Paper 4
Model question paper 20006
NATIONAL CERTIFICATION EXAMINATION
FOR
ENERGY AUDITORS
PAPER – 4: ENERGY PERFORMANCE ASSESSMENT FOR EQUIPMENT AND
UTILITY SYSTEMS
Duration: 2 HRS
Section - I:
Max. Marks: 100
SHORT DESCRIPTIVE QUESTIONS
Marks: 10 x 1 = 10
(i) Answer all Ten questions
(ii) Each question carries One marks
(iii) Answer should not exceed 50 words
S-1
How boiler capacity rating is specified and why..
Boilers capacity is always specified in terms of equivalent evaporation
(kg of steam / hour at 100°C). The amount of steam generation - "from
and at" 100°C. It is easy to compare the ratings of different boilers from
different manufactures.
S-2
List any two losses that are difficult to measure while determining the
furnace efficiency evaluation.
The losses that are difficult to measure in the furnace efficiency
evaluation are
heat storage loss, loss of furnace gases around
charging door and opening, loss of heat by conduction through hearth
and loss due to formation of scales.
S-3
What is the relation between effectiveness and length of the heat
exchanger for the same duty?
Ratio of the cold fluid temperature range to that of the inlet temperature
difference of the hot and cold fluid. Higher the ratio lesser will be
requirement of heat transfer surface and there by length of heat
exchanger
S-4
Which method is used to find the calorific value of unburned contents in the fly
ash and bottom ash
Gravimetric analysis is used to find calorific value of fly ash and bottom ash
NPC-AIP-2006
1
Paper 4
S-5
How to determine the Friction & wind age loss alone in electric motor
efficiency evolution.?
To separate core and F & W losses, no load test should be repeated at variable
voltages. By plotting no-load input kW versus Voltage, the intercept will indicate F &
W kW loss component.
S-6
Explain the characteristic of power factor in electric motor with VSD and
without VSD drives?
Variable frequency drives should also offer a true system power factor of 0.95
or better across the operational speed range, to save on demand charges, and to protect
the equipment (especially motors). In conventional motor, PF will redce at part load.
S-7
What is the main difference in the system characteristics curves between fans
and pumps distribution network.
There is no static head in FAN system characteristic curves whereas pump
has static head
S-8
S-9
How to ensure the accurate measurements in squire duct while measuring the
airflow with pit tot tubes.

When in use, the pitot tube shall be connected by means of airtight tubes to
a pressure measuring instrument.

For rectangular duct, let us calculate the traverse points. 16 points are to be
measured. In small ducts or where traverse operations are otherwise
impossible, an accuracy of ± 5% can frequently be achieved by placing
Pitot in center of duct.
Which of the following is better choice for maximum power generation and
why?
Site A: wind speed of 25 KM per hour & capacity factor 20%
Site B: wind speed of 29 KM per hour & capacity factor 19%
Site-A is better choice because power generation is high for any scapacity of
wind mill
S-10
Compare the advantages between Direct burning process and Biogas
generation foe effective utilization of Cow dung.
Bio gas route has many benefits like energy conversion efficiency is high ,
useful energy is high per kg of Dung plus free manure .
-------- End of Section - I ---------
NPC-AIP-2006
2
Paper 4
Section - II:
LONG DESCRIPTIVE QUESTIONS
Marks: 2 x 5 = 10
(i) Answer all Two questions
(ii) Each question carries Five marks
L-1
(i) What are the methods of recycling of waste minimization and Give two
example each.
i)
On-site Recovery and Reuse - Reuse of wasted materials in the same
process or for another useful application within the industry.
ii)
Production of Useful by-product - Modification of the waste generation
process in order to transform the wasted material into a material that can be reused
or recycled for another application within or outside the company.
L-2
In an industrial office building lighting system, the ILER ratio has been
improved from 0.6 to 0.8 for reducing lighting power consumption by modifying
fittings layout. The initial lighting circuit load was 5KW. Calculate the
percentage of waste reduction and annual energy savings if operating period is
10 hours and 360 days /year.
Before modification
Annual energy wastage = (1 - ILER) x K.Watts x no. of operating hours
= (1 - 0.6) x 5 x 10 hrs/day x 360 days
= 7200 kWh/year
After modification
= (1 - 0.8) x 5 x 10 hrs/day x 360 days
= 3600 kWh/year
i)
% waste reduction 50%
ii)
Annual energy savings= 3600 kwh/year
-------- End of Section - II ---------
Section - III:
Numerical Questions
Marks: 4 x 20 = 80
(i) Answer all Four questions
NPC-AIP-2006
3
Paper 4
(ii) Each question carries Twenty marks
N -1
A process industry has planned to install a Cogeneration plant with capital
investment of Rs. 18 Crores. The estimated annual energy savings is Rs 6
Crores for the 6 years life cycle. The annual operating cost would be 20 % of
the annual savings. The minimum expected internal rate of return is 18%
considering their capital resources.
I. Find out whether the investment will meet the company expectations.
II. Perform the sensitivity analyses for the following two scenario changes and
suggest the project feasibility.
a. If Annual savings is increased by 20% than expected
b. If annual saving is increased by 20% and operating cost also
increased from the existing 20 to 30 %
Solution :
Investment
Annual Savings
Annual Operating Cost
Annual Net Cash Flow
Expected
IRR
by
Company
(i ) NPV
= 18 Crores
= 6 Crores
= 6x0.2 = 1.2 Crores
= 6-1.2 = 4.8 Crores
the = 18.0%
= CF0
CF1
------ + ----- …
(1+r)0 (1+r)1
CF6
….+ -----(1+r)6
1
1
1
1
------ + ------ + --- .. +...----}
(1.18)1 (1.18)2 (1.18)3 (1.18)6
= -18 + 4.8 (3.5573)
= -18 + 17.07
= (-) 0.925
As NPV is negative, it is not meeting the
Company requirement. Therefore project is
not feasible.
= -18+ 4.8 {
(ii) Sensitivity Analysis
Scenario-1
Annual Saving with 20 % raise = 6 Crores x 1.2 = 7.2 Crores
Annual Operating Cost
= 20% of Annual Savings
= 7.2 x 1.2 = 1.44
Annual net cash flow
= 7.2 – 1.44 = 5.76
NPV
NPC-AIP-2006
= -18 + 5.76 (3.5573)
= -18 – 20.490
4
Paper 4
= (+) 2.49 (Positive)
Scenario-2
Annual Saving with 20% raise
Operating Cost
Annual net cash flow
NPV
= 6.0 x 1.22 = 7.2 Crores
= 30% of Annual Savings
= 7.2 x 1.3 = 2.16 Crores
= 7.2 – 2.16 = 5.04 Crores
= -18+5.04 (3.5573)
= (-) 0.712
In this case, NPV is Negative
As with sensitivity factor NPV becomes
negative. Hence this project is not feasible.
N -2
A In a five star hotel building air conditioning system, Cold air at 23oc is
supplied from air handling unit. The cold air flow rate is 20,000 M3/hr at
a density of 1.2 Kg/m3 .The inlet and outlet enthalpy of the air are 105
KJ/Kg and 80 KJ/Kg. The COP of the system is 3.75. Hotel
management wants to install Double effective VAR SYSTEM .The
saturated steam at 5kg/cm2 will be supplied either from 500 KVA DG
Sets exhaust gas boiler or from the existing LDO Fuel fired boiler. The
plant is operating for 8760 hr. The investment VAR system is 20 lacs.
The investment for waste heat boiler is 6 lacs. Power cost is Rs 4/KWH.
As an energy auditor which option can be recommended to the hotel
management?
Option1: Supply steam from the existing LDO fired boiler to VAR
system and avoid the investment of waste heat boiler
Option2 - Supply steam from the waste heat boiler, which needed
investment of waste heat boiler in addition to VAR system
The steam consumption per TR will be 4.7 Kg/TR at 5kg/cm2 pressure.
The cost of LDO is Rs.16,800 Ton
Solution :
Existing Base Case VCR System
TR Rating
COP
NPC-AIP-2006
= (20,000m3/hr x 1.2kg/m3) (105-80)KJ/kg
-----------------------------------------------------3024 x 4.187
= 47.38 TR
= 3.75
5
Paper 4
COP
Power Input
Annual Energy Consumption
Annual cost in VCR system (Base
Case)
= Refrigeration effect
Kcal/hr
----------------------------------------Power Input
Kcal/hr
= 47.38 x 3024 Kcal
-----------------------------3.75 x 860
= 44.43 KW
= 44.43 x 8760 = 3.892 Lakhs KWH
=3.892 x4= Rs. 15.56 Lakhs
Option – 1 : VAR System with Steam Supply from Existing Boiler
Steam Consumption /TR
= 4.7 kg/TR
Steam Consumption per hr
= 4.7 x 47.38 = 222.7 kg/hr
Evaporation Ratio IN THE
EXISTING BOILER
1ton of steam cost
= 14
= Rs.16800
-------------- = Rs.1200 /Ton of steam
14
Investment for VAR system
= Rs.20.00 Lakhs
Electricity cost saving per hr
= 44.43 x 4 =Rs. 178
Steam cost per hr
= 222.7 x 1.2 = 277
Cost Savings /hr
= 178-277 = negative
Hence this option of project is not feasible
Option-2: With VAR & steam supply from WHR steam boiler of DG set
Total Investment
= Rs. 20.00 + 6.00 = 26.00 Lakhs
Savings/hr
= Rs. 178 – Free steamt = Rs.178
Annual Savings
= Rs.178x8760 = Rs.15.60 lakhs
Simple Pay back period
= 26/15.6 = 1.67 years.
Solution : Option 2 should be selected
N-3 An FAD test was carried out on a a1000 cfm compressor using nozzle
method .The following are the measurements and results of the test.
Standard Nozzle Diameter selected for 1000cfm capcity : 0.08 m
Flow coefficient of the Nozzle : 0.95
Receiver Pressure maintained = 4.2 kg / cm2 (a)
Inlet Pressure of the - 1.03 kg / cm2(a)
Inlet air temperature 35oC
Pressure before nozzle maintained= 1.06 kg / cm2 a
Temperature before the nozzle = 48oC
Pressure drop across the nozzle = 0.052 kg / cm2
Gas constant
: 287 Joules / kg K
NPC-AIP-2006
6
Paper 4
After the test the compressor is taken for maintenance and overhaul. The isothermal
efficiency increased thereafter by 5 percent and the shaft power of compressor was
measured to be 115 kW. The expenditure was Rs. 1.5 lacs. What is the payback period
for the investment if the operating hours are 8000/annum and the energy cost is
Rs.4/kWh
Compressor capacity ,1000 cfm
1700 m3/hr
Nozzle size
0.080 m
1/ 2
T  2( P3  P4 )( P3 x Ra ) 

Free air delivered , Q f (m / sec)  k x x d x 1 x 
4
P1 
T3


s
 0.95 x

4
2
1/ 2
x (0.080) 2 x
308  2 x 0.052 x (1.06 x 287) 
x

1.03 
321

= 0.44 m3/s
= 1614 m3/hr
Isothermal power (kW ) 
P1 x Q f x log e r
36.7
Isothermal power
1.03 x 1614 x ln (4.2/1.03)
36.7
63.7 kW
New power drawn
New isothermal efficiency
115 kW
=
63.7 / 115 = 55.4 %
Old isothermal efficiency
= 55.4 / 1.05 = 52.76
Shaft power drawn earlier
63.7 / 0.5276
120.7 kW
Annual savings
= (120.7 – 115) x 8000 x 4
Rs. 1,82,400
Payback period
NPC-AIP-2006
1.5/1.824=10 months
7
Paper 4
N-4
A 10 Ton per hour heat treatment furnace is operating for 20 days
continuously in month. An air preheater was installed to preheat the
combustion air temperature. The measurements were taken before and
after WHR implementation
Before Installation of Air preheater
Fuel oil Consumption
= 620 litre/hr
Specific Gravity
= 0.88
Furnace Operating Temperature
= 1000oC
Oxygen % at outlet of furnace
= 4.5%
Measurement after Installation of Air preheater
%O2after Waste Heat Recovery
= 4.5%
Investment for WHR
= Rs. 100.00 lakhs
Flue gas temperature after WHR = 600 oC
Ambient temperature of air
= 30oC
Fuel oil cost
= Rs.17,000/ KL
GCV of fuel oil
= 10200 kCal/kg
Find out pay back period for the investment?
Solution :
Existing loss in flue gas :
02 in Flue Gas before WHR
% Excess Air
Fuel Consumption
Dry flue gas loss %
= 4.5%
= 4.5 x 100
21-4.5
= 27.3%
= 620 lit. x 0.88 = 545.6kg/hr
= [14x1.273)+ 1] x .24(1000-30) x100
10200
= 40.6%
After WHR installation
After WHR Installation % O2 in flue gas = 4.5%
= 4..5 x 100 = 27.3 %
21-4.5
Dry flue gas loss
NPC-AIP-2006
= [ (14x1.27.3) +1] x0.24 (600-30) x100
10200
= 25.24%
8
Paper 4
% Reduction in the dry flue gas loss
Fuel oil saving/hr
Operating hours
Annual fuel oil saving
Annual cost saving
Simple Pay back period
= 40.6-25.24
= 15.36%
= 620 lit/hr x 15.36 = 95.22 lit/hr
100
= 20 days x 24 hrsx12months =5760 hrs
= 5760 x 95.22 = 548467 lit =548.5KL /Year
= 548.5 KLx 17000 = 93.245 Lakhs
= 100 / 93.245 = 1.07 YEARS
End
NPC-AIP-2006
9