Submitted By : Padma Dhar Garg 2011PME5258

ENERGY AUDIT OF PUMP

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

MAIN PUMP COMPONENTS

Pumps

Prime movers: electric motor

Piping to carry fluid

Valves to control flow in system

PUMP LAYOUT

PUMP CURVES

Pump Operating Point

System head curve Pump Head – Flow curve

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

DATA COLLECTION

Specifications and design details

DATA COLLECTION

Specifications and design details contd…

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

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

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,

EFFICIENCY AND PERFORMANCE EVALUATION OF THE PUMPSPerformance parameters for water pumps

EFFICIENCY AND PERFORMANCE EVALUATION OF THE PUMPSPerformance parameters for water pumps contd..

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

FLOW MEASUREMENT, Q

The following are the methods for flow measurements:

• Ultrasonic flow measurement • Tank filling method • Installation of an on-line flow meter

RATING OF MOTOR

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

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

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

CONTD..

Pump efficiency, Pump = Hydraulic power / Pump shaft power

= 0.8754 / 3.73

= 0.24

= 24%

 

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.

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.�

A VFD IN A BLOCK DIAGRAM

AFFINITY LAWS FOR PUMP

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

CONTD..

Total =5750.9 Rs.Annual saving = 10269.60 – 5750.90 =4519Rs.

Pay back period = 2.27yr

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

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

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

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.

THANK YOU