Worked Example 1 - Tipperary Energy...
Transcript of Worked Example 1 - Tipperary Energy...
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 125 10.2 was chosen with the 270mm
diameter impellor. The motor is a 92% efficient 110kw motor, Good practise C. 1994.
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
Answers:
1. 320m3/hour or 7680 per24 hrs
2. 81%
3. 6720/7680 = 21
4. 81kw/ 92% = 88
5. 88/320 = 0.275 kWh/m3
6. 0.275*365*6720*.15= € 101,228
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 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 m3
6. The total annual running cost.
Answers:
1. 200m3 / hour ( 400 total) and 9600 total per day
2. 72%
3. 9120/9600 * 24 = 22.8 hours
4. 134kw/.92 = 146
5. 146/400 = 0.364 kWh/m3
6. 0.364*365*9120*.15= € 181,800
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 125
10.1 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
Answers:
1. 400m3 / hour
2. 79%
3. 9120/10320 * 24 = 21.2 hours
4. 125kw/.962 = 130
5. 130/400 = 0.325 kWh/m3
6. 0.325*365*9120*.15= € 162,200k
Calculate the annual savings for the pump replacement?
€181,800 – €162,200 = 19,600k
What would the expected cost & Payback be (assuming it cost 40k installed)?
Simple payback of 2.1 years
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 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 m3
6. The total annual running cost.
7. Savings due to controls
8. Payback for the VSD
Answers:
1. (380*88/367) / 120 = 80%
2. 24 hours
3. 115kw/.962 = 120
4. 120/380 = 0.315 kWh/m3
5. 0.315*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
Baureihe--Größe
Type--Size
Modèle
Tipo
Serie
Tipo
Nenndrehzahl
Nom. speed
Vitesse nom.
Velocità di rotazione nom.
Nominaal toerental
Revolutiones nom.
Laufrad--ø
Impeller Dia.
Diamètre de roue
ø Girante
ø Waaier
ø Rodete
Projekt
Project
Projet
Progetto
Projekt
Proyecto
Pos.--Nr.
Item No.
No. de pos.
Pos.--Nr.
Positienr.
Pos.--Nr.
Angebots--Nr.
Project No.
No. de l’offre
Offerta--No.
Offertenr.
Offerta--No.Division Pompes Industrie et Eau
KSB S. A.Allée de Sagan -- B.P. 18936004 Châteauroux
Multitec 125 10.1
1.4408 2900 1/min
K1777.452/10/1
0 500 1000 1500 2000 2300US.gpm
0 500 1000 1500 1900IM.gpm
0 50 100 150l/s
0 100 200 300 400 500 540m /h3
0 100 200 300 400 500 540m /h3
Förderstrom/Flow/Débit/Portata/Kapaciteit/Caudal
200
300
400
ft
60
80
100
120
45
135
m
Förderhöhe
TDH
Hauteur
Prevalenza
Opvoerhoogte
Altura
50
100
150
200
hp
50
100
150
30
kW
Leistungsbedarf
Power Input
Puiss. abs.
Potenza ass.
Opgenomen
vermogen
Potencia nec.
20
40
ft
5
10
2
13
mNPSHR
♣ ξΒζ
ø274
78.9
2020 30 40 5060
70
70
75
75
78
78
ø274
Axial
ø305
80.9
20 30 40 505060
70
70
75
75
78
78
80
80
ø305
Radial
Werkstoff/Material/MatériauMateriale/Materiaal/Material
22, 23, 30
Laufradaustrittsbreite/Impeller outlet width/Largeur à la sortie de la roueLuce della girante/Waaier uittredebreedte/Anchura de salida rodete
20,0 mm20,0 mm
86
Baureihe--Größe
Type--Size
Modèle
Tipo
Serie
Tipo
Nenndrehzahl
Nom. speed
Vitesse nom.
Velocità di rotazione nom.
Nominaal toerental
Revolutiones nom.
Laufrad--ø
Impeller Dia.
Diamètre de roue
ø Girante
ø Waaier
ø Rodete
Projekt
Project
Projet
Progetto
Projekt
Proyecto
Pos.--Nr.
Item No.
No. de pos.
Pos.--Nr.
Positienr.
Pos.--Nr.
Angebots--Nr.
Project No.
No. de l’offre
Offerta--No.
Offertenr.
Offerta--No.Division Pompes Industrie et Eau
KSB S. A.Allée de Sagan -- B.P. 18936004 Châteauroux
Multitec 125 10.2
1.4408 2900 1/min
K1777.452/11/1
0 500 1000 1500 2000 2300US.gpm
0 500 1000 1500 1900IM.gpm
0 50 100 150l/s
0 100 200 300 400 500 540m /h3
0 100 200 300 400 500 540m /h3
Förderstrom/Flow/Débit/Portata/Kapaciteit/Caudal
100
200
300
ft
20
40
60
80
100
m
Förderhöhe
TDH
Hauteur
Prevalenza
Opvoerhoogte
Altura
50
100
hp
40
60
80
25
90
kW
Leistungsbedarf
Power Input
Puiss. abs.
Potenza ass.
Opgenomen
vermogen
Potencia nec.
20
40
ft
5
10
2
13
mNPSHR
♣ ξΒζ
ø243
79.5
2020 30 40 50 60 6570
70
75
75
78
78
ø243
Axial
ø270
81.5
20 30 40 50 60 65
65
70
70
75
75
78
78
81
81
ø270
Radial
Werkstoff/Material/Matériau
Materiale/Materiaal/Material22, 23, 30
Laufradaustrittsbreite/Impeller outlet width/Largeur à la sortie de la roueLuce della girante/Waaier uittredebreedte/Anchura de salida rodete
20,0 mm20,0 mm
88
© TEA. www.tea.ie 1
Water Services Energy Training
William Moynihan, Manager, Kerry Energy Agency
Paul Kenny, Senior Consultant, Tipperary Energy Agency
© TEA. www.tea.ie 2
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. www.tea.ie 3
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 125
10.2 was chosen with the 270mm diameter impellor. The motor is a 92%
efficient 110kw motor, Good practise C. 1994.
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. www.tea.ie 6
Q3: Operating hours per day:
= Required flow
Max Flow per hour *24
= 6720 * 24
320*24
= 21 hours per day
Example 1
© TEA. www.tea.ie 8
Q4. Power = shaft power (from graph)/ motor eff= 81/0.92
= 88kW
Q5. kWh/ m3 = motor power/ flow
= 88/320
= 0.275 kWh/m3
Q6. Annual cost
= kWh/m3 x Flow per annum x cost /kwh
= 0.275 * 365 * 6720 * 0.15
= €101,228
Example 1
© TEA. www.tea.ie 9
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
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 m3
6. The total annual running cost.
Example 2
© TEA. www.tea.ie 11
Q3: Operating hours per day:
= Required flow Or Average flow
Flow per hour *24 flow per hour
= 9120 * 24
9600
= 22.8
Example 2
© TEA. www.tea.ie 13
Q4. Power = shaft power (from graph)/ motor eff= 134/0.92
= 146kW
Q5. kWh/ m3 = motor power/ flow
= 146/400
= 0.364 kWh/m3
Q6. Annual cost
= kWh/m3 x Flow per annum x cost /kwh
= 0.364 * 365 * 9120 * 0.15
= €181,800
Example 2
© TEA. www.tea.ie 14
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 125 10.1 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
Example 3
© TEA. www.tea.ie 16
Q3: Operating hours per day:
= Required flow Or Average flow
Flow per hour *24 flow per hour
= 9120
400
= 22.8
Example 3
© TEA. www.tea.ie 18
Q4. Power = shaft power (from graph)/ motor eff= 125/0.962
= 130kW
Q5. kWh/ m3 = motor power/ flow
= 130/ 400
= 0.325 kWh/m3
Q6. Annual cost
= kWh/m3 x Flow per annum x cost /kwh
= 0.325 * 365 * 9120 * 0.15
= €162,200
Q7. Annual savings for pump replacement
= 182,600 – 162,200 = 20,800
Q8. Simple payback
= 40,000/20,800 = 2 years
Example 3
© TEA. www.tea.ie 19
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 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 m3
6. The total annual running cost.
7. Savings due to controls
8. Payback for the VSD
Example 4
© TEA. www.tea.ie 21
Q3: Operating hours per day:
= Required flow Or Average flow
Flow per hour *24 flow per hour
= 9120 * 24
9120
= 24
Example 4
© TEA. www.tea.ie 23
Q4. Power = shaft power (from graph)/ motor eff= 115/0.962
= 120kW
Q5. kWh/ m3 = motor power/ flow
= 120/380
= 0.315 kWh/m3
Q6. Annual cost
= kWh/m3 x Flow per annum x cost /kwh
= 0.315 * 365 * 9120 * 0.15
= €157,285
Q7. Cost and payback 162,200- 157,285 = 4,915
Q8. 8,000/ 4,915 = < 1.62 years
Example 4