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									                                                                            Paper 3 –Set B Solutions

Regn No: _________________
Name: ___________________
(To be written by the candidates)


        PAPER – 3:                  Energy Efficiency in Electrical Utilities

        Date: 23.04.2006           Timings: 0930-1230 HRS        Duration: 3 HRS         Max. Marks: 150

General instructions:
    o     Please check that this question paper contains 9 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.             A fan with 25 cm pulley diameter is driven by a 1470 rpm motor through a v-belt system.
                    If the motor pulley is reduced from 20 cm to 15 cm at the same motor rpm and fan pulley
                    diameter, the fan speed will reduce by

                    a) 882 rpm            b) 294 rpm           c) 1176 rpm          d) none of the above
     2.             For centrifugal fans, the relation between Power (kW) and Speed (N) is given by

                                     2                     3
                          kW1 N 1                 kW1 N 1                 kW1 N 1
                    a)       =               b)      =               c)      =             d) none of the above
                          kW2 N 2 2               kW2 N 2 3               kW2 N 2

     3.             The pressure flow characteristic curve of a centrifugal fan changes with the following flow
                    control method

                    a) inlet damper        b) outlet damper        c) inlet guide vane      d) none of the above
     4.             The hydraulic power of a motor pump set is 6.9 kW. If the power drawn by the motor is
                    14.0 kW at a 88% efficiency, the pump efficiency is given by

                    a) 49.3%             b) 56%           c) 43.4%           d) none of the above

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Bureau of Energy Efficiency
                                                                       Paper 3 –Set B Solutions

     5.      Which of the following is not true of air receivers?

             a) increases the pressure of air             b) smoothen pulsating air output
             c) storage of large volumes of air           d) a source for draining of moisture
                                                                                   o            o
     6.      If inlet and outlet water temperatures of a cooling tower are 42 C and 36 C respectively
                                                      o          o
             and atmospheric DBT and WBT are 39 C and 32 C respectively, then the effectiveness
             of cooling tower is

             a) 60%                b) 75%           c) 85.7%              d) none of the above
     7.      The efficiency of forward curved fans compared to backward-inclined fans is__

             a) lower                   b) same
             c) higher                  d) none of the above
     8.      The motor efficiency is 0.9 and pump efficiency is 0.6. The power transmitted to the
             water is 15.11 kW. The input power to the motor driving the pump is about

             a) 16.8 kW             b) 25.2 kW            c) 28.0 kW             d) none of the above
     9.      What is the impact on flow and head when the impeller of a pump is trimmed?

             a) both head and flow decreases                 b) flow decreases and pump head increases
             c) both flow and pump head increases            d) none of the above
     10.     In a pumping system the static head is 10 m and the dynamic head is 15 m. If the pump
             speed is doubled, then the total head will be

             a) 70 m          b) 40 m           c) 50 m             d) none of the above
     11.     Typical exit flue gas temperature of a 5 MW DG set operating above 80% load is in the
             range of

             a) 250 to 280 deg C                b) 200 to 230 deg C
             c) 340 to 370 deg C                d) none of the above
     12.     In a DG set, the generator is consuming 150 litre per hour diesel oil. If the specific fuel
             consumption of this DG set is 0.25 litres/ kWh at that load, then what is the kVA loading
             of the set at 0.6 PF?

             a) 600 kVA             b) 1000 kVA              c) 300 kVA                d) none of the above
     13.     The factors affecting Waste Heat Recovery from exhaust flue gases of DG set are:

             a) Temperature of exhaust flue gases after turbo charger
             b) Back pressure on the DG set
             c) DG set loading                                                         d) all of the above
                                                                             3                           o
     14.     The jacket cooling water in a diesel engine flows at 12.9 m /hr with a range of 10 C and
             accounts for 30% of the engine input energy. The power output of the engine will be

             a) 500 kW           b) 387 kW           c) 430 kW              d) none of the above
     15.     The maximum back pressure drop (mmWC) allowed in a DG set is

             a) 100-150            b) 150-200               c) 250-300            d) none of the above

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                                                                   Paper 3 –Set B Solutions

     16.     The electronic ballast in lighting application does not have one of the following

             a) low temperature rise
             b) lower operational losses than conventional ballasts
             c) tuned circuit to deliver power at 28-32 KHz
             d) requiring a mechanical switch (starter)
     17.     Energy savings potential of variable torque applications in comparison to constant torque
             application is:

             a) equal           b) lower             c) higher            d) none of the above
     18.     The function of soft starter includes

             a) provides smooth, stepless acceleration and deceleration.
             b) extension of motor life
             c) delivers a controlled release of power to the motor
             d) all of the above
     19.     The occupancy sensors in a lighting installation are best suited for

             a) entrances of offices/buildings          b) large production shops/hangars
             c) conference halls                        d) street lighting
     20.     The blowdown quantity required in cooling towers is given by

             a) evaporation loss/ (cycle of concentration +1)
             b) evaporation loss/ (cycle of concentration –1)
             c) (cycle of concentration –1)/ evaporation loss
             d) evaporation loss/ (1 - cycle of concentration)
     21.     A 7.5 kW, 415 V, 14.5 A, 1460 RPM, 3 phase rated induction motor with full load
             efficiency of 88% draws 10.1 A and 5.1 kW of input power. The percentage loading of
             the motor is about

             a) 70 %                b) 50%              c) 60 %             d) none of the above
     22.     The approximate size of the capacitor selected for the PF improvement and its
             installation at the induction motor terminals may be equal to

             a) 90% of no-load kVAr of motor             b) full load kVAr of motor
             c) No-load kVAr of motor                    d) none of the above
     23.     The nearest kVAr compensation required for improving the power factor of a 1000 kW
             load from 0.8 lagging power factor to unity power factor is

             a) 750 kVAr           b) 1000 kVAr           c) 500 kVAr            d) none of the above
     24.     Power factor is the ratio of
                           2      2 1/2                    2      2 1/2
             a) kVAr/ (kW +kVAr )                b) kW/ (kW +kVAr )
                   2      2 1/2
             c) (kW +kVAr ) /kW                  d) kVAr/kW

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                                                                   Paper 3 –Set B Solutions

     25.     The actual speed of a 2 pole induction motor operating at 49.8 Hz and at a slip of 1.5 %
             is given by

             a) 2980 RPM           b) 2943 RPM              c) 2955 RPM             d) none of the above
     26.     A 10 kVAr, 415 V rated power factor capacitor was found to be having terminal supply
             voltage of 370 V. The capacity of the power factor capacitor at the operating supply
             voltage would be approximately

             a) 9 kVAr          b) 8 kVAr          c) 10 kVAr             d) none of the above
     27.     In a 11 kV feeder, if the voltage is raised from 11 kV to 22 kV for the same loading
             conditions, the voltage drop in the same feeder system would be lower by a factor of

             a) 1/2             b) 1/3             c) 1/4                 d) none of the above
     28.     Select the incorrect statement:

             a) transformers operating near saturation level create harmonics
             b) harmonics occur as spikes at intervals which are multiples of the supply frequency
             c) harmonics are multiples of the fundamental frequency
             d) devices that draw sinusoidal currents when a sinusoidal voltage is applied create
     29.     A pure capacitive load in an alternating current (AC) circuit draws

             a) active power                            b) leading reactive power
             c) lagging reactive power                  d) none of the above
     30.     Select the incorrect statement:
             In system distribution loss optimization, the various options available include

             a) selection of Aluminium Cored Steel Reinforced (ACSR) lines instead of All Aluminium
                Alloy Conductors (AAAC)
             b) re-routing and re-conducting such feeders and lines where the voltage drops are
             c) power factor improvement
             d) optimum loading of transformers in the system
     31.     Which of the following is the best definition of illuminance?

             a) flux density emitted from an object in a given direction
             b) time rate of flow of light energy
             c) luminous flux incident on an object per unit area
             d) flux density emitted from an object without regard for direction
     32.     The ratio of luminous flux emitted by a lamp to the power consumed by the lamp is ___.

             a) Colour Rendering Index                  b) Illuminance
             c) Lux                                     d) Luminous Efficacy
     33.     Which is the most energy efficient lamp?

             a) GLS           b) LPSV          c) HPMV           d) FTL

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                                                                     Paper 3 –Set B Solutions

     34.     If voltage is reduced for gas discharge lamps to its optimum value, it will result in

             a) no change in power consumption
             b) reduced power consumption
             c) increased power consumption
             d) increased light levels
     35.     The unit of lux is

             a) one lumen per square feet               b) 1000 lumens per square feet
             c) 10 lumen per square metre               d) one lumen per square metre
     36.     The lowest theoretical temperature to which water can be cooled in a cooling tower is

             a) difference between DBT and WBT of the atmospheric air
             b) average DBT and WBT of the atmospheric air
             c) WBT of the atmospheric air
             d) DBT of the atmospheric air
     37.     The L/G ratio of a cooling tower does not depend on

             a) dry bulb temperature                    b) range
             c) enthalpy of inlet air                   d) outlet wet bulb temperature
     38.     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 30 C WBT                   b) 40 C DBT and 37 C WBT
                  o            o                             o            o
             c) 38 C DBT and 37 C WBT                   d) 35 C DBT and 29 C WBT
     39.     A water pump is delivering 200 cubic metres per hour at ambient conditions. The
             impeller diameter is trimmed by 10%. The water flow at the changed condition is given
                      3                    3                     3
             a) 160 m /hr         b) 140 m /hr          c) 180 m /hr        d) none of the above
     40.     Lowering the Cycles of Concentration (C.O.C) in circulating water in a cooling tower, the
             blow down quantity will

             a) decrease          b) not change            c) increase           d) none of the above
     41.     Select the wrong statement:

             a) centrifugal compressors are better suited for applications requiring very high
                capacities, typically above 12,000 cfm
             b) for every 4°C rise in the air inlet temperature of an air compressor, the power
                consumption will normally increase by one percentage points for the same output.
             c) after-coolers remove the moisture in the air before it enters the next stage of
                compressor to reduce the work of compression
             d) for every 250 mm WC pressure drop increase across the suction path due to
                choked filters etc., the compressor power consumption increases by about 2 percent
                for the same output
     42.     The reciprocating air compressor efficiency does not depend on

             a) discharge pressure                      b) flow rate
             c) suction pressure                        d) system air leakages

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                                                                    Paper 3 –Set B Solutions

     43.     The leak test results show load time of 10 seconds and unload time of 20 seconds in a
             load-unload reciprocating compressor. If the compressor capacity is 256 cfm, then the
             approximate leakage would be

             a) 128 cfm           b) 85 cfm                  c) 170 cfm        d) 256 cfm
     44.     Dynamic air compressors are mainly of the following type

             a) centrifugal compressors                           b) two stage screw compressors
             c) two stage reciprocating compressors               d) none of the above
     45.     The flow output of which of the following changes with the discharge pressure

             a) centrifugal compressor                   b) reciprocating compressor
             c) screw compressor                         d) none of the above
     46.     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 lower condensing temperature
             d) none of the above
     47.     In water cooled refrigeration systems, condenser cooling water temperature should be
             closest to

             a) ambient wet bulb temperature             b) dew-point temperature
             c) ambient dry bulb temperature             d) none of the above
                                                         3                                  o          o
     48.     The refrigeration load in TR when 20 m /hr of water is cooled from a 13 C to 8 C is

             a) 80.3          b) 39.6          c) 33              d) none of the above
     49.     One ton of refrigeration (TR) is equal to

             a) 3516 W          b) 200 BTU/min           c) 50.4 kcal/min        d) all of the above
     50.     In a vapour compression refrigeration system, the component where the refrigerant fluid
             experiences no heat loss or gain is

             a) evaporator       b) compressor               c) condenser        d) expansion valve

                                 ……. End of Section – I …….

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Bureau of Energy Efficiency
                                                         Paper 3 –Set B Solutions

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

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

S-1    What are the technical aspects of energy efficient motors with respect to insulation
       life, slip and starting torque?

       Lower temperatures in energy efficient motors translate to long lasting
       insulation. Generally, motor life doubles for each 10°C reduction in operating
       The lower the slip, the higher the efficiency. Less slippage in energy efficient
       motors results in speeds about 1% faster than in standard counterparts.

       Starting torque for efficient motors may be lower than for standard motors.

S-2    A pump is delivering 45 m3/hr of water with a discharge pressure of 30 metre. The
       water is drawn from a sump where water level is 5 metre below the pump centerline.
       The power drawn by the motor is 8.0 kW at 89% motor efficiency. Find out the pump

       Hydraulic power Ph = Q (m3/s) xTotal head, hd - hs (m) x  (kg/m3) x g (m/s2) / 1000
       Q = 45/3600 m3/s , hd - hs = 30 – (-5) = 35 m
       Hydraulic power Ph = (45/3600) x 35 x 1000 x 9.81 / 1000
                              = 4.29 kW

       Pump shaft power       = 8 kW x 0.89
                              = 7.12 kW

       Pump efficiency        = hydraulic power / pump shaft power
                              = 4.29 / 7.12
                              = 60 %

S-3    A 180 kVA, 0.80 PF rated DG set has diesel engine rating of 220 BHP. What is the
       maximum power factor which can be maintained at full load on the alternator without
       overloading the DG set? (Assume alternator losses and exciter power requirement as
       5.60 kW and there is no derating of DG set)

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                                                                  Paper 3 –Set B Solutions


       Engine rated Power              = 220 x 0.746 = 164 kW

       Rated power available for alternator = 164 – 5.6 = 158.4 kW

       Maximum power factor possible = 158.4 /180 = 0.88

S-4      Briefly explain the benefit of installing servo stabilizer for lighting feeder


         In many plants, during the non-peaking hours, the voltage levels are on the
         higher side. During this period, voltage can be optimized, without any
         significant drop in the illumination level.

         The servo stabilizer will provide stabilized voltage for the lighting equipment.

         The performance of “gears” such as chokes, ballasts, will also improved due to
         the stabilized voltage.

         Further, servo stabilizer can maintain optimum voltage, which would help in
         saving electricity.

S-5      How do you calculate the velocity of air/gas in a duct using the average differential
         pressure and density of the air/gas?

         Velocity V, m/s = CP x (2 x 9.81 Δp x y)1/2

         Cp = Pitot tube constant, 0.85 (or) as given by the manufacturer
         Δp = Average differential pressure measured by pitot tube by taking

         measurement at number of points over the entire cross section of the duct.
         γ = Density at air/ gas at test condition

S-6      Estimate the cooling tower capacity (TR) and approach with the following parameters
         Water flow rate through CT      = 140 m3/hr
         Specific heat of water          = 1 kcal/kg °C
         Inlet water temperature         = 40 °C
         Outlet water temperature        = 36 °C
         Ambient WBT                     = 31 °C

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Bureau of Energy Efficiency
                                                              Paper 3 –Set B Solutions

Cooling tower capacity (TR)           = (flow rate x density x sp.heat x diff. temp)/ 3024
                                      = 140 x 1000 x 1.0 x (40-36)/ 3024
                                      = 185.2 TR

Approach                              = 36- 31 = 5oC

S-7    The COP of a vapour compression refrigeration system is 3.0. If the compressor
       motor draws power of 8.5 kW at 89% motor efficiency, find out the tonnage of the
       refrigeration system.

       Power input to compressor                 =     0.89 x 8.5
                                                 =     7.57 kW

       Cooling effect                            =     7.57 x 3.0
                                                 =     22.71 kW

           22.71 kW x 860 kcal/kwh               =     19531 kcals/hr

       Refrigeration tones                       =     19531/3024
                                                 =     6.46 Tonnes

S-8    Calculate the free air delivery (FAD) capacity of a compressor in m3/hr for the
       following observed data:
       Receiver capacity:                              0.4 m3
       Initial pressure (with empty receiver):         0 kg/cm2 (g)
       Final pressure:                                 7 kg/cm2 (g)
       Initial air temperature:                        35oC
       Final air temperature:                          55 oC
       Additional holdup volume:                       0.05 m3
       Compressor pump up time:                        4.5 minutes
       Atmospheric pressure:                           1.026 kg/sq. cm absolute

           P2  P1 V  273  t1 
       =                      
             P0    t  273  t 2 
                                


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Bureau of Energy Efficiency
                                                            Paper 3 –Set B Solutions

       = 0.64 m3/min

S-9    Fill in the blanks
       a) Totally-enclosed, fan cooled (TEFC) motors are more efficient than Screen –
          protected, drip-proof (SPDP) induction motors

       b) Induction motor efficiency increases with increase in its rated capacity

       c) Low speed Squirrel cage induction motors are normally less efficient than high
          speed Squirrel cage induction motors

       d) The capacitor requirement for PF improvement at induction motor terminal
          increases with decrease in rated speed of the induction motor

       e) Slip ring induction motors are normally less efficient than squirrel cage induction

S-10   Calculate the transformer total losses for a 100 kVA transformer for an average
       loading of 50%. Assume no load and full load losses as 1.70 kW and 10.50 kW

       Transformer losses =           No load losses + (% loading)2 x Full load losses

                              =       1.70+ (0.5)2 x 10.5
                              =       4.33 kW

                                  ……. End of Section - II …….

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Bureau of Energy Efficiency
                                                            Paper 3 –Set B Solutions

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

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

L-1        Briefly explain the step-by-step approach for conducting Energy Performance
           Assessment of DG set on shopfloor.


Routine energy efficiency assessment of DG sets on shopfloor involves following
typical steps:

1) Ensure reliability of all instruments used for trial.
2) Collect technical literature, characteristics, and specifications of the plant.
3) Conduct a 2 hour trial on the DG set, ensuring a steady load, wherein the following
   measurements are logged at 15 minutes intervals.

      a) Fuel consumption (by dip level or by flow metre)
      b) Amps, volts, PF, kW, kWh
      c) Intake air temperature, Relative Humidity (RH)
      d) Intake cooling water temperature
      e) Cylinder-wise exhaust temperature (as an indication of engine loading)
      f) Turbocharger RPM (as an indication of loading on engine)
      g) Charge air pressure (as an indication of engine loading)
      h) Cooling water temperature before and after charge air cooler (as an indication
         of cooler performance)
      i)   Stack gas temperature before and after turbocharger (as an indication of
           turbocharger performance)

4) The fuel oil/diesel analysis is referred to from an oil company data.

5) Analysis: The trial data is to be analysed with respect to:

      a) Average alternator loading.
      b) Average engine loading.
      c) Percentage loading on alternator.
      d) Percentage loading on engine.
      e) Specific power generation kWh/liter.
      f) Comments on Turbocharger performance based on RPM and gas temperature

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                                                             Paper 3 –Set B Solutions

      g) Comments on charge air cooler performance.
      h) Comments on load distribution among various cylinders (based on exhaust
         temperature, the temperature to be  5% of mean and high/low values indicate
         disturbed condition).
      i)   Comments on housekeeping issues like drip leakages, insulation, vibrations,

L-2        List down any 10 energy conservation opportunities in compressed air system.


              Ensure air intake to compressor is not warm and humid by locating
               compressors in well-ventilated area or by drawing cold air from outside.
               Every 40C rise in air inlet temperature will increase power consumption by 1

              Clean air-inlet filters regularly. Compressor efficiency will be reduced by 2
               percent for every 250 mm WC pressure drop across the filter.

              Keep compressor valves in good condition by removing and inspecting
               once every six months. Worn-out valves can reduce compressor efficiency
               by as much as 50 percent.

              Install manometers across the filter and monitor the pressure drop as a
               guide to replacement of element.

              Minimize low-load compressor operation; if air demand is less than 50
               percent of compressor capacity, consider change over to a smaller
               compressor or reduce compressor speed appropriately (by reducing motor
               pulley size) in case of belt driven compressors.

              Consider the use of regenerative air dryers, which uses the heat of
               compressed air to remove moisture.

              Fouled inter-coolers reduce compressor efficiency and cause more water
               condensation in air receivers and distribution lines resulting in increased
               corrosion. Periodic cleaning of inter-coolers must be ensured.

              Compressor free air delivery test (FAD) must be done periodically to check
               the present operating capacity against its design capacity and corrective
               steps must be taken if required.

              If more than one compressor is feeding to a common header, compressors
               must be operated in such a way that only one small compressor should
               handle the load variations whereas other compressors will operate at full

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                                                        Paper 3 –Set B Solutions

          The possibility of heat recovery from hot compressed air to generate hot air
           or water for process application must be economically analyzed in case of
           large compressors.

          Consideration should be given to two-stage or multistage compressor as it
           consumes less power for the same air output than a single stage

          If pressure requirements for processes are widely different (e.g. 3 bar to 7
           bar), it is advisable to have two separate compressed air systems.

          Reduce compressor delivery pressure, wherever possible, to save energy.

          Provide extra air receivers at points of high cyclic-air demand which permits
           operation without extra compressor capacity.

          Retrofit with variable speed drives in big compressors, say over 100 kW, to
           eliminate the `unloaded’ running condition altogether.

          Keep the minimum possible range between load and unload pressure

          Automatic timer controlled drain traps wastes compressed air every time the
           valve opens. So frequency of drainage should be optimized.

          Check air compressor logs regularly for abnormal readings, especially
           motor current cooling water flow and temperature, inter-stage and discharge
           pressures and temperatures and compressor load-cycle.

          Compressed air leakage of 40- 50 percent is not uncommon. Carry out
           periodic leak tests to estimate the quantity of leakage.

          Install equipment interlocked solenoid cut-off valves in the air system so
           that air supply to a machine can be switched off when not in use.

          Present energy prices justify liberal designs of pipeline sizes to reduce
           pressure drops.

          Compressed air piping layout should be made preferably as a ring main to
           provide desired pressures for all users.

          A smaller dedicated compressor can be installed at load point, located far
           off from the central compressor house, instead of supplying air through
           lengthy pipelines.

          All pneumatic equipment should be properly lubricated, which will reduce
           friction, prevent wear of seals and other rubber parts thus preventing energy
           wastage due to excessive air consumption or leakage.

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             Misuse of compressed air such as for body cleaning, agitation, general floor
              cleaning, and other similar applications must be discouraged in order to
              save compressed air and energy.

             Pneumatic equipment should not be operated above the recommended
              operating pressure as this not only wastes energy bus can also lead to
              excessive wear of equipment’s components which leads to further energy

             Pneumatic transport can be replaced by mechanical system as the former
              consumed about 8 times more energy. Highest possibility of energy savings
              is by reducing compressed air use.

             Pneumatic tools such as drill and grinders consume about 20 times more
              energy than motor driven tools. Hence they have to be used efficiently.
              Wherever possible, they should be replaced with electrically operated tools.

             Where possible welding is a good practice and should be preferred over
              threaded connections.

             On account of high pressure drop, ball or plug or gate valves are preferable
              over globe valves in compressed air lines.

L-3 (a) A fan is delivering 20,000 Nm3/hr. of air at static pressure difference of 60 mm WC. If
        the shat power of the fan is 6.9 kW, find out the static fan efficiency.
      (b) Explain briefly the difference between static and dynamic head of a centrifugal
          pumping system.

(a)      Q = 20,000 Nm3 / hr. , Pst       = 60 mmWC, St = ? , P = 6.9 kW

             = 20,000/3600 = 5.56 m3/sec

               Fan static St    =   Volume in m3/sec x Pst in mmWc
                                       102 x Power input to shaft

                                =      5.56 x 60
                                       102 x 6.9

                                = 47.4 %

(b)      Static head is simply the difference in height of the supply and destination
         reservoirs and it is independent of flow.

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                                                              Paper 3 –Set B Solutions

       Dynamic head is the friction loss, on the liquid being moved, in pipes, valves
       and equipment in the system. The friction losses are proportional to the square
       of the flow rate.

L-4    A 7.5 kW, 415 V, 14.5 A, 4 pole, 50 Hz, 3 phase rated squirrel cage induction motor
       has a full load efficiency and power factor of 88% and 0.87 respectively.
       An energy auditor measures the following operating data of the motor
(a)    Supply voltage         =       410 V
(b)    Current dram           =       9.7 A
(c)    PF                     =       0.8
(d)    Supply frequency       =       50.1 Hz
(e)    RPM                    =       1480

       Find out the following at the motor operating conditions:
1.     Power input in kW
2.     % motor loading
3.     % slip


1.     Power input = 1.7321 x 0.410 x 9.7 x 0.8 = 5.5 kW

2.     % motor loading = power input/ rated input x 100 = 5.5/ (7.5/0.88) = 5.397/ 8.4269
       = 64.5 %

3.     Synchronous RPM at 50.1 Hz,
       NS = 120 f/ P = 120 x 50.1/ 4 = 1503 RPM
       % slip = (Ns – N)/ Ns x 100 = (1503 – 1480) / 1503 = 23/ 1503 x 100 = 1.53%

L-5    The contract demand of a process plant is 5000 kVA with the electricity supply utility
       company. The average monthly recorded maximum demand of the process plant is
       4500 kVA at a power factor of 0.82. The utility bill analysis provides the following tariff
a)     Minimum monthly billing demand is 75% of the contract demand or the actual
       recorded maximum demand whichever is higher.
b)     Monthly maximum demand (MD) charge is Rs. 300 per kVA.
       Find out the optimum limit of power factor capacitor requirement entirely from the view
       of reducing maximum demand so that no excess demand charges are paid to the
       supply company. Also work out the simple payback period, assuming cost of power
       factor capacitor installation along with automatic power factor correction controller is
       as Rs. 500 per kVAr.

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Bureau of Energy Efficiency
                                                         Paper 3 –Set B Solutions

Minimum payable demand = 5000 x 0.75 = 3750 kVA
Margin available for reduction of MD        =    4500 – 3750 = 750 kVA

Present MD in kW      =       4500 x 0.82   =    3690 kW
Desired power factor = 3690/ 3750           =    0.984

Power factor capacitor requirement to achieve the desired power factor
= 3690 [tan (Cos-1 0.82) – tan (Cos-1 0.984)] = 1908 kVAr (say 1900 kVAr)

Cost of power factor capacitor installation = Rs. 500 per kVAr x 1900 kVAr = 9.5 lakhs
Monthly savings due to MD reduction = 750 kVA
Yearly savings = 750 x 300 x 12 = Rs. 27 lakhs
Simple payback period = investment cost / yearly savings = 9.5/ 27 = 0.352 years = 4.2

                               ……. End of Section - III …….

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Bureau of Energy Efficiency

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