PumpsEquation of Mechanical Energy The equation of mechanical energy in terms of Energy per Unit Mass, Energy per Unit Volume and Energy per Unit Weight involving head The Energy Equation is a statement of the first law of thermodynamics. The energy equation involves energy, heat transfer and work. With certain limitations the mechanical energy equation can be compared to the Bernoulli Equation.The Mechanical Energy Equation in Terms of Energy per Unit MassThe mechanical energy equation for a pump or a fan can be written in terms of energy per unit mass:pin / + vin2 / 2 + g hin + wshaft = pout / + vout2 / 2 + g hout + wloss (1)wherep = static pressure = densityv = flow velocityg = acceleration of gravityh = elevation heightwshaft = net shaft energy inn per unit mass for a pump, fan or similarwloss = loss due to frictionThe energy equation is often used for incompressible flow problems and is called the Mechanical Energy Equation or the Extended Bernoulli Equation.The mechanical energy equation for a turbine can be written as:pin / + vin2 / 2 + g hin = pout / + vout2 / 2 + g hout + wshaft + wloss (2)wherewshaft = net shaft energy out per unit mass for a turbine or similarEquation (1) and (2) dimensions are energy per unit mass (ft2/s2 = ft lb/slug or m2/s2 = N m/kg)

EfficiencyAccording to (1) a larger amount of loss - wloss - result in more shaft work required for the same rise of output energy. The efficiency of a pump or fan process can be expressed as: = (wshaft - wloss) / wshaft (3)The efficiency of a turbine process can be expressed as: = wshaft/ (wshaft + wloss) (4)The Mechanical Energy Equation in Terms of Energy per Unit VolumeThe mechanical energy equation for a pump or a fan (1) can also be written in terms of energy per unit volume by multiplying (1) with fluid density - :pin + vin2 / 2 + hin + wshaft = pout + vout2 / 2 + hout + wloss (5)where = g = specific weightThe dimensions of equation (5) are energy per unit volume (ft.lb/ft3 = lb/ft2 or N.m/m3 = N/m2)The Mechanical Energy Equation in Terms of Energy per Unit Weight involves HeadsThe mechanical energy equation for a pump or a fan (1) can also be written in terms of energy per unit weight by dividing with gravity - g:pin / + vin2 / 2 g + hin + hshaft = pout / + vout2 / 2 g + hout + hloss (6)where = g = specific weighthshaft = wshaft / g = net shaft energy head inn per unit mass for a pump, fan or similarhloss = wloss / g = loss head due to frictionThe dimensions of equation (6) are energy per unit weight (ft.lb/lb = ft or N.m/N = m)Head is the energy per unit weight.hshaft can also be expressed as:hshaft = wshaft / g = Wshaft / m g = Wshaft / Q (7)whereWshaft = shaft powerm = mass flow rateQ = volume flow rateExample - Pumping WaterWater is pumped from an open tank at level zero to an open tank at level 10 ft. The pump adds four horsepowers to the water when pumping 2 ft3/s.Since vin = vout = 0, pin = pout = 0 and hin = 0 - equation (6) can be modified to:hshaft = hout + hlossorhloss = hshaft - hout (8)Equation (7) gives:hshaft = Wshaft / Q = (4 hp)(550 ft.lb/s/hp) / (62.4 lb/ft3)(2 ft3/s) = 17.6 ft specific weight of water - 62.4 lb/ft3 1 hp (English horse power) = 550 ft. lb/sCombined with (8):hloss = (17.6 ft ) - (10 ft) = 7.6 ftThe pump efficiency can be calculated from (3) modified for head: = ((17.6 ft) - (7.6 ft)) / (17.6 ft) = 0.58Pumps, Compressors, Blowers and Fans Comparing pumps, compressors, blowers and fans A generic definition of a Pump is"A machine or device for raising, compressing, or transferring fluids."In practice - it is common to differentiate between pumps compressors blowers fansas summarized below:Pumps A pump is a machine for raising a liquid - a relatively incompressible fluid - to a higher level of pressure or head.Compressors A compressor is a machine for raising a gas - a compressible fluid - to a higher level of pressure.Blowers A blower is a machine for moving volumes of a gas with moderate increase of pressure.Fans A fan moves large amounts of gas with low increase in pressure.Specific Work of Turbo Machines - Pumps, Compressors and Fans Specific work of pumps, fans, compressors and turbines Specific work is work per unit weight. Specific work in turbo machines as fans, pumps, compressors or turbines has the SI-units Nm/kg = J/kg = m2/s2Specific Work of a Pump or FanSpecific work of a pump or fan working with an incompressible fluid can be expressed as:w = (p2 - p1) / (1)wherew = specific work (Nm/kg = J/kg = m2/s2)p = pressure (N/m2) = density (kg/m3)Specific Work of a TurbineSpecific work of a turbine with an incompressible fluid can be expressed as:w = (p1 - p2) / (2)Specific Work of a CompressorA compressor works with compressible fluids and the specific work for an isentropic compressor process can be expressed with the help ofp1 v1 = p2 v2 (3)wherev = volume (m3) = cp / cv - ratio of specific heats (J/kg K)Specific work:w = / ( -1) R T1 [( p2 / p1)((-1)/) - 1] (4)whereR = individual gas constant (J/kg K)T = absolute temperature (K)Specific Work of a Gas TurbineA gas turbine expands a compressible fluid and the specific work can be expressed asw = / ( -1) R T1 [1 - ( p2 / p1)((-1)/)] (5)Head in TurbomachinesThe specific work can on basis of the energy equation be expressed with the head as:w = g h (6)whereh = head (m)g = acceleration of gravity (m/s2)Transformed to express head:h = w / g (7)Example - Specific Work of a Water PumpA water pump works between 1 bar (105 N/m2) and 10 bar (10 105 N/m2). The specific work can be calculated with (1):w = (p2 - p1) / = ( (10 105 N/m2) - (105 N/m2) ) / (1000 kg/m3) = 900 Nm/kgDividing by acceleration of gravity the head can be calculated using (7):hwater = (900 Nm/kg) / (9,81 kg/s2) = 91,74 (m) water columnExample - Specific Work of an Air CompressorAn air compressor works with air at 20 oC compressing the air from 1 bar absolute (105 N/m2) to 10 bar (10 105 N/m2). The specific work can be expressed with (4):w = / ( -1) R T1 [( p2 / p1)((-1)/) - 1] = ( (1.4 J/kg K) / ((1.4 J/kg K) - 1 ) ) (286.9 J/kg K) ((273 K) + (20 K)) [( (10 105 N/m2) / (105 N/m2) )(((1.4 J/kg K) - 1)/(1.4 J/kg.K)) - 1 ] = 273826 Nm/kgwhereair = 1.4 (J/kg K) - ratio of specific heat airRair = 286.9 (J/kg K) - individual gas constant airDividing by acceleration of gravity the head can be calculated using (7):hair = (274200 N m/kg) / (9.81 kg/s2) = 27951 (m) air columnHead Rise and the Energy Equation for a Pump or a Fan The energy equation can be used to calculate the actual head rise in a pump or fan Actual Head Rise of a Pump or FanUsing the Energy Equation the head rise through a pump or fan can be expressed as:ha = (p2 - p1) / + h2 - h1 + (v22 - v12) / 2 g (1)whereha = actual head risep = pressureh = elevation height = g = specific weightv = velocityg = acceleration of gravityThe actual head rise can be expressed as:ha = hshaft - hloss (2)wherehshaft = shaft work in the pump or fanhloss = head loss through the pump or fanThe head loss - hloss - through a pump or fan is related to the skin friction in the blade passages and is proportional with volume flow - q2. flow separation impeller blade casing clearance flows other three dimensional flow effectsActual Head Rise for an Inline PumpFor a very common installation - the inline pump or fan - where the inlet velocity and the outlet velocity are the same (v2 = v1), and the inlet and outlet elevation are the same (h2 = h1), the generic equation (1) can be modified to:ha = (p2 - p1) / (3)Specific WorkBy multiplying (3) with acceleration of gravity - g -, specific work - w - of the pump or fan may be calculated:w = ha g (4)wherew = specific work

Example - Head Rise of an Inline PumpAn inline water pump works between the pressure 1 bar (1 105 N/m2) and 10 bar (10 105 N/m2). Density of water is 1000 kg/m3. The actual water head (water column) can be calculated using (3):hwater = (p2 - p1) / = (p2 - p1) / g = ((10 105 N/m2) - (1 105 N/m2)) / (1000 kg/m3) (9.81 m/s2) = 91.7 m - water columnExample - Head Rise of an FanAn inline fan working with hot air with density = 1,06 kg/m3 add a pressure of 400 Pa (N/m2) to the flow.The air head (air column) can be calculated with (3):hair = (p2 - p1) / g = (400 N/m2) / (1.06 kg/m3) (9.81 m/s2) = 38.5 m - air columnThe water head (water column) can be calculated with (3) using the density of water:hwater = (p2 - p1) / g = (400 N/m2) / (1000 kg/m3) (9.81 m/s2) = 0.041 m = 41 mm - water columnMeasuring pressure with water column in an U-tube manometer is common in air distribution applications as ventilation and air condition systems.Alternating Pumps Secure the system and achieve equal wear by alternate pumps in parallel Critical systems should always be equipped with more than one pump. Choosing between the installation of one or more backup pumps depends on the costs of the installation and how critical the operation of the system is - a water supply system for a hospital is more critical than a chilled water cooling systems for homes. the delivery and installation time of a new pump.If two or more pumps are used, the operation between the pumps should be systematically altered to achieve equal wear.Alternatives for Alternating PumpsPumps can be systematically altered by manual alteration - where the operator selects the lead pump and the sequence of the lag pumps. duty alternation - where the lead pump change every time the pump or system is stopped. timed alternation - where the lead pump is switched by a timer or clock. equal run time - where the lead pump is switched to achieve the same operating time for each pump.Note! Running an automatic system for equalizing the wear of the pumps has the advantage of extending the time before repair and reinvestments. has the disadvantage that all pumps may wear out on at the same time - reducing the operation safety for the whole system.Specific Speed - Pumps Specific speed may be determined independent of the pump size Specific speed is a number characterizing the type of impeller in a unique and coherent manner. Specific speed are determined independent of pump size and can be useful comparing different pump designs. The specific speed identifies the geometrically similarity of pumps. Specific speed can be expressed asNs = q1/2 / h3/4 (1)whereNs = specific speed = pump shaft rotational speed (rpm)q = flow rate (m3/h, l/s, l/min, m3/min, US gpm, British gpm) at Best Efficiency Point (BEP)h = head rise (m, ft)Pumps - Specific Suction Speed Specific Suction Speed may be used to determine what general pump design to use for maximum efficiency Suction Specific Speed may be used to determine what pump geometry - radial, mixed flow or axial - to use for maximum efficiency and prevent cavitation. Suction Specific Speed is commonly used as a basis for estimating the safe operating range of a pump.Suction Specific Speed is dimensionless and are expressed asNss = q1/2 / NPSHr3/4 (1)whereNss = Suction Specific Speed = pump shaft rotational speed (rpm)q = flow rate capacity (m3/h, l/s, m3/min, US gpm, British gpm) at Best Efficiency Point BEPNPSHr = pump NPSH required at best efficiency point (m, ft)Specific Suction Speed can be compared with Specific Speed but instead of using the total head of the pump, the available Net Positive Suction Head N.P.S.H.a is used.As a rule of thumb the Specific Suction Speed should be below 8500 (calculated with US gpm) to avoid cavitation.Best Efficiency Point - BEP Maximum designed pump efficiency A pump does not completely convert the kinetic to pressure energy. Some of the energy is always lost internal and external in the pump. Internal losses hydraulic losses - disk friction in the impeller, loss due to rapid change in direction an velocities through the pump volumetric losses - internal recirculation at wear rings and bushesExternal losses mechanical losses - friction in seals and bearings

The efficiency of the pump at the designed point is normally maximum and is called the Best Efficiency Point - BEP It is possible to operate the pump at other points than BEP, but the efficiency of the pump will always be lower than BEP. Suction Head affected by Altitude Pump suction head as affected by altitude Suction head affected by altitude as:Elevation above Sea LevelPressurePractical Suction Lift

(ft)(m)(psig)(ft H2O)(bar)(ft H2O)(m H2O)

0* 0* 14.7133.951226.7

132040214.0232.380.97216.4

264080513.3330.790.92206.1

3960120712.6629.240.87185.5

5280160912.0227.760.83175.2

6600201211.4226.380.79164.9

7920241410.8824.130.75154.6

1056032199.8822.820.68144.3

* Sea LevelVapor Pressure Saturation pressure - exerted by escaping molecules The vapor pressure of a liquid is defined as the pressure exerted by the molecules that escapes from the liquid to form a separate vapor phase above the liquid surface.The pressure exerted by the vapor phase is called the vapor or saturation pressureVapor or saturation pressure depends on temperature. If a fluid consist of more than one component (a solution), components with higher vapor pressure are called light components lower vapor pressure are called heavy components Temperature and Vapor or Saturation Pressure for some common FluidsAt atmospheric pressure saturation temperature of water : 100oC ethyl alcohol : 78.5oC NPSH - Net Positive Suction Head A definition and an introduction to Net Positive Suction Head - NPSH Low pressure at the suction side of a pump can encounter the fluid to start boiling with reduced efficiency cavitation damageof the pump as a result. Boiling starts when the pressure in the liquid is reduced to the vapor pressure of the fluid at the actual temperature.

To characterize the potential for boiling and cavitation, the difference between the total head on the suction side of the pump - close to the impeller, and the liquid vapor pressure at the actual temperature, can be used.Suction HeadBased on the Energy Equation - the suction head in the fluid close to the impeller can be expressed as the sum of the static and the velocity head:hs = ps / + vs2 / 2 g (1)wherehs = suction head close to the impellerps = static pressure in the fluid close to the impeller = specific weight of the fluidvs = velocity of fluidg = acceleration of gravity

Liquids Vapor HeadThe liquids vapor head at the actual temperature can be expressed as:hv = pv / (2)wherehv = vapor headpv = vapor pressureNote! The vapor pressure in fluids depends on temperature. Water, our most common fluid, starts boiling at 20 oC if the absolute pressure in the fluid is 2.3 kN/m2. For an absolute pressure of 47.5 kN/m2, the water starts boiling at 80 oC. At an absolute pressure of 101.3 kN/m2 (normal atmosphere), the boiling starts at 100 oC.Net Positive Suction Head - NPSHThe Net Positive Suction Head - NPSH - can be expressed as the difference between the Suction Head and the Liquids Vapor Head and expressed likeNPSH = hs - hv (3)or, by combining (1) and (2)NPSH = ps / + vs2 / 2 g - pv / (3b)Available NPSH - NPSHa or NPSHAThe Net Positive Suction Head made available the suction system for the pump is often named NPSHa. The NPSHa can be determined during design and construction, or determined experimentally from the actual physical system.

The available NPSHa can be calculated with the Energy Equation. For a common application - where the pump lifts a fluid from an open tank at one level to an other, the energy or head at the surface of the tank is the same as the energy or head before the pump impeller and can be expressed as:h0 = hs + hl (4)whereh0 = head at surfacehs =headbefore the impellerhl =headloss from the surface to impeller - major and minor loss in the suction pipeIn an open tank the head at surface can be expressed as:h0 = p0 / = patm / (4b)For a closed pressurized tank the absolute static pressure inside the tank must be used.The head before the impeller can be expressed as:hs = ps / + vs2 / 2 g + he (4c)wherehe = elevation from surface to pump - positive if pump is above the tank, negative if the pump is below the tankTransforming (4) with (4b) and (4c):patm / = ps / + vs2 / 2 g + he + hl (4d)The head available before the impeller can be expressed as:ps / + vs2 / 2 g = patm / - he - hl (4e)or as the available NPSHa:NPSHa = patm / - he - hl - pv / (4f)Available NPSHa - the Pump is above the TankIf the pump is positioned above the tank, the elevation - he - is positive and the NPSHa decreases when the elevation of the pump increases.At some level the NPSHa will be reduced to zero and the fluid starts to evaporate.Available NPSHa - the Pump is below the TankIf the pump is positioned below the tank, the elevation - he - is negative and the NPSHa increases when the elevation of the pump decreases (lowering the pump).It's always possible to increase the NPSHa by lowering the pump (as long as the major and minor head loss due to a longer pipe don't increase it more). This is important and it is common to lower the pump when pumping fluids close to evaporation temperature.Required NPSH - NPSHr or NPSHRThe NPSHr, called as the Net Suction Head as required by the pump in order to prevent cavitation for safe and reliable operation of the pump.The required NPSHr for a particular pump is in general determined experimentally by the pump manufacturer and a part of the documentation of the pump.

The available NPSHa of the system should always exceeded the required NPSHr of the pump to avoid vaporization and cavitation of the impellers eye. The available NPSHa should in general be significant higher than the required NPSHr to avoid that head loss in the suction pipe and in the pump casing, local velocity accelerations and pressure decreases, start boiling the fluid on the impeller surface.Note that the required NPSHr increases with the square capacity.Pumps with double-suction impellers has lower NPSHr than pumps with single-suction impellers. A pump with a double-suction impeller is considered hydraulically balanced but is susceptible to an uneven flow on both sides with improper pipe-work.Example - Pumping Water from an Open TankWhen increasing the the elevation for a pump located above a tank, the fluid will start to evaporate at a maximum level for the actual temperature.At the maximum elevation NPSHa is zero. The maximum elevation can therefore be expressed by (4f):NPSHa = patm / - he - hl - pv / = 0For optimal theoretical conditions we neglect the major and minor head loss. The elevation head can then be expressed as:he = patm / - pv / (5)The maximum elevation or suction head for an open tank depends on the atmospheric pressure - which in general can be regarded as constant, and the vapor pressure of the fluid - which in general vary with temperature, especially for water.The absolute vapor pressure of water at temperature 20 oC is 2.3 kN/m2. The maximum theoretical elevation height is therefore:he = (101.33 kN/m2) / (9.80 kN/m3) - (2.3 kN/m2) / (9.80 kN/m3) = 10.1 mDue to the head loss in the suction pipe and the local conditions inside the pump - the theoretical maximum elevation is significantly decreased.The maximum theoretical elevation of a pump above an open water tank at different temperatures can be found from the table below.Suction Head as Affected by TemperatureTemperatureVapor PressureMax. elevation

(oC)(oF)(kN/m2)(m)(ft)

0320.610.333.8

5410.910.233.5

10501.210.233.5

15591.710.233.5

20682.310.133.1

25773.210.032.8

30864.39.932.5

35955.69.832.2

401047.79.531.2

451139.69.430.8

5012212.59.129.9

5513115.78.728.5

60140208.327.2

65149257.825.6

7015832.17.123.3

7516738.66.421

8017647.55.518

8518557.84.414.4

90194703.210.5

9520384.51.75.6

100212101.330.00

Pumping HydrocarbonsBe aware that the NPSH specification provided by the manufacturer in general is for use with cold water. For hydrocarbons these values must be lowered to account for the vapor release properties of complex organic liquids.FluidTemperature (oC)Vapor Pressure (kPa abs)

Ethanol205.9

6558.2

Methyl Acetate2022.8

5593.9

Note that the head developed by a pump is independent of the liquid, and that the performance curves for water from the manufacturer can be used for Newtonian liquids like gasoline, diesel or similar. Be aware that required power depends on liquid density and must be adjusted.Power Gained by Fluid from Pumps or Fans The power gained by fluid from operating pumps or fans Power Gained by FluidThe power gained by the fluid from a pump or fan can be expressed as:P = m w (1)whereP = powerm = mass flow ratew = specific workSpecific WorkSpecific work - w - can be expressed:w = g h (2)whereh = headg = acceleration of gravityMass Flow RateMass flow rate - m - can be expressed:m = Q (3)where = densityQ = volume flow rateCombining (1), (2) and (3) the power gained by the fluid from a pump or fan can be expressed as:P = Q g h (4)With specific weight expressed as: = g (5)where = specific weightequation (4) can be modified so the power gained by the fluid from a pump or fan can be expressed as:P = Q h (6)Since head can be expressed ash = (p2 - p1) / (7)equation (4) can be modified so the power gained by the fluid from a pump or fan can be expressed as:P = Q (p2 - p1) (8)Example - Head Rise of a Inline PumpAn inline water pump works between measured pressure 1 bar (1 105 N/m2) and 10 bar (10 105 N/m2). Density of water is 1,000 kg/m3. The volume flow is measured to 1 10-3 m3/s.The actual water head (water column) can be calculated using (7):h = (p2 - p1) / = (p2 - p1) / g = ((10 105 N/m2) - (1 105 N/m2)) / (1,000 kg/m3) (9.81 m/s2) = 91.7 m - water columnThe power gained by the fluid can be calculated using equation (4):P = Q g h = (1,000 kg/m3) (1 10-3 m3/s) (9.81 m/s2) (91.7 m) = 899.6 kg.m2/s3 (W) = 0.9 kWVolume Flow and Temperature Rise in Pumps Calculate temperature rise in pumps No pump is perfect with 100% efficiency. The energy lost in friction and hydraulic losses are transformed to heat - heating up the fluid transported through the pump.

The temperature rise can be calculated as dt = Ps (1 - ) / cp q (1)where dt = temperature rise in the pump (oC)q = volume flow through the pump (m3/s)Ps = brake power (kW)cp = specific heat capacity of the fluid (kJ/kgoC)= pump efficiency = fluid density (kg/m3)Typical relation between the centrifugal pump flow, efficiency and power consumption, is indicated in the figure below:

Example - Temperature rise in water pumpThe temperature rise in a water pump working at normal conditions with flow 6 m3/h (0.0017 m3/s), brake power 0.11 kW and pump efficiency of 28% (0.28), can be calculated as dt = (0.11 kW) (1 - 0.28) / (4.2 kJ/kgoC) (0.0017 m3/s) (1000 kg/m3) = 0.011 oCIf the flow of the pump is reduced by throttling the discharge valve, the temperature rise through the pump will increase. If the flow is reduced to 2 m3/h (0.00056 m3/s), the brake power is slightly reduced to 0.095 kW and pump efficiency reduced to 15% (0.15), the temperature rise can be calculated as dt = (0.095 kW) (1 - 0.15) / (4.2 kJ/kgoC) (0.00056 m3/s) (1000 kg/m3) = 0.035 oCWith the standard documentation provided by a manufacturer it should be possible to express the temperature rise as a function of volume flow as shown in the figure below:

Cavitation - an Introduction Cavitation may occur in fluid flow systems where local static pressure is below the vapor pressure Cavitation is a common problem in pumps and control valves - causing serious wear and tear and damage. Under the wrong conditions, cavitation reduces the component life time dramatically.What is Cavitation?Cavitation may occur when the local static pressure in a fluid reach a level below the vapor pressure of the liquid at the actual temperature.According to the Bernoulli Equation this may happen when the fluid accelerates in a control valve or around a pump impeller.The vaporization itself does not cause the damage - the damage happens when the vapor almost immediately collapses after evaporation when the velocity is decreased and pressure increased.Avoiding CavitationCavitation can in general be avoided by increasing the distance between the actual local static pressure in the fluid - and the vapor pressure of the fluid at the actual temperatureThis can be done by: reengineering components initiating high speed velocities and low static pressures increasing the total or local static pressure in the system reducing the temperature of the fluidReengineering of Components Initiating High Speed Velocity and Low Static PressureCavitation and damage can be avoided by using special components designed for the actual rough conditions. conditions as huge pressure drops can - with limitations - be handled by Multi Stage Control Valves challenging pumping conditions - with fluid temperatures close to the vaporization temperature - can be handled with special pumps - working after other principles than centrifugal pumpsIncreasing the Total or Local Pressure in the SystemBy increasing the total or local pressure in the system the distance between the static pressure and the vaporization pressure is increased and vaporization and cavitation can be avoided.The ratio between static pressure and the vaporization pressure - an indication of the possibility of vaporization, is often expressed by the Cavitation Number.Unfortunately it may not always be possible to increase total static pressure due to systems classifications or other limitations. Local static pressure in components may be increased by lowering the component in the system. Control valves and pumps should in general be positioned in the lowest part of the systems to maximize static head.This is common solution for boiler feeding pumps receiving hot condensate (water close to 100 oC) from condensate receivers.Reducing the Temperature of the FluidThe vaporization pressure depends of fluid temperature. Vapor pressure of Water, our most common fluid, is indicated below:Temperature(oC)Vapor Pressure(kN/m2)

00.6

50.9

101.2

151.7

202.3

253.2

304.3

355.6

407.7

459.6

5012.5

5515.7

6020

6525

7032.1

7538.6

8047.5

8557.8

9070

9584.5

100101.33

Note! - the possibility of evaporation and cavitation increases dramatically with the water temperature.Cavitation can also be avoided by locating components to the coldest part of a system. It is common to locate pumps in heating systems in the "cold" return lines.This is the same for control valves. If it is possible control valves should be located on the cold sides of heat exchangers.System Curve and Pump Performance Curve To select a proper pump for a particular application it is necessary to utilize the system curve and the pump performance curve The System CurveA fluid flow system can in general be characterized with the System Curve - a graphical presentation of the Energy Equation.

The system head visualized in the System Curve is a function of the elevation - the static head in the system, and the major and minor losses and can be expressed as:h = dh + hl (1)whereh = system headdh = h2 - h1 = elevation (static) head - difference between inlet and outlet of the systemhl = head lossA generic expression of head loss is:hl = k q2 (2)whereq = flow ratek = constant describing the total system characteristics - including all major and minor lossesIncreasing the constant - k - by closing some valves, reducing the pipe size or similar - will increase the head loss and move the system curve upwards. The starting point for the curve - at no flow, will be the same.Pump Performance CurveThe pump characteristic is normally described graphically by the manufacturer as a pump performance curve. The pump curve describes the relation between flowrate and head for the actual pump. Other important information for proper pump selection is also included - efficiency curves, NPSHr curve, pump curves for several impeller diameters and different speeds, and power consumption.

Increasing the impeller diameter or speed increases the head and flow rate capacity - and the pump curve moves upwards.The head capacity can be increased by connecting two or more pumps in series, or the flow rate capacity can be increased by connecting two or more pumps in parallel.Selection of PumpA pump can be selected by combining the System Curve and the Pump Curve:

The operating point is where the system curve and the actual pump curve intersect.Best Efficiency Point - BEPThe best operating conditions will in general be close to the best efficiency point - BEP.Special consideration should be taken for applications where the system conditions change frequently during operation. This is often the situation for heating and air conditioning system or water supply systems with variable consumption and modulating valves.Carry OutWhen a pumps operates in the far right of its curve with poor efficiency - the pumps carry out.Shutoff HeadShutoff head is the head produced when the pump operates with fluid but with no flow rate.ChurnA pump is in churn when it operates at shutoff head or no flow.Pumps and Viscosity When a liquid flow through a pump, hydrodynamic losses depends on fluid viscosity When a viscous fluid is handled by a centrifugal pump brake horsepower requirement increases the head generated is reduced capacity is reduced efficiency of pump is reduced and the Best Efficiency Point - BEP - is moved

The head, flow and capacity at other viscosities than used in the original documentation can be modifying with coefficients. Flow qv= cq q (1)where qv = flow compensated for viscosity (m3/h, gpm)cq = viscosity flow coefficient q = original flow according pump curve (m3/h, gpm)Headhv= ch h (2)where hv = head compensated for viscosity (m, ft)ch = viscosity head coefficient h = original head according pump curve (m, ft)Efficiencyv= c (3)where v = effciency compensated for viscosity c = viscosity efficiency coefficient = original efficiency according pump curve Power - SI unitsPv= qv hvv g / (3.6 106 v) (4)where Pv = power compensated for viscosity (kW) v = density of viscous fluid (kg/m3)g = acceleration of gravity (9.81 m/s2)

Discharge Regulation of Centrifugal Pumps Adapting pump capacities to process demand It is often necessary to adapt the pump capacity to a temporary or permanent change in the process demand.The capacity of a centrifugal pump can be regulated either at constant speed, or varying speedCapacity Regulating by Varying Speed

Speed regulating is energy efficient since the energy to the pump is reduced with the decrease of speed.The speed of the pump can be varied with hydraulic/hydrostatic drives - hydraulic coupling between input and output shaft - speed ratio 5 to 1 is controlled by adjusting the volume of oil in the coupling mechanical drives - belt and sheave drive eddy current drive/clutch - magnetic coupling transfer load torque between input and output shaft variable speed drives - inverters - AC drives - adjustable frequency drives - operates by varying the frequency and voltage to the electric motorThe change in power consumption, head and volume rate can be estimated with the affinity laws.Capacity Regulating by Constant SpeedCapacity can be regulated at constant speed by throttling bypassing flow changing impeller diameter modifying the impellerThrottlingThrottling can be carried out by opening and closing a discharge valve.

Throttling is energy inefficient since the energy to the pump is not reduced. Energy is wasted by increasing the dynamic loss.Bypassing FlowThe discharge capacity can be regulated by leading a part of the discharge flow back to the suction side of the pump. Bypassing the flow is energy inefficient since the energy to the pump is not reduced.Changing the Impeller DiameterReducing the impellers diameter is a permanent change and the method can be used where the change in process demand is not temporary. The method may be energy efficient if the motor is changed and the energy consumption reduced.The change in power consumption, head and volume rate can be estimated with the affinity laws.Modifying the ImpellerThe flow rate and the head can be modulated by changing the pitch of the blades. Complicated and seldom used.Pump Regulation and Power ConsumptionPower consumption of some of the regulating methods are compared in the figure below:

Stable and Unstable Centrifugal Pumps Characteristics of stable and unstable centrifugal pumps For stable head-flow characteristics the differential head - h - progressively falls with increasing flow rate - q.

For unstable head-flow characteristics the differential head - h - rises to a maximum and then progressively falls with increasing flow rate - q.

An unstable pump may start to oscillate between the two possible combinations of head-flow. The flow rate will modulate and the pipeline vibrate.It is recommended that centrifugal pumps are designed and operated under stable conditions. Pump Affinity Laws Turbo machines affinity laws are used to calculate volume capacity, head or power consumption in centrifugal pumps when changing speed - rpm - or wheel diameters The Affinity Laws of centrifugal pumps or fans indicates the influence on volume capacity, head (pressure) and/or power consumption of a pump or fan due to change in speed of wheel - revolutions per minute (rpm) geometrically similarity - change in impeller diameterBe aware that there are two sets of affinity laws affinity laws for a specific centrifugal pump - to approximate head, capacity and power curves for different motor speeds and /or different diameter of impellers affinity laws for a family of geometrically similar centrifugal pumps - to approximate head, capacity and power curves for different motor speeds and /or different diameter of impellers

Pump Affinity Laws for a Specific Centrifugal PumpVolume CapacityThe volume capacity of a centrifugal pump can be expressed likeq1 / q2 = (n1 / n2)(d1 / d2) (1)whereq = volume flow capacity (m3/s, gpm, cfm, ..)n = wheel velocity - revolution per minute - (rpm)d = wheel diameterHead or PressureThe head or pressure of a centrifugal pump can be expressed likedp1 / dp2 = (n1 / n2)2 (d1 / d2)2 (2)wheredp = heador pressure (m, ft, Pa, psi, ..)PowerThe power consumption of a centrifugal pump can be expressed asP1 / P2 = (n1 / n2)3 (d1 / d2)3 (3)WhereP = power (W, bhp, ..)Changing the Wheel Velocity

Head or Pressuredp1 / dp2 = (n1 / n2)2 (2a)PowerP1 / P2 = (n1 / n2)3 (3a)

Note! If the speed of a pump is increased with 10% the volume flow increases with 10% the head increases with 21% the power increases with 33 %If we want to increase the volume flow capacity of an existing system with 10% we have to increase the power supply with 33%.Changing the Impeller DiameterIf wheel velocity is constant a change in impeller diameter can simplify the affinity laws toVolume Capacityq1 / q2 = d1 / d2 (1b)Head or Pressuredp1 / dp2 = (d1 / d2)2 (2b)PowerP1 / P2 = (d1 / d2)3 (3b)

Pump and Fan Efficiency Overall pump and fan efficiency is the ratio - power actually gained by the fluid - to shaft power supplied For a fluid flow process involving a pump or fan the overall efficiency is related to the hydraulic mechanical volumetricloss in the pump or fan.Hydraulic Loss and Hydraulic EfficiencyHydraulic loss relates to the construction of the pump or fan, and is caused by the friction between the fluid and the walls, acceleration and retardation of the fluid and the change of the fluid flow direction.The hydraulic efficiency can be expressed as:h = w / (w + wl) (1)whereh = hydraulic efficiencyw = specific work from the pump or fanwl = specific work lost due to hydraulic effectsMechanical Loss and Mechanical EfficiencyMechanical components - as transmission gear and bearings - generates a mechanical loss that reduces the power transferred from the motor shaft to the pump or fan impeller.The mechanical efficiency can be expressed as:m = (P - Pl) / P (2)wherem = mechanical efficiencyP = power transferred from the motor to the shaftPl = power lost in the transmissionVolumetric Loss and Volumetric EfficiencyDue to leakage of fluid between the back surface of the impeller hub plate and the casing, or through other pump components - there is a volumetric loss reducing the pump efficiency.The volumetric efficiency can be expressed as:v = q / (q + ql) (3)wherev = volumetric efficiencyq = volume flow out of the pump or fanql = leakage volume flowTotal Loss and Overall EfficiencyThe overall efficiency is the ratio of power actually gained by the fluid to the shaft power supplied. The overall efficiency can be expressed as: = h m v (4)where = overall efficiencyThe losses in the pump or fan converts to heat transferred to the fluid and the surroundings. As a rule of thumb the temperature increase in a fan transporting air is approximately 1 oC.Example - Hydraulic Efficiency for a PumpAn inline water pump works between pressure 1 bar (1 105 N/m2) and 10 bar (10 105 N/m2). Density of water is 1000 kg/m3. The hydraulic efficiency h = 0.91.The actual water head (water column) can be calculated as:h = (p2 - p1) / = (p2 - p1) / g = ((10 105 N/m2) - (1 105 N/m2)) / (1,000 kg/m3) (9.81 m/s2) = 91.7 m - water columnThe pump must be constructed for the specific work:wc = g h / h = (9.81 m/s2) (91.7 m) / 0.91 = 988.6 m2/s2The construction or design head is:h = wc / g = (988.6 m2/s2) / (9.81 m/s2) = 100.8 m - water columnPumps in Parallel or Serial Connection For pumps in serial - add head, for pumps in parallel - add flowrates Pumps can be arranged in serial or parallel to provide an additional head or flow rate capacity.Pumps in Serial - Head AddedWhen two (or more) pumps are arranged in serial their resulting pump performance curve is obtained by adding theirheads at the same flow rate as indicated in the figure below.

Centrifugal pumps in series are used to overcome larger system head loss than one pump can handle alone. for two identical pumps in series the head will be twice the head of a single pump at the same flow rate - as indicated in point 2.With a constant flowrate the combined head moves from 1 to 2.Note! In practice the combined head and flow rate moves along the system curve to point 3. point 3 is where the system operates with both pumps running point 1 is where the system operates with one pump runningSeries operation of single stage pumps is seldom encountered - more often multistage centrifugal pumps are used.Pumps in Parallel - Flow Rate AddedWhen two or more pumps are arranged in parallel their resulting performance curve is obtained by adding their flowrates at the same headas indicated in the figure below.

Centrifugal pumps in parallel are used to overcome larger volume flows than one pump can handle alone. fortwo identical pumps in parallel, and the head is kept constant, the flowrate doubles as indicated with point 2 compared to a single pumpNote! In practice the combined head and volume flow moves along the system curve as indicated from 1 to 3. point 3 is where the system operates with both pumps running point 1 is where the system operates with one pump runningIn practice, if one of the pumps in parallel or series stops, the operation point moves along the system resistance curve from point 3 to point 1 - the head and flow rate are decreased.Classifications of Pumps Selecting between Centrifugal Pumps and Positive Displacement Pumps Pumps are in general classified as Centrifugal Pumps (or Roto-dynamic pumps) and Positive Displacement Pumps.Centrifugal Pumps (Roto-dynamic pumps)

The centrifugal or roto-dynamic pump produce a head and a flow by increasing the velocity of the liquid through the machine with the help of a rotating vane impeller.Centrifugal pumps include radial, axial and mixed flow units.Centrifugal pumps can further be classified as end suction pumps in-line pumps double suction pumps vertical multistage pumps horizontal multistage pumps submersible pumps self-priming pumps axial-flow pumps regenerative pumpsPositive Displacement PumpsThe positive displacement pump operates by alternating of filling a cavity and then displacing a given volume of liquid. The positive displacement pump delivers a constant volume of liquid for each cycle against varying discharge pressure or head.The positive displacement pump can be classified as: Reciprocating pumps - piston, plunger and diaphragm Power pumps Steam pumps Rotary pumps - gear, lobe, screw, vane, regenerative (peripheral) and progressive cavitySelecting between Centrifugal or Positive Displacement PumpsSelecting between a Centrifugal Pump or a Positive Displacement Pump is not always straight forward.Flow Rate and Pressure HeadThe two types of pumps behave very differently regarding pressure head and flow rate: The Centrifugal Pump has varying flow depending on the system pressure or head The Positive Displacement Pump has more or less a constant flow regardless of the system pressure or head. Positive Displacement pumps generally gives more pressure than Centrifugal Pump's.Capacity and ViscosityAnother major difference between the pump types is the effect of viscosity on the capacity: In the Centrifugal Pump the flow is reduced when the viscosity is increased In the Positive Displacement Pump the flow is increased when viscosity is increasedLiquids with high viscosity fills the clearances of a Positive Displacement Pump causing a higher volumetric efficiency and a Positive Displacement Pump is better suited for high viscosity applications. A Centrifugal Pump becomes very inefficient at even modest viscosity.Mechanical EfficiencyThe pumps behaves different considering mechanical efficiency as well. Changing the system pressure or head has little or no effect on the flow rate in the Positive Displacement Pump Changing the system pressure or head has a dramatic effect on the flow rate in the Centrifugal PumpNet Positive Suction Head - NPSHAnother consideration is the Net Positive Suction Head NPSH. In a Centrifugal Pump, NPSH varies as a function of flow determined by pressure In a Positive Displacement Pump, NPSH varies as a function of flow determined by speed. Reducing the speed of the Positive Displacement Pump pump, reduces the NPSHCentrifugal Pumps An introduction to Centrifugal Pumps A centrifugal pump converts the input power to kinetic energy in the liquid by accelerating the liquid by a revolving device - an impeller. The most common type is the volute pump. Fluid enters the pump through the eye of the impeller which rotates at high speed. The fluid is accelerated radially outward from the pump chasing. A vacuum is created at the impellers eye that continuously draws more fluid into the pump.

The energy created by the pump is kinetic energy according the Bernoulli Equation. The energy transferred to the liquid corresponds to the velocity at the edge or vane tip of the impeller. The faster the impeller revolves or the bigger the impeller is, the higher will the velocity of the liquid energy transferred to the liquid be. This is described by the Affinity Laws.Pressure and HeadIf the discharge of a centrifugal pump is pointed straight up into the air the fluid will pumped to a certain height - or head - called the shut off head. This maximum head is mainly determined by the outside diameter of the pump's impeller and the speed of the rotating shaft. The head will change as the capacity of the pump is altered.The kinetic energy of a liquid coming out of an impeller is obstructed by creating a resistance in the flow. The first resistance is created by the pump casing which catches the liquid and slows it down. When the liquid slows down the kinetic energy is converted to pressure energy. it is the resistance to the pump's flow that is read on a pressure gauge attached to the discharge lineA pump does not create pressure, it only creates flow. Pressure is a measurement of the resistance to flow.In Newtonian fluids (non-viscous liquids like water or gasoline) the term head is used to measure the kinetic energy which a pump creates. Head is a measurement of the height of the liquid column the pump creates from the kinetic energy the pump gives to the liquid. the main reason for using head instead of pressure to measure a centrifugal pump's energy is that the pressure from a pump will change if the specific gravity (weight) of the liquid changes, but the head will notThe pump's performance on any Newtonian fluid can always be described by using the term head.Different Types of Pump Head Total Static Head - Total head when the pump is not running Total Dynamic Head (Total System Head) - Total head when the pump is running Static Suction Head - Head on the suction side, with pump off, if the head is higher than the pump impeller Static Suction Lift - Head on the suction side, with pump off, if the head is lower than the pump impeller Static Discharge Head - Head on discharge side of pump with the pump off Dynamic Suction Head/Lift - Head on suction side of pump with pump on Dynamic Discharge Head - Head on discharge side of pump with pump onThe head is measured in either feet or meters and can be converted to common units for pressure as psi or bar. it is important to understand that the pump will pump all fluids to the same height if the shaft is turning at the same rpmThe only difference between the fluids is the amount of power it takes to get the shaft to the proper rpm. The higher the specific gravity of the fluid the more power is required. Centrifugal Pumps are "constant head machines"Note that the latter is not a constant pressure machine, since pressure is a function of head and density. The head is constant, even if the density (and therefore pressure) changes.The head of a pump in metric units can be expressed in metric units as:h = (p2 - p1)/( g) + v22/(2 g) (1)whereh = total head developed (m)p2 = pressure at outlet(N/m2)p1 = pressure at inlet(N/m2) = density(kg/m3)g = acceleration of gravity(9.81) m/s2v2 = velocity at the outlet(m/s)Head described in simple terms a pump's vertical discharge "pressure-head" is the vertical lift in height - usually measured in feet or m of water - at which a pump can no longer exert enough pressure to move water. At this point, the pump may be said to have reached its "shut-off" head pressure. In the flow curve chart for a pump the "shut-off head" is the point on the graph where the flow rate is zeroPump EfficiencyPump efficiency, (%) is a measure of the efficiency with wich the pump transfers useful work to the fluid. =Pout/ Pin (2)where = efficiency (%)Pin= power input Pout = power output Centrifugal Pump Speed-Torque Curve Centrifugal pumps characteristics - speed torque curve The theoretical characteristic of a centrifugal pump is a parabola starting from the origin and proportional to the square of the speed.

The torque can be expressed as T = k n2 (1)where T = torque (Nm, lbf ft)k = constantn = pump speed or velocity (rpm)With the discharge valve closed the torque amounts to 30 - 50 % of the nominal torque at full speed. Full Load TorqueThe torque at full load can be calculated as T = 30 P / ( n) (2)where T = torque (kN m)P = power (kW)Pump Power Calculator Calculate pump hydraulic and shaft power? Hydraulic Pump PowerThe ideal hydraulic power to drive a pump depends on the mass flow rate, the liquid density and the differential height

- either it is the static lift from one height to an other, or the friction head loss component of the system - can be calculated asPh = q g h / (3.6 106) (1)where Ph = power (kW)q = flow capacity (m3/h) = density of fluid (kg/m3)g = gravity (9.81 m/s2)h = differential head (m) Shaft Pump PowerThe shaft power - the power required transferred from the motor to the shaft of the pump - depends on the efficiency of the pump and can be calculated asPs = Ph / (2)where Ps = shaft power (kW) = pump efficiencyPositive Displacement Pumps An introduction tutorial to the basic operating principles of positive displacement pumps Positive Displacement Pumps has an expanding cavity on the suction side and a decreasing cavity on the discharge side. Liquid flows into the pumps as the cavity on the suction side expands and the liquid flows out of the discharge as the cavity collapses. The volume is constant given each cycle of operation.The positive displacement pumps can be divided in two main classes reciprocating rotaryThe positive displacement principle applies applies whether the pump is a rotary lobe pump progressing cavity pump rotary gear pump piston pump diaphragm pump screw pump gear pump vane pump regenerative (peripheral) pump peristalticPositive Displacement Pumps, unlike a Centrifugal or Roto-dynamic Pumps, will produce the same flow at a given speed (RPM) no matter the discharge pressure. Positive Displacement Pumps are "constant flow machines"A Positive Displacement Pump must not be operated against a closed valve on the discharge side of the pump because it has no shut-off head like Centrifugal Pumps. A Positive Displacement Pump operating against a closed discharge valve, will continue to produce flow until the pressure in the discharge line are increased until the line bursts or the pump is severely damaged - or both.A relief or safety valve on the discharge side of the Positive Displacement Pump is therefore absolute necessary. The relief valve can be internal or external. The pump manufacturer has normally the option to supply internal relief or safety valves. The internal valve should in general only be used as a safety precaution, an external relief valve installed in the discharge line with a return line back to the suction line or supply tank is recommended.Reciprocating PumpsTypical reciprocating pumps are plunger pumps diaphragm pumpsPlunger pumps comprise of a cylinder with a reciprocating plunger in it. In the head of the cylinder the suction and discharge valves are mounted. In the suction stroke the plunger retracts and the suction valves opens causing suction of fluid into the cylinder. In the forward stroke the plunger push the liquid out the discharge valve. With only one cylinder the fluid flow varies between maximum flow when the plunger moves through the middle positions, and zero flow when the plunger is in the end positions. A lot of energy is wasted when the fluid is accelerated in the piping system. Vibration and "water hammers" may be a serious problem. In general the problems are compensated by using two or more cylinders not working in phase with each other.In diaphragm pumps the plunger pressurizes hydraulic oil which is used to flex a diaphragm in the pumping cylinder. Diaphragm valves are used to pump hazardous and toxic fluids. Rotary PumpsTypical rotary pumps are gear pumps lobe pumps vane pumps progressive cavity pumps peripheral pumps screw pumpsIn gear pumps the liquid is trapped by the opening between the gear teeth of two identical gears and the chasing of the pump on the suction side. On the pressure side the fluid is squeezed out when the teeth of the two gears are rotated against each other. The motor provides the drive for one gear. The lobe pumps operates similar to the gear pump, but with two lobes driven by external timing gears. The lobes do not make contact. Progressive cavity pumps consist of a metal rotor rotating within an elastomer-lined or elastic stator. When the rotor turns progressive chambers from suction end to discharge end are formed between the rotor and stator, moving the fluid. Viscosity Converting Chart A converting table between viscosity units like Centiposes, milliPascal, CentiStokes and SSU The viscosity of a fluid is its resistance to shear or flow, and is a measure of the fluids adhesive/cohesive or frictional properties. This arises because of the internal molecular friction within the fluid producing the frictional drag effect. There are two related measures of fluid viscosity which are known as dynamic and kinematic viscosity.

Dynamic viscosity is also termed "absolute viscosity" and is the tangential force per unit area required to move one horizontal plane with respect to the other at unit velocity when maintained a unit distance apart by the fluid.Centipoise (CPS) Millipascal seconds (mPas)Poise (P)Centistokes (cSt)Stokes (S)Saybolt Seconds Universal (SSU)

10.0110.0131

20.0220.0234

40.0440.0438

70.0770.0747

100.1100.160

150.15150.1580

200.2200.2100

250.24250.24130

300.3300.3160

400.4400.4210

500.5500.5260

600.6600.6320

700.7700.7370

800.8800.8430

900.9900.9480

10011001530

1201.21201.2580

1401.41401.4690

1601.61601.6790

1801.81801.8900

200220021000

2202.22202.21100

2402.42402.41200

2602.62602.61280

2802.82802.81380

300330031475

3203.23203.21530

3403.43403.41630

3603.63603.61730

3803.83803.81850

400440041950

4204.24204.22050

4404.44404.42160

4604.64604.62270

4804.84804.82380

500550052480

5505.55505.52660

600660062900

700770073380

800880083880

900990094300

1000101000104600

1100111100115200

1200121200125620

1300131300136100

1400141400146480

1500151500157000

1600161600167500

1700171700178000

1800181800188500

1900191900199000

2000202000209400

2100212100219850

22002222002210300

23002323002310750

24002424002411200

25002525002511600

30003030003014500

35003535003516500

40004040004018500

45004545004521000

50005050005023500

55005555005526000

60006060006028000

65006565006530000

70007070007032500

75007575007535000

80008080008037000

85008585008539500

90009090009041080

95009595009543000

150001501500015069400

200002002000020092500

3000030030000300138500

4000040040000400185000

5000050050000500231000

6000060060000600277500

7000070070000700323500

8000080080000800370000

9000090090000900415500

10000010001000001000462000

12500012501250001250578000

15000015001500001500694000

17500017501750001750810000

20000020002000002000925000

Note! The viscosities are based on materials with a specific gravity of one (1).

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