ENERGY EFFICIENT MOTOR

Compiled By: Savan Changela IU1241060008 Mit Kavathiya IU1241060020 Shubham Logad IU1241060022 Ajaysinh Zala IU1241060062

Guided By: Prof. Parth Patel.

Index

• Design features• Energy “efficient” motor• Calculations of “efficient” motor• Comparison between standard and efficient

motor(theoretical)• Testing of standard motor

Design Features:

1. Constructional Features:-a. Main Dimensionb. Stator Designc. Rotor Design

2. Performance Calculations:-a. No load current b. Loss Componentc. Leakage reactanced. Loss and efficiency

Energy “efficient” motor An energy “efficient” motor produces the same

shaft output power (hp), but uses less input power (kw) than a standard motor.

Energy efficient motor require less maintenance and have longer life because of lower motor operating temperature.

Contd.• Energy efficient motors are manufactured using same frame as

a standard frame motor, but have difference in following parameter:

a. Steel Lamination.b. Copper Windings.c. Air Gap.d. Fan Losses.e. Length.

Calculation Of Energy Efficient Motor:

• Design of 1 H.P, 415 V, 3Φ, 4 pole, 50 Hz star connected, squirrel cage induction motor

• Soln: Assume the following data:-

Specific Magnetic loading, Bav = 0.4546 Tesla Specific Electric loading, ac = 23246 ac/m Full load Efficiency, η = 0.825 Full load Power factor, cosΦ = 0.80 Winding Factor, Kw =0.955 Stacking Factor, Ks=0.9

(i) Main Dimensions: We have from output equation: Q = KW/ η cosΦ = 0.750/( 0.825 x 0.80 ) = 1.136 KVA

Co = 11 Bav ac Kw cosΦ x 10-3

= 11x 0.4546 x 23246 x 0.955 x 10-3 = 111.01

ns = 2f/p = 25 rps

D2 * L = Q/ (Co ns ) = 1.136/(111.01x25) = 4.094x10-4 m3

•

(ii) No. of stator turns Φ = (πDL/p) Bav = (π x 0.08 x 0.09/ 4) x 0.4546 = 2.56 mwb

Assuming Eph =Vph = 415 volts Tph = Eph / 4.44fΦkw = 415/(√3x4.44 x 50 x 2.56x10-3 x 0.955)

= 440(iii) No. of stator slots Assuming no of slot/pole/phase =3 Total no. of slots Ss= 3 x 3 x 4 = 36

(iv) Slot Pitch Yss = π *D/Ss = π x 0.08/36 = 6.97 mm

(v) No of conductors /slot Total no of stator conductors = 6Ts = 6 x 440 = 2646 No. of conductors /slot = 2646/36 = 74(vi) Coil Span = 36/4 = 9 Distribution factor kd = 0.96 Pitch Factor kp = 0.985 Winding Factor = 0.94 (vi) Conductor Size:- Stator Current (line) = = 1.58 A

Current Density = 5.5 A/mm2 Wire gauge = 23.5 SWG Diameter = 0.58 mm Area = 0.2679 mm2

(vii) Slot Dimension:- Space required for bar = Zss x as conductor in a slot = 74 x 0.2679 = 19.82 mm2

Area of each slot = 49 mm2

Max. Allowable flux density = 1.7 T (Wts)min. = = 2.065 mm

Therefore,Slot Width = 4.3 mmSlot Depth = 13.5 mm

Lmts = 2L + 2.3T + 0.24 = 0.56 m Flux density in stator teeth = = 0.8166 T Flux in core = = 1.28x10-3 wbMax. Allowable flux density in core is 1.65 T

Acs = Flux in core / 1.65 = 0.776 x 10-3 m2

Depth of stator core dcs = (0.776 x 10-3) / Li = 9.57 x 10-3 m

•

(xi) Rotor Bars :- Rotor Bar Current Ib = (2 x m x kws x Ts x Is cos ) / 28 = 112 A. Current density = 6 A/mm2

Area of each rotor bar = 18.67 mm2. (xii) Dimension of rotor slots :- Slot area = 33.16 mm2

Width of rotor slot = 3 mm Depth of rotor slot = 10.75 mm Length of each bar Lb = 90 + 30 +10 = 130 mm Resistance of each bar Rb = = 82.32 x 10-6 ohm Copper loss in bar = Sr x I2

b x Rb

= 28.91 W

•

•

•

(xvi) Friction and Windage losses :- 1) Reduce the friction and Windage losses, we have to change the fan design design for better cooling and quieter operation. 2) And use of better quality bearing and lubricating material.

by this we can optimize F&W losses to 20 W approx. (xvii) Stray load losses of 1hp motor is about 5 W as per data.

• LOSSES W Stator copper loss 85.51Rotor copper loss 33.84Iron loss 27.82Friction and Windage loss 20

Stray load loss 5 TOTAL 172.17 W

Comparison of parameters Sr No.

Parameter Standard motor Energy efficient motor

Ratings

1. Full load output 763.04W 737.79W2. Efficiency 73.2% 81.07%3. Power Factor 0.75 0.804. KVA input 1.39KVA 1.136KVA5. Full load line Current 1.85A 1.56A6. Specific Magnetic

Loading0.45 Wb/m2 0.4546 Wb/m2

7. Specific Electric Loading

23000A/m 23246A/m

8. Output co-efficient 108.73 111.01

Sr No.

Parameter Standard motor Energy efficient motor

Main Dimension1. Stator bore 0.08 m 0.08 m2. Gross iron length 0.07 m 0.09 m3. Net iron length 0.063 m 0.081 m4. Lamination Thickness 0.65 mm C5

coating0.5 mm C3 coating

Material

1. Silicon Content 1.1% 2.1%

2. Lamination CRNGO – M47Grade

CRNGO – M45Grade

3. Stator Copper Gauge 24swg/0.55mm 23.5swg/0.58mm

Contd.

Contd.

Sr No.

Parameter Standard motor Energy efficient motor

Stator1. Type of winding Single Layer Double Layer2. Flux per pole 1.97 Wb 2.56 Wb3. Turns per Phase 492 4404. Conductor per Slot 82 745. Conductor Resistance 24ohm/phase 22.9ohm/phase6. Length of mean turn 0.52 0.56

Rotor1. Length Of Air Gap 0.3 0.252. Outer Diameter 79.4mm 79.5mm3. Rotor bar current 167.35A 112A4. Current Density In

Bar4.32 A/mm2 4 A/mm2

5. End ring current 319.79A 249.68A

Contd. Sr No.

Parameter Standard motor Energy efficient motor

Losses1. Stator copper loss 144.05W 96.66W2. Rotor copper loss 49.04W 38.99W3. Iron loss 37.1W 27.82W4. Friction and Windage

loss41.25W 20W

5. Stray load loss 5.55W 5W6. Total 276.92W 172.6W7. Input = Output + losses 1026.14W 922.17W

Efficiency 73.26% 81.01%

Testing of standard motor• No load test (open circuit):

• Short-Circuit Test(blocked rotor test):

• High voltage withstand test: Applying 2 KV for 20 seconds, Phase to phase - OK Phase to Earth - OK

Full-load Test(Temperature rise test):

EFFICIENCY CALCULATION ACCORDING TO PRACTICAL TEST RESULT:-  R per phase = 24/2 = 12 ohm R75 = * Rph Where, T = ambient temperature = 30º C R75 = 14.03 ohm Fixed losses = No load losses – 3(No load current)2 * R75

= 136.2 – 3(1.17)2 * 14.03 = 78.58 W  Stator Copper Losses = 3 * (1.85)2 * 14.03 = 144.05 W  Rotor Copper Losses = (1040 – 78.58 – 144.05) * 0.06 = 49.04 W   Stray Losses = = 1040/200 = 5.25 W   Motor Output = 1040 – 5.25 – 49.04 – 144.05 – 78.58 = 763.04 W So, Efficiency η = * 100 % = 73.2 % 

Testing result of energy efficient motor

• NO LOAD TEST (Open circuit):-

SHORT CIRCUIT TEST (Blocked rotor test):-

• HIGH VOLTAGE WITHSTAND TEST:-  Applying 2KV for 20 seconds, Phase to phase – OK Phase to Earth – OK 

Volts. Amps. Watts N(rpm) Hz

415 0.85 61.1 1492 50

Volts Amps. Watts Torque

230 3.8 1022 1.204

PRACTICAL IMAGES

• EFFICIENCY CALCULATION ACCORDING TO PRACTICAL TEST RESULT:-

R per phase = 22.9/2 = 11.45 ohm R75 = * Rph Where, T = ambient temperature = 33º C R75 = 13.24 ohm Fixed losses = No load losses – 3(No load current)2 * R75

= 61.1 – 3(0.85)2 * 13.24 = 32.4 W  Stator Copper Losses = 3 * (1.56)2 * 13.24 = 96.66 W  Rotor Copper Losses = (909 – 96.66 – 32.4) * 0.05 = 38.99 W   Stray Losses = = 909/200 = 4.55 W   Motor Output = 909 – 4.55 – 39 – 96.66 – 32.4 = 736.45 W

So, Efficiency η = * 100 % = 81.01 % 

PAYBACK PERIOD CALCULATION: The efficiency of standard motor is 73.3% and EEM is 81.03%. The total losses in standard motor are 276.92 W and in EEM is 172.6 W. So, we compensate the losses = 104.32 W = 0.104 KW

If load cycle of motor is 8 hours, So, KWH saving = 0.834 units Consider industrial tariff = 7.16 Rs./unit Per day saving = 7.16*0.834 = 5.97 Rs. Now, Costing defined by Megha motors, Standard motors = 3040 Rs. Energy efficient motor = 3745 Rs. Extra cost =705 Rs. So, Payback period in days = Extra cost / per day savings = 705 / 5.97 = 119 Days So, payback period of motor is approximately 4 months.

Generally, motor life is about 10 years. Total saving in 10 year, except payback period Saving = 116*30*5.97 = Rs. 20,775.5

If small industry having motor units around 20, which has average rating of 1hp and if all motors are replaced by EEM then, Saving energy in a day’s = 0.834*20 = 16.68 units Energy saving in Rs. = 16.68*7.16 = 119.42 Rs. Per day And its average life 10 years, Total saving in 10 years = 119.42*116*30 = Rs. 4, 15,581.6

 

Future Work

• On theory basis all calculations are being completed of efficient motor, now for future work we will wind the stator and build the motor and carry out all the practical tests and the efficiency calculations.