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
o
o
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Please check that this question paper contains 4 printed pages
Please check that this question paper contains 16 questions
The question paper is divided into three sections
All questions in all three sections are compulsory
All parts of a question should be answered at one place
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. may be given mark
S-2
What are the disadvantages of heating the charge above the optimum
temperature in steel re-rolling furnaces?
1. Increased energy consumption
rates
S-3
2. Increase in scale losses 3. High rejection
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
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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
S-6
Static fan efficiency =
(m3/s) x pressure gain in Pascal
. 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
Static Fan Efficiency % 
Volume in m 3 / Sec x tota l pressure in mmwc
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.…
0.25 – 0.30
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Paper 4 – Energy Auditor – Set A Solutions
-------- 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),
(hd - hs) = Total head in metres
hd - hs
=
g = acceleration due to gravity (m/s2),
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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Paper 4 – Energy Auditor – Set A Solutions
(550/800) = (N1/1485)
N1 = 1021
The pump shaft power would be:
3
=
L-2
1021
1485
x 114
= 37 kW
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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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
Boiler efficiency
- 18 bar (g), 310OC
- 81%
Investment required
Generated power
- Rs. 20 crores
= 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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Paper 4 – Energy Auditor – Set A Solutions
Rs. 0.52 crores
Maximum demand charges per
annum
1000 x 12 x Rs.300
Rs. 0.36 crores
Total cost per year
Savings
8.5 + 0.52 + 0.36 = 9.38 crores
16.8 – 9.38 = Rs. 7.42 crores
=
Investment
20 = 7.42
1
(1+i)1
Rs. 20 crores
1
(1+i)2
+
+
1
(1+i)3
+
1
(1+i)4
+
1
(1+i)5
+
1
(1+i)6
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
Cp of hot fluid
Cp of cold fluid
(i)
= 100 kg/hr
= 1.1 kcal/kg°C
= 0.95 kcal/kg°C
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)
b)
LMTD Parallel flow
t1  t 2
t1
In
t 2
LMTD
=
t1
=
210°C
t2
=
40°C
LMTD
=
210  40
210
ln
40
= 102.5°C
LMTD Counter current flow
t1
=
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Bureau of Energy Efficiency
145°C
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Paper 4 – Energy Auditor – Set A Solutions
t2
=
LMTD
=
=
105°C
t1  t 2
t1
ln
t 2
145 105
145
ln
105
= 123.9°C
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
N-3
=
=
100 x 1.1 x (220  115) x 0.95
0.95 x (75  10)
=
177.7 kg/hr
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
Rated efficiency
Actual steam generation pressure
Feed water temperature
= 10 TPH (F&A 100oC)
= 84%
= 7 kg/cm2 (g) saturated
= 45oC
Boiler was found to be operating at rated steam pressure and flow conditions
Furnace Oil Data:
Furnace oil consumption
GCV of oil
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Bureau of Energy Efficiency
= 600 litre per hour
= 10200 kcal/kg
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Paper 4 – Energy Auditor – Set A Solutions
Specific gravity of oil
% Carbon
% Hydrogen
% Sulphur
% Oxygen
% Nitrogen
Cost
= 0.92
= 84%
= 12%
= 3%
= Nil
= 1%
= Rs. 20/kg
Flue Gas Data:
% O2 in flue gas
CO in flue gas
Flue gas temperature
Specific heat of flue gas
= 5.5% by volume
= Nil
= 240oC
= 0.24 kcal/kgoC
Moisture in ambient air
Ambient air temperature
= 0.03 kg/kg of air
= 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 ( H2  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
Actual air supplied (AAS) = 1 + EA
100
=
35.48%
x Theoretical air
1 + 35.48 x 14.05=19.03 kg of air/kg of oil
100
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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
(ii)
Oil Saving per hour
= 600 x 0.8145 = 581.79 lit./hr
0.84
= 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
Density of air at 0oC
Diameter of the discharge air duct
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40oC
1.293 kg/m3
1m
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Paper 4 – Energy Auditor – Set A Solutions
Velocity pressure measured by Pitot tube in
discharge duct
Pitot tube coefficient
Static pressure at fan inlet
Static pressure at fan outlet
Power drawn by the motor coupled with the fan
Belt transmission efficiency
Motor efficiency at the operating load
47 mmWC
0.9
- 22 mmWC
188 mmWC
72 kW
95%
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
Diameter of the discharge air duct
Velocity pressure measured by Pitot tube
Static pressure at fan inlet
Static pressure at fan outlet
Power drawn by the motor
Transmission efficiency
Motor efficiency
Area of the discharge duct
Pitot tube coefficient
Corrected gas density
Air velocity
40oC
1m
47 mmWC
- 22 mmWC
188 mmWC
72 kW
95%
90 %
3.14 x 1 x 1/4
0.785 m2
0.9
(273 x 1.293) / (273 + 40) = 1.127
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
Power input to the shaft
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Bureau of Energy Efficiency
25.7 x 0.785
20.17 m3/s
72 x 0.95 x 0.9
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Paper 4 – Energy Auditor – Set A Solutions
61.6 kW
Static Fan Efficiency % 
Volume in m3 / Sec x total static pressure in mmwc
102 x Power input to the shaft in (kW)
20.17 x (188 – (-22)
102 x 61.6
67 %
Fan static Efficiency
ii)
New flow
Fan shaft Power drawn due to flow
reduction to 50 % by damper closing
Power drawn by the motor
Energy savings
20.17/2 = 10.1
10.1 x 333
102 x 0.61
54 kW
54/(0.95 x 0.9) =63
(72 – 63) x 8000 x Rs.4.50
Rs.3,24,000/annum
iii) various energy conservation measures
i.
ii.
iii.
iv.
Pulley change
Impeller trimming
New fan
New motor
-------- End of Section - III ---------
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