Q: Work out the Pump Efficiency, Power consumption by the pump and the motor.

Transcription

1 Worked Example 1 Q: Work out the Pump Efficiency, Power consumption by the pump and the motor. In 1994 the system was designed to pump at a Total dynamic head of 75m and an average flow rate of 6720m3 per day. The KSB Multitek was chosen with the 270mm diameter impellor. The motor is a 92% efficient 110kw motor, Good practise C Determine: 1. The pump flow rate at that head. 2. The Pump efficiency at that point. 3. Calculate the operating hours per day assuming it operates to a reservoir on an onoff control using a level sensor. 4. The electrical load on the motor (shaft power/ motor eff) 5. The kwh per m3 6. The total annual running cost. 1 kwh = 15cents, head is 75m Answers: m3/hour or 7680 per24 hrs 2. 81% /7680 = kw/ 92% = /320 = kwh/m *365*6720*.15= 101,228

2 Worked example 2 demand change over time. Over the last 15 years, the demand has risen substantially to 9120 m3 / day and the station is using the standby pump to deliver the additional flow into the reservoir. The increased head losses in the pipe due to the extra flow has added a further 12m of headloss (now 87m) when the second pump is running. Assuming two identical pumps, what is the new operating point of the pump when the second pump is now working? Determine: 1. Each pump flow rate at that head. 2. The Pump efficiency at that point. 3. Calculate the operating hours per day assuming it operates to a reservoir on an onoff control using a level sensor. (to keep calculations simple - both pumps come on together) 4. The load on the motor 5. The kwh per m3 6. The total annual running cost. Answers: m3 / hour ( 400 total) and 9600 total per day 2. 72% /9600 * 24 = 22.8 hours kw/.92 = /400 = kwh/m *365*9120*.15= 181,800

3 Worked Example 3 Select a new pump Over the last 15 years, the demand has risen substantially to 9120 m3 / day and you want to be able to supply (a new commercial park being planned) for the foreseeable future 9600m3/ day using one pump (92m head). You have selected the The KSB Multitek with a 305mm dia impellor. The Motor is a Brook Crompton Premium efficiency IEC 3 motor with an efficiency of 96.2%. Determine: 1. Pump flow rate at that head. 2. The Pump efficiency at that point. 3. Calculate the operating hours per day assuming it operates to a reservoir on an onoff control using a level sensor. The load on the motor 4. The kwh per m3 5. The total annual running cost. 6. Annual Savings versus example 2 7. Simple payback Assumed cost 40k Answers: m3 / hour 2. 79% /10320 * 24 = 21.2 hours kw/.962 = /400 = kwh/m *365*9120*.15= 162,200k Calculate the annual savings for the pump replacement? 181, ,200 = 19,600k What would the expected cost & Payback be (assuming it cost 40k installed)? Simple payback of 2.1 years

4 Worked example 4 VSD controls Given the above pump can operate on a VSD at 48Hz at a flow rate of 9120 m3 per day, (head of 87m) draws (shaft) 120kW, calculate the parameters again. Determine: 1. Pump flow rate at that head. 2. The Pump efficiency at that point. 3. Calculate the operating hours per day assuming it operates to a reservoir on an onoff control using a level sensor. (to keep calculations simple - both pumps come on together) 4. The load on the motor 5. The kwh per m3 6. The total annual running cost. 7. Savings due to controls 8. Payback for the VSD Answers: 1. (380*88/367) / 120 = 80% hours kw/.962 = /380 = kwh/m *365*9120*.15= 162,825 How much did the extra controls having the pump match demand versus on/ off save? 4915 What is the simple payback assuming a 8,000 increase for a VSD versus a soft start? 1.62 Years

5 Baureihe-Größe Type-Size Modèle Projekt Project Projet Tipo Serie Tipo Progetto Projekt Proyecto Nenndrehzahl Nom. speed Vitesse nom. Multitec /min Angebots-Nr. Project No. No. de l offre Velocità di rotazione nom. Nominaal toerental Revolutiones nom. Offerta-No. Offertenr. Offerta-No. Laufrad-ø Impeller Dia. Diamètre de roue Pos.-Nr. Item No. No.depos. øgirante øwaaier ørodete Pos.-Nr. Positienr. Pos.-Nr. Division Pompes Industrie et Eau KSB S.A. AlléedeSagan -B.P Châteauroux US.gpm IM.gpm ξβζ Werkstoff/Material/Matériau Materiale/Materiaal/Material 22,23, Förderhöhe TDH Hauteur Prevalenza Opvoerhoogte Altura 100 m ø ft ø m/h l/s Radial Axial 40 ft NPSH R m ø Leistungsbedarf Power Input Puiss. abs. Potenza ass. Opgenomen vermogen Potencia nec. kw 100 ø hp m/h Förderstrom/Flow/Débit/Portata/Kapaciteit/Caudal K /10/1 50 Laufradaustrittsbreite/Impeller outlet width/largeur à la sortie de la roue Luce della girante/waaier uittredebreedte/anchura de salida rodete 86 20,0 mm 20,0 mm

6 Baureihe-Größe Type-Size Modèle Tipo Serie Tipo Nenndrehzahl Nom. speed Vitesse nom. Multitec /min Projekt Project Projet Progetto Projekt Proyecto Angebots-Nr. Project No. No. de l offre Velocità di rotazione nom. Nominaal toerental Revolutiones nom. Offerta-No. Offertenr. Offerta-No. Laufrad-ø Impeller Dia. Diamètre de roue Pos.-Nr. Item No. No.depos. øgirante øwaaier ørodete Pos.-Nr. Positienr. Pos.-Nr. Division Pompes Industrie et Eau KSB S.A. AlléedeSagan -B.P Châteauroux US.gpm Förderhöhe TDH Hauteur Prevalenza Opvoerhoogte Altura m IM.gpm m/h ξβζ Werkstoff/Material/Matériau Materiale/Materiaal/Material 22,23, ø ø ft l/s Radial Axial 40 ft NPSH R m ø Leistungsbedarf Power Input Puiss. abs. Potenza ass. Opgenomen vermogen Potencia nec. 60 kw hp 40 ø m/h Förderstrom/Flow/Débit/Portata/Kapaciteit/Caudal K /11/1 Laufradaustrittsbreite/Impeller outlet width/largeur à la sortie de la roue Luce della girante/waaier uittredebreedte/anchura de salida rodete 88 20,0 mm 20,0 mm

7 Water Services Energy Training William Moynihan, Manager, Kerry Energy Agency Paul Kenny, Senior Consultant, Tipperary Energy Agency TEA. 1

8 Examples 4 examples: 1. Basic operating point of a given system (as designed) 2. Over time, demand has increased, 2nd pump operating on system, work out the new operating point 3. Select a new point for the maximum that will be required in the future 4. Given a Variable speed drive curve, work out the new operating point for current demand TEA. 2

9 Example 1 Q: Work out the Pump Efficiency, Power consumption by the pump and the motor. In 1994 the system was designed to pump at a Total dynamic head of 75m and an average flow rate of 6720m3 per day. The KSB Multitek was chosen with the 270mm diameter impellor. The motor is a 92% efficient 110kw motor, Good practise C Determine: 1. The pump flow rate at that head. 2. The Pump efficiency at that point. 3. Calculate the operating hours per day assuming it operates to a reservoir on an on-off control using a level sensor. 4. The electrical load on the motor (shaft power/ motor eff) 5. The kwh per m3 6. The total annual running cost. 1 kwh = 15cents, head is 75m TEA. 3

10 Example 1 TEA. 4

11 Example 1 Q2: Pump Eff = 81% Q1: Pump Flow Rate = 320m 3 /Hour TEA. 5

12 Example 1 Q3: Operating hours per day: = Required flow Max Flow per hour *24 = 6720 * *24 = 21 hours per day TEA. 6

13 Example 1 Q4: Shaft power = 81kW TEA. 7

14 Example 1 Q4. Power = shaft power (from graph)/ motor eff = 81/0.92 = 88kW Q5. kwh/ m3 = motor power/ flow = 88/320 = kwh/m 3 Q6. Annual cost = kwh/m 3 x Flow per annum x cost /kwh = * 365 * 6720 * 0.15 = 101,228 TEA. 8

15 Example 2 Worked example 2 demand change over time. Over the last 15 years, the demand has risen substantially to 9120 m 3 / day and the station is using the standby pump to deliver the additional flow into the reservoir. The increased head losses in the pipe due to the extra flow has added a further 12m of headloss (now 87m) when the second pump is running. Assuming two identical pumps, what is the new operating point of the pump when the second pump is now working? Determine: 1. Each pump flow rate at that head. 2. The Pump efficiency at that point. 3. Calculate the operating hours per day assuming it operates to a reservoir on an on-off control using a level sensor. (to keep calculations simple - both pumps come on together) 4. The load on the motor 5. The kwh per m 3 6. The total annual running cost. TEA. 9

16 Example 2 Q2: Pump Eff = 72% Q1: Pump Flow Rate = 200m 3 /Hour TEA. 10

17 Example 2 Q3: Operating hours per day: = Required flow Or Average flow Flow per hour *24 flow per hour = 9120 * = 22.8 TEA. 11

18 Example 2 Q4: Shaft power = 67kW (134kW for 2) TEA. 12

19 Example 2 Q4. Power = shaft power (from graph)/ motor eff = 134/0.92 = 146kW Q5. kwh/ m3 = motor power/ flow = 146/400 = kwh/m3 Q6. Annual cost = kwh/m3 x Flow per annum x cost /kwh = * 365 * 9120 * 0.15 = 181,800 TEA. 13

20 Example 3 Over the last 15 years, the demand has risen substantially to 9120 m3 / day and you want to be able to supply (a new commercial park being planned) for the foreseeable future 9600m3/ day using one pump (92m head). You have selected the The KSB Multitek with a 305mm dia impellor. The Motor is a Brook Crompton Premium efficiency IEC 3 motor with an efficiency of 96.2%. Determine: 1. Pump flow rate at that head. 2. The Pump efficiency at that point. 3. Calculate the operating hours per day assuming it operates to a reservoir on an on-off control using a level sensor. The load on the motor 4. The kwh per m3 5. The total annual running cost. 6. Annual Savings versus example 2 7. Simple payback Assumed cost 40k TEA. 14

21 Example 3 Q2: Pump Eff = 79% Q1: Pump Flow Rate = 400m 3 /Hour TEA. 15

22 Example 3 Q3: Operating hours per day: = Required flow Or Average flow Flow per hour *24 flow per hour = = 22.8 TEA. 16

23 Example 3 Q4: Shaft power = 125kW TEA. 17

24 Example 3 Q4. Power = = shaft power (from graph)/ motor eff 125/0.962 = 130kW Q5. kwh/ m3 = motor power/ flow = 130/ 400 = kwh/m3 Q6. Annual cost = kwh/m3 x Flow per annum x cost /kwh = * 365 * 9120 * 0.15 = 162,200 Q7. Annual savings for pump replacement = 182, ,200 = 20,800 Q8. Simple payback = 40,000/20,800 = 2 years TEA. 18

25 Example 4 Given the above pump can operate on a VSD at 48Hz at a flow rate of 9120 m 3 per day, (head of 87m) draws (shaft) 120kW, calculate the parameters again. Determine: 1. Pump flow rate at that head. 2. The Pump efficiency at that point. 3. Calculate the operating hours per day assuming it operates to a reservoir on an on-off control using a level sensor. (to keep calculations simple - both pumps come on together) 4. The load on the motor 5. The kwh per m 3 6. The total annual running cost. 7. Savings due to controls 8. Payback for the VSD TEA. 19

26 Example 4 Q2: Pump Eff = 80% Q1: Pump Flow Rate = 380m 3 /Hour TEA. 20

27 Example 4 Q3: Operating hours per day: = Required flow Or Average flow Flow per hour *24 flow per hour = 9120 * = 24 TEA. 21

28 Example 4 Q4: Shaft power = 115kW TEA. 22

29 Example 4 Q4. Power = = shaft power (from graph)/ motor eff 115/0.962 = 120kW Q5. kwh/ m3 = motor power/ flow = 120/380 = kwh/m 3 Q6. Annual cost = kwh/m 3 x Flow per annum x cost /kwh = * 365 * 9120 * 0.15 = 157,285 Q7. Cost and payback 162, ,285 = 4,915 Q8. 8,000/ 4,915 = < 1.62 years TEA. 23