Rajiv Gandhi Proudyogiki Vishwavidyalaya

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					 Management is required when there is a
 1. Crisis
 2. Regulation
 3. Competition
 4. Waste

Implementation of energy demand side management
  is to
1. Eliminate the waste;
2. Minimize the losses.
                   Why Important?

1. Energy Conservation and its Management has become a
   prime factor for the nation, society and individual due to
   high cost and non availability of energy.

2. Non awareness of latest technologies and methods, non-
   conventional energy sources and renewable energy
   sources and how to run the plant and equipment in an
   energy efficient manner.
             Our Concern
1. Though, energy losses in power distribution
   equipment including switch gear, transformer,
   transmission and distribution lines etc are also of
   our concern, but they are smaller amount as
   compared to losses in electrical motors used for any

2. Electrical motors when not selected properly can
   cause energy losses as high as 25% to 30%.

3. The motors designed in lower frame might not be
   properly selected and have lower margins when
   operated at abnormal operating conditions run with
   very low power factor and efficiency and result in to
   high losses and consumption
                      Electrical Drives
The general meaning of a Drive is the system, which is driven by
some energy. The source of energy may be any thing like wind,
water, oil, steam, solar or electricity etc.
When the source of energy is electricity, the drive is called Electric
Drive. In any drive system, we take some output in terms of energy
or work done.
Utilization of electricity for driving the mechanical system employs
the use of Electric Motors, which gives an output in terms of
Mechanical Energy. These electric motors are DC Motors,
Synchronous motors or Induction Motors.
Many industrial applications requiring rotating electric drives are
normally capable of speed control and often require an equipment to
attain a versatile and smooth speed control and make the motor to
operate at a desired specific speed torque characteristic. These
drives are characterized by the nature of speed torque characteristic
such as constant torque drives and constant power drives. These are
sometimes characterized by the type of motor used in the drive i.e.
dc and ac drives making use of dc and ac motors respectively.
                    Type of Drives

 The various types of electric drives used in industries
may be divided into three types:

 1. Individual motor drive;
 2. Group drive;
 3. Multi-motor drive.
                   Individual motor drive
In individual drive, a single electric motor is used to drive one
individual machine.

The machines can be placed in any desired position and can be
moved very easily. The machines can be built as an integral part of
the complete system, which results in a good appearance, cleanliness
and safety. For the purposes where constancy of speed and flexibility
of control is required, such as in paper mills and textile industry,
individual drive is essential.
                                Group Drive
By group drive is meant a drive in which a single electric motor
drives a line shaft by means of which an entire group of working
machines may be operated.
It is also sometimes called the line shaft drive. The line shaft is fitted with multi-
stepped pulleys and belts that connect these pulleys and the shafts of the driven
machines serve to vary their speed.

This drive is economical in consideration of the first cost of the
motors and control gear. A single motor of large capacity costs less
than the total cost of a number of small motors of the same total

The efficiency and power factor of a large group drive motor will be
higher, provided it is operated fairly 10% overload when being
driven by group drive.
   This form of drive has become obsolete now-a-days because
  of its following draw-backs and objectionable features, and
  the modern trend is to employ individual and multi-motor
 1.In group drive, speed control of individual machine is very cumbersome
  using stepped pulleys, belts etc.

2.Owing to use of line shafting pulleys and belts group drive does not give
  good appearance and is also less safe to operate.

3.In group drive since machines have to be installed to suit the layout of the
  line shafting, as such flexibility of layout of the various machines is lost.
  Also it is not possible to install any machine at a desired place.

 4.The possibility of installation of additional machines in an existing
  industry is limited.

5.If, at any time, all operations are not required, the main motor will work
  at low capacity and, therefore, operation efficiency will be low.
                   Multi-motor Drive
It consists of several individual drives each of which serves to
operate one of many working members or mechanisms in some
production unit.
Such drive is essential in complicated metal-cutting machine tools,
paper making machines, rolling mills, and similar types of
machinery. The use of multi-motor drive is continuously expanding
in modern industry as their advantages outweigh the increase in
capital cost as compared to the group drives.
The use of individual and multi-motor
drives has enable the introduction of
automation in production processes,
which in turn has considerably increased
the productivity of various industrial
organizations. Complete or partial
automation helps to operate various
mechanisms at optimum conditions and
to increase reliability and safety of
Various Types of Driven Equipments
1. Pump;
2. Fan;
3. Compressors;
4. Conveyers;
5. Mill;
6. Crusher;
7. Crane;
8. Hoist;
9. Traction;
Amongst motor used in any modern plant and industry,
the maximum number of motors used is 3-phase Induction
motor because of their cheapness, robust construction and
satisfactory performance. Their maintenance in service is
also easier as compared to other types of electrical motors.

The Torque – speed characteristic, inertia and duty cycle of load will
mainly determine the electrical characteristic and rating
requirements of the driving motor.
The proper selection of motor rating and design will result
in a minimum motor cost for a specified motor life
expectancy, torque-speed characteristic, inertia and duty
cycle of load.

The selection of under size motor for low motor cost
however may result in overloads and a consequent
reduction in motor life.
                                             Additionally, in actual
                                             practice, motors are
                                             subjected to abnormal
                                             operating conditions
                                             because of system
                                             inability to maintain the
                                             normal operating
                                             conditions when some
                                             accidents or faults take
Any form of abnormal operating conditions may affect the
performance, high losses and life expectancy of the motor. The
degree of deterioration depends upon the magnitude and nature of
abnormal conditions. Therefore, the abnormal operating condition
may also be taken in to account in design stage so that its withstand
capability is increased. Proper protection shall also be employed to
save motor against severe damage.
                    Energy Management
The energy management is the work for energy manager assigned by the management
who will exclusively look about energy matter such as

1. A detailed energy monitoring system.
2. Comparison of specific energy consumption values on a
   monthly and yearly basis.
3. Exploring possibility of improvement in energy

The Energy Audit is carried out to critically examine each of
the major energy consuming units to determine whether
there exists any unwanted use of energy, losses,
idle/redundant running etc. All efforts should be made to run
the machine at full/optimum capacity.
        Energy Conservation Plan
Specific energy consumption value is the index to determine
how effectively the plant and machinery are utilized in any
industrial process. The KWH/ton or KWH/unit of production is
calculated in each month and energy consumption indices
are worked out separately for major equipment and process.
These are then compared on a monthly and yearly basis
regularly to detect any deviation from the norm (targeted
value) and to take necessary correcting steps.

On identification of areas where electrical energy is not
efficiently utilized, remedial measures are to be taken to
either replace the old equipment with energy efficient or to
implement with energy efficient equipment or to implement
modifications to make them more energy – efficient.
 Implementation of energy conservation
The energy conservation measures include:

  1. Method of installation i.e. recycling (i.e. using scrap),
     retrofitting and changing process ( from existing to
     more efficient one);
  2. Method of heat use (i.e. installation of equipment for
     waste heat recovery and utilization, waste material
     utilization and process efficiency improvement);
  3. Changing the equipments with energy efficient i.e.
     energy efficient motors and drives;
  4. Improving the power factor of the system;
  5. Utilizing the energy during off peak period;
Energy conservation in Mechanical systems
  Various Mechanical drives are pumps, fans, compressors, mill,
  crushers, hoists, tractions system etc. In all these drives, there are
  mechanical losses like friction and windage losses.

1. In the bearings, friction loss occurs. In order to reduce the frictional loss the
   surface are made smooth and bearings are lubricated. The size and grade of
   bearing is decided based on the torque on shaft. Small and medium range
   drives are provided with ball and roller bearings while Higher capacity and
   higher speed drive are always provided with sleeve bearing with pedestals.
   These bearings are forced oil lubrication. The oil is cooled by water
   separately. The quantity of oil will depend on the heat developed in the
2. The pump and fan has the blades in rotation. While rotating, there is friction
   windage loss in fan or blower, which is minimized by proper angle of blade
   design. Similarly, the blade angle of pumps is designed to minimize the
   friction loss of water in the pump. Similarly, for compressor, the friction
   occurs because of movement of piston in the cylinder, the friction loss is
   minimized to have proper lubrication and gap between piston and cylinder.
   There may be other losses in pulley drive; belt drive and gear drive
   mechanism which is designed properly to have minimum mechanical losses.
       Energy conservation during design
Induction Motor represents the majority of all electric motors
used in industries world- wide. Therefore, there is
tremendous amount of energy consumed in operating these
The important design features worth mentioning are:

1. More copper section in stator and rotor thus reducing I2R loss;

2. Better quality of stampings and lamination to reduce the iron loss;

3. Enhanced cooling by adopting improved design of fans and ventilation

4. Designing for lower flux density for improving power factor;
A case study for 750KW, 6.6KV, 6 Pole Squirrel cage Induction motor which is optimally
design for the specification as given below:
Starting torque      – 90%
Starting current - 600%
Maximum torque - 230%
Temperature rise - 70 0 C
Out of various design, two close designs were selected with the following parameters as shown in Table

                 Description                     1st Design                   2nd Design
           1     Frame size                      1LA7 636-6                   1LA7 710-6
           2     Temperature rise (oC) 69.1                                   54.5
           3     Starting current                6.027                        6.12
           4     Starting torque                 1.037                        0.894
           5     Efficiency at 100% 93.33                                     95.95
           6     Power factor at 100% 0.827                                   0.864
           7     Cost of machine                 6.0 Lakhs                    8.0 Lakhs
           8     Total loss                      53.6 kW                      31.66 kW
1. From the above table, it can be seen that the efficiency and power
   factor of Frame 1LA7 710-6 is more than the frame 1LA7 636-6.
2. The saving in power loss by using 1LA7 710-6 motor is ( 53.6 – 31.66)
   kW = 21.94 kW. If the motors are running continuously, then the
   number of unit saved is 24x 21.94 = 526.56 kWh per day per motor or
   192194 kWh per year per motor.
3. Taking the unit charge as Rs 4 per unit; the total saving is Rs 192194 x
   4 = Rs 768776.
4. Let us take 15% interest and depreciation. The extra expenditure per
   year on the cost is 2x0.15 Lakh = Rs 30000. The pay back period is
   30000 x12/768776 months = 0.468. It is less than 15 days only. It may be
   noted that the working hours were taken 24 hours which is not
   realistic. Let us take 8 hrs per day. The amount saved per year is Rs
   768776x8/24 = 256259/-. Therefore payback period is 0.468x3 = 1.4
5. If, there are 1x105 ( 1 Lakhs) motors operating every day, we can save
   power 21.94 x 1x105 kW = 2194 MW.
                                                                                    Variable Speed Drives

                                                  Q-H characteristics of pump                    In pumps and fans using constant speed
                                    600                                                          motor with conventional control, variation
                                                                                                 in flow is achieved by means of throttling
      Pressure (H) in mm of water

                                                                                                 valve or damper as shown in Fig.
                                                                                                 The Power consumed
                                    300                                                    Hr1         P  Q H/ (m xp)
                                    200                                                          Where Q = Delivery of pump in m3/sec
                                                                                                          H = pressure head (m)
                                                                                                         m = motor efficiency

                                          0   1       2        3        4       5      6
                                                                                                         p = pump efficiency
                                                     Discharge(Q) in m3/sec                               s = m p

Q1 = 4; H1 = 200
Q2 = 3.6; H2 = 240
Let us assume that there is no change in system efficiency. Then
  P1            Q1 H1              4 x 200
          =                 =                 = 0.9259
  P2            Q2 H2             3.6 x 240

Therefore P2 = 1.08 P1
                                        Q-H Curve with variable speed motor            Fig.: Pump curve at different
                                                                                       speed with constant resistive
                                                                                       curve for lower quantity
Pressure (H) in mm of water

                              500                                                      Q1 = 4; H1 = 200
                                                                                       Q2 = 3.6; H2 = 160
                                                                                  H1   P1         4 x 200
                              300                                                 H2         =              = 1.39
                                                                                       P2         3.6 x 160
                                                                                            Therefore, P2 = 0.72 P1

                                    0    1       2        3        4     5    6
                                                Discharge(Q) in m3/sec

         It means, the power required by variable speed motor is only 72%. It is also
         possible to keep the system efficiency same as original one. Even if the system
         efficiency is down by 1%, the power required by pump system will be 72x85/84 =
         72.85%. It can be seen that there is an advantage of having the variable speed
         motor. We can also see that in case of still lower delivery, the power reduction in
         percentage will be more because of lower head.
Energy conservation by Adjustable speed control

1. DC Motor
DC motor with variable speed by rheostatic control in field and armature would give
speed control. Some losses also occur in the resistance. DC motor with Thyristor
converter is widely used for efficient and precise speed control in steel mills, Cement
Plant, paper and textile mills. With fully controlled bridge, regenerative braking can be
2. Induction Motor
In case of squirrel cage motor multispeed winding may help the operation at different
speed while speed change can be achieved with the help of addition of rotor resistance.
Energy conservation in Slip ring Induction Motor with Slip recovering is one of best
method. In conventional SRIM, introducing resistance in the rotor circuit varies motors
speed. Since power is absorbed by rotor resistance, efficiency of the motor drops as the
speed decreases. The power otherwise wasted in rotor resistance can be fed back to
the system by using a static converter, which converts the slip frequency power to DC
and then converting in to three phase which would be fed to main supply for controlling
the speed. This is fed back to the line with a line-commutated inverter. The variable
voltage and variable frequency is also adopted for variable speed in Induction motor.
 3. Synchronous motor
 The speed change is achieved by application of Variable frequency variable voltage.
 Fans – a case study
 To have similarity of measurements, two 210 MW units at Vijaywada Thermal power
 station of Andhra Pradesh State Electricity Board were selected for carrying out site
 measurements and subsequent energy consumption comparisons. Unit 3 & 4 are 210
 units and having tower type boilers supplied by the same source.
 Fixed speed induction motors drive unit 3 ID fans. The fan is coupled to induction
 motor through a hydraulic coupling so that the fan speed could be varied by scoop
 tube control.
 Unit 4 ID fans are driven by Variable Frequency Drive. The fan is coupled to the motor
 through a flexible coupling. A synchronous motor fed from Load Commutated Inverter
 (LCI) was used as a variable Frequency drive.

The study conclusively proves that introduction of Variable Frequency Drive system for
flow control application has a definite advantage in terms of substantial energy savings
as shown in Table.

Variable Frequency Drive has got the following advantages in addition to power savings:
(1) Increase in life of equipment due to soft start(2)Unlimited number of starts (3)
Assimilation of plant automation system for higher productivity.
                                      Unit 3               Unit 4

 Type of motor                        Sq. cage             Synchronous Motor
                                      Ind. Motor           Brushless
 Motor frame                          1LA7 902             1DQ3735
 Rating Kw                            1600                 1368
 Voltage                              6.6KV                1.2 Kv
 Frequency                            50 Hz                Up to 47.33 Hz
 Method of speed variation            Hydraulic coupling   VVVF
 Power consumption for the same load 2319 Kw               1845 Kw
 factor (207 MW)
 Power saving due to VFD at 207 Mw                         474 Kw
 Energy saving for Generation time                         2527 MwHr
 of 5333 Hr per year
 Total Energy saving in Rs per year                        Rs 81.62 (taking    Rs
                                                           1000/- per MWhr)

Taking the cost of Control Rs 3 to 4 Crores, Payback period
can be of the order of 4-5 years only.
Energy conservation by Improving the power factor
in case of Induction motor
For improving the power factor, there should be reduction in reactive power. For,
inductive load, leading reactive power is required and for capacitive load, lagging
reactive power is required.
I1 = current before pf improvement;
I2 = current after pf improvement;
1 = pf angle before pf improment;
2 = pf angle after pf improvement.
P = Power consumed = V I1Cos 1= V I2Cos 2


                       1 I2

The important disadvantages of low power factor are:

1.  Higher currents require larger size cable, switchgears,
   transformer and alternators etc. Thus the capital cost of
   the equipment is increased. This is, uneconomical from
   the supplier's point of view.
2. Higher currents give rise to higher copper losses in the
   system and therefore, the efficiency of the system is
   reduced. Also, the cost of energy loss (that is running
   cost) in the system increased.
3. Higher currents produce larger voltage drop in cables and
   other apparatus. This results in poor voltage regulation.
Optimization of Power factor correction when Power is same

If x is the annual cost per KVA of maximum demand then
annual saving in the KVA demand charges

 = x ( S1 – S2) = x P(Sec 1 – Sec 2)

If y is the annual cost per KVAr of the power factor
correction equipment then annual cost of the power factor
correction equipment.
CPF = y Qc = y P ( Tan1 – Tan 2)
The total annual saving ,

Cs = CD - CPF = x P(Sec 1 – Sec 2) - y P ( Tan1 – Tan 2)
Condition for optimization is

Sin 2 = y/x
   Optimization of Power factor correction when KVA demand is same
For the same kVA, the power can be conserved or productivity can be increased.

Let us assumed that z is the annual cost per KW of the installation, then the annual
saving due to increased power out would be
 = z S ( Cos 2 – Cos 1)

Let y is the annual cost per KVAr of the pf correction equipment then the annual cost of
the power factor correction equipment is given by
CPF = y Qc = y S (Sin 1 – Sin 2)

The net saving would be
Cs = z S ( Cos 2 – Cos 1) – y S ( Sin 1 – Sin 2)

For maximum annual saving

Tan 2 = y/Z
Power Factor Improvement Using Synchronous Condensers

When the KVAR requirement is small, it can be met through static capacitors.
   However when requirements exceed 10,000 KVAR, it is generally more
   economical to use the synchronous condensers.
A synchronous condenser is essentially an over excited synchronous motor.
   Generally, it does not supply any active mechanical power. The excitation of
   the machine is varied to provide the necessary amount of the leading KVAR.


1. By the use of synchronous condenser a finer control is possible than by the
    use of static capacitors.
2. A synchronous condenser can be overloaded for short periods but a static
    capacitor cannot be overloaded.
3. A momentary drop in voltage causes the synchronous condenser to supply
    greater KVAR to the system whereas in the case of static capacitor, the KVAR
    supplied is reduced.
4. The inertia of the synchronous condenser improves the system stability and
    reduces the effect of sudden changes in load.