁnes the shape of the impeller, low speciﬁc speed means low ﬂow rate and high pressurerise (radial impeller) while high speciﬁc speed occurs when the ﬂow rate is high and the pressure rise is low,see Fig. 2.9. Figure 2.9: Speciﬁc speed and shape of the impeller.2.4.1 ProblemsProblem 2.4.11The revolution number of a pump is 1450 rpm, the head and ﬂow rate at the best-eﬃciency point are 17mand 0.03 m3 /s. Find the speciﬁc speed. Find the diameter of the impeller if, based on industrial experience,the pressure number at the best-eﬃciency point should be ψ = 1. Find the ﬂow number ϕ. Find the head andﬂow rate at 970rpm. (Solution: nq = 30, D2 = 240mm, ϕ = 0.036, Q970rpm = 0.02m3 /s, H970rpm = 7.61m)Problem 2.4.12
Hydraulic Machines 17Find the speciﬁc speed of the pump given by 2.10, if the revolution number is 3000 rpm. Make a sketch ofthe impeller. (Solution: nq = 94, mixed impeller.) Figure 2.10: Performance chart for Problem 2.4.12.
Hydraulic Machines 18Problem 2.4.13The performance curve of a pump at 1450 rpm is given by H = 100 − 30000 Q2 and the eﬃciency is givenby η = −78000 Q2 + 4500 Q. Find the head and ﬂow rate of the best-eﬃciency point. Find the performancecurve at 1740 rpm. (Hopt = 76m, Qopt = 0.02855m3 /s, etam ax = 64.9%, H1740rpm = 144 − 30000 Q2 .)2.5 Forces on the impeller2.5.1 Radial forceTODO2.5.2 Axial forceThe axial force results from two components: • Momentum force • Pressure distribution on the hub and shroud.The momentum force is Fm = mv = ρr1 πc2 , ˙ 2 1 (2.19) Figure 2.11: Pressure distribution on the hub.The pressure distribution is ρ ρ 2 2 p(r) = r2 ωf + K 2 p(R2 ) = p2 − ∆p2 → p(r) = p2 − ∆p2 − ωf r2 − r2 . (2.20) 2 2The axial force becomes, e.g. on the hub (back of the impeller) r2 2 ρ 2 r2 − rs Fhub = 2rπp(r)dr = · · · = r2 − rs π p2 − ∆p2 − ωf 2 2 2 . (2.21) rs 2 2A similar result is obtained for the shroud (front of the impeller) with replacing rs by r1 . The overall axialforce is Fax = Fhub − Fshroud − Fm , (2.22)
Hydraulic Machines 19and its direction is towards the suction side (the axial force tries to ’pull down’ the impeller from the shaft).TODO: Extend explanations.Problem 2.5.14Find the axial force on the back of the impeller, whose outer diameter is D2 = 300mm, the shaft diameteris Ds = 50mm, the outlet pressure is 2.3bar and the revolution number is 1470rpm. The average angularvelocity of the ﬂuid is 85% of that of the impeller. (Solution: F = 9.36kN)
Chapter 3Hydraulic Systems3.1 Analysis3.1.1 ExercisesProblem 16Calculate the head loss of the pipe depicted in the ﬁgure below as a function of the volume ﬂow rate!Parameters: ζA = 1.5, ζB,D = 0.26, ζC = 0.35, ζF = 0.36, λ = 0.0155, Ds = Dp = D = 0.6[m] andQ = 0.4[m3 /s]. zC zD zF zB zASolution: • Static (geodetic) head + dynamic (friction) losses of the pipe: Hpipe = Hstat + Hf riction • Volume ﬂow rate: Q = cs(uction) Apipe,suction = cp(ressure) Apipe,pressure = cApipe • The ’extra’ 1 in the pressure side (...ζD + 1) represents the outﬂow losses. • Hstat = 8 + 4 = 12[m], Ls = 7 + 6 = 14[m], Lp = 12 + 20 + 8 = 40[m] 20
Hydraulic Machines 21 Ls c2 s Lp c2 p Hpipe = Hstat + KQ2 = Hstat + λ + ζA + ζB + λ + ζF + ζC + ζD + 1 = Ds 2g Dp 2g = 12[m] + 3.25[s2 /m5 ] × Q2 [m3 /s]2Problem 17The artiﬁcal fountain Beneath the St. Gellert is fed by two pipelines of 30m length. The height distancebetween the pump and the fountain is 22m. The diameter of the pipes is D1 = 100mm and D2 = 70mm,the friction coeﬃcient of the straight segments is λ = 0.02 and the friction coeﬃcient of the other segments(bends, etc.) is ζ = 0.5. Assuming that the ﬂow velocity in the ﬁrst pipe is 1.5m/s, calculate the the requiredhead. Calculate the ﬂow velocity in the second pipe and the overall ﬂow rate of the common pump feedingthe two pipes.Assuming 65% overall (pump+motor) eﬃciency, calculate the energy demand for 100 days andthe cost of the operation if the energy tariﬀ is 32HU F/kW h.Solution:Without the bypass line: H = 22.826m, Q = 0.01678[m3 /s], P = 5.78[kW ] and Cost = 443691HU F .Problem 18A pump delivers Q = 1200[dm3 /min] water from an open-surface well, whose water level is 25[m] below thedefault level. The pressure side ends 5[m] above the default level and the water ﬂows into an open-surfaceswimming pool. The diameter of the pipe on the suction side is Ds = 120[mm] and Dp = 100[mm] on thepressure side. The loss coeﬃcients are ζs = 3.6 and ζp = 14 (without the outﬂow losses). Calculate therequired pump head!Solution: Hp,req. = 35.7[m]3.2 Control3.2.1 ExercisesProblem 19A pump running at 1470[rpm] with Hpump = 45 − 2781Q2 head delivers water into a pipeline with Hpipe =20 + 1125Q2 . Calculate the required revolution number for the reduced ﬂow rate Q = 0.05[m3 /s].
Hydraulic Machines 22 Solution: H pump head at 1470 rpm • The actual working point is given pipeline by the solution of Hpump = Hpipe , central parabola which gives Q = 0.08[m3 /s] and H = 27.2[m]. pump head at ? rpm • Aﬃnity states that while vary- ing the revolutionary speed, H/n2 and Q/n remain constant. Thus, also H/Q2 remains constant, let’s denote this constant by a. So, while varying the revolutionary speed, the working point moves Q’ Q* Q Q along the central parabola (see ﬁg- ure), given by Hap = a Q2 .However, as Q is given and we also know that this point has to be located on the pipeline characteristic, weknow that H = 20 + 1125 × 0.052 = 22.81[m]. Thus, the parameter of the aﬃne parabola is a = H /Q 2 =9125.Q∗ is given by the intersection of the aﬃne parabola and the original pump characteristic: Hap (Q∗ ) =Hpump (Q∗ ), which gives Q∗ = 0.06148[m3 /s] with H ∗ = 34.5[m].Now we can employ aﬃnity between Q∗ and Q : Q 0.05 n = n∗ = 1470 × = 1195.5[rpm] Q∗ 0.06148and just for checking the calculation 2 n 1195.52 H = H∗ = 34.5 × = 22.81[m]. n∗ 14702Problem 20Solve the previous control problem (pump: Hpump = 45 − 2781Q2 , pipeline: Hpipe = 20 + 1125Q2 , desiredﬂow rate: Q = 0.05[m3 /s]) using a throttle at the pressure side of the pump and also with a bypass line.Compare the resulting operations in terms of power loss!Problem 21A pump, whose characteristic curve is given by Hpump = 70 − 90000[s2 /m5 ]Q2 , works together with twoparallel pipes. The main pipe is given by H1 = 30 + 100000[s2 /m5 ]. Calculate the head-ﬂow relationshipH2 (Q) of the side pipe, whose opening results in a ﬂow rate of 480[l/min] in the main pipe. The static headof the second side pipe is 25[m].Solution:Head of the main pipe at the prescribed ﬂow rate: Q1 = 480[l/min] = 0.008[m3 /s] → H1 (Q1 ) = 36.4[m]The head is the same, so the ﬂow rate of the pump is 70−36.4Hp (Qp ) = H1 (Q1 ) → Qp = 90000 = 0.0193[m3 /s]Thus, the ﬂow rate on the side pipe is Q2 = Qp − Q1 = 0.0193 − 0.008 = 0.0113[m3 /s]
Hydraulic Machines 23 36.4−25The actual characteristic of the side pipe: H2 (Q2 ) = 25 + aQ2 = 36.4[m] 2 → a= 0.01132 = 89279The solution is H2 (Q2 ) = 25 + 89279Q2 .
Chapter 4Positive displacement pumps4.1 ExercisesProblem 22Calculate the hydraulic power of the double-acting piston pump, which delivers water from an open-surfacetank into a closed one with 500[kP a] gauge pressure (i.e. relative pressure) located 50[m] above the suctiontank. Diameter of the piston is D = 120[mm], the stroke is 150[mm] and the driving motor runs at 120[rpm].Solution: 0.122 π 3Qmean = 2 × Apiston × s × n = 2 × 4 × 0.15 × 120 60 = 6.78 × 10−3 [ m ] s∆p = ptank,abs. − p0 + ρgH = ptank,rel. + ρgH = 991[kP a]P = Q∆p = 6.72[kW ] 24
Chapter 5Laboratory measurements5.1 Getting prepared to the measurements • The measurement descriptions are to be read and fully understood. Students failing to explain the experimental set up, the steps of the measurement, etc. will not be allowed to attend. • Table for measured data has to be prepared in advance. • Millimetre paper, pocket calculator, pencil and rubber are needed.5.2 Report preparation and submissionThe report can be prepared by hand or by computer as long as it is readable, clean and the ﬁgures andtables look nice. The sections of the reports are: • the aim of measurement, • description of the system (with ﬁgures) and steps of the measurement, • calculations (the equations used for data processing) and results, • derivation of error propagation, • measured data, computed quantities and error estimation, • graphs • and short summary.Tables have header with denomination and unit of the quantities, the measured and computed columns areseparated consistently as it is illustrated in Table 5.2.Requirements of diagram are shown in Figure 5.1. Measured Computed No. a b c d e f g h [unit1] [unit2] [unit3] [unit4] [unit5] [unit6] [unit7] [unit8] 1 2 25
Hydraulic Machines 26 Title of diagram H [m] × Quantity 1 + Quantity 2 Quantities and Legend has to be inserted units on the axes escpecially when more quantities are represented Error estimation has to be +σQ _ performed and represented } } }_ }+ σH @n=2000 f/min Parameter value at which measurement was performed has to be shown Measured points are marked by × or + Q [m³/h] Figure 5.1: Format of diagram.
Hydraulic Machines 275.3 Error estimation5.3.1 IntroductionIn technical practice we usually need to measure certain physical quantities directly (e.g. mass, length) orindirectly (e.g. eﬃciency). The measuring devices are built with ﬁnite range of operation and ﬁnite precision(reading precision). For example in case of a U-tube manometer (see Figure 5.2) we approximate the watercolumn displacement with a discrete value of the scale (called nominal value). However, as it is illustrated inFigure 5.2 all the displacements between 4.5mm and 5.5mm are read as 5mm. Thus if we want to deﬁne theread value precisely we say that h = hn ± σ, where hn is the nominal value, σ is the deviation. The deviationcan be approximated to be the half of the absolute error (E). In our case the absolute error Eh = 1mm,thus the deviation is σ = ±Eh /2 = ±0.5mm, the read value of the water column displacement is 5 ± 0.5mm.Relative error is also used to classify measuring devices. The relative error is the rate of the absolute errorand the measured value (eh = Eh /h = 1mm/5mm = 0.2 = 20%). 11 11 10 10 9 9 8 8 7 7 6 6 }Eh=1mm } +σh 5 5 } -σh 4 4 3 3 2 2 1 1 0 0 Figure 5.2: U-tube manometer.5.3.2 Example for error propagationWe have a pump-system represented in Figure 5.3. The pressure rise of the pump is measured with a U-tubemanometer to determine the ﬂow rate we employ metering oriﬁce. The mechanical power (input power) ofthe pump is measured with the driving balancing electric motor. The working medium is water while themeasuring medium in the manometers is mercury.Usual characteristics of the pump are η = f1 (Q) and Pi = f2 (Q). Since these computed quantities are derivedfrom measured ones, the reading errors summarized are present in their value (called error propagation).For this reason error estimation has to be given for each quantity. To compute the eﬃciency the followingevaluations have to be done:Eﬃciency of a machine is the fraction of useful and input power Pu(sef ul) η (Pu , Pi ) = . (5.1) Pi(nput)
Hydraulic Machines 28 Metering Orifice Throttle valve Δh~Δp ΔH~Δpmo Balancing motor M ω Pump Figure 5.3: Sketch of the pump test rig.Useful power of a pump is the product of the ﬂow rate and performed pressure rise. Pu (Q, ∆p) = Q ∆p (5.2)The ﬂow rate is measured with metering oriﬁce and computed as follows Q (∆pmo ) = αA 2∆pmo /ρ, ∆pmo (H1 , H2 ) = (ρHg − ρw )g(H1 − H2 ), (5.3)while the pressure rise is measured simple with U-tube manometer ∆p (h1 , h2 ) = (ρHg − ρw )g(h1 − h2 ). (5.4)The input power of the pump is the mechanical power on the shaft that is the product of the torque neededto balancing the motor and the angular velocity. Pi (M, n) = M ω = M 2πn, M (m) = (m − m0 )gl. (5.5)In general the deviation of a derived quantities (w = f (x, y, z)) is computed as follows 2 2 2 2 ∂f 2 ∂f 2 ∂f 2 σw = σx + σy + σz (5.6) ∂x ∂y ∂zIn our example the eﬃciency is the function of useful and input power (η = f (Pu , Pi )). To ﬁnd the deviationof eﬃciency we need not only the partial derivatives evaluated in operating point but the deviation of theuseful and input powers. These quantities depend on additional quantities thus we need derive each deviationuntil they can be expressed with the absolute error of a quantity that was measured directly. (For examplethe ﬂow rate is computed from the pressure drop on the metering oriﬁce that is measured with U-tubemanometer.) These quantities are underlined in the next section.
Hydraulic Machines 29 2 2 2 ∂η 2 ∂η 2 ∂η 1 ∂η Pu ση = σPu + σPi , = , =− 2 ∂Pu ∂Pi ∂Pu Pi ∂Pi Pi 2 2 2 ∂Pu 2 ∂Pu 2 ∂Pu ∂Pu σPu = σQ + σ∆p , = ∆p, =Q ∂Q ∂∆p ∂Q ∂∆p 2 2 ∂Q 2 ∂Q αA σQ = σ∆pmo , = ∂∆pmo ∂∆pmo ρw 2∆pmo /ρw 2 2 2 ∂∆pmo 2 ∂∆pmo 2 ∂∆pmo ∂∆pmo σ∆pmo = σH1 + σH2 , = (ρHg −ρw )g, = −(ρHg −ρw )g ∂H1 ∂H2 ∂H1 ∂H2 2 2 2 E H1 2 EH2 σH1 = , σH2 = 2 2 2 2 2 ∂∆p 2 ∂∆p 2 ∂∆p ∂∆p σ∆p = σh1 + σh 2 , = (ρHg −ρw )g, = −(ρHg −ρw )g ∂h1 ∂h2 ∂h1 ∂h2 2 2 2 Eh1 2 Eh2 σh1 = , σh2 = 2 2 2 2 2 ∂Pi 2 ∂Pi 2 ∂Pi ∂Pi σPi = σM + σn = 2πn, = M 2π ∂M ∂n ∂M ∂n 2 2 ∂M 2 ∂M σM = σm , = gl ∂m ∂m 2 2 Em σm = 2 2 2 En σn = (5.7) 2Known quantities and parameter are listed in Table 5.3.2. The absolute error of each device is the readingprecision. However, it is worth noting that random error can cause also deviation in measured value. Notation Value Inner diameter of the oriﬁce plate d = 70mm Inner diameter of the pipeline D = 100mm Discharge coeﬃcient of the oriﬁce plate α = 0.62 Balancing mass for unloaded operation m = 0.03kg Armlength of the balancing motor l = 0.96m Density of water ρw = 1000kg/m3 Absolute error of h1 and h2 Eh1,2 = 1mm Absolute error of H1 and H2 EH1,2 = 1mm Absolute error of n En = 5rpm Absolute error of m Em = 10g Table 5.1: Table of known quantitiesTable of measured and computed quantities are in Table 5.3.2.Error propagation evaluated in the second operation point. First we collect the known deviation and compute
Hydraulic Machines 30 Measured Computed No. h1 h2 H1 H2 m n ∆p ∆pmo M Q Pu Pi η [mm] [mm] [mm] [mm] [kg] [rpm] [P a] [P a] [N m] [m3 /s] [W ] [W ] [-] 1 2 820 180 412 588 1.08 1790 79107 21755 9.89 0.01574 1245.1 1853.6 0.672 3 Table 5.2: Table of measured and computed quantitiesrelative error. 2 2 2 2 Eh1,2 2 σh1 = σh2 = σh1,2 = = (0.5mm) , → eh1 = 0.12%, eh2 = 0.56% (5.8) 2 2 2 2 2 EH1,2 2 σH1 = σH2 = σH1,2 = = (0.5mm) , → eH1 = 0.24%, eH2 = 0.17% (5.9) 2 2 2 Em 2 σm = = (5g) , → em = 0.93% (5.10) 2 2 2 En 2 σn = = (2.5rpm) , → en = 0.28%. (5.11) 2Then we express all the unknown deviation. 2 2 σ∆p = ((ρHg −ρw )g) σh1,2 + (−(ρHg −ρw )g) σh1,2 = (87.4P a)2 , 2 2 2 → e∆p = 0.22% (5.12) 2 2 2 2 2 2 σ∆pmo = ((ρHg −ρw )g) σH1,2 + (−(ρHg −ρw )g) σH1,2 = (87.4P a) , → e∆pmo = 0.80% (5.13) 2 αA 2 σQ = σ∆pmo = (3.162 · 10−5 m3 /s)2 , 2 → eQ = 0.40% (5.14) ρw 2∆pmo /ρw σPu = ∆p2 σQ + Q2 σ∆p = (2.85W )2 , 2 2 2 → ePu = 0.46% (5.15) σM = (gl)2 σm = (0.047N m)2 , 2 2 → eM = 0.95% (5.16) 2 σPi= (2πn)2 σM + (M 2π)2 σn = (13.5W )2 , → ePi = 1.46% 2 2 (5.17) 12 2 Pu 2 2 2 ση = σPu + 2 σPi = (5.13 · 10−3 )2 , → eη = 1.53% (5.18) Pi PiThen we can say that the eﬃciency is η ± ση = 67.2 ± 0.51%. The evaluation has to be done for eachoperating point and in case of drawing diagram the error width (deviation) has to be represented.