SGChapt11 - XI. Turbomachinery This chapter considers the...

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Unformatted text preview: XI. Turbomachinery This chapter considers the theory and performance characteristics of the mechanical devices associated with the fluid circulation. General Classification: Turbomachine - A device which adds or extracts energy from a fluid. Adds energy: Extracts energy: Pump Turbine In this context, a pump is a generic classification that includes any device that adds energy to a fluid, e.g. fans, blowers, compressors. We can classify pumps by operating concept: 1. Positive displacement 2. Dynamic (momentum change) General Performance Characteristics Positive Displacement Pumps 1. Delivers pulsating or periodic flow (cavity opens, fluid enters, cavity closes, decreasing volume forces fluid out exit opening. 2. Not sensitive to wide viscosity changes. 3. Delivers a moderate flow rate. 4. Produces a high pressure rise. 5. Small range of flow rate operation (fixed pump speed). Dynamic Pumps 1. Typically higher flow rates than PD’s. 2. Comparatively steady discharge. 3. Moderate to low pressure rise. 4. Large range of flow rate operation. 5. Very sensitive to fluid viscosity. | v v XI - 1 | e-Text Main Menu | Textbook Table of Contents | Study Guide Typical Performance Curves (at fixed impeller speed) Fig. 11.2 Performance curves for dynamic and positive displacement pumps Centrifugal Pumps Most common turbomachine used in industry. Includes the general categories of (a) liquid pumps, (b) fans, (c) blowers, etc. They are momentum change devices and thus fall within the dynamic classification. Typical schematic shown as Fig. 11.3 Cutaway schematic of a typical centrifugal pump | v v XI - 2 | e-Text Main Menu | Textbook Table of Contents | Study Guide Writing the energy equation across the device and solving for hp – hf ,we have P − P1 V2 − V1 H = hp − hf = 2 + + Z 2 − Z1 ρg 2g 2 2 where H is the net useful head delivered to the fluid, the head that results in pressure, velocity, and static elevation change. Since for most pumps (not all), V1 = V2 and ∆Z is small, we can write H≅ ∆P ρg Since friction losses have already been subtracted, this is the ideal head delivered to the fluid. Note that velocity head has been neglected and can be significant at large flow rates where pressure head is small. Pw = ρ Q g H The ideal power to the fluid is given by The pump efficiency is given by η= ρQgH ρQgH Pw = = BHP BHP ωT where BHP = shaft power necessary to drive the pump ω = angular speed of shaft T = torque delivered to pump shaft Note that from the efficiency equation, pump efficiency is zero at zero flow rate Q and at zero pump head,H. | v v XI - 3 | e-Text Main Menu | Textbook Table of Contents | Study Guide Basic Pump Theory Development of basic pump theory begins with application of the integral conservation equation for moment-of-momentum previously presented in Ch. III. Applying this equation to a centrifugal pump with one inlet, one exit, and uniform properties at each inlet and exit, we obtain & & T = me r x Ve − mi r x Vi where T is the shaft torque needed to drive the pump Vi , Ve are the absolute velocities at the inlet and exit of the pump This is used to determine the change of angular momentum across the device. Fig. 11.4 Inlet and exit velocity diagrams for an idealized impeller Since the velocity diagram is key to the analysis of the device, we will discuss the elements in detail. | v v XI - 4 | e-Text Main Menu | Textbook Table of Contents | Study Guide 1. At the inner radius r1 have two velocity components: a. the circumferential velocity due to the impeller rotation u1 = r1 ω V 1 w1 blade tip speed at inner radius β1 b. relative flow velocity tangent to the blade w1 tangent to the blade angle α1 Vn 1 u1 Vt β1 1 These combine to yield the absolute inlet velocity V1 at angle α1 The absolute velocity can be resolved into two absolute velocity components: 1. Normal ( radial ) component: Vn1 = V1 sin α1 = w1 sin β1 Note that for ideal pump design, Vn1 = V1 and α 1 = 90 o 2. Absolute tangential velocity: Vt1 = V1 cos α 1 = u 1 - w1 cos β1 again, ideally Vt1 = 0 It is also important to note that Vn1 is use to determine the inlet flow rate, i.e., Q = A1 Vn1 = 2 π r1 b1 Vn1 where b1 is the inlet blade width | v v XI - 5 | e-Text Main Menu | Textbook Table of Contents | Study Guide Likewise for the outer radius r2 we have the following: a. the circumferential velocity due to the impeller rotation u 2 = r2 ω blade tip speed at outer radius b. relative flow velocity tangent to the blade w2 tangent to the blade angle V2 w2 β2 Vn2 α2 Vt u2 2 β2 These again combine to yield the absolute outlet velocity V2 at angle α2 The exit absolute velocity can also be resolved into two absolute velocity components: 1. Normal ( radial ) component: Vn2 = V2 sin α 2 = w2 sin β2 = Q 2 π r2 b2 Note that Q is the same as for the inlet flow rate 2. Absolute tangential velocity: Vt 2 = V2 cos α 2 = u 2 - w2 cos β2 Vt 2 = u 2 - Vn2 tan β 2 = u2 - Q 2 π r2 b 2 tan β2 Q = A1 Vn1 = 2 π r1 b1 Vn1 = A 2 Vn2 = 2 π r2 b 2 Vn 2 where Again, each of the above expressions follows easily from the velocity diagram, and the student should draw and use the diagram with each pump theory problem. | v v XI - 6 | e-Text Main Menu | Textbook Table of Contents | Study Guide We can now apply moment - of – momentum equation. { T = ρ Q r2 * Vt 2 − r1 * Vt1 } (again Vt1 is zero for the ideal design) For a sign convention, we have assumed that Vt1 and Vt2 are positive in the direction of impeller rotation. The “ ideal” power supplied to the fluid is given by { Pw = ω T = ρ Q ω r2 Vt2 − ω r1 Vt1 or { } } Pw = ω T = ρ Q u 2 Vt 2 − u1 Vt1 = ρ Qg H Since these are ideal values, the shaft power required to drive a non-ideal pump is given by BHP = Pw ηp The head delivered to the fluid is H= ρ Q{u 2 Vt2 − u1 Vt1 } ρQg {u = 2 Vt 2 − u 1 Vt1 } g For the special case of purely radial inlet flow H* = u 2 Vt 2 g | v v XI - 7 | e-Text Main Menu | Textbook Table of Contents | Study Guide From the exit velocity diagram, substituting for Vt2 we can show that u ωQ H= 2 − g 2 π b 2 g tan β 2 2 2 has the form C1 - C2 Q shutoff head, the head produced at zero flow, Q = 0 u2 where: C1 = g Example: A centrifugal water pump operates at the following conditions: speed = 1440 rpm, r1 = 4 in, r2 = 7 in, β1 = 30o, β2 = 20o, b1 = b2 = 1.75 in Assuming the inlet flow enters normal to the impeller (zero absolute tangential velocity): find: (a) Q, (b) T, (c) Wp, (d) hp, (e) ∆P ω = 1440 rev 2 π rad = 150.8 min 60 s Calculate blade tip velocities: u1 = r1 ω = 4 rad ft ft150.8 = 50.3 12 s s u 2 = r2 ω = Since design is ideal, at inlet 7 rad ft ft150.8 = 88 12 s s V = V1 n 1 w1 α1 = 90 , Vt1 = 0 o Vn1 = U1 tan 300 = 50.3 tan 30o = 29.04 ft/s 30Þ 90Þ Q = 2 π r1 b1 Vn1 r1 • | v v XI - 8 | e-Text Main Menu | Textbook Table of Contents | Study Guide 30Þ u1 3 ft ft 4 Q = 2 π ft1.75 ft 29.04 = 8.87 s 12 s 3 ft gal gal s Q = 8.87 60 7.48 3 = 3981 ft min s min Repeat for the outlet: ft 3 8.87 Q s Vn2 = = 2 π r2 b 2 2 π 7 ft 1.75 ft 12 12 ft Vn2 = 16.6 s Vn2 16.6 ft/s ft w2 = = = 48.54 s sin 20 o sin 20 o V 2 w2 α2 20Þ 20Þ u2 r2 • Vt 2 = u 2 - w 2 cos β2 = 88 − 48.54 cos 20 o = 42.4 ft s We are now able to determine the pump performance parameters. Since for the centrifugal pump, the moment arm r1 at the inlet is zero, the momentum equation becomes Ideal moment of momentum delivered to the fluid: { T = ρ Q r2 * Vt2 } 3 slug ft 7 ft = 1.938 3 8.87 ft 42.4 = 425.1ft − lbf ft s s 12 Ideal power delivered to the fluid: P = ω T = 150.8 rad ft − lbf 425.1ft − lbf = 64,103 = 116.5 hp s s | v v XI - 9 | e-Text Main Menu | Textbook Table of Contents | Study Guide Head produced by the pump (ideal): H= P 64,103 ft − lbf/s = = 115.9 ft lbf ft 3 ρ gQ 62.4 3 8.87 ft s Pressure increase produced by the pump: 3 ft ∆ P = ρ g H = 62.4 115.9 ft = 7226 psf = 50.2 psi s Pump Performance Curves and Similarity Laws Pump performance results are typically obtained from an experimental test of the given pump and are presented graphically for each performance parameter. • Basic independent variable - Q {usually gpm or cfm } • Dependent variables typically – head pressure rise, in some cases ∆P H BHP – input power requirements (motor size) η – pump efficiency • These typically presented at fixed pump speed and impeller diameter Typical performance curves appear as | v v XI - 10 | e-Text Main Menu | Textbook Table of Contents | Study Guide Fig. 11.6 Typical Centrifugal Pump Performance Curves at Fixed Pump Speed and diameter These curves are observed to have the following characteristics: 1. hp is approximately constant at low flow rate. 2. hp = 0 at Qmax. 3. BHP is not equal to 0 at Q = 0. 4. BHP increases monotonically with the increase in Q. 5. ηp = 0 at Q = 0 and at Qmax. 6. Maximum pump efficiency occurs at approximately Q* = 0.6 Qmax . This is the best efficiency point BEP. At any other operating point, efficiency is less, pump head can be higher or lower, and BHP can be higher or lower. 7. At the BEP, Q = Q*, hp = hp*, BHP = BHP*. Measured Performance Data Actual pump performance data will typically be presented graphically as shown in Fig. 11.7. Each graph will usually have curves representing the pump head vs flow rate for two or more impeller diameters for a given class/model of pumps having a similar design. The graphs will also show curves of constant efficiency and constant pump power (BHP) for the impeller diameters shown. All curves will be for a fixed pump impeller speed. | v v XI - 11 | e-Text Main Menu | Textbook Table of Contents | Study Guide Fig. 11.7 Measured performance curves for two models of a centrifugal water pump | v v XI - 12 | e-Text Main Menu | Textbook Table of Contents | Study Guide How to Read Pump Performance Curves Care must be taken to correctly read the performance data from pump curves. This should be done as follows: (1) For a given flow rate Q (2) Read vertically to a point on the pump head curve h for the impeller diameter D of interest. (3) All remaining parameters ( efficiency & BHP) are read at this point; i.e., graphically interpolate between adjacent curves for BHP to obtain the pump power at this point. Note that the resulting values are valid only for the conditions of these curves: (1) pump model and design, (2) pump speed – N, (3) impeller size – D, (4) fluid (typically water) Thus for the pump shown in Fig. 11.7a with an impeller diameter D = 32 in, we obtain the following performance at Q = 20,000 gpm: Q = 20,000 gpm, D = 32 in, N = 1170 rpm H ≅ 385 ft, BHP ≅ 2300 bhp, ηp ≅ 86.3 % Note that points that are not on an h vs. Q curve are not valid operating points. Thus for Fig. 11.7b, the conditions Q = 22,000 gpm, BHP = 1500 bhp, hp = 250 ft do not correspond to a valid operating point because they do not fall on one of the given impeller diameter curves. However, for the same figure, the point Q = 20,000 gpm, BHP = 1250 bhp is a valid point because it coincidentally also falls on the D = 38 in impeller curve at hp = 227 ft. | v v XI - 13 | e-Text Main Menu | Textbook Table of Contents | Study Guide Net Positive Suction Head - NPH One additional parameter is typically shown on pump performance curves: NPSH = head required at the pump inlet to keep the fluid from cavitating. NPSH is defined as follows: 2 P NPSH = i + ρg Vi P −v 2g ρg where Pi = pump inlet pressure Pv = vapor pressure of fluid Pump inlet Considering the adjacent figure, write the energy equation between the fluid surface and the pump inlet to obtain the following: P NPSH = i + ρg zi Pa Pi z=0 2 Vi P P P − v = a − Z i − h f,a − i − v 2g ρg ρ g ρg For a pump installation with this configuration to operate as intended, the righthand-side of the above equation must be > the NPSH value for the operating flow rate for the pump. Example: A water supply tank and pump are connected as shown. Pa = 13.6 psia and the water is at 20 o C with Pv = 0.34 psia. The system has a friction loss of 4.34 ft. Will the NPSH of the pump of Fig. 11.7a at 20,000 gpm work? a 10 ft i | v v XI - 14 | e-Text Main Menu | Textbook Table of Contents | Study Guide Applying the previous equation we obtain NPSH = Pa P − Z i − h f,a − i − v ρg ρg (13.6 − 0.34) lbf/in2 *144 in2 /ft 2 − (−10 ft) − 4.34 ft NPSH = 3 62.4 lbf/ft NPSH = 36.26 ft The pump will work because the system NPSH as shown in Fig. 11.7a is 30 ft which provides a 6.3 ft safety margin. Conversely, the pump could be located as close as 3.7 ft below the water surface and meet NPSH requirements. Pump Similarity Laws Application of the dimensional analysis procedures of Ch. V will yield the following three dimensionless performance parameters: Dimensionless flow coefficient: CQ = Q ω D3 Dimensionless head coefficient: CH = gH ω 2 D2 Dimensionless power coefficient: CP = BHP ρω 3 D5 where ω is the pump speed in radians/time and other symbols are standard design and operating parameters with units that make the coefficients dimensionless. How are these used? These terms can be used to estimate design and performance changes between two pumps of similar design. | v v XI - 15 | e-Text Main Menu | Textbook Table of Contents | Study Guide Stated in another way: If pumps 1 and 2 are from the same geometric design family and are operating at similar operating conditions, the flow rates, pump head, and pump power for the two pumps will be related according to the following expressions: Q2 N2 = Q1 N1 D2 D1 2 3 H 2 N2 D2 = H1 N1 D1 Use to predict the new flow rate for a design change in pump speed N and impeller diameter D. 2 Used to predict the new pump head H for a design change in pump speed, N and impeller diameter D. 3 BHP2 ρ2 N 2 D2 = BHP1 ρ1 N1 D1 5 Used to predict the new pump power BHP for a design change in fluid, ρ, pump speed N and impeller diameter D. Example It is desired to modify the operating conditions for the 38 in diameter impeller pump of Fig. 11.7b to a new pump speed of 900 rpm and a larger impeller diameter of 40 in. • H(ft) • BEP1 Determine the new pump head and power for the new pump speed at the BEP. Q(gpm) | v v XI - 16 | e-Text Main Menu | Textbook Table of Contents | Study Guide BEP 2 For the D = 38 in impeller of Fig. 11.7b operating at 710 rpm, we read the best efficiency point (BEP) values as Q* = 20,000 gpm, H* = 225 ft, BHP * = 1250 hp Applying the similarity laws for N2 = 900 rpm and D2 = D1 = 38 in, we obtain 3 3 Q2 N2 D2 900 40 = = 1.478 = 710 38 Q1 N1 D1 Q2 = 20,000*1.478 = 29,570 gpm 2 ans. 2 2 2 H 2 N2 D2 900 40 = 1.78 = = 710 38 H1 N1 D1 H2 = 225*1.78 = 400.5 ft 3 ans. 5 3 5 BHP2 ρ2 N 2 D2 900 40 = 2.632 = = (1) 710 38 BHP1 ρ1 N1 D1 BHP2 = 3290 hp ans. Thus, even small changes in the speed and size of a pump can result in significant changes in flow rate, head, and power. It is noted that every point on the original 38 in diameter performance curve exhibits a similar translation to a new operating condition. The similarity laws are obviously useful to predict changes in the performance characteristics of an existing pump or to estimate the performance of a modified pump design prior to the construction of a prototype. | v v XI - 17 | e-Text Main Menu | Textbook Table of Contents | Study Guide Matching a Pump to System Characteristics The typical design/sizing requirement for a pump is to select a pump which has a pump head which matches the required system head at the design/operating flow rate for the piping system. Key Point hp = hsys at Qdes. It is noted that pump selection should occur such that the operating point of the selected pump should occur on the pump curve near or at the BEP. From the energy equation in Ch. VI, the system head is typically expressed as 2 h sys 2 2 P2 − P1 V2 − V1 f L + ∑ K V = + + Z 2 − Z1 + i D 2g ρg 2g Thus the selection of a pump for a piping system design should result in a pump for which the pump head hp at the design flow rate Qdes is equal ( or very close) to the head η p hp Hdes requirements hsys of the piping system at the same flow rate, and this should occur at or near the point of maximum efficiency for the chosen pump. • hsys Q(gpm) Qdes Other operating and performance requirements (such as NPSH) are obviously also a part of the selection criteria for a pump. | v v XI - 18 | e-Text Main Menu | Textbook Table of Contents | Study Guide Pumping Systems: Parallel and Series Configurations For some piping system designs, it may be desirable to consider a multiple pump system to meet the design requirements. Two typical options include parallel and series configurations of pumps. Specific performance criteria must be met when considering these options. Given a piping system which has a known design flow rate and head requirements, Qdes, hdes. The following pump selection criteria apply. Pumps in Parallel: Assuming that the pumps are identical, each pump must provide the following: Q(pump) = 0.5 Qdes h(pump) = hdes Pumps in Series: Assuming that the pumps are identical, each pump must provide the following: Q (pump) = Qdes h(pump) = 0.5 hdes For example, if the design point for a given piping system were Qdes = 600 gpm, and hsys = 270 ft, the following pump selection criteria would apply: 1. Single pump system Q(pump) = 600 gpm, hp = 270 ft 2. Parallel pump system Q(pump) = 300 gpm, hp = 270 ft for each of the two pumps 3. Series pump system Q(pump) = 600 gpm, hp = 135 ft for each of the two pumps | v v XI - 19 | e-Text Main Menu | Textbook Table of Contents | Study Guide ...
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This note was uploaded on 10/27/2009 for the course MAE 101a taught by Professor Sakar during the Spring '08 term at UCSD.

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