Electrical motor efficiency

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  • 1. Electrical Motor EfficiencyElectrical motor efficiency is the ratio between the shaft output power - and the electricalinput power.Electrical Motor Efficiency when Shaft Output is measured in WattIf power output is measured in Watt (W), efficiency can be expressed as:ηm = Pout / Pin (1)whereηm = motor efficiencyPout = shaft power out (Watt, W)Pin = electric power in to the motor (Watt, W)Electrical Motor Efficiency when Shaft Output is measured in HorsepowerIf power output is measured in horsepower (hp), efficiency can be expressed as:ηm = Pout 746 / Pin (2)wherePout = shaft power out (horsepower, hp)Pin = electric power in to the motor (Watt, W)Primary and Secondary Resistance LossesThe electrical power lost in the primary rotor and secondary stator winding resistanceare also called copper losses. The copper loss varies with the load in proportion to thecurrent squared - and can be expressed asPcl = R I2 (3)wherePcl = stator winding - copper loss (W)R = resistance (Ω)I = current (Amp)
  • 2. Iron LossesThese losses are the result of magnetic energy dissipated when when the motorsmagnetic field is applied to the stator core.Stray LossesStray losses are the losses that remains after primary copper and secondary losses, ironlosses and mechanical losses. The largest contribution to the stray losses is harmonicenergies generated when the motor operates under load. These energies are dissipatedas currents in the copper windings, harmonic flux components in the iron parts, leakagein the laminate core.Mechanical LossesMechanical losses includes friction in the motor bearings and the fan for air cooling.NEMA Design B Electrical MotorsElectrical motors constructed according NEMA Design B must meet the efficienciesbelow: Power Minimum Nominal (hp) Efficiency1) 1-4 78.8 5-9 84.0 10 - 19 85.5 20 - 49 88.5 50 - 99 90.2 100 - 124 91.7 > 125 92.41) NEMA Design B, Single Speed 1200, 1800, 3600 RPM. Open Drip Proof (ODP) orTotally Enclosed Fan Cooled (TEFC) motors 1 hp and larger that operate more than 500hours per year.
  • 3. Options:- Useful Formulas Formulas- Transformer FormulasCalculating Motor Speed:A squirrel cage induction motor is a constant speed device. It cannot operate for anylength of time at speeds below those shown on the nameplate without danger of burningout.To Calculate the speed of a induction motor, apply this formula: Srpm = 120 x F PSrpm = synchronous revolutions per minute.120 = constantF = supply frequency (in cycles/sec)P = number of motor winding polesExample: What is the synchronous of a motor having 4 poles connected to a 60 hz powersupply? Srpm = 120 x F P Srpm = 120 x 60 4 Srpm = 7200 4 Srpm = 1800 rpmCalculating Braking Torque:Full-load motor torque is calculated to determine the required braking torque of a motor.To Determine braking torque of a motor, apply this formula: T = 5252 x HP rpm
  • 4. T = full-load motor torque (in lb-ft)5252 = constant (33,000 divided by 3.14 x 2 = 5252)HP = motor horsepowerrpm = speed of motor shaftExample: What is the braking torque of a 60 HP, 240V motor rotating at 1725 rpm? T = 5252 x HP rpm T = 5252 x 60 1725 T = 315,120 1725 T = 182.7 lb-ftCalculating Work:Work is applying a force over a distance. Force is any cause that changes the position,motion, direction, or shape of an object. Work is done when a force overcomes aresistance. Resistance is any force that tends to hinder the movement of an object.If anapplied force does not cause motion the no work is produced.To calculate the amount of work produced, apply this formula: W=FxDW = work (in lb-ft)F = force (in lb)D = distance (in ft)Example: How much work is required to carry a 25 lb bag of groceries vertically fromstreet level to the 4th floor of a building 30 above street level? W=FxD W = 25 x 30 W = 750 -lbCalculating Torque:Torque is the force that produces rotation. It causes an object to rotate. Torque consist ofa force acting on distance. Torque, like work, is measured is pound-feet (lb-ft). However,torque, unlike work, may exist even though no movement occurs.To calculate torque, apply this formula:
  • 5. T=FxDT = torque (in lb-ft)F = force (in lb)D = distance (in ft)Example: What is the torque produced by a 60 lb force pushing on a 3 lever arm? T=FxD T = 60 x 3 T = 180 lb ftCalculating Full-load Torque:Full-load torque is the torque to produce the rated power at full speed of the motor. Theamount of torque a motor produces at rated power and full speed can be found by using ahorsepower-to-torque conversion chart. When using the conversion chart, place a straightedge along the two known quantities and read the unknown quantity on the third line.To calculate motor full-load torque, apply this formula: T = HP x 5252 rpmT = torque (in lb-ft)HP = horsepower5252 = constantrpm = revolutions per minuteExample: What is the FLT (Full-load torque) of a 30HP motor operating at 1725 rpm? T = HP x 5252 rpm T = 30 x 5252 1725 T = 157,560 1725 T = 91.34 lb-ftCalculating Horsepower:Electrical power is rated in horsepower or watts. A horsepower is a unit of power equal to746 watts or 33,0000 lb-ft per minute (550 lb-ft per second). A watt is a unit of measureequal to the power produced by a current of 1 amp across the potential difference of 1
  • 6. volt. It is 1/746 of 1 horsepower. The watt is the base unit of electrical power. Motorpower is rated in horsepower and watts.Horsepower is used to measure the energy produced by an electric motor while doingwork.To calculate the horsepower of a motor when current and efficiency, and voltage areknown, apply this formula: HP = V x I x Eff 746HP = horsepowerV = voltageI = curent (amps)Eff. = efficiencyExample: What is the horsepower of a 230v motor pulling 4 amps and having 82%efficiency? HP = V x I x Eff 746 HP = 230 x 4 x .82 746 HP = 754.4 746 HP = 1 Hp Eff = efficiency / HP = horsepower / V = volts / A = amps / PF = power factor Horsepower Formulas Example To Find Use Formula Given Find Solution HP = 240V x 20A x 85% HP = I X E X Eff. HP 240V, 20A, 85% Eff. HP 746 746 HP=5.5 I = 10HP x 746 I = HP x 746 10HP, 240V, I I 240V x 90% x 88% E X Eff x PF 90% Eff., 88% PF I = 39 ATo calculate the horsepower of a motor when the speed and torque are known,apply this formula: HP = rpm x T(torque) 5252(constant)
  • 7. Example: What is the horsepower of a 1725 rpm motor with a FLT 3.1 lb-ft? HP = rpm x T 5252 HP = 1725 x 3.1 5252 HP = 5347.5 5252 HP = 1 hpCalculating Synchronous Speed:AC motors are considered constant speed motors. This is because the synchronous speedof an induction motor is based on the supply frequency and the number of poles in themotor winding. Motor are designed for 60 hz use have synchronous speeds of 3600,1800, 1200, 900, 720, 600, 514, and 450 rpm.To calculate synchronous speed of an induction motor, apply this formula: rpmsyn = 120 x f Nprpmsyn = synchronous speed (in rpm)f = supply frequency in (cycles/sec)Np = number of motor polesExample: What is the synchronous speed of a four pole motor operating at 50 hz.? rpmsyn = 120 x f Np rpmsyn = 120 x 50 4 rpmsyn = 6000 4 rpmsyn = 1500 rpm
  • 8. E = Voltage / I = Amps /W = Watts / PF = Power Factor / Eff = Efficiency / HP = Horsepower AC/DC FormulasTo Find Direct Current AC / 1phase AC / 1phase AC 3 phase 115v or 120v 208,230, or 240v All VoltagesAmps when HP x 746 HP x 746 HP x 746 HP x 746Horsepower is Known E x Eff E x Eff X PF E x Eff x PF 1.73 x E x Eff x PFAmps when kW x 1000 kW x 1000 kW x 1000 kW x 1000Kilowatts is known E E x PF E x PF 1.73 x E x PFAmps when kVA x 1000 kVA x 1000 kVA x 1000kVA is known E E 1.73 x EKilowatts IxE I x E x PF I x E x PF I x E x 1.73 PF 1000 1000 1000 1000Kilovolt-Amps IxE IxE I x E x 1.73 1000 1000 1000Horsepower I x E x Eff I x E x Eff x PF I x E x Eff x PF I x E x Eff x 1.73 x PF(output) 746 746 746 746 Three Phase Values For 208 volts x 1.732, use 360 For 230 volts x 1.732, use 398 For 240 volts x 1.732, use 416 For 440 volts x 1.732, use 762 For 460 volts x 1.732, use 797 For 480 Volts x 1.732, use 831 E = Voltage / I = Amps /W = Watts / PF = Power Factor / Eff = Efficiency / HP = Horsepower AC Efficiency and Power Factor Formulas To Find Single Phase Three Phase 746 x HP 746 x HP Efficiency E x I x PF E x I x PF x 1.732 Input Watts Input Watts Power Factor VxA E x I x 1.732 Power - DC Circuits Watts = E xI
  • 9. Amps = W / E Ohms Law / Power Formulas P = watts I = amps R = ohms E = Volts Voltage Drop Formulas 2 x K x I x L K = ohms per mil foot VD = Single Phase CM (2 or 3 wire) 2K x L x I (Copper = 12.9 at 75°) CM= VD (Alum = 21.2 at 75°) 1.73 x K x I x L Note: K value changes with temperature. See Code chapter 9, VD= CM Table 8 L = Length of conductor in feet Three Phase 1.73 x K x L x I CM= VD I = Current in conductor (amperes) CM = Circular mil area of conductor (Download This Button Today!) Home > Forums > Mechanical Engineers > Activities >Member Feedback HVAC/R engineering Forum "...I just wanted to say THANKS for theforum. The knowledge I gain from your site is invaluable..." Fan Horsepower equation
  • 10. More...Geography derivationWhere in the world do Eng-Tips members come from? thread403-82316 Click Here To Find Out! Share ThisEng-Tips Shirts! Forum Searc FAQ Link Job Whitepaper MVP h s s s s s Check Out Our Whitepaper Library. Click Here . Get your Eng-Tips Forums wilg (Mechanical) 23 Dec 03 SWAG here! 13:28 The equation for Fan Horsepower is BHP = (cfm x static friction x specific gravity)/(6356 x motor efficiency)....How is the 6356 derived ? KHilgefort (Mechanical) 24 Dec 03 9:33 I usually see this figure in terms of eff = (volume flowrate in cfm) X (pressure in inches of water) / 6356 / (brake horsepower). R.A. Wallis in "Axial Flow Fans" similarly puts the equation as Shaft h.p. = (5.2) X (in. water) X (cu. ft/min) / (33,000 X efficiency). If you convert the horsepower units to ft-lb/min (33,013 per hp) and convert the pressure to lb/ft^2 (5.202 per inch H2O), you get 33013/5.202 = 6346. This at least gets you to a dimensionless value for the efficiency. lilliput1 (Mechanical) 24 Dec 03 13:35 The efficiency to get bhp is fan efficiency, not motor efficiency. A good apporoximation for fan efficiency is 0.65 For pumps bhp = (gpm x ft wg TDH)/(3960 x eff)
  • 11. Here alse 0.65 is a good approximation for pumpefficiency.On projects we HVAC engineers have to give electricalloads to the Elect Engineer on the project. We use theabove formula + guick estimate from experience ofpressure drops to come up with motor hp. We want to besafe at the initial stage so we always pick mhpconservatively. Later on as plans are developed, actualpressure drops are calculated & fans/pumps are selectedon actual fan/pump curves. quark (Mechanical) 24 Dec 03 23:56There are two mistakes in your equation. First one, assuggested by lilliput, it is fan efficiency you have to use toget bhp. Secondly you can omit specific gravity as this isinbuilt in the equation.1HP = 33000 ft-lbf/min = 33000 (cuft/min)x(lbf/sq.ft)and 1 inch. wc = 5.192 lbf/sq.ft (this is where sg isincluded)therefore, 1 HP = 33000 (cu.ft/min)x inches wc/5.192which gives, 1HP = 6356 cfm x inches wcSo fan BHP = cfmx dp across fan/(6356xeff. blower)PS: I HP is the amount of power required to lift a weightof 76 kgs to a height of 1 meter in 1 second. I dont knowexact definition in IP units but you can convert to get theabove said value.
  • 12. Basic Motor Formulas And CalculationsThe formulas and calculations which appear below should be used for estimatingpurposes only. It is the responsibility of the customer to specify the required motor Hp,Torque, and accelerating time for his application. The salesman may wish to check thecustomers specified values with the formulas in this section, however, if there is seriousdoubt concerning the customers application or if the customer requires guaranteedmotor/application performance, the Product Department Customer Service group shouldbe contacted.Rules Of Thumb (Approximation)At 1800 rpm, a motor develops a 3 lb.ft. per hpAt 1200 rpm, a motor develops a 4.5 lb.ft. per hpAt 575 volts, a 3-phase motor draws 1 amp per hpAt 460 volts, a 3-phase motor draws 1.25 amp per hpAt 230 volts a 3-phase motor draws 2.5 amp per hpAt 230 volts, a single-phase motor draws 5 amp per hpAt 115 volts, a single-phase motor draws 10 amp per hpMechanical Formulas Torque x Torque in lb.ft. HP x 5250 rpm 120 x Frequency HP = rpm = = rpm No. of Poles 5250Temperature ConversionDeg C = (Deg F - 32) x 5/9Deg F = (Deg C x 9/5) + 32High Inertia Loads WK2 x rpmt= WK2 = inertia in lb.ft.2 308 x T av. t = accelerating time in sec. WK2 x rpm T = Av. accelerating torqueT= lb.ft.. 308 x t inertia reflected to motor = Load Inertia Load rpm
  • 13. Motor rpm 2Synchronous Speed, Frequency And Number Of Poles Of AC Motors 120 x f P x ns 120 x fns = f= P= P 120 nsRelation Between Horsepower, Torque, And Speed Txn 5250 HP 5250 HPHP = T= n= 5250 n TMotor Slip ns - n x% Slip = 100 nsCode KVA/HP Code KVA/HP Code KVA/HP Code KVA/HP A 0-3.14 F 5.0 -5.59 L 9.0-9.99 S 16.0-17.99 B 3.15-3.54 G 5.6 -6.29 M 10.0-11.19 T 18.0-19.99 C 3.55-3.99 H 6.3 -7.09 N 11.2-12.49 U 20.0-22.39 D 4.0 -4.49 I 7.1 -7.99 P 12.5-13.99 V 22.4 & Up E 4.5 -4.99 K 8.0 -8.99 R 14.0-15.99SymbolsI = current in amperesE = voltage in voltsKW = power in kilowattsKVA = apparent power in kilo-volt-amperesHP = output power in horsepowern = motor speed in revolutions per minute (RPM) synchronous speed in revolutions per minutens = (RPM)P = number of polesf = frequency in cycles per second (CPS)T = torque in pound-feetEFF = efficiency as a decimalPF = power factor as a decimalEquivalent Inertia
  • 14. In mechanical systems, all rotating parts do not usually operate at the same speed. Thus,we need to determine the "equivalent inertia" of each moving part at a particular speed ofthe prime mover.The total equivalent WK2 for a system is the sum of the WK2 of each part, referenced toprime mover speed.The equation says: WK2EQ = WK2part Npart Nprime mover 2This equation becomes a common denominator on which other calculations can be based.For variable-speed devices, inertia should be calculated first at low speed.Lets look at a simple system which has a prime mover (PM), a reducer and a load. WK2 = 900 lb.ft.2 WK2 = 100 lb.ft.2 WK2 = 27,000 lb.ft.2 (as seen at output shaft) PRIME MOVER 3:1 GEAR REDUCER LOADThe formula states that the system WK2 equivalent is equal to the sum of WK2parts at theprime movers RPM, or in this case: WK2EQ = WK2pm + WK2Red. Red. RPM + WK2Load Load RPM PM RPM 2 PM RPM 2Note: reducer RPM = Load RPM WK2EQ = WK2pm + WK2Red. 1 + WK2Load 1 3 2 3 2The WK2 equivalent is equal to the WK2 of the prime mover, plus the WK2 of the load.This is equal to the WK2 of the prime mover, plus the WK2 of the reducer times (1/3)2,plus the WK2 of the load times (1/3)2.This relationship of the reducer to the driven load is expressed by the formula givenearlier: WK2EQ = WK2part Npart Nprime mover 2
  • 15. In other words, when a part is rotating at a speed (N) different from the prime mover, theWK2EQ is equal to the WK2 of the parts speed ratio squared.In the example, the result can be obtained as follows:The WK2 equivalent is equal to: WK2EQ = 100 lb.ft.2 + 900 lb.ft.2 1 + 27,000 lb.ft.2 1 3 2 3 2Finally: WK2EQ = lb.ft.2pm + 100 lb.ft.2Red + 3,000 lb.ft2Load WK2EQ = 3200 lb.ft.2The total WK2 equivalent is that WK2 seen by the prime mover at its speed.Electrical Formulas Alternating Current To Find Single-Phase Three-Phase HP x 746 HP x 746Amperes when horsepower is known E x Eff x pf 1.73 x E x Eff x pf Kw x 1000 Kw x 1000Amperes when kilowatts are known E x pf 1.73 x E x pf Kva x 1000 Kva x 1000Amperes when kva are known E 1.73 x E I x E x pf 1.73 x I x E x pfKilowatts 1000 1000 IxE 1.73 x I x EKva 1000 1000 I x E x Eff x pf 1.73 x I x E x Eff x pffHorsepower = (Output) 746 746
  • 16. I = Amperes; E = Volts; Eff = Efficiency; pf = Power Factor; Kva = Kilovolt-amperes;Kw = KilowattsLocked Rotor Current (IL) From Nameplate Data 577 x HP x KVA/HPThree Phase: IL = E See: KVA/HP Chart 1000 x HP x KVA/HPSingle Phase: IL = EEXAMPLE:Motor nameplate indicates 10 HP, 3 Phase, 460 Volts, Code F. 577 x 10 x (5.6 or IL = 6.29) 460 70.25 or 78.9 Amperes (possible IL = range)Effect Of Line Voltage On Locked Rotor Current (I L) (Approx.) ELINEIL @ ELINE = IL @ EN/P x EN/PEXAMPLE:Motor has a locked rotor current (inrush of 100 Amperes (IL) at the rated nameplate voltage (EN/P) of 230 volts. What is IL with 245 volts (ELINE) applied to this motor? IL @ 245 V. = 100 x 254V/230V IL @ 245V. = 107 AmperesBasic Horsepower CalculationsHorsepower is work done per unit of time. One HP equals 33,000 ft-lb of work perminute. When work is done by a source of torque (T) to produce (M) rotations about anaxis, the work done is: radius x 2 x rpm x lb. or 2 TMWhen rotation is at the rate N rpm, the HP delivered is:
  • 17. radius x 2 x rpm x TN HP = lb. = 5,250 33,000For vertical or hoisting motion: HP WxS = 33,000 x EWhere: W = total weight in lbs. to be raised by motor S = hoisting speed in feet per minute overall mechanical efficiency of hoist and gearing. For purposes of E = estimating E = .65 for eff. of hoist and connected gear.For fans and blowers: Volume (cfm) x Head (inches of HP = water) 6356 x Mechanical Efficiency of FanOr Volume (cfm) x Pressure (lb. Per sq. ft.) HP = 3300 x Mechanical Efficiency of FanOr Volume (cfm) x Pressure (lb. Per sq. HP = in.) 229 x Mechanical Efficiency of FanFor purpose of estimating, the eff. of a fan or blower may be assumed to be 0.65.Note:Air Capacity (cfm) varies directly with fan speed. Developed Pressure varies with square of fan speed. Hp varies with cube of fan speed.For pumps:
  • 18. GPM x Pressure in lb. Per sq. in. x Specific HP = Grav. 1713 x Mechanical Efficiency of PumpOr GPM x Total Dynamic Head in Feet x HP = S.G. 3960 x Mechanical Efficiency of Pump where Total Dynamic Head = Static Head + Friction HeadFor estimating, pump efficiency may be assumed at 0.70.Accelerating TorqueThe equivalent inertia of an adjustable speed drive indicates the energy required to keepthe system running. However, starting or accelerating the system requires extra energy.The torque required to accelerate a body is equal to the WK2 of the body, times thechange in RPM, divided by 308 times the interval (in seconds) in which this accelerationtakes place: WK2N (in lb.ft.) ACCELERATING TORQUE = 308tWhere: N = Change in RPM W = Weight in Lbs. K = Radius of gyration Time of acceleration t= (secs.) WK2 = Equivalent Inertia 308 = Constant of proportionalityOr TAcc = WK2N
  • 19. 308tThe constant (308) is derived by transferring linear motion to angular motion, andconsidering acceleration due to gravity. If, for example, we have simply a prime moverand a load with no speed adjustment:Example 1 PRIME LOADER LOAD WK2 = 200 lb.ft.2 WK2 = 800 lb.ft.2The WK2EQ is determined as before: WK2EQ = WK2pm + WK2Load WK2EQ = 200 + 800 WK2EQ = 1000 ft.lb.2If we want to accelerate this load to 1800 RPM in 1 minute, enough information isavailable to find the amount of torque necessary to accelerate the load.The formula states: 1000 x WK2EQN 1800000 TAcc = or 1800 or 308t 18480 308 x 60 TAcc = 97.4 lb.ft.In other words, 97.4 lb.ft. of torque must be applied to get this load turning at 1800 RPM,in 60 seconds.Note that TAcc is an average value of accelerating torque during the speed change underconsideration. If a more accurate calculation is desired, the following example may behelpful.Example 2The time that it takes to accelerate an induction motor from one speed to another may befound from the following equation:
  • 20. WR2 x change in rpm t= 308 x TWhere: T = Average value of accelerating torque during the speed change under consideration. t = Time the motor takes to accelerate from the initial speed to the final speed. WR2 = Flywheel effect, or moment of inertia, for the driven machinery plus the motor rotor in lb.ft.2 (WR2 of driven machinery must be referred to the motor shaft).The Application of the above formula will now be considered by means of an example.Figure A shows the speed-torque curves of a squirrel-cage induction motor and a blowerwhich it drives. At any speed of the blower, the difference between the torque which themotor can deliver at its shaft and the torque required by the blower is the torque availablefor acceleration. Reference to Figure A shows that the accelerating torque may varygreatly with speed. When the speed-torque curves for the motor and blower intersectthere is no torque available for acceleration. The motor then drives the blower at constantspeed and just delivers the torque required by the load.In order to find the total time required to accelerate the motor and blower, the areabetween the motor speed-torque curve and the blower speed-torque curve is divided intostrips, the ends of which approximate straight lines. Each strip corresponds to a speedincrement which takes place within a definite time interval. The solid horizontal lines inFigure A represent the boundaries of strips; the lengths of the broken lines the averageaccelerating torques for the selected speed intervals. In order to calculate the totalacceleration time for the motor and the direct-coupled blower it is necessary to find thetime required to accelerate the motor from the beginning of one speed interval to thebeginning of the next interval and add up the incremental times for all intervals to arriveat the total acceleration time. If the WR2 of the motor whose speed-torque curve is givenin Figure A is 3.26 ft.lb.2 and the WR2 of the blower referred to the motor shaft is 15ft.lb.2, the total WR2 is: 15 + 3.26 = 18.26 ft.lb.2,And the total time of acceleration is: WR2 rpm1 rpm2 rpm3 +--------- rpm9 + + 308 T1 T2 T3 + T9Or t = 18.2 150 + 150 + 300 + 300 + 200 + 200 + 300 + 100 + 40 6
  • 21. 46 48 47 43.8 39.8 36.4 32.8 29.6 11 308 t = 2.75 sec.Figure ACurves used to determine time required to accelerate induction motor and blower Accelerating Torques T1 = 46 lb.ft. T4 = 43.8 lb.ft. T7 = 32.8 lb.ft. T2 = 48 lb.ft. T5 = 39.8 lb.ft. T8 = 29.6 lb.ft. T3 = 47 lb.ft. T6 = 36.4 lb.ft. T9 = 11 lb.ft.Duty CyclesSales Orders are often entered with a note under special features such as: "Suitable for 10 starts per hour"Or "Suitable for 3 reverses per minute"Or "Motor to be capable of accelerating 350 lb.ft.2"Or "Suitable for 5 starts and stops per hour"Orders with notes such as these can not be processed for two reasons.
  • 22. 1. The appropriate product group must first be consulted to see if a design is available that will perform the required duty cycle and, if not, to determine if the type of design required falls within our present product line. 2. None of the above notes contain enough information to make the necessary duty cycle calculation. In order for a duty cycle to be checked out, the duty cycle information must include the following: a. Inertia reflected to the motor shaft. b. Torque load on the motor during all portions of the duty cycle including starts, running time, stops or reversals. c. Accurate timing of each portion of the cycle. d. Information on how each step of the cycle is accomplished. For example, a stop can be by coasting, mechanical braking, DC dynamic braking or plugging. A reversal can be accomplished by plugging, or the motor may be stopped by some means then re-started in the opposite direction. e. When the motor is multi-speed, the cycle for each speed must be completely defined, including the method of changing from one speed to another. f. Any special mechanical problems, features or limitations.Obtaining this information and checking with the product group before the order isentered can save much time, expense and correspondence.Duty cycle refers to the detailed description of a work cycle that repeats in a specific timeperiod. This cycle may include frequent starts, plugging stops, reversals or stalls. Thesecharacteristics are usually involved in batch-type processes and may include tumblingbarrels, certain cranes, shovels and draglines, dampers, gate- or plow-positioning drives,drawbridges, freight and personnel elevators, press-type extractors, some feeders,pressesof certain types, hoists, indexers, boring machines,cinder block machines, keyseating,kneading, car-pulling, shakers (foundry or car), swaging and washing machines, andcertain freight and passenger vehicles. The list is not all-inclusive. The drives for theseloads must be capable of absorbing the heat generated during the duty cycles. Adequatethermal capacity would be required in slip couplings, clutches or motors to accelerate orplug-stop these drives or to withstand stalls. It is the product of the slip speed and thetorque absorbed by the load per unit of time which generates heat in these drivecomponents. All the events which occur during the duty cycle generate heat which thedrive components must dissipate.Because of the complexity of the Duty Cycle Calculations and the extensive engineeringdata per specific motor design and rating required for the calculations, it is necessary forthe sales engineer to refer to the Product Department for motor sizing with a duty cycleapplication.Last Updated SeptemberChapter 10: Fans and Drives10.1. Theory and Applications
  • 23. 10.2. Commissioning Fans and Drives 10.2.1. Functional Testing Field Tips Key Commissioning Test Requirements Key Preparations and Cautions Time Required to Test 10.2.2. Design Issues Overview10.3. Typical Problems 10.3.1. Static pressure requirements in excess of design 10.3.2. Improper belt drive system adjustment10.4. Testing Guidance and Sample Test Forms10.5. Supplemental Information10.1. Theory and ApplicationsThe fan is the heart of the air handling system since it is one of the most significant energy usersin a building. Commissioning and re-commissioning fans and drives is a key factor for ensuringthat a building�s efficiency goals are met over the life of the building.There are both indirect and direct components to a fan�s energy consumption. The indirectcomponent relates to the system the fan serves. The fan must impart enough energy to the airstream to overcome the system�s resistance to flow. This energy consumption can besignificantly altered by:� Fan installation considerations like system effect� Duct and fitting design and their related pressure drops� Component pressure drops� Duct system leakage� Duct system thermal lossThese topics are discussed in Chapter 11: Distribution, and Chapter 13: Return, Relief andExhaust System.The direct fan energy component relates to how efficiently the fan can covert the energy goinginto its prime mover (usually electricity into a motor) into air flow and pressure in the fan system.This energy consumption is a function of the following items:� Fan efficiency� Motor efficiency� Drive system efficiency and adjustmentThe fan horsepower equation (Equation 10.1) is a function of several fundamental components:flow rate, static pressure, fan efficiency, and motor efficiency. Application of this equation to fansystem analysis is discussed in detail in Appendix C: Calculations. Commissioning efforts shouldbe targeted at these factors to ensure system efficiency, performance, and reliability.Equation 10.1 Fan Horsepower
  • 24. The fundamental technology associated with the fans, coils, casings and other major systemcomponents is well established. Most of the advances in technology that improve theperformance of these components are related to the drive systems and control systems, not withthe components themselves. Drive and control systems can be readily upgraded as technologicalimprovements warrant. This easy upgrade is in direct contrast with a more complex machine likea chiller where technology changes are continually improving performance, and where thistechnology is generally an integral part of the machine�s package.If one examines a reasonably well-maintained 50-year-old airfoil centrifugal fan and a similarunit right off of the production line, one will probably see only modest performance differences.Over time, the shaft and/or bearings in the older fan may have required replacement, and thewheel periodic cleaning. However, it is likely that the fan is capable of moving air as efficientlyas the newer fan. Through attention to proper maintenance and equipment location (indoors ratherthan outdoors), fans can be a lasting component of the air handling system.10.2. Commissioning Fans and DrivesThe following sections present benefits, practical tips, and design issues associated withcommissioning an air handler�s fans and drives.10.2.1. Functional Testing Field TipsKey Commissioning Test Requirements lists practical considerations for functional testing. KeyPreparations and Cautions address potential problems that may occur during functional testingand ways to prevent them.
  • 25. Key Commissioning Test RequirementsFan energy constitutes a significant portion of a buildings overall energy consumption. Evenminor improvements in efficiency can have a major impact on the energy consumption patternassociated with a building. A well executed commissioning plan for the fans and theirassociated drive systems ensures that the systems are set up for peak efficiency and that thisefficiency will persist. The fan and drive control should reliably integrate with the overallsystem control strategy in a manner that provides the intended function and level ofperformance.1 Verify the fan size and capacity. Capacity tests results should be evaluated in the light of the accuracy of in instrumentation and the actual conditions at the time of the test.2 Backdraft dampers need to be tested for proper operation. Non-motorized dampers must open and close freely without binding. Motorized dampers must be connected to the DDC control system and verified that they are commanded open prior to fan operation.3 Verify that network failures do not result in unsafe operating modes. The recovery from the failure should safely return the drive to the network.4 Verify that drive settings and adjustments provide for safe and reliable system operation at peak efficiency levels in all operating modes.Key Preparations and CautionsCautions1 Applicable cautions as outlined in Functional Testing Basics should be observed.2 Safety and interlock testing, verification of some of the drive settings, and loop tuning efforts will place the system at risk. Appropriate precautions and procedures should be in place to protect the personnel and machinery involved in the test process, including plans for quickly aborting the test.3 Any belt drives have been adjusted and aligned.5 All safeties, interlocks, and alarms are programmed (or hard-wired, if applicable) and function correctly, regardless of VFD operating position (i.e. hand, auto, by-pass).6 If necessary, the motor shaft is grounded.7 Distribution system pressure drops do not exceed design expectations. This is typically performed while conducting construction observation. If changes increase distribution system pressure drop, ensure all equipment still receives design flow rate.8 Verify all VFD operating parameters are correct for the application, including acceleration and deceleration times and minimum speed setting.Test Conditions1 Tests that are targeted at verifying design parameters and settings for the fan and its enclosure can generally be performed after the assembly of the air handling unit but prior to its start-up.2 Other tests targeted at the interlocks and fundamental control functions, loop testing and tuning, and capacity testing will require that the air handling system be operational and moving the design volume of air, but not necessarily fully under control. Safety systems should be operational to protect the machinery and occupants in the event of a problem
  • 26. 10.2.2. Design Issues OverviewThe Design Issues Overview presents issues that can be addressed during the design phase toimprove system performance, safety, and energy efficiency. These design issues are essential forcommissioning providers to understand, even if design phase commissioning is not a part of theirscope, since these issues are often the root cause of problems identified during testing.Does the unit have good access for control installation, maintenance, andcomponent replacement?Access to the fan and its related components is critical for ensuring the persistence of energyefficiency and other commissioning related benefits.1 Piping should be arranged to ensure that access panels are not blocked, service routes remain open, and components such as coils and fan shafts can be removed and replaced without shutting down adjacent systems or central plant equipment.2 Fan scrolls should be provided with access doors to allow the wheel to be inspected and cleaned.3 Coils should be provided with space between then and access to that space to facility inspection and cleaning and allow for the installation of control elements in their proper location. For example, space is required between a preheat coil and the next coil downstream to allow the freezestat to be installed downstream of the preheat coil (which by design will see subfreezing entering air temperatures and should be capable of handling them safely).Have variable speed drive installation and operation requirements been taken intoaccount?1 Most VFD manufactures have some specific requirements regarding the length, routing, and general configuration of the power circuit from the drive to the motor. Failure to pay attention to the requirements can cause operational problems in the electrical system and in severe cases, cause failures in switchgear, drives, and transfer switches.2 Many VFDs can be damaged if they start against a reverse spinning motor. This condition is likely to occur in parallel fan systems, even if they are equipped with backdraft dampers. No damper is 100% leak proof, and it does not take much reverse flow to set a fan wheel in motion. Most drives also have a feature to handle reverse flow, usually called DC injection braking. The process pulses the motor with a DC signal before starting and accelerating it. The DC signal brakes the rotating armature. Usually there are adjustments that need to be made to tailor this feature to the load served in addition to activating it. Verifying this feature is properly set and functioning should be part of the commissioning process both during the pre-start checks as well as the functional tests.3 Many drives are supplied with bypass contactors that allow the motor to run at full speed if the drive fails. In some cases, the system could be damaged by full speed fan operation when the loads were configured for minimum flow conditions.4 The drive should be configured and wired to ensure that all safety interlocks are effective in all possible selector switch configurations (local, auto, hand, inverter, bypass, etc.). Some drives are arranged to allow the safety interlocks to be effective when the drive is operating but not effective if the drive is bypass. Some drives can also be configured so that if they are placed in the local mode, any external interlock (external to the drive circuit board) will be ignored. This feature may be desirable in process applications, but it is highly undesirable in most HVAC applications. Verifying that the drive is properly set and functioning should be
  • 27. part of the commissioning process both during the pre-start checks as well as the functional tests.Are the VFDs and the motors compatible?Motors that are not rated for VFD applications may have a reduced life if used with a VFD. Inretrofits, it is desirable to evaluate the motor�s capabilities relative to the drive. Even it budgetconstraints prevent a motor replacement when the drive is installed, the potential for a futureproblem and early failure can be anticipated. In new installations, the drives and motors should becoordinated to be compatible with each other.Does the VFD shaft need to be grounded?The variable voltages, magnetic fields and harmonics associated with VFD operation can inducecurrents in the motor shaft that have no path to ground other than through the bearings for mostconventional motors. Evidence suggests that these eddy currents can lead to premature bearingfailures, perhaps in a matter of years on some motors. Shaft grounding kits installed on the motorprovide a direct path from the shaft to ground via a brush system.Is the drive arrangement suitable for the application?Given the wide array of drive options available, it is important to tailor the selection to theapplication.1 If direct drives are applied, then fan speed adjustment for balancing purposes will have to rely on less efficient approaches like discharge dampers, or will require that a variable speed drive be included as a part of the package.2 A variable speed drive on a constant volume fan may represent false economy. While it does minimize balancing efforts and eliminate the need for a final sheave change or adjustment to set the fan speed, the drive results in a loss in fan system efficiency that increases in magnitude as the speed is reduced (see Chapter 10: Fans and Drives Supplemental Information). The drive also introduces operating complexity, first cost, potential electrical system harmonic problems, and multiple failure mode possibilities into the system. These issues coupled with the efficiency reduction will probably outweigh any modest savings in balancing costs achieved.3 Variable pitch sheaves provide flexibility and a good intermediate stepping stone between start-up and final balanced speed as a system is brought on line. But some of their disadvantages may make the installation of a fixed pitch sheave as the final step in the balancing effort a desirable feature to include in the project.4 During design review, verify that the fan and drive capacity is properly sized so that the VFD will operate near 100% speed at full load (do not use the VFD as a throttling device).Could the fan motor run in the wrong direction?For most axial fans, if the impeller were to run in the opposite direction, it would move air in theopposite direction. With centrifugal fans, running the impeller backwards will still provide flowin the correct direction, but the performance will be degraded significantly.Reverse flow or back-draft through most fan wheels will cause them to spin in the reversedirection. Forward curved fan wheels will spin in the wrong direction if air is blown through themin the right direction but they are not energized. For most single phase motors, if the motor isspinning in the wrong direction when power is applied, the fan will simply run in the wrongdirection. The rotational direction of most three phase motors used for HVAC applications is tied
  • 28. to the phase rotation established by the way the windings are connected to the distributionsystem. Thus, if the motor is spinning backwards when voltage is applied, it will reverse and runin the proper direction. Problems can occur with variable speed drives when they attempt to startagainst a reverse rotating motor.Systems with operating conditions that could cause backflow should be designed and installed tosafely and reliably deal with any problems. Both normal and failure modes need to be considered.Common examples of situations where backflow potential exists include:1 Systems with parallel fans or air handling units. Don�t forget that parallel fan terminal units have fans that are essentially in parallel with the supply fan.2 Systems with series fans: the supply and exhaust fans associated with a 100% outdoor air systems and the fans in series powered fan terminal boxes relative to the supply fan.Does the air handler specification include desirable options?Most fans and air handling units are available with an array of options, some of which aredesirable in most installations and others of which are only required for special installations.Examples include:1 Access doors in casings and fan scrolls.2 Lubrication lines extended to be accessible from the exterior of the unit.3 Baseline vibration characteristics measured at the factory.4 Premium efficiency motors.5 Special vibration isolation provisions.6 Scroll drains (essential for exhaust fans located outdoors and discharging in the up-blast configuration)7 Factory installed back draft dampers.8 Non- sparking or explosion proof construction for hazardous locations.9 Special coatings for handling abrasive or corrosive fluid streams.10.3. Typical ProblemsThe following problems are frequently encountered with fans and drives.10.3.1. Static pressure requirements in excess of designA typical problem found during commissioning or retro-commissioning is high static pressure inthe fan system. In creating excess static pressure that is not required to operate the system, a fanwastes significant amounts of energy. This problem arises because fan selections often fall into arange where there is a difference between the design brake-horsepower (bhp) requirement and theactual motor horsepower installed due to the standard horsepower ratings available in motorproduct lines. The difference between available sizes can become quite significant for larger fans.For example, a fan with an 82 bhp motor requirement would probably come with a 100 hp motor.If the fan was unable to deliver design flow against the installed system static requirement, thenthere would be a lot of margin for speeding the fan up to achieve the design requirement withoutoverloading the motor (assuming the operating point did not end up in a different fan classrequirement). This safety net may be desirable, as the excess motor capacity allows problems tobe solved in the field. But, the added energy consumed by the fan beyond that intended by thedesign will become an energy burden that will persist for the life of the system.
  • 29. Extra diligence during design and construction can prevent conditions that add unanticipatedstatic pressure to the system, thus averting the need to run the fan at an operating point in excessof design. If the balancing team discovers that they have excess system static pressure, there areways to lower static pressure that will allow the system to function at or near its intended designpoint rather than adding on ongoing energy burden to the project by simply throwing energy atthe problem. An example of such a situation is contained in Example 2 in Appendix C :Calculations.10.3.2. Improper belt drive system adjustmentWhile simple in concept, there are some critical parameters associated with the installation andadjustment of this belt drive systems that are often ignored, resulting in belt failures, poorperformance, noise, reduced equipment life and energy waste.Alignment of the drive and motor sheaves is a critical step in the belt installation process.Without proper alignment, belts will run less efficiently, wear out more quickly, and, in extremecases, be thrown off the drive sheaves.Over-tensioning the belts can cause problems with bearings and shafts due to the excessive loadsimposed. In addition, new belts will stretch during the first 8 to 24 hours of operation; belts thathave been properly set initially will require re-tensioning after they have run. This contingency isoften overlooked to the detriment of the drive system efficiency.Multiple belt drives will function best if factory matched belt sets are installed. This ensures thatthe drive loads are equally distributed between all of the belts, equalizing wear and life.The T.B. Woods Company offers a very good guide to proper belt drive selection and adjustmenton their web site.xx10.4. Testing Guidance and Sample Test FormsClick the button below to access all publicly-available prefunctional checklists, functional testprocedures, and test guidance documents referenced in the Testing Guidance and Sample TestForms table of the Air Handler system module.AHU Testing Guidance and Sample Test Formsxx10.5. Supplemental InformationSupplemental information for fans and drives has been developed to provide necessarybackground information for functional testing.
  • 30. Chapter 10: Fans and Drives Supplemental Information10.1. Fans10.2. Capacity control strategies10.3. Drive systems and arrangementsFiguresFigure 10.1: Typical VFD Efficiency vs. SpeedFigure 10.2: Motor and Drive Arrangement Block Access10.1. FansAlthough fans come in a wide variety of designs, shapes, sizes, and configurations. Theygenerally, they fall into two categories:� Centrifugal fans This type of fan imparts kinetic energy to the air primarily by centrifugal force. In essence, the air is drawn into the center of the fan wheel where it is captured and contained by blades. These parcels of air are then �flung� to the periphery of the wheel.[1] The wheel itself can have an inlet on one side (Single Width, Single Inlet or SWSI) or an inlet on both sides (Double Width, Double Inlet or DWDI). The design of the blades on the wheel can have a significant impact on efficiency, performance and cost. Common designs are forward curved, backward included, airfoil, and radial. [2]� Axial fans This type of fan uses aerodynamic effects to impart velocity to the air as it passes through the impeller. Generally, the air travels along the axis of the fan and impeller as compared the centrifugal design where the air enters the impeller by flowing parallel to the shaft, but exits the impeller radially relative to the shaft. Generally, the impeller for this type of fan will be resemblance to an airplane propeller, but with many more blades.10.2. Capacity control strategiesRegardless of the design, the rotating nature of the fan wheel can create significant structuralloads on the shaft, wheel, bearings, and housing. Issues related to these factors are accounted forin the fan class rating. A fan with a wheel that is rated Class II has a higher speed and pressurecapability than the same fan with a wheel that is rated Class I. Therefore, some caution must beused when changing fan speeds in the field to be sure that the new operating point is still withinthe fan�s class rating.There are a variety of techniques used to control fan capacity. The most common include:� Discharge dampers Dampers located on the outlet of the fan can simply throttle the fan. Basically, the discharge damper increases/makes worse, the system effect associated with the fan outlet, thereby degrading its performance. Generally, this is probably the least expensive but also the least desirable approach due to the efficiency implications of a damper on the fan discharge. It can also be quite noisy.
  • 31. � Inlet vanes Inlet vanes modify the performance of the fan by �pre-swirling� the air as it enters the eye of the fan. This has the effect of changing the shape of the fan performance curve as can be seen in Figure 16 of Chapter 13 of the 2000 ASHRAE Systems and Equipment Handbook. This approach is much more desirable than a discharge damper, but not as desirable as a variable speed approach. The emergence of affordable and reliable variable speed drive technology has displaced this approach, but when VAV systems first emerged, it was a common means of achieving the required capacity control and is still found on many existing systems or on systems where the variable speed drives have been eliminated by a value engineering effort.� Blade pitch Varying blade pitch is a efficient but mechanically complex approach to controlling capacity on axial fans. The effect is very similar to a speed change as can be seen from Figure 17 in Chapter 13 of the 2000 ASHRAE Systems and Equipment Handbook. Most fans that use this approach require additional maintenance in the form of periodic lubrication, inspection, and over-haul of the mechanical vane pitch control system.� Variable speed Currently, this is probably the most common approach to controlling fan capacity due to its efficiency, mechanical simplicity, and steadily improving first cost. Commonly referred to as VSD technology (for Variable Speed Drive), it is not necessarily mean VFD technology, which is a subset. Before modern electronic technology made semi- conductor based Variable Frequency Drives a practical and affordable reality, there were a variety of more exotic approaches used including: � Variable speed DC motors These were complex and costly and were usually found only on industrial or very large commercial applications. � Hydro-mechanical clutches This technology employed hydraulics and a clutch system to vary the speed of the output shaft relative to the input shaft. They too were not common on commercial HVAC systems and tended to have relatively high mechanical losses. � Variable pulley systems Often termed �pulley pincher� drives[3], these systems did find somewhat wide application on commercial HVAC systems. The devices functioned by moving the sides of an adjustable drive pulley towards or away from each other. This changed the effective pitch diameter of the pulley, and thus, the output speed. While capable of modulating speeds, the devices tended to be hard on belts and had relatively high mechanical losses. � Solid-state variable frequency drives Typically called VFDs or invertors[4], current technology drives of this type provide a nearly ideal solution to the fan capacity control problem. In and of themselves, they tend to be more efficiency than some of the other approaches (see Figure 10.1 for a typical efficiency plat), but they also tend to maintain the fan efficiency at or near the selected efficiency as they vary its capacity by changing speed. However, this is not without its complications, but paying careful attention to design and commissioning issues can readily overcome any problems and the advantages typically outweigh the disadvantages.Figure 10.1: Typical VFD Efficiency vs. Speed
  • 32. While efficiency does decay with load, these drives will generally deliver better efficiency andless decay than some of the other alternatives like variable pulley systems.Regardless of the technique used, capacity control systems will subject the fan and itscomponents to a wide array of continuously varying performance conditions. The interaction ofthe multiple operating points with the fan components, system components, and building can leadto a number of surprising and unanticipated problems, especially for larger fans with a lot ofpower. Examples include:� On one late 1990�s project the resonance between the large air handling unit fans and the building resulted in vibration in the building�s structural system under certain operating conditions.� In a semiconductor facility, resonance between process exhaust fans and sensitive machinery in the process clean room caused quality control problems.� Over the years, there have been multiple occurrences of fan failure related to resonate frequency problems including axial fans shedding blades and centrifugal fan wheels disintegrating.These problems can be difficult to predict and often show up as commissioning issues. Often, themost viable approach to solving them is to make sure the design incorporates features that willallow you to solve the problem if it occurs. For example, avoiding operation at the triggeringcondition can solve most of these types of problems. And, most current technology variable speeddrives will allow you to program in multiple frequency ranges that the drive will �jump over�
  • 33. as it is commanded through its speed range. Thus, ensuring that the drives that will be suppliedfor your project include this feature can give the start-up and commissioning team the tools theyneed to solve this type of problem when if it crops up. Another desirable feature to include in theproject is vibration analysis and documentation under a variety of operating modes for large fans,especially if they will be operating at variable capacities and speeds. It is also possible to do testson the building structure to determine it�s resonate frequency and then use that information forsetting up the drive systems. The project structural engineer may also be able to predict the rangeof resonate frequencies anticipated for the structure and this information can be reviewed by therest of the team in light of the anticipated operating parameters for the system to allow potentialproblems to be identified and addressed during design.10.3. Drive systems and arrangementsIn all but direct drive applications, some sort of sheave or pulley and belt system will typically beassociated with a fan and its motor. It is not uncommon for one of these pulleys to be supplied inan adjustable configuration to allow the speed of the fan to be easily adjusted by the balancingcontractor in the field. While desirable from this standpoint, there are several draw backs toadjustable pulleys or sheaves:� Belt service life Most V-belts will provide the best service life if they run with their outside perimeter (the flat part at the open end of the V cross section) slightly above the edge of the sheaves they are installed on. If an adjustable pitch sheave requires significant adjustment, it is not uncommon for the outside perimeter of the belt to run below the top of the sheave sides. This results in extra wear on the belt and can reduce service live significantly.� Loss of setting Despite its advantages, adjustability can also be the downfall of adjustable pitch sheaves. It is not uncommon for the balanced setting of the sheave to be lost inadvertently when the belts are replaced, especially if the mechanic performing the work has not been trained regarding adjustable sheaves and mistakenly thinks that the adjustability feature is a convenient way to tension the belt(s) or make the set of belts that they happen to have with them fit. As a result, the once balanced system ends up out of balance and performance suffers. If the new setting delivers less air than was intended, then capacity problems may show up at a later date when design loads show up on the system. If the settings deliver more air than was intended, energy can be wasted, especially if the system is one of the constant volume reheat systems frequently found in hospital or process environments. Both problems can lead to pressure relationship problems if the misadjusted fan happens to be an exhaust fan. If the exhaust is hazardous, a loss of airflow can create a dangerous condition in the area served by the fan that may not be immediately detected.Fans and there prime movers come in a variety of mounting arrangements. AMCA Standard 99-86 illustrates these along with other standards related to fans and air handling units includingdimensioning, motor positions, etc. This information can also be found in most manufacturers fancatalogs. Usually, one or more of the following considerations will dictate the specificarrangement:� The needs of the HVAC process, prime mover and drive system Some HVAC applications may be sensitive to potential by-products from the drive system and there-for, may which to place the entire drive assembly outside of the conditioned air steam. Similarly, certain HVAC process may be a hostile environment for belts or motors and installing them outside of the air stream will improve their serviceability and service life. This can be particularly important for exhaust systems handling hazardous and/or explosive or flammable materials where a motor in the air stream could be a source of ignition.
  • 34. � The arrangement of the fan By their nature, the arrangement of some fans precludes some of the drive arrangements. In addition, physical constraints of the fan installed location may place limitations on the type of drive arrangement that might be used.Figure 10.3: Motor and Drive Arrangement Block AccessOn this new construction project, access to the inlet side of this SWSI fan, which was difficult to begin with,will be totally blocked by the belts and belt guard between the motor and shaft (red circles). It was too lateto solve the problem on this project but a different arrangement may have prevented it. On this project, themaintenance staff will need to remove the belt and drives to inspect the fan wheel.� Service requirements Some arrangements may make service of the motor or fan wheel impossible in the installed location or may block access to some other component in the fan room (see Figure 10.3)� Balancing A belt and pulley system provides a convenient way to adjust fan speed for balancing purposes. Direct drive fans do not have this option and require other methods to adjust for final balance such as adjustable blade pitch or a variable speed drive. Adjustable blades to not have to be automated but are labor intensive to set as compared to a sheave change. Variable speed drives are attractive from an ease of use standpoint, but add unnecessary cost, complexity, and failure modes to a constant volume system.� Heat gains Because they are doing work on the air stream and air is compressible, all fans will show a temperature rise across them, even if the motor is not in the air stream. This temperature rise is called fan heat and can be calculated by converting the fan brake horsepower into btu�s per hour and then solving the following equation for the fan temperature rise: If the motor is located in the air stream, then the motor efficiency losses will also show up as a part of the temperature rise.[5] For large fans with large motors, this can be a significant load on the system that could be avoided by locating the motor outside of the air stream. These
  • 35. advantages have to be weighed against the complications this can introduce for some arrangements in terms of sealing the drive shaft where it penetrates the casing and vibration isolation. � Vibration and sound isolation The method by which vibration and sound isolation will be accomplished can also affect arrangement selection. Mounting the entire fan and drive on an isolation mount will allow the assembly to be further soundproofed by locating it inside an acoustically treated fan casing at the cost of placing the motor in the air stream. By their nature, direct drive fan usually have their vibration isolation problems addressed by the motor mounting arrangement. A hidden but sometimes significant aspect of the vibration isolation technique relates to how the equipment will be seismically restrained (see Functional Testing Basics: Supplemental Information for details).It is becoming increasingly common for manufactures to provide two parallel fans in packagedequipment. Usually space constraints, redundancy requirements, or both drive this design. Whenemployed, there are several issues that need to be considered. � Backdraft Even if the intent of the design is to always run two fans, it is quite likely that at some point in time a failure in the power system, drive system or fan itself will result in one fan needing to operate while the other sits idle. Backdraft dampers are commonly employed to prevent air from the active fan from re-circulating into the inactive fan. However, if not carefully applied, there can be some operational difficulties that will show up during the commissioning process. � Surge When two identical fans are operated in parallel, there is a potential for surge to occur between the two fans.[6] This is because it is very difficult to create two fans that are exactly identical and then get them operating at exactly the same point on their performance curve. Since the fans are coupled to the same system and but that system places them at slightly different points on their operating curves, pressure fluctuations can occur as the fans shift around and interact, trying to find a mutually agreeable operating point. The effects from this can range from unnoticeable to noise to (in rare cases) fan damage.Chapter 18 of the 2000 ASHRAE Systems and Equipment Handbook, AMCA publications 99-86,200, 201, 202-88, and 203, and the Trane Fan Engineering Handbook are all excellent resourcesfor additional detailed information regarding the topics outlined above.[1] It�s the same effect you experienced as a child on the merry-go-round at the playground.[2] While less common than the other designs, radial blade fans are sometimes found in exhaust systems, especially exhaust systems that handle materials like dust or other particulate matter or where high pressures are required.[3] For those who are wood workers, this is basically the same approach as is used for varying speed on a Shopsmith� multipurpose tool.[4] This is a reference to the electronic process going on in most drives; basically the drives take alternating current, rectify it to direct current perform their �magic�, and then invert the direct current to create an alternating current out put with the desired frequency and other electrical characteristics necessary to control the motor.[5] This temperature rise can be calculated in the same manner as the fan heat but the motor horsepower (vs. fan brake horsepower) at the current operating condition is used.[6] This should not be confused with the surge that can occur in a single fan if it is operated at a point on its curve where the pressure difference across it fights with the fans ability to generate that pressure difference causing sporadic flow reversals through the impeller.