View stunning SlideShares in full-screen with the new iOS app!Introducing SlideShare for AndroidExplore all your favorite topics in the SlideShare appGet the SlideShare app to Save for Later — even offline
View stunning SlideShares in full-screen with the new Android app!View stunning SlideShares in full-screen with the new iOS app!
1.
PUMPS CHAPTER –11For more chemical engineering eBooks and solution manuals visit here www.chemicallibrary.blogspot.com
2.
INTRODUCTION DESIGNING OF ANY FLUID FLOWING SYSTEM REQUIRES; 1. Design of system through which fluid will flow 2. Calculation of losses that will occur when the fluid flows 3. Selection of suitable device which will deliver enough energy to the fluid to overcome these lossesDevices: Deliver Energy To Liquids/Gases: Pumps/CompressorsDevices: Extracts Energy From Fluids: Turbines TYPES OF PUMPS POSITIVE DISPLACEMENT PUMPS DYNAMIC PUMPS RECIPROCATING PUMPS CENTRIFUGAL ROTARY PUMPS PUMPS
3.
POSITIVE DISPLACEMENT PUMPS, (PDP’S) WORKING PRINCIPLE AND FEATURES;1. Fixed volume cavity opens2. Fluid trapped in the cavity through an inlet3. Cavity closes, fluid squeezed through an outlet4. A direct force is applied to the confined liquid5. Flow rate is related to the speed of the moving parts of the pump6. The fluid flow rates are controlled by the drive speed of the pump7. In each cycle the fluid pumped equals the volume of the cavity8. Pulsating or Periodic flow9. Allows transport of highly viscous fluids10. Performance almost independent of fluid viscosity11. Develop immense pressures if outlet is shut for any reason, HENCE 1. Sturdy construction is required 2. Pressure-relief valves are required (avoid damage from complete shutoff conditions)
4.
PDP’S, contd. RECIPROCATING TYPE PDPS Piston OR Plunger pumps Diaphragm pumpsSingle acting piston pump Single diaphragm pumpDouble acting Simplex pump Double diaphragm pumpDouble acting Duplex pump
5.
ROTARY TYPE PDPS SINGLE ROTOR MULTIPLE ROTORS Sliding vane pump Gear PumpFlexible tube or lining 2 Lobe Pump Screw pump 3 Lobe Pump Radial Pump AND MANY MORE
6.
DYNAMIC PUMPSWORKING PRINCIPLE AND FEATURES1. Add somehow momentum to the fluid (through vanes, impellers or some special design2. Do not have a fixed closed volume3. Fluid with high momentum passes through open passages and converts its high velocity into pressure TYPES OF DYNAMIC PUMPS ROTARY PUMPS SPECIAL PUMPS Centrifugal Pumps Jet pump or ejector Axial Flow Pumps Electromagnetic pumps for liquid metals Mixed Flow Pumps Fluid-actuated: gas-lift or hydraulic-ram
8.
COMPARISON OF PDPS AND DYNAMIC PUMPSCRITERIA PDPS DYNAMIC PUMPSFlow rate Low, typically 100 gpm As high as 300,000 gpm Pressure As high as 300 atm Moderate, few atm Priming Very rarely AlwaysFlow Type Pulsating Steady Constant flow rate for virtually Head varies with any pressure flow rate Constant OR OR RPM Flow rate cannot be changed Flow rate changes with without changing RPM head for same RPM Hence used for metering Viscosity Virtually no effect Strong effects
9.
CENTRIFUGAL PUMPS Centrifugal Pumps: Construction Details and Working1. A very simple machine Illustration-12. Two main parts 1. A rotary element, IMPELLER Illustration-2 2. A stationary element, VOLUTE3. Filled with fluid & impeller rotated4. Fluid rotates & leaves with high velocity Impeller-1 Impeller-55. Outward flow reduces pressure at inlet, (EYE OF THE IMPELLER), more fluid Impeller-2 Impeller-6 comes in. Impeller-36. Outward fluid enters an increasing area region. Velocity converts to pressure Impeller-4 Impeller Impart Energy/Velocity By Rotating Fluid Volute Converts Velocity To Pressure
10.
CENTRIFUGAL PUMPS, contd.Centrifugal Pumps: Working Principal1. Swinging pale generates centrifugal force → holds water in pale2. Make a bore in hole → water is thrown out3. Distance the water stream travels tangent to the circle = f(Vr)4. Volume flow from hole = f(Vr)5. In centrifugal pumps, flow rate & pressure = f(Vr) (tip velocity)A freely falling body achieves a velocity V = (2gh)1/2 ORA body will move a distance h = V2/2g, having an initial velocity VFind diameter that will generate ‘V’ to get required ‘h’ for given ‘N’
11.
CENTRIFUGAL PUMPS, contd.Q. FOR AN 1800 RPM PUMP FIND THE DIAMETER OF IMPELLER TO GENERATE A HEAD OF 200 FT. Find first initial velocity V = (2gh)1/2 = 113 ft/sec Convert RPM to linear distance per rotation 1800 RPM = 30 RPS → V/RPS = 113/30 = 3.77 ft/rotation3.77 = circumference of impeller → diameter = 1.2 ft = 14.4 inches CONCLUSIONFLOW THROUGH A CENTRIFUGAL PUMP FOLLOWS THE SAME RULES OF FREELY FALLING BODIES DO WE GET THE SAME DIAMETER OR HEAD OR FLOW RATE AS PREDICTED BY THESE IDEAL RULES
12.
CENTRIFUGAL PUMPS, contd. BASIC PERFORMANCE PARAMETERS The Energy Equation for This Case & & & V12 V2 2 Q − W shaft − W vis = − m1 h1 + & + gz1 + m 2 h2 + & + gz 2 2 2 Assumptions:• No heat generation V2 2 V12 & W shaft = m h2 + & + gz 2 − h1 + + gz1 • No viscous work. 2 2 • Mass in = mass out
13.
CENTRIFUGAL PUMPS, contd.What would be the difference in ‘z’, can we assume z2-z1≈0 V2 2 V1 2 Hence & W shaft = m h2 + & − h1 + 2 2 p2 V2 2 p1 V12 &W shaft = m & + u2 + − + u1 + ρ 2 ρ 2 p 2 V2 2 p1 V1 2 Thermodynamically, u = u(T) &W shaft = m & + − + ρ 2 ρ 2 only and Tin ≈ Tout p 2 V2 2 p1 V1 2 & W shaft = ρ Q + − + ρ 2 ρ 2
14.
CENTRIFUGAL PUMPS, contd. ρ V2 2 ρ V12 Pw = ρ gHQ = W shaft & = Q p2 + − p1 + 2 2 Where Pw = water power Pw 1 ρ V 2 2 ρ V12 H = = ( p 2 − p1 ) + − ρ gQ ρ g 2 2 Generally V1 and V2 are of same order of magnitude If the inlet and outlet diameters are same Pw 1 H = ≅ ( p 2 − p1 ) ρ gQ ρ g
15.
CENTRIFUGAL PUMPS, contd. The power required to drive the pump; bhp The power required to turn the pump shaft at certain RPM bhp = ω T T = torque required to turn shaftThe actual power required to drive the pump depends upon efficiency Pw ρ gQH η= = bhp ωT η = η vη hη m Efficiency has three components; Volumetric Mechanical Hydraulic• casing leakages 1. Losses in bearings • Shock 2. Packing glands etc • friction, Q • re-circulationηv = Pf hf Q + QL ηm = 1 − ηv = 1 − bhp hs
16.
CENTRIFUGAL PUMPS, contd.Torque estimation ⇒ 1D flow assumption1-D angular momentum balance gives T = ρ Q ( r2Vt 2 − rVt1 ) 1 Vt1 and Vt2 absolute circumferential or tangential velocity components Pw = ωT = ωρ Q ( r2Vt 2 − rVt1 ) = ρ Q ( u2Vt 2 − u1Vt1 ) 1 Pw ρ Q ( u2Vt 2 − u1Vt1 ) 1 DOH= = = ( u2Vt 2 − u1Vt1 ) DETAILS ρ gQ ρ gQ g IN TUTORIALEuler turbo- Torque, Power and Ideal Head depends on,machinery Impeller tip velocities ‘u’ & abs. tangential velocities Vtequations; Independent of fluid axial velocity if any
17.
CENTRIFUGAL PUMPS, contd. Doing some trigonometric and algebraic manipulationH= 1 (V22 − V12 ) + ( u2 − u12 ) + ( w2 − w12 ) 2 2 p w2 r 2ω 2 +z+ − = const 2g ρg 2g 2g BERNOULLI EQUATION IN ROTATING COORDINATES Applicable to 1, 2 and 3D Ideal Incompressible Fluids One Can Also Relate the Pump Power With Fluid Radial Velocity Pw = ρ Q ( u2Vn 2 cot α 2 − u1Vn1 cot α1 ) DO Q Q EX. 11.1 Vn 2 = and Vn1 = 2π r2b2 2π r1b1 IN TUTORIALWith known b1, b2, r1, r2, β1, β2 and ω one can find centrifugal pump’s ideal power and ideal head as a function of Discharge ‘Q’
18.
CENTRIFUGAL PUMPS, contd. EFFECT OF BLADE ANGLES β1, β2 ON PUMP PERFORMANCE Pw 1 H= = ( u2Vt 2 − u1Vt1 ) ρ gQ gAngular Angular >>momentum out momentum in QVn 2 = Vt 2 = u2 − Vn 2 cot β 2 2π r2b2Doing all this leads to if β < 90, backward curve blades, stable op if β = 90, straight radial blades, stable op u2 u2 cot β 2 2 If β > 90, forward curve blades, unstable opH≈ − Q g 2π r2b2 g
19.
CENTRIFUGAL PUMPS, CHARACTERISTICS1. Whatever discussed earlier is qualitative due to assumptions.2. Actual performance of centrifugal pump → extensive testing3. The presentation of performance data is exactly same for 1. Centrifugal pumps 2. Axial flow pumps 3. Mixed flow pumps 4. Compressors4. The graphical representation of pumps performance data obtained experimentally is called “PUMP CHARACTERSTICS” OR “PUMP CHARACTERSTIC CURVES” 1. This representation is almost always for constant shaft speed ‘N’ 2. Q (gpm) discharge is the independent variable (LIQUIDS) 3. H (head developed), P (power), η (efficiency) and NPSH (net positive suction head) are the dependent variables (LIQUIDS) 4. Q (ft3/m3/min), discharge is the independent variable (GASES) 5. H (head developed), P (power), η (efficiency) are the dependent variables (GASES)
21.
CENTRIFUGAL PUMPS, CHARACTERISTICS, contd.General Features of Characteristic Curves of Centrifugal Pumps1. ‘H’ is almost constant at low flow rates2. Maximum ‘H’(shut off head) is at zero flow rate3. Head drops to zero at Qmax4. ‘Q’ is not greater than Qmax → ‘N’ and/or impeller size is changed5. Efficiency is always zero at Q = 0 and Q = Qmax6. η is not an independent parameter → P ρ gHQ η= w = P P7. η = ηmax at roughly Q=0.6Qmax to 0.93Qmax8. η = ηmax is called the BEST EFFICIENCY POINT (BEP)9. All the parameters corresponding to ηmax are called the design points, Q*, H*, P*10. Pumps design should be such that the efficiency curve should be as flat as possible around ηmax11. ‘P’ rises almost linearly with flow rate
22.
CENTRIFUGAL PUMPS, CHARACTERISTICS, contd. (a ) basic casing with three (b) 20 percent larger casing with three impeller sizes larger impellers at slower speed Typical Characteristic Curves of Commercial Centrifugal Pumps1. Having same casing size but different impeller diameters2. Rotating at different rpm3. For power requirement and efficiency one needs to interpolate
23.
CENTRIFUGAL PUMPS, CHARACTERISTICS, contd. Calculate the ideal Head to be developed by the pump shown in last figure (1170 × 2π / 60 rad / s ) ( 36.75 / 2 ×12 ft ) 2 2 ω 2 r22 H o (ideal ) = = 2 = 1093 ft g 32.2 ft / sActual Head = 670 ft or 61% of Ho(ideal) at Q=0Differences are due to1. Impeller recirculations, important at low flow rates2. Frictional losses3. Shock losses due to mismatch of blade angle and flow inlet important at high flow rates
24.
CENTRIFUGAL PUMPS, CHARACTERISTICS, contd. IMPORTANT POINTS TO REMEMBER1. EFFECT OF DENSITY 1. Pump head reported in ‘ft’ or ‘m’ of that fluid → ρ important 2. These characteristic curves, valid only for the liquid reported 3. Same pump used to pump a different liquid → H and η would be almost same. OR. A centrifugal pump will always develop the same head in feet of that liquid regardless of the fluid density 4. However P will change. Brake HP will vary directly with the liquid density2. EFFECT OF VISCOSITY 1. Viscous liquids tend to decrease the pump Head, Discharge and efficiency → tends to steepen the H-Q curve with η ↓ 2. Viscous liquids tend to increase the pump BHP
27.
CENTRIFUGAL PUMPS, CHARACTERISTICS, contd.µ ≥ 300µwor µ > 2000 SSU µ PDP’s are preferredµ ≤ 10µw or µ < 50 SSU µCentrifugal pumps are preferred
28.
SUCTION HEAD AND SUCTION LIFT• A centrifugal pump cannot pull or suck liquids• Suction in centrifugal pump → creation of partial vacuum at pump’s inlet as compared to the pressure at the other end of liquid• Hence, pressure difference in liquid → drives liquid through pump• How one can increase this pressure difference – Increasing the pressure at the other end • Equal to 1 atm for reservoirs open to atmosphere • > or < 1 atm for closed vessels – Decreasing the pressure at the pump inlet • Must be > liquid vapor pressure → temperature very important • By increasing the capacity → Bernoullis equation
29.
SUCTION HEAD AND SUCTION LIFT MAXIMUM SUCTION DEPENDS UPON• Pressure applied at liquid surface at liquid source, hence – Maximum suction decreases as this pressure decreases• Vapor pressure of liquid at pumping temperature – Maximum suction decreases as vapor pressure increases• Capacity at which the pump is operating CASE OF OPEN RESERVOIRS• Maximum suction varies inversely with altitude Table-1 CASE OF HOT LIQUIDS• Maximum suction varies inversely with temp. Table-2 CASE OF INCREASING CAPACITY• Maximum suction varies inversely with capacity Table-3
30.
NET POSITIVE SUCTION HEAD• Problem of Cavitation –The lowest pressure occurs at the pump’s inlet –Pressure at pump inlet < liquid vapor pressure → cavitation occurs –What are the effects of cavitation • Lot of noise and vibrations are generated • Sharp decrease in pump’s ‘H’ and ‘Q’ • Pitting of impeller occurs due to bubble collapse • May occur before actual boiling in case of dissolved gases / low boiling mixtures of hydrocarbons• Hence ‘P’ at pump’s inlet should greater than the Pvp• This extra pressure above Pvp available at pump’s inlet is called Net Positive Suction Head ‘NPSH’ P Vi 2 Pvp• Mathematically → NPSH = 1 + − ρg 2 ρg
31.
NET POSITIVE SUCTION HEAD, contd.• NPSH calculated from this equation is the ‘REQUIRED NPSH’ specified by manufacturer → “PUMP’S CHARACTERISTIC”• The NPSH actually available at the pump’s inlet is called ‘AVAILABLE NPSH’ → “SYSTEM’S CHARACTERISTIC”• ‘AVAILABLE NPSH’ must be ≥‘REQUIRED NPSH’• Rule of thumb for design ‘AVAILABLE NPSH’ ≥ (2+‘REQUIRED NPSH’ NPSH’) ft of liquidHOW TO CALCULATE AVAILABLE NPSHWrite Energy Equation between the free surface of fluid reservoirand pump inlet Psurface Pvp NPSH available = − Z i − h fi − ρg ρgThus Zi can be important parameter in designers hand to ensure thatcavitation does not occur for a given Psurface and temperature
32.
NET POSITIVE SUCTION HEAD, contd. EFFECT OF VARYING HEIGHT Psurface Pvp Given, Psurface, Pvp and hfi , Zi can NPSHA = − Z i − h fi − ≥ NPSHR ρg ρg be varied to avoid cavitation An Example The 32-in pump of Fig. 11.7a is to pump 24,000 gpm of water at 1170 rpm from a reservoir whose surface is at 14.7 psia. If head loss from reservoir to pump inlet is 6 ft, where should the pump inlet be placed to avoid cavitation for water at (a) 60°F, pvp0.26 psia, SG 1.0 and (b) 200°F, pvp 11.52 psia, SG 0.9635? NPSHR = 40 ≤ Psurface Pvp − Z i − h fi − = (14.7 − 0.26 ) − Z − 6 Z i ≤− 12.7ρ g = 62.4 ρg ρ g 62.4 (144 )−1 i Pump must be placed at least 12.7 ft below the reservoir surface to avoid cavitation. ρ g = 62.4 × .9653 = 60.1 Z i ≤ −38.4 Pump must now be placed at least 38.4 ft below the reservoir surface, to avoid cavitation
33.
NET POSITIVE SUCTION HEAD, contd.TYPICAL EXAMPLEA pump installed at an altitude of 2500 ft and has a suction lift of 13 ftwhile pumping 50 degree water. What is NPSHA? Ignore friction Psurface Pvp NPSH available = − Z i − h fi − = 31 − 13 − 0 − .41 = 17.59 ft ρg ρg Actual NPSHA = 17.59 – 2 = 15.59 ftTYPICAL EXAMPLEWe have a pump that requires 8 ft of NPSH at I20 gpm. If the pump isinstalled at an altitude of 5000 ft and is pumping cold water at 60oF,what is the maximum suction lift it can attain? Ignore friction Psurface Pvp NPSHA = NPSHR + 2 = 8 + 2 = − Z i − h fi − = 28.2 − Z i − 0 − .59 = 17.59 ft ρg ρg
34.
DIMENSIONLESS PUMP PERFORMANCE-1 PERFORMANCE- EVERY PUMP HAS THREE PERFORMANCE PARAMETERS1. Head ‘H’ (or pressure difference ∆P-recall that ∆P= ρgH)2. Volume Flow Rate ‘Q’3. Power ‘P’ TWO "GEOMETRIC" PARAMETERS: 1. D diameter 2. n (or ω) rotational speed THREE FLUID FLOW PARAMETERS:1. ρ density2. viscosity3. ε roughness Above parameters involve only three dimensions, M-L-T
35.
DIMENSIONLESS PUMP PERFORMANCE-2 PERFORMANCE-Buckingham π Theorem suggests7 -3 = 4 π’s to represent the physical phenomena in a pump. Any pump’s performance parameters are 1. Head H (or gH ) → gH = f1 ( Q, D, n, ρ , µ , ε ) 2. Power P → P = f 2 ( Q, D, n, ρ , µ , ε ) Hence The Two π Groups Are gH Q ρ nD 2 ε P Q ρ nD 2 ε = g1 3 , , = g2 3 , , n2 D 2 nD µ D ρn D 3 5 nD µ D WHEREε = relative roughnessD gH ρ nD 2 ρ ( nD ) D 2 2 = CH = Head Coefficient = = Re. Number n D µ µ Q P 3 = CQ = Capacity Coefficient 3 5 = CP = Power Coefficient nD ρn D
36.
DIMENSIONLESS PUMP PERFORMANCE-3 PERFORMANCE-Reynolds number inside a centrifugal pump Hence, we may write:1. ≈ 0.80 to 1.5x107)2. Flow always turbulent CH = CH ( CQ )3. Effect of Re, almost constant4. May take it out of the functions g1and g2 CP = CP ( CQ )5. Same is true for ε/D For geometrically similar pumps, Head and Power coefficients should be (almost) unique functions of the capacity coefficients.In real life, however:-manufacturers use the same case for different rotors(violating geometrical similarity)-larger pumps have smaller ratios of roughness and clearances-the fluid viscosity is the same, while Re changes with diameters.
37.
DIMENSIONLESS PUMP PERFORMANCE-4 PERFORMANCE-CH, CP and CQ combined to give a coefficient having practical meaning CH CQ η= = η ( CQ ) CP Similarly one can also define the CNPSH the NPSH coefficient as g ⋅ NPSH C NPSH = 2 2 = C NPSH ( CQ ) n D
38.
DIMENSIONLESS PUMP PERFORMANCE-5 PERFORMANCE- Representing the pump performance data in dimensionless form Pump data Results in graphical form•Choose two geometricallysimilar pumps•32 in impeller in pump (a) & 38in in pump (b)•Pump (b) casing 20% > pump(a) casing.•Hence same diameter to casingratios DISCRIPENCIES•A few % in η and CH•pumps not truly dynamically similar•Larger pump has smaller roughness ratio•Larger pump has larger Re. number
39.
DIMENSIONLESS PUMP PERFORMANCE-6 PERFORMANCE-The BEP lies at η=0.88, corresponding to,CQ* ≈ 0.115 CP* ≈ 0.65 CH* ≈ 5.0 CNPSH* ≈ 0.37A unique set of values• Valid for all pumps of this geometrically similar family• Used to estimate the performance of this family pumps at BEP Comparison of Values Discharge Head Power D, ft n, r/s nD3, ft3/s n2D2/g, ft n3D5/550, hpFig. 11.7a 32/12 1170/60 370 84 3527Fig. 11.7b 38/12 710/60 376 44 1861Ratio - - 1.02 0.52 0.53
40.
SIMILARITY RULES/AFFINITY LAWS-1 LAWS-If two pumps are geometrically similar, then1. Ratio of the corresponding coefficients =12. This leads to estimation of performance of one based on the performance of the other MATHEMATICALLY THIS CONCEPT LEADS TO Q2 gH 2 P2CQ2 3 n2 D2 CH 2 2 2 n2 D2 CP2 ρ 2 n2 D2 3 5 = =1 = =1 = =1 CQ1 Q1 CH1 gH1 CP1 P1 n1 D13 n12 D12 ρ1n13 D15 3 2 2 3 5 Q2 n2 D2 H 2 n2 D2 P2 ρ 2 n2 D2 = = = Q1 n1 D1 H1 n1 D1 P ρ1 n1 D1 1 THESE ARE CALLLED SIMILARITY RULES
41.
SIMILARITY RULES/AFFINITY LAWS-1 LAWS-The similarity rules are used to estimate the effect of1. Changing the fluid2. Changing the speed3. Changing the sizeVALID ONLY AND ONLY FORGeometrically similar family of any dynamic turbo machine pump/compressor/turbine Effect of changes in size and speed on homologous pump performance (a) 20 percent change in speed at constant size (b) 20 percent change in size at constant speed
42.
SIMILARITY RULES/AFFINITY LAWS-1 LAWS-For Perfect Geometric Similarity η1 = η2, butLarger pumps are more efficient due to 1. Higher Reynolds Number 2. Lower roughness ratios 3. Lower clearance ratiosEmpirical correlations are availableTo estimate efficiencies in geometrically similar family of pumps 1 1 − η2 D2 4Moody’s Correlation ≈ Based on size changes 1 − η1 D1 0.33Anderson’s Correlation 0.94 − η2 Q2 ≈ Based on flow rate changes 0.94 − η1 Q1
43.
Concept of Specific Speed-1 Speed- A confusing exampleWe want to use a centrifugal pump from the family of Fig. 11.8 todeliver 100,000 gal/min of water at 60°F with a head of 25 ft. Whatshould be (a) the pump size and speed and (b) brake horsepower,assuming operation at best efficiency?H* = 25 ft = (CH n2 D2)/g = (5 × n2 D2)/32.2Q* = 100000 gpm = 222.8 ft3/m = CQ n D3 = 0.115 × n D3Bhp* = Cpρ n3 D5 = 720 hpSolving simultaneously gives, D = 12.4 ft, n = 62 rpm
44.
Concept of Specific Speed-1 Speed-The type of applications for which centrifugal pumps are required are;1. High head low flow rate2. Moderate head and moderate flow rate3. Low head and high flow rateQ. Would a general design of the centrifugal pump will do all the three jobs?Ans. NoQ. What should be the design features to accomplish the three specified jobs? PHYSICS FOR OUR RESCUE1. Answer to this question lies in the basic concept of centrifugal pump working principle.2. Vanes are used to impart momentum to the fluid by applying the centrifugal force to the fluid.
45.
Concept of Specific Speed-2 Speed-3. More the diameter of the vane more will be the centrifugal force4. More will be the diameter more will be the radial component of velocity and lesser will be the axial component5. More will be the radial velocity more will be the head developed6. Hence to get more head you need longer vanes and vice versa7. More will be the clearance between the impeller and casing more will the flow rate & also more will be the axial component8. These simple physics principles lead us to the variation in impeller design to accomplish the three jobs mentioned
46.
Concept of Specific Speed-3 Speed- POINT TO PONDER• We represent the performance of a family of geometrically similar pumps by a single set of dimensionless curves• Can we use even a smaller amount of information or even a single number to represent the same information?• We have a huge variety of pumps each with a different diameter impeller, shape of impeller and running at certain rpm• Impeller shape ultimately dictates the type of application• RPM is not related to the pump design however it effects its performance• Hence the biggest problem is to avoid diameter in the pump performance informationAgain dimensional analysis comes to rescue, a combination of π’s is also a π, giving the same information in a different form
47.
Concept of Specific Speed-4 Speed- REARRANGE THE THREE COEFFICIENTS INTO A NEW COEFFICIENT SUCH THAT DIAMETER IS ELIMINATED 1 CQ 2 n (Q ) 1 2 N s = 17182 N s/ N = / = ( gH ) s 3 3 CH 4 4 Points to remember 1. Ns refers only to BEP 2. Directly related to most efficientRigorous form, dimensionless pump design 3. Low Ns means low Q, High H ( RPM )( GPM ) 1 2 Ns = 4. High Ns means High Q, Low H ( H , ft ) 3 4 5. Ns leads to specific pump applications 6. Low Ns means high head pump Lazy but common form, 7. High Ns means high Q pump Not dimensionless Experimental data suggests, pump is inSimilarly one can define Nss danger of cavitation, based on NPSH If Nss ≥ 8100
48.
Concept of Specific Speed-5 Speed- GEOMETRICAL VARIATION OF SPECIFIC SPEEDDetailed shapes
49.
Concept of Specific Speed-5 Speed-Specific speed is an indicator of Pump performance Pump efficiencyThe Q is a rough indicator of Pump size Pump Reynolds Number THE PUMP CURVES
50.
Concept of Specific Speed-5 Speed-Note How The Head, Power and Efficiency curves change as specific speed changes
51.
Revisit of Confusing Example-1 Example- Dimensionless performance curves for a typical axial- flow pump. Ns = 12.000. Constructed from data for a 14-in pump at 690 rpm. CQ* =0.55, CH*=1.07, Cp*=0.70,ηmax= 0.84. Ns = 12000 D = 14 in, n = 690 rpm, Q* = 4400 gpm.
52.
Revisit of Confusing Example-2 Example-Can this propeller pump family provide a 25-ft head & 100,000 gpmdischarge Since we know the Ns and Dimensionless coefficients then using similarity rules let us calculate the Diameter and RPM D = 48 in and n = 430 r/min, with bhp = 750: a much more reasonable design solution
53.
Pump vs System Characteristics• Any piping systems has the following components in its total head which the selected pump would have to supply1. Static head due to elevation2. The head due to velocity head, the fictional head loss3. Minor head losses H sys = ( z2 − z1 ) = a h f ,la min ar = 128µ LQ πρ gD 4 H sys = ( z2 − z1 ) + h f ,la min ar = a + bQ Mathematically, 3 possibilities h f ,turbulent = Through Moody s Method V2 fL H sys = ( z2 − z1 ) + ∑ + ∑ K = a + cQ 2 2g D
54.
Pump vs System Characteristics, contd• Graphical Representation Of The Three Curves
55.
Match between pump & system•In industrial situation the resistance often varies for variousreasons•If the resistance factor increases, the slope of the systemcurve (Resistance vs flow) increases & intersect thecharacteristic curve at a lower flow.•The designed operating points are chosen as close to thehighest efficiency point as possible.•Large industrial systems requiring different flow rates oftenchange the flow rate by changing the characteristic curve withchange in blade pitch or RPM
57.
Pump in Parallel or Series•To increase flow at a given head 1. Reduce system resistance factor with valve 2. Use small capacity fan/pumps in parallel.Some loss in flow rate may occur when operatingin parallel•To increase the head at a given flow 1. Reduce system resistance by valve 2. Use two smaller head pumps/fans in series.But some head loss may occur.
60.
Unstable operation (Hunting)If the characteristic issuch that the systemfinds two flow rates fora given head it cannotdecide where to stay.The pump couldoscillate betweenpoints. It is calledhunting.
Views
Actions
Embeds 0
Report content