NATIONAL CERTIFICATION EXAMINATION by 4DSkbV

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									                                 Paper 4 – Energy Auditor – Set A Solutions

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

              NATIONAL CERTIFICATION EXAMINATION 2006
                                FOR
                         ENERGY AUDITORS

  PAPER – 4:            Energy Performance Assessment for Equipment and Utility
                        Systems

  Date: 23.04.2006      Timings: 1400-1600 HRS      Duration: 2 HRS    Max. Marks: 100


General instructions:
      o   Please check that this question paper contains 4 printed pages
      o   Please check that this question paper contains 16 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
      o   Open book examination



Section - I:       SHORT DESCRIPTIVE QUESTIONS                           Marks: 10 x 1 = 10

          (i) Answer all Ten questions
          (ii) Each question carries One mark
          (iii) Answer should not exceed 50 words


S-1       State two causes for rise in exit flue gas temperature in a boiler.

          1. Scale deposition on water side     2. Soot deposition on gas side

             Any other relevant cause such as reduction in excess air levels, problems with
               Air preheater, economizer etc. are also the cause for the rise in exit flue gas
                                                                                 temperature

S-2       What are the disadvantages of heating the charge above the optimum
          temperature in steel re-rolling furnaces?

          1. Increased energy consumption        2. Increase in scale losses 3. High rejection
          rates


S-3       State the impact of fouling factor on the overall heat transfer coefficient.

          They are inversely proportional. Increase in fouling factor will result in decreased
          overall heat transfer coefficient.


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Bureau of Energy Efficiency
                                           Paper 4 – Energy Auditor – Set A Solutions


S-4        List the basic parameters required for assessing refrigeration capacity.

         Mass of circulating fluid, its specific heat and temperature difference


S-5        While using Pitot tube for airflow measurement in large ducts, series of traverse
           measurements are recommended. Why?

           Because the velocity is not uniform across the duct cross section


                                               (m3/s) x pressure gain in Pascal
S-6       Static fan efficiency =                                               . Right or wrong?
                                                 Power input to shaft in Watt

           Justify your answer.

         Right

          The units are balanced in both SI and MKS system.

           For example, if Volume=10 m3/s, Pressure gain is 500 Pa and power consumed
          is 10,000 W, then the efficiency as per SI units given in the problem

          (10 x 500 / 10,000) x 100 = 50%

          As per MKS the fan efficiency is
                                  Volume in m 3 / Sec x tota l pressure in mmwc
      Static Fan Efficiency % 
                                      102 x Power input to the shaft in (kW)      X 100

   1mmwc = 9.81 Pa
    Volume = 10 m3/s, Pressure gain in mmWC = 500/9.81 = 51 mmwc,
    Power consumed = 10 kW

           (10 x 51) x100 /(102 x 10) = 50%


S-7        State two methods of non-intrusive water flow measurements in a pipe.

          1. Ultrasonic / Doppler 2. Tracer


S-8        What is meant by compression ratio for air compressor?

           Ratio of discharge to suction pressures

S-9        How many volt-amperes (VA) does a 60 Watt incandescent light require?

          60 VA

S-10       A reasonable range of capacity factors for wind electric generators is.…


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                                   Paper 4 – Energy Auditor – Set A Solutions

          0.25 – 0.30


                               -------- End of Section - I ---------



Section - II:        LONG DESCRIPTIVE QUESTIONS                               Marks: 2 x 5 = 10

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

L-1       A centrifugal clear water pump rated for 800 m3/hr was found to be operating at
          576 m3/hr with discharge valve throttled. The pumps speed is 1485 RPM. The
          discharge pressure of the pump before the throttle valve is 2 kg/cm 2g. The pump
          draws the water from a sump 4 metres below the centerline of the pump. The
          input power drawn by the motor is 124 kW at a motor efficiency of 92%.
(i)       Find out the efficiency of the pump.
(ii)      If the normal required water flow rate is 500 m3/hr to 700 m3/hr, what in your
          opinion should be the most energy efficient option to get the required flow rate
          variation?
(iii)     And what would be the pump shaft power for that most energy efficient option if
          the pump is delivering the flow rate of 550 m3/hr.

Ans:

(i) Hydraulic power, Ph (kW)         = Q x (hd - hs) x r x g / 1000

Q = Volume flow rate (m3/s), r = density of the fluid (kg/m3),    g = acceleration due to gravity (m/s 2),
(hd - hs) = Total head in metres


hd - hs          =        20 – ( - 4 ) = 24 m


hydraulic power = (576 x 24 x 1000 x 9.81) / (3600 x 1000)
                      = 37.67 kw


Input power to pump                = 124 kW x 0.92 = 114 kW


Efficiency of the pump             = (37.67 /114) x 100 = 33 %


 (ii)     Since the pump discharge requirement varies from 500 m3/h to 700 m3/h,
          the ideal option would be to operate with a VSD (variable frequency drive,
          hydraulic coupling)



(iii) For a flow rate 550 m3/h, the reduced speed of pump would be:
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Bureau of Energy Efficiency
                                 Paper 4 – Energy Auditor – Set A Solutions



                                          (550/800) = (N1/1485)
                                                N1 = 1021
The pump shaft power would be:
                                             3
                                      1021
                                  =              x 114        = 37 kW
                                      1485




L-2    A 30 kW four pole induction motor operating at 50 Hz and rated for 415 V and
       1440 RPM, the actual measured speed is 1460 RPM. Find out the percentage
       loading of the motor if the voltage applied is 425 V.

Ans:
                       % Loading =                  Slip          x 100%
                                          (Ss – Sr) x (Vr / V)2


                       Synchronous speed = 120 x 50 / 4 = 1500 rpm


                       Slip = Synchronous Speed – Measured speed in rpm.
                       = 1500 – 1460 = 40 rpm.


                        % Loading =                      40             x 100% = 69.9%
                                         (1500 - 1440) x (415/425)2




                              -------- End of Section - II ---------

Section - III:   Numerical Questions                                       Marks: 4 x 20 = 80

       (i)     Answer all Four questions
       (ii) Each question carries Twenty marks

N -1   A process plant requires 28 tons of steam per hour and 2250 kW of electric
       power. The plant operates for 8000 hours per annum. Steam is generated at 2
       bar (g) in a coal fired boiler with an efficiency of 75%. The feed water
       temperature is 80OC. The calorific value of coal is 4000 kcal/kg. The cost of coal
       is Rs. 2000/ton. Power is drawn from the grid at Rs. 4/kWh. The contract
       demand is 3000 kVA with the electricity supply company and the plant is
       charged for 100% of the contract demand at Rs. 300/kVA/month. The plant has
       never exceeded its contract demand in the past.
       The plant is planning for a back pressure cogeneration system using the same
       coal with the following parameters. The power and steam demand are to be fully
       met by the cogeneration plant and a contract demand of 1000 kVA with the grid
       is to be kept for emergency purposes.
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Bureau of Energy Efficiency
                              Paper 4 – Energy Auditor – Set A Solutions


       Find out the IRR over a project life cycle of 6 years for the proposed
       cogeneration system
           Cogeneration System data:

           Boiler generation pressure                - 18 bar (g), 310OC
           Boiler efficiency                         - 81%
           Investment required                       - Rs. 20 crores
           Generated power                           = 2250 kW
           Steam enthalpy data:

           Total enthalpy at 2 bar (g)               = 647.13 kcal/kg
           Total enthalpy at 18 bar (g), 310oC       = 730.28 kcal/kg

Ans:
       Existing condition
       Coal consumption                          28,000 x [(647.13) – 80]
                                                           0.75 x 4000

                                                 5293 kg/hr

       Fuel cost per annum                       5.293 x 2000 x 8000

                                                 Rs. 8.5 crores (approx)


       Annual electrical energy charges          2250 x 8000 x Rs.4

                                                 Rs. 7.2 crores

       Maximum demand charges                    3000 x 12 x 300

                                                 Rs. 1.1 crores (approx)

       Total electricity bill per annum          7.2 + 1.1 = Rs. 8.3 crores



       Total energy bill per annum               8.5 + 8.3 = Rs. 16.8 crores


       With cogeneration plant



       Coal consumption                          28,000 x [(730.28) – 80]
                                                        0.81 x 4000

                                                 5620 kg/hr

       Incremental coal consumption              5620 – 5293 = 327 kg/hr

       Incremental fuel cost per annum           0.327 x 8000 x Rs. 2000
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Bureau of Energy Efficiency
                                           Paper 4 – Energy Auditor – Set A Solutions


                                                                  Rs. 0.52 crores

        Maximum demand charges per                                1000 x 12 x Rs.300
        annum                                                     Rs. 0.36 crores

        Total cost per year                                       8.5 + 0.52 + 0.36 = 9.38 crores
        Savings                                 =                 16.8 – 9.38 = Rs. 7.42 crores

        Investment                                                Rs. 20 crores

                 1                 1                  1             1             1             1
20 = 7.42            1   +             2    +             3   +         4   +         5   +         6
               (1+i)             (1+i)              (1+i)         (1+i)         (1+i)         (1+i)

        IRR = 29 to 30%




N -2    In a double pipe heat exchanger hot fluid is entering at 220°C and leaving at
        115°C. Cold fluid enters at 10oC and leaves at 75°C. The following data is
        provided for hot and cold fluids.
              Mass flow rate of hot fluid                     = 100 kg/hr
              Cp of hot fluid                                 = 1.1 kcal/kg°C
              Cp of cold fluid                                = 0.95 kcal/kg°C
(i)     Calculate LMTD
        a) For parallel flow
        b) For counter current flow
(ii)    Which flow arrangement is preferable and why?
(iii)   Find the mass flow rate of cold fluid if the heat loss during the exchange is 5%.


Ans:
a)      LMTD Parallel flow
                                           t1  t 2
        LMTD                 =
                                               t1
                                            In
                                               t 2
        t1                  =             210°C
        t2                  =             40°C
                                           210  40
        LMTD                 =                                = 102.5°C
                                               210
                                            ln
                                               40


b)      LMTD Counter current flow
        t1                  =             145°C

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Bureau of Energy Efficiency
                                  Paper 4 – Energy Auditor – Set A Solutions

       t2              =         105°C
                                  t1  t 2
       LMTD             =
                                      t1
                                   ln
                                      t 2
                                  145105
                        =                      = 123.9°C
                                      145
                                   ln
                                      105


2)     Counter flow is preferred since the LMTD is more, the area of the heat
       exchanger will be less


3)     Mass flow rate of cold fluid
       Data:
       m. Mass flow rate of hot fluid          =     100 kg/hr
       cph.                                    =     1.1 kcal/kg°C
       cpC                                     =     0.95 kcal/kg°C
       Cold fluid inlet temperature            =     10°C
       Cold fluid outlet temperature           =     75°C
       Hot fluid inlet temperature             =     220°C
       Hot fluid outlet temperature            =     115°C
       Q = m.hxCPh X th x 0.95                =     mc x CPC x tc


Mass flow rate of cold fluid mc                =



                                                     100 x 1.1 x (220  115 ) x 0.95
                                               =
                                                            0.95 x (75  10 )
                                               =     177.7 kg/hr

N-3    An efficiency trial was conducted in furnace oil fired boiler during the conduct of
       energy audit study and the following data were collected.
              Boiler Data:
              Rated capacity                         = 10 TPH (F&A 100oC)
              Rated efficiency                       = 84%
              Actual steam generation pressure       = 7 kg/cm2 (g) saturated
              Feed water temperature                 = 45oC
       Boiler was found to be operating at rated steam pressure and flow conditions
              Furnace Oil Data:
              Furnace oil consumption                = 600 litre per hour
              GCV of oil                             = 10200 kcal/kg
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Bureau of Energy Efficiency
                                   Paper 4 – Energy Auditor – Set A Solutions

             Specific gravity of oil                          = 0.92
             % Carbon                                         = 84%
             % Hydrogen                                       = 12%
             % Sulphur                                        = 3%
             % Oxygen                                         = Nil
             % Nitrogen                                       = 1%
             Cost                                             = Rs. 20/kg


        Flue Gas Data:
             % O2 in flue gas                                 = 5.5% by volume
             CO in flue gas                                   = Nil
             Flue gas temperature                             = 240oC
             Specific heat of flue gas                        = 0.24 kcal/kgoC
             Moisture in ambient air                          = 0.03 kg/kg of air
             Ambient air temperature                          = 40oC
             Assume surface heat and unaccounted losses = 2%
        Determine the following:
(i)     Boiler efficiency by indirect method
(ii)    Find out the annual savings in Rs per year if the boiler was operating at its rated
        efficiency.
(iii)   Also suggest possible measures to improve the efficiency of the boiler.

Ans:

Theoritical air Requirement for Furnace Oil
[(11 .6 x C )  {34 .8 x ( H 2  O2 / 8)}  (4.35 x S )] / 100 kg/kg of oil

Theoritical air required

        11.6 x 84 + ((34.8 x (12-0 )) + 4.35 x 3
                         100

        = 11.6 x 84 + 34.8 x 12 + 4.35 x 3 = 14.05 kg/kg of fuel

                        100

        Excess air Supplied (EA) =              5.5 x 100
                                                  21-5.5

        Excess air for 5.5 % O2 in flue gas              =    35.48%




Actual air supplied (AAS) = 1 + EA                x Theoretical air
                                100

                                     1 + 35.48 x 14.05=19.03 kg of air/kg of oil
                                         100

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Bureau of Energy Efficiency
                              Paper 4 – Energy Auditor – Set A Solutions

Mass of Dry Flue gas = .84 x 44 + 19.03x 77 + .01 + (19.03 - 14.05) x 23
                          12            100                          100

                       + 0.03 x 64/32
                     = 3.08 + 14.65 + 0.01 + 1.14 + 0.06
                     = 18.94 kg of air / kg of oil



        Calculation of All Losses :
1.      Dry flue gas loss          = 18.94 x 0.24 (240-40) x 100
                                            10200

                                        = 8.91%

2.      Loss due to Hydrogen in the Fuel = 9x 0.12 (584 + 0.45 (240-40) x 100
                                                      10200
                                         = 7.14%

3.      Loss due to Moisture vapour
        present in combustion air             = 19.03 x .03 x .45 (240-40) x 100
                                                                 10200
                                                     = 0.50%



4.      Boiler surface heat loss and unaccounted losses (given) = 2%

(i)     Boiler Efficiency               = 100 – (8.91 + 7.14 + 0.5 +2%)

                                        = 100-18.55 = 81.45%

        Efficiency Improvement from Existing to Rated = 2.55%
        Fuel Input after improvement
        in boiler efficiency                  = 600 x 0.8145 = 581.79 lit./hr
                                                 0.84
(ii)    Oil Saving per hour      = 600-581.79       = 18.21 litre/hr.

               (18.21 x 0.92) x Rs. 20 x 8000 = Rs. 26.80 lakhs

(iii)   1. Reduce excess air
        2. Reduce flue gas temperature (by cleaning the boiler tubes to get rid of
        scales or soot )


N-4     A V-belt centrifugal fan is supplying air to a chemical process. The performance
        test on the fan gave the following parameters.
         Ambient temperature                                        40oC
         Density of air at 0oC                                      1.293 kg/m3
         Diameter of the discharge air duct                         1m
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Bureau of Energy Efficiency
                               Paper 4 – Energy Auditor – Set A Solutions

         Velocity pressure measured by Pitot tube in              47 mmWC
         discharge duct
         Pitot tube coefficient                                   0.9
         Static pressure at fan inlet                             - 22 mmWC
         Static pressure at fan outlet                            188 mmWC
         Power drawn by the motor coupled with the fan            72 kW
         Belt transmission efficiency                             95%
         Motor efficiency at the operating load                   90 %

(i)     Find out the efficiency of the fan.
(ii)    Due to modification in the chemical process, only half of the operating flow will
        be required in future. This is to be effected by damper control method. The fan
        characteristic curve shows that the total static head developed by the fan will be
        333 mmWC and static fan efficiency will be 61% by damper control method.
        Find out what will be the annual savings at 8000 hours of operation per year and
        an energy cost of Rs. 4.50 /kWh. Assume that the motor efficiency and belt
        transmission efficiency remains same.
(iii)   List down the various energy conservation options to achieve the modified flow
        rate.

Ans:
i)
Ambient temperature                             40oC
Diameter of the discharge air duct              1m
Velocity pressure measured by Pitot tube        47 mmWC
Static pressure at fan inlet                    - 22 mmWC
Static pressure at fan outlet                   188 mmWC
Power drawn by the motor                        72 kW
Transmission efficiency                         95%
Motor efficiency                                90 %
Area of the discharge duct                      3.14 x 1 x 1/4
                                                0.785 m2
Pitot tube coefficient                          0.9
Corrected gas density                           (273 x 1.293) / (273 + 40) = 1.127

Air velocity                                     Cp x  2 x 9.81 x  p x      
                                                               

                                                0.9 x x Sq rt.(2 x 9.81 x 47 x 1.127)
                                                                 1.127
                                                25.7 m/s

Volume                                          25.7 x 0.785
                                                20.17 m3/s
Power input to the shaft                        72 x 0.95 x 0.9

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Bureau of Energy Efficiency
                                       Paper 4 – Energy Auditor – Set A Solutions

                                                             61.6 kW

                                     Volume in m3 / Sec x total static pressure in mmwc
         Static Fan Efficiency % 
                                           102 x Power input to the shaft in (kW)


Fan static Efficiency                                        20.17 x (188 – (-22)
                                                             102 x 61.6
                                                             67 %



ii)
New flow                                                     20.17/2 = 10.1
Fan shaft Power drawn due to flow                            10.1 x 333
reduction to 50 % by damper closing                          102 x 0.61
                                                             54 kW
Power drawn by the motor                                     54/(0.95 x 0.9) =63
Energy savings                                               (72 – 63) x 8000 x Rs.4.50
                                                             Rs.3,24,000/annum



iii) various energy conservation measures

      i.     Pulley change
      ii.    Impeller trimming
      iii.   New fan
      iv.    New motor



                                -------- End of Section - III ---------




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Bureau of Energy Efficiency

								
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