Paper EA4 - Energy Auditor - Set B Solutions


                                  ENERGY AUDITORS

PAPER – EA4: Energy Performance Assessment for Equipment and Utility Systems

Date: 23.05.2004        Timings: 1400 – 1600 HRS    Duration: 2 HRS.     Max. Marks: 100

General instructions:
      o   Please check that this question paper contains 3 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 Questions                                           Marks: 10 x 1 = 10

            (i)    Answer all Ten questions
            (ii)   Each question carries One mark

S-1       Why does an energy auditor prefer to establish the efficiency of a steam boiler
          based on the indirect method?

(i) The indirect method derives an efficiency number by identifying the nature
and quantities of the losses occurring in the system. The direct method arrives
only at a number without identifying and quantifying losses.

(ii) The direct method is also too inaccurate for testing solid fuel fired boilers
because fuel flow must be measured introducing a large inaccuracy

S-2       Explain why a project with a high IRR is not necessarily more attractive then a
          project with a lower IRR.

The IRR cannot distinguish between lending and borrowing and hence a high
IRR is not necessarily a desirable feature.

Also the NPV is the increase in shareholders or companies wealth through the
project. It therefore can happen that a project A has a very high IRR but low
NPV, while another project B has a low IRR but high NPV. In this case B should
be selected.

S-3       What are the two most important rules to improve measurement accuracy
          when measuring airflow with a pitot tube in a large duct?

(i) To insure accurate velocity pressure readings, the Pitot tube tip must be
pointed directly into (parallel with) the air stream.

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Bureau of Energy Efficiency
                          Paper EA4 - Energy Auditor - Set B Solutions

(ii) To obtain the average total velocity in ducts of 100 mm diameter or larger, a
series of velocity pressure readings must be taken at points of equal area.

Even in a turbulent velocity profile the air velocity decreases towards the duct
walls. Consequently measuring the velocity in the middle of a duct is

S-4    Name the two most common types of extended surface heat exchangers.

1. Plate – fin heat exchanger
2. Tube - fin heat exchanger

S-5    Define volumetric efficiency for a compressor. State the units.

(i) The ratio of free air delivered to compressor swept volume.
(ii) Units are (m3/min)/(m3/min). Multiplied by 100 results in % efficiency.

S-6    Is there a conceptual difference between COP and EER? Explain how they

(i) There is no conceptual difference. Both parameters are calculating with
different units the ratio “refrigeration effect/ power input.”.

(ii) The relation is COP = 0.293 EER.

S-7    Define color rendering index.

Color Rendering Index (CRI) is a measure of the effect of light on the perceived
color of objects.

S-8    Define the term “overall plant fuel rate”. State the units.

                                    Fuel consumption in kg / hr
Overall plant fuel rate kg/kWh =
                                       power output in kW

The fuel consumption refers to the fuel to generate the electrical power as well
as the steam of a cogeneration system

S-9    Which three variables need to be measured to calculate the efficiency of a
       pump. State which units apply in an SI system?

The variables to be measured (i) volumetric flow in cubic meter per second, (ii)
pressure difference across the pump in Pascal and (iii) electrical power input to
the motor driving the pump in Watt.

S-10   Under what circumstances does cogeneration not make any sense?

Cogeneration does not make any sense whenever the demand for process
steam does not coincide with demand for electrical power, because neither
steam nor electricity can be stored in large amounts.

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

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                           Paper EA4 - Energy Auditor - Set B Solutions

Section - II:    Long Questions                                        Marks: 2 x 5 = 10

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

L-1     Assume that the boiler efficiency is calculated by the direct method using
        either NCV or GCV.
       (i) What is the difference in the efficiency calculation using NCV or
       (ii) State an example where boiler efficiency is more than 100%, if not,
            explain why this is not possible.

Within the framework of Book 4 the energy efficiency of a boiler is defined as

                                       useful energy output
                              fuel energy input as either NCV or GCV

a) The value of useful energy output does not change and is independent of whether we
use NCV or GCV in the calculation.
b) The value of NCV is always smaller than the value of GCV because GCV contains the
heat of vaporization of the water in the fuel and the water generated by the hydrogen in
the fuel. The relation is

                           9 Hydrogen  w
NCV  GCV  584                           in kCal/kg of fuel
where Hydrogen is the percentage of hydrogen in the fuel and w is the percentage of
water in the fuel. The constant 584 is the latent heat of vaporization of water in kCal/kg

(i) Answer: The difference in the calculation is that the efficiency calculated by
using NCV is always larger than the efficiency calculated by using GCV for the
same boiler and the same operating conditions because using NCV one divides
by a smaller number but useful energy output is the same.

(ii) Answer: All systems where NCV is used in the efficiency equation but flue
gas temperature is reduced below 100 C in a heat exchanger. This means the
heat of vaporization is recovered and useful energy output is therefore larger
then the energy input resulting in efficiencies larger 100%.

L-2     Which is one of the first essential steps in determining the suitability of a
        variable speed drive in a pump system? Explain why!

Answer: The first essential step is to observe the load pattern of the pump and
drawing a graph by plotting percent operating hours on the Y axes versus
percent rated flow on the X-axes, or vice versa.

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                                Paper EA4 - Energy Auditor - Set B Solutions

                                -------- 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      Determine the simple payback period of the incremental investment for two
         transformers with the following details

                                                          Option A                      Option B
         Capacity                                         315 kVA                       315 kVA
         Efficiency at rated capacity                     98%                           99%
         Capital cost                                     Rs. 2.2 lakhs                 Rs. 3.0 lakhs

         Assume the following for both the transformers
         Operating PF at rated capacity                                       = 0.9
         No load losses                                                       = same
         Energy charge                                                        = Rs. 4.50/kWh

         For the analysis consider two cases for the length of time during which the
         transformers are used at rated capacity
         (a) 10 hours/day and 250 days/year of operation
         (b) 16 hours/day and 300 days/year of operation

Two solutions are possible depending on whether the kVa is taken as transformer
input or output.

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                          Paper EA4 - Energy Auditor - Set B Solutions

Solution 1 (Taking kVA as output)
                    315 kVA 0.9
kW drawn case A                  289.286

                    315 kVA 0.9
kW drawn case B                  286.364

Reduction in kW by buying transformer B instead of A is
289.286 – 286.364 = 2.922 kW
                    1        1 
Short cut is 315               0.9  2.922 kW
                    0.98 0.99 

Scenario a) Annual cost savings are = 10 x 250 x 2.922 x 4.50 = Rs 32,873
and simply payback period is (300,000 – 220,000)/32,873 = 2.43 years

Scenario b) Annual cost savings are = 16 x 300 x 2.922 x 4.50 = Rs 63,115
and simply payback period is (300,000 – 220,000)/63,115 = 1.27 years

Solution 2 (Taking kVA as input)

Based on the above, the solution to the problem can also be derived as below

Total demand                                 = 315 kVA

Power factor                                        = 0.9

Power demand                                        = 315 x 0.9 = 283.5 kW

Losses with transformer A (98 %)                    = 283.5 x 0.02 = 5.67 kW

Losses with transformer B (98.5 %)           = 283.5 x 0.01   = 2.835 kW

Difference in losses                         = 5.67 – 2.835 = 2.835 kW

Scenario a) Annual cost savings are = 10 x 250 x 2.835 x 4.50 = Rs 31893.75
and simply payback period is (300,000 – 220,000)/31893.75 = 2.5 years

Scenario b) Annual cost savings are = 16 x 300 x 2.835 x 4.50 = Rs 61,236
and simply payback period is (300,000 – 220,000)/61,236 = 1.3 years

N-2    A reciprocating single stage compressor coupled with an electric motor has a
       mechanical shaft power requirement of 75 kW at a discharge pressure of 600
       kPa. Determine the energy cost savings if the discharge pressure is reduced to
       500 kPa of both isothermal and adiabatic compression processes. Assume
       the following for the existing and modified pressure conditions:
       Intake air pressure = 1 atmosphere,  motor operating efficiency = 95%
       Average load factor = 75%,           operating hours = 7000 hours/year
       Average energy charge = Rs. 4.50/kWh No change in remaining parameters

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Bureau of Energy Efficiency
                          Paper EA4 - Energy Auditor - Set B Solutions

Use equation for isothermal power (kW) on page 110 Book 4.

Scenario (i) Isothermal process

a)      kW = P1 x Qf x ln (r) / 36.7
b)      Reduction in percent = 100 x (kW@600 – kW@500)/kW@600

        = 100 x (ln(5.922) –ln(4.935)) /ln(5.922) = 10.25%
c)      Annual savings = 75 x 0.75 x 7000 x 0.1025 x 4.5/0.95 =            Rs 191,176

Adiabatic compression efficiency solution may be ignored.

N-3     It is proposed to install at the beginning of the year a heat recovery equipment
        in a food processing industry. The capital cost of the equipment is Rs 20,000/-.
        The savings accrued by the unit are constant and Rs 5,000/- annually. The
        discount rate is 25%.
        (i)     Calculate the Net Present Value (NPV) for 5 years.
        (ii)    Is the investment recovered after 5 years? Explain!
        (iii)   Is the investment recovered after 7 years? Explain!
        (iv)    Estimate the IRR for this investment after 7 years if the salvage value
                of the equipment is Rs 2,000 at the end of 7th year.

The fastest solution is to use the equation as shown in Book 1, or use table
format as on page 140 in Book 4.

(I)     NPV over 5 years at 25% = -20,000 + 5,000/ 1.25 + 5,000/ 1.252 + 5,000/
        1.253 + 5,000/ 1.254 + 5,000/ 1.255 = MINUS 6554 Rs.

(II) The investment is not recovered after 5 years because the NPV is negative.

(III)   NPV = Result of (i) + 5,000/ 1.256 + 5,000/ 1.257 = Minus Rs 4195.
        The investment is still not recovered because NPV is still negative

(IV)    Use discount rate of 17.4% which results in NPV ~ 0 over 7 years.
        Therefore IRR is estimated to be 17.4% .

N -4    The following are the data collected for a boiler using furnace oil as the fuel.
        Determine the boiler efficiency based on GCV by indirect method ignoring
        radiation and convection losses.
        Ultimate chemical analysis (% weight) :       Carbon : 84, Hydrogen : 12,
        Nitrogen : 0.5, Oxygen : 1.5, Sulphur: 1.5, Moisture : 0.5, NCV of fuel 9,763
        kCal/kg and humidity 0.015 kg moisture/kg of dry air.
        Flue gas analysis: CO2 : 10.5% volume, flue gas temperature : 180oC and
        ambient temperature : 20oC


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                          Paper EA4 - Energy Auditor - Set B Solutions

Calculation is an almost exact copy of page 14 in Book 4 where coal would be
replaced by fuel oil, or page 19.

a) Theoretical Air = 11.43 x 0.84 + 34.5 x (0.12 – 0.015/8) + 4.32 x 0.015 = 13.74


  Theoretical Air = 11.6 x 0.84 + 34.8x (0.12 – 0.015/8) + 4.35 x 0.015 = 13.92

b) Theoretical % CO2 = Moles of C/(Moles of N2 + Moles of C + Moles of Sulphur)
= 100 x (0.84/12)/(13.74 x 0.77/28 + 0.005/28 +0.84/12 + 0.015/32) =

Theoretical % CO2     = Moles of C/(Moles of N2 + Moles of C + Moles of Sulphur)
= 100 x (0.84/12)/(13.92 x 0.77/28 + 0.005/28 +0.84/12 + 0.015/32) = 15.44%

Depending on the variation in the above values (a and b) the subsequent
calculations will also have a minor variation in the end result, which can be

c) Excess air supplied = 7900 x(15.62 -10.5)/(10.5(100-15.62)) =           45.65%

d) Actual mass of air supplied = (1+0.4565) x 13.74 =              20.01 kg/kg of oil.

e) Actual mass of dry flue gas =
0.84 x 44/12 + 0.005 + 20.01 x 0.77 + 0.015 x 64/32 +(20.01-13.74) x 0.23 =
                                                          19.96 kg/kg oil

At this point it is required to derive GCV from NCV

f) GCV = 9,763 + 584 x (9 x 0.12 + 0.005) =                        10,397 kcal/kg

g) % loss by dry flue gas = 19.96 x 0.23 x (180-20)/10,397 = 0.07065 or    7.06%

h) % loss by formation of water from Hydrogen in the fuel
9 x 0.12 x (584 + 0.45 (180-20))/10,397 = 0.681 or                         6.81%

i) % loss due to moisture in fuel = 0.005 x (584 +0.45 (180-20))/10,397 = 0.03%

j) % loss due to moisture in air = 20.01 x 0.015 x 0.45 (180-20)/10,397 = 0.21%

k) Boiler efficiency = 100- 7.06 – 6.81 – 0.03 – 0.21 =                    85.89%.

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

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