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NATIONAL CERTIFICATION EXAMINATION

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                            Model Question Paper - 2006

                   NATIONAL CERTIFICATION EXAMINATION
                                    FOR
                   ENERGY MANAGERS AND ENERGY AUDITORS

            PAPER –3 : ENERGY EFFICIENCY IN ELECTRICAL UTILITIES

 Section – 1 : OBJECTIVE TYPE



1.    Power generating capacity in India, from coal based thermal power plants is
      about –
      b) 70%
2.    Power loss in a line is the product of -
      a) Resistance and Square of the current
3.    Power factor is called as – of the angle between kW and kVA
       a) Cosine
4.    If the fundamental frequency of the power system is 50 Hz, 5th harmonic will
      have a frequency of -- Hz
      b) 250
5.    MD recorder is recording ---
      b) time-integrated demand over a predefined recording cycle
6.    For a synchronous speed of 1500 rpm, at a given mains frequency of 50 Hz, the
      motor will have the number of poles as ---
      b) 4
7.    Squirrel cage motors are normally --- efficient than Slip ring motors
      a) more
8.    Core losses in a motor accounts for --- of its total losses
      b) 20-25%
9.    For efficient overall operation in a Plant, it is desirable to have –
      a) High value of efficiency and high power factor
10.   Fixed losses in a motor consist of ---
      a) Magnetic core loss, friction and Windage losses



11.   For every 250 mm Wc pressure drop increase across at the suction path due to
      choked filter etc., the compressor power consumption increases by about ----
      percent for the same output
      b) 2
12.   The likely estimate on equivalent power wastage for a leakage from 7 bar
      compressed air system through 1.6 mm orifice size is ---kW
      c) 0.8
13.   Vertical type reciprocating compressors are used in the capacity range of ---
      cfm
      a) 50-150


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14.   The compression ratios for axial flow compressors are ---
      b) higher
15.   The temperature at which moisture condenses is called as ---
      b) Dew point
16.   The driving force for refrigeration in vapour absorption refrigeration plants is –
      b) mechanical energy
17.   In general, designed chill water temperature drop across the chillers is ---0 C
      a) 5
18.   The compressor that has recently become practical in the Market is—
      compressor
      b) Scroll
19.   Screw compressor is also called as – compressor
      d) helical rotary
20.   The output of a reciprocating compressor is ---
      a) nearly constant
21.   Specific ratio is used to define fans, blowers and compressors as per ---
      c) ASME
22.   System curve of a fan depicts the relation between ---
      a) air flow and pressure
23.   Axial flow fans are equipped with ---blades
      a) variable pitch
24.   Peak efficiency of a radial fan is ---
      c) 69-75%
25.   The most efficient axial fan is of – type
      c) vane axial
26.   Frictional losses in a pumping system is proportional to –
      b) Q-2
27.   The intersection point of the pump curve and the system curve is called ---
      b) best efficiency point
28.   Increased suction lift from open wells, the delivery flow rate ---
      b) decreases
29.   For large capacity of centrifugal pumps, design efficiencies are in the range of
      around ---
      a) 85%
30.   Throttling the delivery valve of a pump results in increased ---
      c) both (a) and (b)
31.   Natural draft cooling towers are mainly used in –
      b) Power stations
32.   Cooling capacity of cooling tower is ---
      a) the heat rejected
33.   Which one of the following has the maximum effect on cooling tower
      performance?
      c) fill media
34.   This fill material is more energy efficient
      b) film fill
35.   Some of the components of cooling tower are:


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       c) both (a) and (b)
36.    Colour rending index is measured in the scale of ---
       a) 1-100
37.    Which of the following lighting source has the least life?
       a) incandescent
38.    The average rated life of CFL is – hr
       b) 10,000
39.    This device aids initial voltage build-up for a Tube light
       a) Choke
40.    For better street lighting, ---type of lamp is recommended
       a) HPSV
41.    Power requirement of DG set is determined by:
       b) minimum load
42.    Designed power factor of a DG set is –
       b) 0.8
43.    The radiator cooling temperature for DG sets should be in the range of – 0 C
       d) 80-90
44.    High pressure gas injection is resorted to in the case of ---
       b) slow speed dual-fuel engines
45.    Water-cooled DG sets are ---air- cooled DG sets in performance
       c) more efficient
46.    Hydrodynamic principle of speed control is used in ---
       b) Fluid coupling
47.    MD controller is used to switch off ---loads, in logical sequence
       b) non-essential
48.    Identify the ―odd one‖ out from the following:
       d) capacitor based sensor
49.    Eddy current Drive is used to ---
       b) vary the output speed
50.       Amorphous core transformers are available at present up to –KVA capacity
        d) 1600



             Section ii : SHORT DESCRIPTIVE QUESTIONS 10 X 5 =50 MARKS



S-1 “One Unit saved in Industry is equal to Two Units generated in the Power Station”-
      Justify this Statement.

The generated power from the Power station is transmitted and distributed to consumers.
Though the T & D losses are said to be around 17% in India and hence efficiency being 87%, all
these losses may not constitute technical losses as, un-metered consumption and pilferage are
also accounted in this loss.
Before the generated power reaches Industry, it passes thro‘ transformers having 95%
efficiency and in Industry the motor efficiency is about 90%. Another 30% is lost in mechanical
system that includes coupling/driving system and driven equipments like pumps, valves etc. Thus
the overall efficiency becomes 0.83 x 0.95 x 0.9 x 0.70 = 0.50 i.e. 50% efficiency.



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Hence, one unit saved in the Industry is equal to two units generated in the Power station.

S-2 What is an energy-efficient motor?

An energy-efficient motor is the one in which, design improvements are incorporated to
increase operating efficiency over motors of standard design. These improvements are in
reducing motor losses, use of lower-loss silicon-steel, a longer core, thicker wires to reduce
resistance, thinner laminations, smaller air gap between stator and rotor, copper instead of
aluminium bars in the rotor, superior bearings and a smaller fan. As per the stipulations of BIS,
energy-efficient motors are designed to operate with out loss in efficiency at loads between
75% and 100% of rated capacity.

S-3 What are the major compressed air system components?

Compressed air system consists of the major components as stated below:
      Intake Air Filters, preventing dust entering compressor
      Inter-stage Coolers, reducing the temperature of air before it enters the next stage
        to reduce the work of compression and increase efficiency
      After Coolers, removing moisture in the air by reducing the temperature in a water-
        cooled heat exchanger
      Air-dryers, removing the remaining traces of moisture
      Moisture Drain Traps, used for the removal of moisture in the compressed air
      Receivers, provided to store and smoothen the pulsating air output.

S-4 What are the several heat transfer loops in a refrigeration system?

The several heat transfer loops in a refrigeration system are:
    Indoor air loop: indoor air is driven in the leftmost loop by the supply air fan through a
       cooling coil, where it transfers its heat to chilled water. The cool air then cools the
       building
    Chilled water loop: It is driven by the chilled water pump and water returns from the
       cooling coil to the chiller‘s evaporator to be re-cooled
    Refrigerant loop: Using a phase-change refrigerant, the chiller‘s compressor pumps heat
       from the chilled water to the condenser water
    Condenser water loop: Water absorbs heat from the chiller‘s condenser and the
       condenser water pump sends it to the cooling tower
    Cooling tower loop: The cooling tower‘s fan drives air across an open flow of the hot
       condenser water, transferring the heat to the outdoors.



   S-5 What are the design and selection criteria in respect of fan?

   Accurate determination of air-flow and required outlet pressure are most important
   in the proper selection of fan type and size. The required air flow depends on the process
   requirements, normally determined from heat transfer rates or combustion air or flue gas
   quantity to be handled. System pressure requirement is to be computed, though it is
   difficult. Pressure drop across the length, bends, contractions and expansions in the ducting
   system, pressure drop across filters, drop in branch lines etc. A very conservative approach
   is adopted allocating large safety margins, resulting in over-sizing of fans, operating at flow
   rates much below their design values and consequently of poor efficiency.




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S-6 What are the affinity laws relating to a rotodynamic pump performance?

The equations relating to rotodynamic pump performance of flow, head, and power absorbed,
to speed are known as ―Affinity Laws‖. They are:
1. Q is proportional to N
2. H is proportional to N2
3. P is proportional to N3
Where,
Q = Flow rate
H = Head
P = Power absorbed
N = Rotating speed

In other words,
Flow is proportional to speed
Head is proportional to the square of speed
Power is proportional to the cube of speed

S-7 What is meant by Cycles of concentration and how is it related to cooling tower
Blow down?

Cycles of Concentration (COC) is the ratio of dissolved solids in circulating water to the
dissolved solids in make up water.
Blow down losses depend up on Cycles of concentration and evaporation losses and is related
as:
Blow down = Evaporation loss / (COC – 1)

S-9 What is „lamp efficacy‟? How savings in Industrial sector can be achieved by using
high efficiency lamps?

Lamp efficiency is the ratio of light output in lumens to power input to lamps in watts.
High efficacy gas discharge lamps which are suitable for different types of applications,
offer appreciable scope for energy conservation.
In Industrial sector, instead of HPMV lamps of 250 W low efficacy, if changed to HPSV
lamps of 150 W, there will be a saving of 100W in power saving will be 37%.
For GLS lamps of 13 W existing, 9W CFL lamps, if used saves 31% power.

S-9 Briefly explain two important factors, to decide the type of DG set.

The two most important factors to decide the type of DG set are: power and speed of the
engine.
The power requirement is determined by the maximum load. The engine power rating should
be 10-20% more than the power demand by the end use.
Speed of the engine should be such of a value that the fuel efficiency at that speed is the
greatest. The DG sets should be run at this rated speed to avoid poor efficiency and to
prevent build up of engine deposits due to incomplete combustion, leading to higher
maintenance costs.

S-10 Why Amorphous core transformers are preferred over conventional transformers?




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   Amorphous core transformers provide excellent opportunities to conserve energy, due to
   low core loss. The reduction in energy loss over the conventional transformers is roughly
   around 70%. Amorphous core transformers are of less weight and hence easier to handle.



Section-iii: LONG DESCRIPTIVE QUESTIONS Marks: 5 x 10 =50


L-1.   The maximum demand approved by a utility is 5500 kVA and tariff provides for
minimum billing demand of 80% of approved. Review of past 12 months records of bills
reveals that the monthly maximum demand recorded is around 4200 kVA.

Will there be any benefits in surrendering part of contract demand? If so what is the kVA
that you recommend for surrendering?

Give the costs savings by surrendering demand, if unit rate for kVA is Rs. 200.

ANS: The maximum demand approved by a utility is 5500 kVA and tariff provides for minimum
billing demand of 80% of approved. Review of past 12 months records of bills reveals that the
monthly maximum demand recorded is around 4200 kVA.

Will there be any benefits in surrendering part of contract demand? If so what is the kVA that
you recommend for surrendering?

Give the costs savings by surrendering demand, if unit rate for kVA demand is Rs 200.



Approved maximum demand                  : 5500 kVA
Minimum billing demand (MBD) : 4400 kVA
Maximum demand recorded                  : 4200 kVA

The approved maximum demand is selected such a way that the gap between MD recorded and
minimum billing demand are narrowed down.

       A different scenario is to be created.

 Contract demand                  5500       5400     5300     5200
 (Minimum Billing Demand) 80% of contract demand 4400          4320 4240     4116
 Monthly demand recorded          4200       4200     4200     4200

From above, it would be advantage to surrender ‗300 kVA‘ demand and set new approved demand
at 5200 kVA.

The actual savings possible is calculated below:

New contract (approved) demand           :5200 kVA

New minimum billing demand               :4160kVA)

Existing MD recorded                     :4200kVA

Present billing demand                   :4200 (since recorded kVA is more than MBD)



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Original minimum billing demand        : 4400kVA (Before surrendering demand)

Reduction demand value taken for billing: 200 kVA

Savings in demand charges/month        Rs 200 * 200kVA

                                       Rs40,000



L-2. What are the factors to be considered while selecting a motor? Explain in detail



ANS:

A.     Torque Requirement

The primary consideration defining the motor choice for any particular application is the torque
required by the load. The relationship between the maximum torque generated by the motor
(break-down torque) and the torque requirements for start-up (locked rotor torque) and during
acceleration periods is very important. The thermal loading on the motor is determined by the
duty/load cycle. One important consideration with totally enclosed fan cooled (TEFC) motors is
that the cooling may be insufficient when the motor is operated at speeds lower than its rated
speed.

B.     Sizing to Variable Load

Industrial motors frequently operate under varying load conditions due to process requirements.
A common practice in cases where such variable loads are found is to select a motor based on
the highest anticipated load. In many instances, an alternative approach is typically less costly,
more efficient and provides equally satisfactory operation. With this approach, the optimum
rating for the motor is selected on the basis of the load duration curve for the particular
application. Thus, rather than selecting a motor of high rather than selecting a motor of high
rating that would operate at full capacity for only a short period, a motor would be selected
with a rating slightly lower than the peak anticipated load and would operate at overload for a
short period of time. Since operating within the thermal capacity of the motor insulation is of
greatest concern in a motor operating at higher than its rated load, the motor rating is selected
as that which would result in the same temperature rise under continuous full-load operation as
the weighted average temperature rise over the actual operating cycle.


L-3. List out different basic capacity control methods for the fans and blowers



Different basic capacity (volume) control methods adopted in fans and blowers are as follows:

       1.   Changing the rotational speed is the most efficient. If the volume requirement is
            constant, it can be achieved by selecting appropriate pulley sizes. If the volume
            varies with the process, adjustable-speed drives can be used.

       2. Changing the blade angle is a method used with some vane-axial fans.




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        3. Restricting the air flow is accomplished with dampers or valves which close off the
           air flow at the inlet or outlet. Inlet vanes, which swirl the air entering the
           centrifugal fan or blower, are more efficient than dampers or butterfly valves.

        4. Venting the high-pressure air, or recirculating it to the inlet, is often used with
           positive-displacement blowers. It is sometimes used with fan systems, but is the
           least efficient method as there is no reduction in the air being moved.




L-4. Explain the methodology of refrigeration plant energy audit.




ANS: The cooling effect produced is quantified as tons of refrigeration.

1 ton of refrigeration = 3024 kCal/hr heat rejected.

The specific power consumption kW/TR is a useful indicator of the performance of
refrigeration system. By measuring refrigeration duty performed in TR and the Kilo Watt
inputs measured, kW/TR is used as a reference energy performance indicator.

The refrigeration TR is assessed as TR = Q Cp  (Ti – To) / 3024

Where           Q is mass flow rate of coolant in kg/hr

Cp is coolant specific heat in kCal /kg deg C

Ti is inlet, temperature of coolant to evaporator (chiller) in 0C

To is outlet temperature of coolant from evaporator (chiller) in 0C.

The above TR is also called as chiller tonnage. In a centralized chilled water system, apart from
the compressor unit, power is also consumed by the chilled water (secondary) coolant pump as
well condenser water (for heat rejection to cooling tower) pump and cooling tower fan in the
cooling tower fan. Effectively, the overall energy consumption would be towards ;

           Compressor kW

           Chilled water pump kW

           Condenser water pump kW

           Cooling tower fan kW, for induced / forced draft towers

The specific power consumption for certain TR output would therefore have to include :

           Compressor kW/TR

           Chilled water pump kW/TR

           Condenser water pump kW/TR




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             Cooling tower fan kW/TR

and overall kW/TR as a sum of the above.

In case of air conditioning units, the air flow at the Fan Coil Units (FCU) or the Air Handling
Units (AHU) can be measured with an anemometer. Dry bulb and wet bulb temperatures are
measured at the inlet and outlet of AHU or the FCU and the refrigeration load in TR is assessed
as:

                                              Q  ρ  h in  h out 
                                      TR 
                                                     3024

Where Q is the air flow in CMH

         is density of air kg/m3

h in      is enthalpy of inlet air kCal/kg

h out     is enthalpy of outlet air kCal/kg

Use of handy psychometric charts can help to calculate h in and hout from dry bulb, wet bulb
temperature values which are, in-turn measured, during trials,by a whirling psychrometer.




L-5. Describe the step by step methodology of lighting system audit in a plant?




ANS:

Improvement option in lighting at any facility would involve following step by step approach:

a.     Step-1: Inventorise the lighting system elements with respect to device rating, population &
       use profile.

b. Step-2: Measure and document lux levels at various plan locations at working place, at day
   time and night times w.r.t lamps ON or OFF during the said period.

c.     Step-3: Use a portable load analyzer to measure and document the voltage and power
       consumption profile at various lighting load distribution panels

d. Step-4: Compare the measured lux values with standard values and identify locations of
   under-lit and over-lit areas

e.     Step-5: Analysis of failure rates of lamps, ballasts and actual life expectancy levels from
       the past data.

f.     Step-6: Based on above careful assessment has to be carried out along with energy saving
       potential, investment required and payback calculations.




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