2. INTRODUCTION
What is a Pump?
A pump is a device used to convert mechanical energy to
pressure energy for the movement of fluids, such as liquids,
gases or slurries.
Objective
• Transfer fluid from source to destination
• Circulate fluid around a system
3. MAIN PUMP COMPONENTS
Pumps
Prime movers: electric motor
Piping to carry fluid
Valves to control flow in system
6. PUMP TERMINOLOGY
Friction Head :- Friction head is the measure of resistance to flow provided by the
pipe , valve etc.
Static Head:-The static head is the amount of feet of elevation the pump must lift
the water regardless of flow.
Operating Point:- It is a point where system curve and pump curve intersect .
Static Suction Lift :- The vertical distance from the water line to the centerline of
the impeller.
Static Discharge Head :- The vertical distance from the discharge outlet to the
point of discharge or liquid level when discharging into the bottom of a water tank
9. INSTRUMENTS REQUIRED
• Power Analyzer: Used for measuring electrical parameters such as kW,
kVA, pf, V, A and Hz
•Stroboscope: To measure the speed of the driven equipment and motor
• Ultra sonic flow meter or online flow meter
•The above instruments can be used in addition to the calibrated online /
plant instruments
10. PARAMETERS TO BE MEASURED
• Energy consumption pattern of pumps (daily / monthly /yearly
consumption)
• Motor electrical parameters (kW, kVA, pf, A, V, Hz) for individual
pumps
11. Contd..
Pump operating parameters to be monitored for each pump
Discharge Flow,
Head (suction & discharge),
Load variation,
Pumps operating hours and operating schedule,
Pump /Motor speed,
14. EFFICIENCYAND PERFORMANCE
EVALUATION OF THE PUMPS
Pump hydraulic power can be calculated by the formula:
Hydraulic kW =
Q x Total Head, (hd – hs) x x g
1000
Parameter Details Unit
Q Water flow rate m3/s
Total head Difference between discharge head, hd & suction head, hs m
Density of water or fluid being pumped Kg/m3
g Acceleration due to gravity m2/s
Pump efficiency, Pump =
Hydraulic power
Pump shaft power
Pump shaft power = Hydraulic power x Motor
15. FLOW MEASUREMENT, Q
The following are the methods for flow measurements:
• Ultrasonic flow measurement
• Tank filling method
• Installation of an on-line flow meter
17. DATA TABLE
Sr. No. Description UOM Present Pump Proposed Pump
1 Name
2 Head Mtrs. 6.70
3 Discharge m3/S 0.013
4 Input Power motor KW 3.89
4 Power consumption daily kWh 11.67
5 Efficiency % 92
6 Specific Power Consumption kWh/m3
7 Running Hrs/day Hrs. 3
8 Running Hrs./ Annum Hrs. 330
9 Power consumption/Annum kWh 1283.70
10 Unit Rate Rs 8.00
11 Total expenditure/Annum Rs. 10269.60
13 Estimated Saving Rs. 4519.00
14 Cost of proposed VFD Rs.
15 Simple Pay Back Period (year) 2.27
18. MEASURED DATA
Flow rate (m3/s) 0.013
HD (mtrs.) 9.14
HS (mtrs.) 2.44
HD - HS (mtrs.) 6.7
Motor voltage (V) 393
Motor current (A) 10
Motor input power (kW) 3.89
19. CALCULATION
Hydraulic power = (0.013 × 6.7 × 1000 × 9.8)/1000
= 0.8754 kW
Motor air gap power PG = 3.89 kW
Motor shaft power = Pump shaft power
Motor shaft power = PG - s PG
Motor slip ‘s’ = (NS – NR)/ NS
NS = 120 f / P
= (120×50) /4
= 1500
s = (1500 – 1440)/1500
= 0.04
Pump shaft power = 3.89 – (0.04 × 3.89)
= 3.73
21. CALCULATION
Running Hrs/day = 3 Hrs
Working days = 110 days/sem
Running Hrs./ Annum = 330 Hrs
Power consumption daily = 3.00 3.89
= 11.67kWh
Power consumption/Annum = 330 3.89
= 1283.7 kWh
Unit Rate = 8 Rs.
Total expenditure/Annum = 8 1283.7
= 10269.6 Rs.
22. Affinity Laws of Centrifugal Loads
ƒ FLOW is proportional to motor speed.
ƒ PRESSURE is proportional to the motor speed squared.
ƒ POWER is proportional to the motor speed cubed.
25. CONTD..
Assuming the Pump does not need to run at full speed all of the day,
we will use an example of:
Running full speed (100%) for 25% of the day
80% speed for 50% of the day
60% speed for the remaining 25% of the day
Cost of running with an AC drive controlling the motor:
• 5.2 hp x (1)3 x 0.746 kW/hp x 82.5 hours x 8/kWh = 2567.40
• 5.2 hp x (0.8)3 x 0.746 kW/hp x 165 hours 8/kWh=2629.00
• 5.2hp x (0.6)3 x 0.746 kW/hp x 82.5 hours x 8/kWh =554.50
26. CONTD..
Total =5750.9 Rs.
Annual saving = 10269.60 – 5750.90
=4519Rs.
Pay back period = 2.27yr
27. EFFECT OF VFD ON FLOW RATE
F N
F =k N
At 100% speed ie. 1440 rpm
0.013= k 1440
K = 9.03 10-6
• At 80% speed ie. 1152
F= 9.03 10-6 1152
= 0.014m3/s
• At 60% speed ie. 864
F= 9.03 10-6 1864
= 0.0071m3/s
864
1152
1440
0
200
400
600
800
1000
1200
1400
1600
0.007 0.01 0.013
speed
flow rate
28. ENERGY CONSERVATION
OPPORTUNITIES
• Compare the actual values with the design / performance test values if any deviation
is found, list the factors with the details and suggestions to over come.
• Compare the specific energy consumption with the best achievable value
(considering the different alternatives). Investigations to be carried out for problematic
areas..
• Enlist scope of improvement with extensive physical checks / observations. Based
on the actual operating parameters, enlist recommendations for action to be taken for
improvement, if applicable such as:
• Replacement of pumps
• Impeller replacement
• Impeller trimming
• Variable speed drive application, etc
29. Avoiding over sizing of pump
ENERGY CONSERVATION
OPPORTUNITIES
Head
Head
Partially
closed valve
Const. Speed
A
B
C
Meters
Pump Efficiency 24%
Pump Curve at
Full open valve
System Curves
Operating Points
14401152
7.14 m
9.14 m
Static
5.14 m
Flow (m3/s)
Oversize Pump
Required Pump
30. ENERGY CONSERVATION
POSSIBILITIES- SUMMARY
• Improvement of systems and drives.
• Use of energy efficient pumps
• Replacement of inefficient pumps
• Trimming of impellers
• Correcting inaccuracies of the Pump sizing
• Use of high efficiency motors
• Integration of variable speed drives into pumps: The integration of adjustable speed
drives (ASD) into compressors could lead to energy efficiency
improvements, depending on load characteristics.
• High Performance Lubricants: The low temperature fluidity and high temperature
stability of high performance lubricants can increase energy efficiency by reducing
frictional losses.
