5.energy audit of pumps
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Transcript of 5.energy audit of pumps
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Submitted By : Padma Dhar Garg 2011PME5258
ENERGY AUDIT OF PUMP
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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
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MAIN PUMP COMPONENTS
Pumps
Prime movers: electric motor
Piping to carry fluid
Valves to control flow in system
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PUMP LAYOUT
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PUMP CURVES
Pump Operating Point
System head curve Pump Head – Flow curve
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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
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DATA COLLECTION
Specifications and design details
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DATA COLLECTION
Specifications and design details contd…
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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
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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
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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,
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EFFICIENCY AND PERFORMANCE EVALUATION OF THE PUMPSPerformance parameters for water pumps
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EFFICIENCY AND PERFORMANCE EVALUATION OF THE PUMPSPerformance parameters for water pumps contd..
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EFFICIENCY AND 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
1000Parameter 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
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FLOW MEASUREMENT, Q
The following are the methods for flow measurements:
• Ultrasonic flow measurement • Tank filling method • Installation of an on-line flow meter
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RATING OF MOTOR
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DATA TABLE
Sr. No. Description UOM Present Pump Proposed Pump
1 Name 2 Head Mtrs. 6.703 Discharge m3/S 0.0134 Input Power motor KW 3.894 Power consumption daily kWh 11.675 Efficiency % 92
6 Specific Power Consumption kWh/m3
7 Running Hrs/day Hrs. 38 Running Hrs./ Annum Hrs. 330
9 Power consumption/Annum kWh 1283.70
10 Unit Rate Rs 8.0011 Total expenditure/Annum Rs. 10269.6013 Estimated Saving Rs. 4519.0014 Cost of proposed VFD Rs.15 Simple Pay Back Period (year) 2.27
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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
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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 powerMotor 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
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CONTD..
Pump efficiency, Pump = Hydraulic power / Pump shaft power
= 0.8754 / 3.73
= 0.24
= 24%
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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.
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Affinity Laws of Centrifugal Loads
FLOW is proportional to motor speed.�
PRESSURE is proportional to the motor speed squared.�
POWER is proportional to the motor speed cubed.�
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A VFD IN A BLOCK DIAGRAM
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AFFINITY LAWS FOR PUMP
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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 day60% speed for the remaining 25% of the dayCost 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
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CONTD..
Total =5750.9 Rs.Annual saving = 10269.60 – 5750.90 =4519Rs.
Pay back period = 2.27yr
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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
0.00700000000000002
0.01 0.0130
200
400
600
800
1000
1200
1400
1600
864
1152
1440
speed flow rate
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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
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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 valveSystem Curves
Operating Points
14401152
7.14 m
9.14 m
Static
5.14 m
Flow (m3/s)
Oversize Pump
Required Pump
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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.
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THANK YOU