Energy Efficiency Measures in Electrical Motors

الكهربائية المحركات المحركات الكهربائية نوعان, محركات

التيار المستمر و محركات التيار المتناوب.

,في األنشطة الصناعية و التجاريةتمثل النظم المدارة بالمحركات أكثر

من استهالك الكهرباء بهذه % 70من المنشآت.

كما و تمثل نظم المحركات الكهربائيةفي المنشآت الصناعية االستهالك

األكبر للطاقة الكهربائية يليها اإلنارة, بينما في المؤسسات التجارية و

التعليمية تمثل نظم اإلضاءة االستهالك األكبر يليها المحركات الكهربائية.

غالبا ما يتم استخدام المحركاتالكهربائية بدون النظر لكفاءتها حيث

يتم تركيبها و تشغيلها حتى يجد مشاكل في تشغيلها.

و قد تستخدم المحركات الكهربائيةبأحجام غير مناسبة أو قد تكون غير

مناسبة للتطبيق المستخدمة من أجله. المحركات الكهربائية أجهزة ذات كفاءة

عالية و لكن كفاءتها تقل مع الزمن خاصة إذا استخدمت في ظروف قاسية

أو تّم تشغيلها و صيانتها بطريقة غير صحيحة.

إذا كان جهد التغذية منخفض جدا و الحمل ثابت فإن التيار المار يكون عالي جدا و يحدث ارتفاع كبير في درجة الحرارة, إذا كان جهد المصدر كبير جدا يحدث إجهادات كهربائية على

العزل الكهربائي و قد تؤدي إلى تدمير ملفات المحرك

TYPES OF ACMOTORS

AlternatingCurrent Motors

ThreePhase

SinglePhase

InductionSynchronousInductionSynchronous

wound RotorSquirrel Rotorwound RotorSquirrel Rotor

التحريضية • المحركات هي أكثر المحركات الكهربائية استخداما في

مجال الصناعة و أكثرها انتشارا في الحياة العامة.

تعمل في حالة عامل استط[اعة متأخر و تكونقيمته منخفضة في حالة الالحمل و يزداد من

في[ حالة الحمل الكامل.% 90إلى % 80 يمكن الحصول من خالل المحركات التحريضية

ثالثية الطور على سرعة ثابتة و منتظ[مة, و فيها تغيرات الجهد و التردد تعط[ي تأثيرات مميزة على

خصائص تشغيل المحرك.

Motor Efficiency Parameters

Efficiency = Mech. Energy Delivered / Electrical Energy Input

Power Factor = Real Power (KW) / Apparent Power (KVA)

Induction Motor Efficiency

Power Out Shaft output

Power Input

Copper Loss

Iron Loss

Ventilation Loss

Friction Loss

Motor Losses

Efficiency=(Input-Losses)/Electrical Energy Input

A- No-load losses (fixed)1- friction losses2- Windage losses3- Core losses ; (Iron Losses)4- Heating losses (no load current)

Losses in Induction Motors

B- Load dependent losses (variable)

1- Resistive (I2R ) or copper losses2- Stray losses caused by components of stray flux ; they include:

• Saturation effects in rotor & stator iron• Leakage flux• Harmonics

Friction & Windage LossesIron Losses

0 40 80 120

12

9

6

3

0

Full

Lad

Pow

er L

oss (

%)

Power Loss in Induction Motors

Load (%)

I2R Losses

Stray Losses

Motor Efficiency

1. Stray Load Loss10% 2. Friction & Windage Losses 5% 3. Core Loss 20% 4. Rotor Loss 25%

5. Stator loss 40%

1 2 3

4

5

losses

Distribution of Losses in Induction Motor

Losses of a 5 HP, 4-pole, 3-phase std. motor, fully loaded

الكهربائية اآلالت في الطاقة كفاءة

مع زيادة الطلب على الطاقة الكهربائية و مع زيادة االعتبارات حول مصادر الطاقة و الكلفة

المرتبطة بذلك فإنه أتى بالمقابل اعتبار الكفاءة كعامل أساسي و مهم جدا, و المحركات الكهربائية هي من أكثر محوالت الطاقة كفاءة و

.الموجودة في االستخدام العام هذه األيامو يؤدي استخدام محركات الطاقة عالية

% من تكاليف 4-2 إلى وفر ( HEM’s)الكفاءة. (STD’s )تشغيل المحركات التقليدية

يمكن أن تستهلك المحركات الكهربائية قدرة زائدة في الحاالت التالية

المحاور (Bearing wear) احتكاك

المناسب غير )Part load( التحميل

التغذية جهود توازن Unbalance( عدمvoltage phases(

Motor Driven Loads

Types of Loads Served by Motor

• Constant speed, constant torque Constant volume air supply fans & circulation pumps

• Constant speed, Variable torque Sawmill blades, escalator motors

Variable speed, variable torque Offer largest potential for energy savings

Decreasing torque loads- vehicle.

Linear-law loads (Reciprocating Compressor, positive displacement pump) Shaft power varies linearly with shaft speed, reducing speed by ½ Cuts the rate of energy consumption by ½

Square-law loads (torque of such loads is prop to speed; power varies with the square of speed, conveyors, hoists,winches)

Variable speed, variable torque

Cube-law loads

The power such loads require increases with The tube of their speed. Doubling the speed will increase Power demand by 8 folds (2)3

(centrifugal fans, centrifugal pumps ) Conversely, reducing its speed by 20% resultsIn 50% drop in power requirement

Energy Conservation Measures in Motor

• Reduce load/switching off the motor• Reduce under loading• Sizing to variable load• Power factor correction• Energy saving starters• Reduce voltage imbalance• Reduce speed

• Flat belts • Energy efficient motors• Replacing multiple motors with one large motors

Energy Saving Opportunity in Common Motor Application

• Fan system• Compressed air• Refrigeration

• Pumping

Energy consumption by induction motors up to 300 kw in industry

Process & Conveyors 15%Other Compressor 14%

Pumps 32%

Air Compressor 8%Others 8%

Fans 23%

• Maintain the Pump

Pumping

• Reduce sharp bends

• Select Efficient Pump

• Use Variable speed drive (VSD)• Remodeling Piping System to Reduce

Friction- Loss and Leakage-Loss

Case study (Cement factory)

The water is cooled with two cooling towers and circulated by 8 pumps. Measurements were made of the water flows and pressures

and electrical inputs to the motors. The operation time 24h/day.

Pump no.3 study

Collected data: motor: TGL, pf=.86, 380V/660V, 141/81amp, 1480rpm,50 cycleMeasured data: motor: average values 391.4V, 95.9amp, pf=.8, rpm=1489Load by slip=.45 full load, load by amps=.585, output=.585 x 75kW=43.9kWMotor efficiency=43.9/52=84.3%

Pump data measured: 17 Sep 2002 flow 434, head estimated=15m, 18 Sep flow 378, head=15mWP=9.8 x .106 x 15=15.59kW (average flow=406cu. m/h) Pump efficiency=WP/motor output=15.59/43.9=35.5%

Wire to water efficiency=.843 x 355 x 100=30%

Replace pump and motor with new 95% efficient motor and 75% efficient pump.Existing pump and motor electricity use=52kW x 8760h/y=455520New pump and motor electricity use=21.9kW x 8760=191844Kwh/yKWh saved=263676kWh/y Cost saved=263676x2.4=632822 s.p Pump and motor cost $8000x50=400000 s.p ,Payback=400000/632822=0.63 y = 0.63x12=7.5 month.

Fan system

• Maintain the Fan

• Reduce sharp bends

• Select Efficient Fan

• Use Variable speed drive (VSD)

 Case study

Replace cooling tower fan motorThe existing cooling tower fan motor is rated at 75kW. Measured test show the load on the motor to be 29.25kW, with an efficiency of 78%. Existing electricity used=29.25kW x 8760h/y=256,230kWh.yr A new 40kW (92%) motor will save=(1/0.78-1/0.92) x 29.25kW x 8760h/y=35,872kWh/y

Cost saved=35,872 x 2.4=86092 s.p=$1793, Motor cost=$3500 x50=175000s.p

Payback=175000/86092=2y

Compressed air

• Air leakage losses• Pressure reduction• Piping resistance losses• Waste heat recovery• Control systems• Miss use of air• Maintenance

Case study

The existing compressed air system consists of a central system with six compressors. Potential savings of 5% - 10 % due to elimination of leaks and waste are possible. The compressors operate with 1-1120Kw, 2-720Kw,3-720Kw compressors running 24h/d 365day/yThe compressors were checked to determine the load on the motors.It is estimated that savings of 10% could be achieved by stopping the waste of air and repairing leaks.Measurements were made to determine the load on each compressor. The compressors are operating at 34% to 51% to .42% of full load.

This would save: (1120kWx.34+720Kwx.51+720kWx.42)x8760h/y x.10 =920,150 kW/yCost saved: 920,150 x 2.4=2208360 s.p/y,

Cost to stop leaks estimated to be=1250000 s.p

Payback=1250000/2208360=.56y

Refrigeration

• Maintenance

• Reducing load by using good insulations• Use the correct size of the motor (not using over sizing motor)

•100

•95

•90

•85

•80

•75•0.75 •1.1 •3.7 •5.5 •11 •22 •30 •55 •75 •110

•Motor Rating (kW )

•H.E.M.

•Std. Motor

• Eff

icie

ncy

(% )

at 8

0% L

oadi

ng

Comparison of HEM and Standard Motor Efficiencies

Large motorsHEM 2-4% more efficient than std.

Motors less than 5.5 kWHEM 4-7% more efficient than std.

•Power factor

•Efficiency

0 25 50 75 100 125Eff

icie

ncy

& P

ower

Fac

tor

(%)

01020304050607080

90

Load (%)

MOTOR OVER-SIZING

Reasons for over-sizing 1. Uncertainty in motor loading

2. Safety margin

Outcomes: 1. Higher motor purchase price 2. Poor PF 3. Higher running costs due to running on low efficiency 4. Increased cost of initial electrical supply equipment

VOLTAGE SUPPLY &MOTOR PERFORMANCE

Voltage Imbalance· Overloading certain phases resulting in loss increase.· Increasing motor vibration and shorten motor life.

Over-Voltage 10 to 15% over-voltage reduce significantly efficiency & PF

Under Voltage Reduction of voltage below 10% increases copper losses

ESTIMATING MOTOR EFFICIENCY

Efficiency.% = 100 * Motor Output (kW) Motor Input (kW)

Output (kW) = Motor loading %*Nameplate kW

Loading% = 100 * [RPMsync - RPMrotor] [RPMsync - RPMnameplate]

Example

The synchronous speed of a 45 kW motor is 1,500. Using a tachometer and kW meter, the rotor rotation speed and motor input power was found to be rpmroto = 1,463 and 38.5 kW respectively. The motor nameplate rotation speed is rpmnameplate = 1,450.

What is the estimated efficiency of the motor at the given loading conditions?

Motor loading% = 100 * [rpmsync - rpmrotor] / [rpmsync - rpmnameplate]= 100 * [1,500 - 1,463] / [1,500 - 1450]= 74%

Motor power output (kW) = Motor loading % * Nameplate kW= 74% * 45 kW= 33.3 kW

Motor Efficiency % = 100 * Motor Power Output / Motor Power Input= 100 * 33.3 kW/38.5 kW= 86.5%

% Rated Load

Effic

ienc

y %

0

10

20

30

40

50

60

70

80

90

100

0 25 50 75 100 125 150 175

75 kW7.5 kW0.75 kW

Motor Efficiency and Loading

Motor loading plays a key role in motor efficiency. Therefore, it is important that the motor efficiency is to be associated with the motor loading conditions.

REWINDING ELECTRIC MOTORS

Rewinding reduces motor efficiency due to:

i. Excessive heating (during fault and rewinding)

ii. Using a thinner gauge wire.

Rewound motor efficiency is reduced by:1.5 - 2.5%

POLICY FOR USING H.E.Ms

1. Motors operating more than 3,000 hrs/year are good candidates for replacing with HEM when they fail.2. Use of HEM with new equipment should be evaluated.3. Failed small Motors (<10 kW) are more economical to be replaced with new one.4. Rewinding or replacement of failed large Motors should be economically evaluated.

5. For centrifugal pump and fans use HEM with

same speed as the original one.

6. Operating Std. motors should not be replaced

with HEM unless economical is viable

Thanks for joining us

MSEE. Nazeh Tannous •E-Mail: syrlove2001@excite.com