NATIONAL CERTIFICATION EXAMINATION 2004 - DOC

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```					                                     Paper EA3 – Energy Auditor – Set A Solutions

NATIONAL CERTIFICATION EXAMINATION 2004
FOR
ENERGY AUDITORS

PAPER – EA3:        Energy Efficiency in Electrical Utilities

Date: 23.05.2004        Timings: 0930-1230 HRS            Duration: 3 HRS        Max. Marks: 150

General instructions:
o      Please check that this question paper contains 7 printed pages
o      Please check that this question paper contains 65 questions
o      The question paper is divided into three sections
o      All questions in all three sections are compulsory
o      All parts of a question should be answered at one place

Section – I: OBJECTIVE TYPE                                                             Marks: 50 x 1 = 50

(ii)    Each question carries one mark
(iii)   Put a () tick mark in the appropriate box in the answer book

1.         If the voltage level of the electricity distribution system is raised from 11 kV to 33 kV
for the same loading conditions, the distribution losses are reduced by a factor of

a) 1/9               b) 1/3            c) 1/6                 d) none of the above
2.         In electricity distribution, if the voltage is raised from 11 kV to 33 kV for the same
loading conditions, the voltage drop in the distribution system would be lower by a
factor of

a) 1/4               b) 1/2             c) 1/3                d) none of the above
3.         If the reactive power drawn by a particular load is zero, it means the load is operating
at

a) lagging power factor          b) leading power factor
c) unity power factor              d) none of the above
4.         Select the location of installing capacitor bank, which will reduce the electricity
distribution losses to the maximum extent

a) main sub-station bus bars                b) motor terminals
c) motor control centre                     d) distribution board bus bars
5.         A pure inductive load draws

a) leading reactive power                        b) active power
c) lagging reactive power                        d) none of the above

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Bureau of Energy Efficiency
Paper EA3 – Energy Auditor – Set A Solutions

6.     The nearest kVAr compensation required for improving the power factor of a 100 kW
load from 0.8 lag to unity power factor is

a) 50 kVAr         b) 75 kVAr             c) 100 kVAr               d) none of the above
7.     The percentage increase in power consumption of a compressor with suction side air
filter pressure drop of 250 mmWC is closest to

a) 0.5%            b) 2%                  c) 3%                     d) 4%
8.     A power factor capacitor designed for 10 kVAR at 415 V was found to be operating at
400 V. The effective capacity of the capacitor would be

a) 9.3 kVAR          b) 10 kVAR             c) 10.8 kVAR            d) none of the above
9.     A four pole induction motor operating at 50 Hz, with 1% slip will run at an actual
speed of

a) 1500 RPM          b) 1515 RPM            c) 1485 RPM         d) none of the above
10.    With decrease in design speed of induction motors the required capacitive kVAr for
reactive power compensation for the same capacity range will

a) increase          b) decrease          c) not change             d) none of the above
11.    kW rating indicated on the name plate of an induction motor indicates

a) rated input of the motor                      b) rated output of the motor
c) maximum input power which the motor can draw
d) maximum instantaneous input power of the motor
12.    For every 4°C reduction in the air inlet temperature of an air compressor, the power
consumption will normally decrease by….. percentage points for the same output.

a) 1              b) 2             c) 3              d) 4
13.    The acceptable pressure drop in mains header at the farthest point of an industrial
compressed air network is

a) 0.3 bar        b) 0.5 bar       c) 1.0 bar        d) 2 bar
14.    PF capacitor installed at the motor starter location will improve

a) motor design power factor
b) motor operating power factor from the starter to the power supply side
c) motor operating power factor from the starter to the motor terminals side
d) all of the above.
seconds respectively during a compressed air leakage test. The air leakage in the
compressed air system would be

a) 20.3 cfm              b) 42.1 cfm              c) 66.6 cfm         d) 132.8 cfm
16.    Higher chiller COP can be achieved with

a) lower evaporator temperature and higher condensing temperature
b) lower evaporator temperature and lower condensing temperature
c) higher evaporator temperature and higher condensing temperature
d) none of the above

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Bureau of Energy Efficiency
Paper EA3 – Energy Auditor – Set A Solutions

17.    Vertical type reciprocating compressors are used in the capacity range of

a) 50 – 150 cfm                b) 200 – 500 cfm
c) 500 - 1000 cfm              d) above 1000 cfm
18.    One ton of refrigeration (TR) is equal to

a) 3.51 kW           b) 3024 kcal/hr          c) 12,000 BTU/hr          d) all of the above
19.    Approximate percentage reduction in power consumption with 1 °C rise in evaporator
temperature in refrigerating systems is

a) 1%               b) 2%                    c) 3%                 d) 4%
20.    The refrigerant used in vapour absorption systems is

a) steam            b) pure water            c) freon                d) lithium bromide
21.    Li – Br water absorption refrigeration systems have a COP in the range of

a) 0.4 – 0.5        b) 0.65 – 0.70           c) 0.75 – 0.80          d) none of the above
22.    Slip ring induction motors, in general, have a …… design efficiency in comparison
with the squirrel cage induction motors for similar ratings

a) lower              b) higher             c) same              d) none of the above
23.    System resistance in water pumping system varies with

a) square of flow rate                        b) cube of flow rate
c) square root of flow rate                   d) none of the above
24.    The outer tube connection of the pitot tube is used to measure …… in the fan system

a) static pressure            b) velocity pressure
c) total pressure             d) dynamic pressure
25.    If the speed of a pump is doubled, pump shaft power goes up by

a) 2 times           b) 6 times               c) 8 times              d) 4 times
26.    If the speed of a pump is doubled, the pump head goes up by

a) 4 times            b) 2 times              c) 8 times               d) 16 times
27.    Friction loss in a piping system carrying fluid is proportional to

a) fluid flow   b) (fluid flow)
2                  1        c)           d)         1
2
fluid flow                      (fluid flow)
28.    Shaft power of the motor driving a pump is 30 kW. The motor efficiency is 0.9 and
pump efficiency is 0.6. The power transmitted to the water is

a) 16.2 kW              b) 18.0 kW           c) 27.0 kW            d) none of the above

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Bureau of Energy Efficiency
Paper EA3 – Energy Auditor – Set A Solutions

29.    For fans, the relation between flow discharge Q and speed N is
2                       3
Q1   N1                     Q1 N 1                Q1 N 1
a)      =               b)         =              c)     =                d) none of the above
Q2 N 2                      Q2 N22                Q2 N 23
o          o
30.    If inlet and outlet water temperatures of a cooling tower are 40 C and 32 C
o        o
respectively and atmospheric DBT and WBT are 35 C and 28 C respectively then
the approach of cooling tower is
o                       o                           o                     o
a) 3 C                b) 4 C                      c) 5 C                    d) 7 C
31.    Cooling tower effectiveness is

a) approach / (range + approach)              b) range/ (range + approach)
c) approach / range                           d) none of the above
32.    The lowest theoretical temperature to which water can be cooled in a cooling tower is

a) DBT of the atmospheric air                  b) WBT of the atmospheric air
c) average DBT and WBT of the atmospheric air
d) difference between DBT and WBT of the atmospheric air
33.    Which of the following ambient conditions will evaporate maximum amount of water
in a cooling tower
o             o                          o                  o
a) 35 C DBT and 25 C WBT                      b) 40 C DBT and 38 C WBT
o            o                                o            o
c) 35 C DBT and 28 C WBT                     d) 38 C DBT and 37 C WBT
34.    In general, design chilled water temperature drop across chillers is approximately
o                           o                           o                 o
a) 5 C                  b) 1 C                    c) 10 C                 d) 15 C
35.    Normally a manufacturer’s guaranteed best approach of a cooling tower is
o                               o                 o                        o
a) 5 C                  b) 12 C                  c) 8 C                   d) 2.8 C
36.    GLS lamp is

a) general lighting service lamp                  b) general lighting source lamp
c) glow light source lamp                      d) glow light service lamp
37.    The unit of illuminance is

a) lux            b) luminaire               c) lumens                   d) none of the above
38.    Luminous efficacy of which of the following is the highest?

a) CFL             b) HPMV                    c) HPSV                     d) LPSV
39.    If voltage is reduced from 230 V to 200 V for a fluorescent tube light, it will result in

a) reduced power consumption b) increased power consumption
c) increased light levels   d) no change in power consumption and light levels
40.    What is the typical frequency of a high frequency electronic ballast?

a) 50 Hz            b) 10 kHz                c) 30 kHz                   d) 50 kHz

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Bureau of Energy Efficiency
Paper EA3 – Energy Auditor – Set A Solutions

41.    The compression ratio in diesel engines is in the range of

a) 5:1 to 10:1        b) 10:1 to 13:1         c) 14:1 to 25:1       d) none of the above
42.    The rated efficiency of a diesel generator captive power plant has a range of

a) 43% – 45%             b) 50% – 60%              c) 60% – 70%            d) above 70%
43.    The maximum unbalanced load between phases should not exceed …… % of the
capacity of the DG set

a) 1                  b) 5                c) 10                    d) none of the above
44.    The exhaust gas waste heat recovery potential of a turbo charged genset at 500 kW
o
o
(Assume exit gas temperature of 180 C and 8 kg gas/ kWh generated)

a) 1.6 lakh kCal/hr     b) 2.2 lakh kCal/hr       c) 3.0 lakh kCal/hr   d) 3.5 lakh
kCal/hr
45.    The operating efficiency of a DG set also depends on

a) turbo charger performance            b) inlet air temperature
46.    The core losses of a transformer are the least if the core is made up of

a) silicon alloyed iron (grain oriented)                     b) copper
c) amorphous core – metallic glass alloy                     d) none of the above
47.    The basic functions of an electronic ballast fitted to a fluorescent tube light exclude
one of the following

a) to ignite the lamp                                b) to stabilize the gas discharge
c) to supply power to the lamp at supply frequency
d) to supply power to the lamp at very high frequency
48.    Modern electronic soft starters are used for motors to

a) achieve variable speed                    b) provide smooth start and stop
49.    The nearest kVA rating required for a DG set with 1000 kW connected load, with
diversity factor of 1.5 and 84% loading and 0.8 power factor is

a) 500 kVA             b) 1000 kVA
c) 1500 kVA            d) 2000 kVA
50.    Maximum demand controller is used to

a) switch off non-essential loads in a logical sequence
b) switch off essential loads in a logical sequence
c) controls the power factor of the plant
d) all of the above.

……. End of Section – I …….

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Bureau of Energy Efficiency
Paper EA3 – Energy Auditor – Set A Solutions

Section – II: SHORT DESCRIPTIVE QUESTIONS                           Marks: 10 x 5 = 50

(ii)   Each question carries Five marks

and full load losses as 3 kW and 25 kW respectively.

=      3 + (0.6)2 x 25
=      12 kW

S-2.   (a) What is synchronous speed of an induction motor?

The speed of a motor is the number of revolutions in a given time frame, typically
revolutions per minute (RPM). The speed of an AC motor depends on the frequency of
the input power and the number of poles for which the motor is wound. The
synchronous speed in RPM is given by the following equation, where the frequency is
in hertz or cycles per second:
120  Frequency
Synchronou Speed (RPM) 
s
No. of Poles
(b) How the % slip of an induction motor is measured?

The actual speed, with which the motor operates, will be less than the synchronous speed.
The difference between synchronous and full load speed is called slip and is measured in
percent. It is calculated using this equation:

Synchronous Speed - Full Load Speed
Slip (%)                                         100
Synchronous Speed

S-3.   How does power factor of an induction motor reduce with the reduction of the applied
motor.
kW
The power factor of the motor is given as: Power Factor  Cos  
kVA
As the load on the motor comes down, the magnitude of the active current or active
power reduces. However, there is no corresponding reduction in the magnetizing
current or reactive power, which is proportional to supply voltage. With the result, the
apparent current or apparent power does not reduce in the same proportion to that of
the active current or active power. Therefore, the motor power factor reduces, with a

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Bureau of Energy Efficiency
Paper EA3 – Energy Auditor – Set A Solutions

S-4.    What are the parameters required to be measured while estimating the chiller
performance in kW/TR?

Q : mass flow rate of coolant in kg/hr
Ti : inlet, temperature of coolant to evaporator (chiller) in 0C
To : outlet temperature of coolant from evaporator (chiller) in 0C.
Actual power drawn : by compressor, chilled water pump, condenser water pump and
cooling tower fan

S-5.    A fan is operating at 900 RPM developing a flow of 3000 Nm 3/hr. at a static pressure
of 600 mmWC. What will be the flow and static pressure if the speed is reduced to
600 RPM.

Q1 = 3000 Nm3 / hr. Q2 = ?

N1 = 900 rpm                     N2   = 600 rpm

PSp1 = 600 mmWc      PSp2 = ?

N1 = Q1
N2   Q2

Q2 = N2 x Q1 = 600 x 3000 = 2000 Nm3 / hr.
N1         900

SP1 = N12
SP2   N22

SP2 = SP1 x N22/N12 =       600 x 6002
900 2
SP2 = 266.6 mmWc

S-6.    What are the various methods of flow control in centrifugal pumps?

1.   Pump control by varying speed.
2.   Parallel operation of pumps
3.   Stop/start control
4.   Flow control valve
5.   By-pass control
6.   Impeller trimming
7.   Variable speed drives

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Bureau of Energy Efficiency
Paper EA3 – Energy Auditor – Set A Solutions

S-7.   In a cooling tower, the Cycle of Concentration (C.O.C.) is 3 and evaporation losses
are 1%. The circulation rate is 1200 m3 /min. Find out the blow down quantity
required for maintaining the desired level of dissolved solids in the cooling water.

Blow down         =    Evaporation loss / (COC – 1)

= 0.01 x 1200
__________       = 12      = 6 m3/min
3-1           2

S-8.   Whether it is advisable to install a servo transformer for controlling the operating

Most of the problems faced by the lighting equipment and the gears are due to
the supply voltage fluctuations. Hence, the lighting equipments have to be
isolated from the power feeders. This is carried out by installing lighting voltage
transformers, which regulate the voltage exclusively for lighting circuits. This
reduces the voltage related problems in turn increasing the efficiency of the
lighting system. This also results in energy saving during the periods when the
supply voltage levels are on higher side. In addition, it increases the life and
performance of the control gear units of the lamps due to better voltage
regulation.

S-9.   A 5 MW DG set with an average load of 3 MW running in parallel with the grid was
found to be exporting 100 kVAr. Without calculating, explain what could be the
possible reasons for the export of reactive power to the grid. List

The DG set besides meeting the reactive power requirement of the load was
generating excessive 100 kVAR . The PF of the DG set must be lower than the
load PF. Since the DG set is in parallel operation with grid, it will be in a position
to export reactive power.
The plant can improve its utility PF, provided it is worked out on the basis of kWH
and kVARh readings. In this process, some of the lagging kVARs imported from
the grid can be nullified by the export of kVARs. The PF penalty imposed or
improved PF benefits on the imported power can be accrued by the plant.
1. The excitation power required by the DG set will increase .
2. I2r losses of the alternator will increase.

S-10. How are energy savings achieved through electronic ballast in a fluorescent tube light
in comparison to the conventional magnetic ballast?

1. The low internal core loss of electronic blast, in comparison to conventional magnetic
ballasts.
2. Increased light output of the lamp fitted with electronic ballast, due to the excitation
of the lamp phosphors with high frequency.

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Bureau of Energy Efficiency
Paper EA3 – Energy Auditor – Set A Solutions

New high frequency (28-32 kHz) electronic ballasts have the following advantages over the

     Energy savings up to 35%
     Less heat dissipation, which reduces the air conditioning load
     Lights instantly
     Improved power factor
     Operates in low voltage load
     Less in weight
     Increases the life of lamp

……. End of Section - II …….

Section – III: LONG DESCRIPTIVE QUESTIONS                             Marks: 5 x 10 = 50

(ii)   Each question carries Ten marks

L-1.   A plant has 2 identical 500 kVA transformers, each with a no load loss of 0.84 kW and
full load copper loss of 5.7 kW. The plant average load is 300 kVA and has never
exceeded 450 kVA in the past.
(a) Compare the transformer losses when single transformer is in operation and when
both transformers are in parallel operation.

Transformer Losses ( when 2 transformers are in parallel operation )
No load loss = 0.84 kW /transformer
F.L loss      = 5.7 kW / transformer
F.L loss at operating load = (150/500)2 x 5.7 = 0.513 kW / transformer
Operating loss per transformer = 0.84+0.513=1.353 kW / transformer
Operating loss for both the transformers =1.353 x 2 = 2.706 kW
Transformer Losses ( when 1 transformer is in operation )
No load loss = 0.84 kW
F.L loss      = 5.7 kW

F.L loss at operating load = (300/500)2 x 5.7 = 2.052kW
Operating loss per transformer = 0.84+2.052=2.892 kW

Savings in transformer losses (when 2 transformers are in operation )= 2.892 -2.706=0.186 kW

(b) What would you like to advise to the plant’s management on transformer operation
keeping in view the energy saving potential, reliability and safety of the system.

The plant should operate both the transformers, as there is a power saving potential of
0.186 kW. Moreover in this case the reliability and safety of the system will be much
higher in comparison to single transformer operation.

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Bureau of Energy Efficiency
Paper EA3 – Energy Auditor – Set A Solutions

L-2.   A compressed air leakage test was conducted in an engineering industry, which
employs a 500 cfm reciprocating compressor. The compressed air system is
respectively. The following was observed for a period of 15 minutes trial:
On load time = 40 secs
Subsequently some of the air leakage points were attended and the leakage test was
On load time = 20 secs

The average power drawn during the above 2 trials was observed as 70 kW during
load and 15 kW during unload condition. Calculate the annual cost savings for
5000 hr/ year of compressor operation. Assume energy charge of Rs. 5.00 per kWh.

= (40x 500) / (40+120) = 125cfm

= (20 x 500) / (20+140) = 62.5 cfm

Specific power consumption = 70kWh / (500 x 60 cfh) = 0.00233 kW/ ft3

Reduction in leakage quantity =125 – 62.5 = 62.5 cfm = 3750 cfh

Energy savings per hour =3750 x 0.0023 = 8.7375kWh

Cost savings @ Rs.5/kWh @5000hrs/annum

=8.7375kWh x 5000hrs/annum x Rs.5 /kWh= Rs.2,18,437/ annum

L-3.   A centrifugal pump is delivering 30 m3/s of water at a discharge pressure of 3
kg/cm2g. The pump suction is 1 meter below the pump center line. Find out the
power drawn by the motor if the pump efficiency is 60% and motor efficiency is 92%.

Q = 30 m3 / Sec.
P1 = 3 Kg/Cm2 = 30 m
Pump = 60%
Motor = 92 %
(The pump suction is I m below the pump centerline. Hence it should be –1m.
So the total developed head would be 30 – (-1) = 31 m)

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Bureau of Energy Efficiency
Paper EA3 – Energy Auditor – Set A Solutions

Hydraulic Power(kW) = Q (m3/s) x Total Head(m) x ρ x g / 1000

= 30 x 31 x 1000 x 9.81
1000

= 9123.3 kW

Shaft Power = Hydraulic Power = 9123.3 kW / 0.60 = 15205.5.5 kW
Pump efficiency

Shaft power
Motor Power                           ==     15205.5     =   16527.7 kW
Motor efficiency            0.9 2

L-4.    Briefly explain the step by step approach for the conduct of energy audit of vapour
compression refrigeration plants.

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

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Bureau of Energy Efficiency
Paper EA3 – Energy Auditor – Set A Solutions

           Condenser water pump kW/TR
           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 hin and hout from dry bulb, wet
bulb temperature values which are, in-turn measured, during trials, by a
whirling psychrometer.

L-5.    List 10 energy saving opportunities in a cooling tower.

1. Follow manufacturer’s recommended clearances around cooling towers and
relocate or modify structures that interfere with the air intake or exhaust.
2. Optimise cooling tower fan blade angle on a seasonal and/or load basis.
3. Correct excessive and/or uneven fan blade tip clearance and poor fan balance.
4. On old counter-flow cooling towers, replace old spray type nozzles with new
square spray ABS practically non-clogging nozzles.
5. Replace splash bars with self-extinguishing PVC cellular film fill.
6. Install new nozzles to obtain a more uniform water pattern
7. Periodically clean plugged cooling tower distribution nozzles.
8. Balance flow to cooling tower hot water basins.
9. Cover hot water basins to minimise algae growth that contributes to fouling.
10. Optimise blow down flow rate, as per COC limit.
11. Replace slat type drift eliminators with low pressure drop, self extinguishing, PVC
cellular units.
12. Restrict flows through large loads to design values.
13. Segregate high heat loads like furnaces, air compressors, DG sets, and isolate
cooling towers for sensitive applications like A/C plants, condensers of captive
power plant etc. A 1oC cooling water temperature increase may increase A/C
compressor kW by 2.7%. A 1oC drop in cooling water temperature can give a heat
rate saving of 5 kCal/kWh in a thermal power plant.

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Bureau of Energy Efficiency
Paper EA3 – Energy Auditor – Set A Solutions

14. Monitor L/G ratio, CW flow rates w.r.t. design as well as seasonal variations. It
would help to increase water load during summer and times when approach is high
and increase air flow during monsoon times and when approach is narrow.
15. Monitor approach, effectiveness and cooling capacity for continuous optimisation
efforts, as per seasonal variations as well as load side variations.
16. Consider COC improvement measures for water savings.

……. End of Section – III …….

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Bureau of Energy Efficiency

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