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					                                     Paper EA3 – Energy Auditor – Set A Solutions

                                          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

            (i)     Answer all 50 questions
            (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

                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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                                     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.
     15.    A 200 cfm compressor has a loading and unloading period of 10 seconds and 20
            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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                                   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
                                                            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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                                  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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                                   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
            loading and 480 C exhaust gas is closest to ……
            (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
     45.    The operating efficiency of a DG set also depends on

            a) turbo charger performance            b) inlet air temperature
            c) % loading                            d) all of the above
     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
            c) improve the loading                       d) none of the above
     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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                                Paper EA3 – Energy Auditor – Set A Solutions

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

         (i)    Answer all Ten questions
         (ii)   Each question carries Five marks

S-1.   Calculate the transformer total losses for an average loading of 60%. Assume no load
       and full load losses as 3 kW and 25 kW respectively.

Transformer losses =           No load losses + (% loading)2 x Full load losses
                        =      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) 
                                         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
       load on the motor? Draw a curve depicting power factor vs percentage loading on the
The power factor of the motor is given as: Power Factor  Cos  
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
reduction in the applied load.

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                               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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                              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
       voltage of the lighting circuit? Justify your answer.

       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

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
       advantages/disadvantages of the above situation.

      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
 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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                                Paper EA3 – Energy Auditor – Set A Solutions

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

            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

           (i)    Answer all Five questions
           (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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                              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
       maintained at the normal loading-unloading settings of 6.5 kg/cm2g and 7 kg/ cm2g
       respectively. The following was observed for a period of 15 minutes trial:
       On load time = 40 secs
       Unload time = 120 secs.
       Subsequently some of the air leakage points were attended and the leakage test was
       repeated. The following was observed while maintaining the same loading &
       unloading pressure settings and trial period:
       On load time = 20 secs
       Unload time = 140 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.

Leakage quantity when loading is 40 secs and unloading is 120 secs
= (40x 500) / (40+120) = 125cfm

Leakage quantity when loading is 20 secs
and unloading is 140 secs.

= (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 %
       Total Head = Deliver head - Suction head = 30- (-1) = 31 metres.
               (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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                              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

                  = 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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                                  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 
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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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