Paper 3 –Set B Solutions
Regn No: _________________
Name: ___________________
(To be written by the candidates)
NATIONAL CERTIFICATION EXAMINATION 2006
FOR
ENERGY MANAGERS & ENERGY AUDITORS
PAPER – 3:
Energy Efficiency in Electrical Utilities
Date: 23.04.2006
Timings: 0930-1230 HRS
Duration: 3 HRS
Max. Marks: 150
General instructions:
o
o
o
o
o
Please check that this question paper contains 9 printed pages
Please check that this question paper contains 65 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
Section – I: OBJECTIVE TYPE
(i)
(ii)
(iii)
1.
Answer all 50 questions
Each question carries one mark
Put a () tick mark in the appropriate box in the answer book
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
2.
b) 294 rpm
2
kW1 N 1
=
kW2 N 2 2
d) none of the above
3
b)
kW1 N 1
=
kW2 N 2 3
c)
kW1 N 1
=
kW2 N 2
d) none of the above
The pressure flow characteristic curve of a centrifugal fan changes with the following flow
control method
a) inlet damper
4.
c) 1176 rpm
For centrifugal fans, the relation between Power (kW) and Speed (N) is given by
a)
3.
Marks: 50 x 1 = 50
b) outlet damper
c) inlet guide vane
d) none of the above
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%
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Bureau of Energy Efficiency
b) 56%
c) 43.4%
d) none of the above
1
Paper 3 –Set B Solutions
5.
Which of the following is not true of air receivers?
a) increases the pressure of air
c) storage of large volumes of air
6.
If inlet and outlet water temperatures of a cooling tower are 42 oC and 36oC respectively
and atmospheric DBT and WBT are 39 oC and 32 oC respectively, then the effectiveness
of cooling tower is
a) 60%
7.
b) 75%
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
b) 25.2 kW
c) 28.0 kW
b) 40 m
d) none of the above
b) 200 to 230 deg C
d) none of the above
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
13.
c) 50 m
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
c) 340 to 370 deg C
12.
b) flow decreases and pump head increases
d) none of the above
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
11.
d) none of the above
What is the impact on flow and head when the impeller of a pump is trimmed?
a) both head and flow decreases
c) both flow and pump head increases
10.
d) none of the above
b) same
d) none of the above
a) 16.8 kW
9.
c) 85.7%
The efficiency of forward curved fans compared to backward-inclined fans is__
a) lower
c) higher
8.
b) smoothen pulsating air output
d) a source for draining of moisture
b) 1000 kVA
c) 300 kVA
d) none of the above
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
14.
The jacket cooling water in a diesel engine flows at 12.9
with a range of 10oC and
accounts for 30% of the engine input energy. The power output of the engine will be
a) 500 kW
15.
d) all of the above
m 3/hr
b) 387 kW
c) 430 kW
d) none of the above
The maximum back pressure drop (mmWC) allowed in a DG set is
a) 100-150
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Bureau of Energy Efficiency
b) 150-200
c) 250-300
d) none of the above
2
Paper 3 –Set B Solutions
16.
The electronic ballast in lighting application does not have one of the following
characteristics
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
18.
b) lower
c) higher
d) none of the above
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
c) conference halls
20.
b) large production shops/hangars
d) street lighting
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 %
22.
b) 50%
c) 60 %
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
c) No-load kVAr of motor
23.
b) full load kVAr of motor
d) none of the above
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
24.
d) none of the above
b) 1000 kVAr
c) 500 kVAr
d) none of the above
Power factor is the ratio of
a) kVAr/ (kW 2+kVAr2)1/2
c) (kW 2+kVAr2)1/2/kW
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Bureau of Energy Efficiency
b) kW/ (kW 2+kVAr2)1/2
d) kVAr/kW
3
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
26.
d) none of the above
b) 8 kVAr
c) 10 kVAr
d) none of the above
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
28.
c) 2955 RPM
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
27.
b) 2943 RPM
b) 1/3
c) 1/4
d) none of the above
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
harmonics
29.
A pure capacitive load in an alternating current (AC) circuit draws
a) active power
c) lagging reactive power
30.
b) leading reactive power
d) none of the above
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
higher
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
c) Lux
33.
b) Illuminance
d) Luminous Efficacy
Which is the most energy efficient lamp?
a) GLS
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Bureau of Energy Efficiency
b) LPSV
c) HPMV
d) FTL
4
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
c) 10 lumen per square metre
36.
b) 1000 lumens per square feet
d) one lumen per square metre
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
c) enthalpy of inlet air
38.
Which of the following ambient conditions will evaporate maximum amount of water in a
cooling tower
a) 35oC DBT and 30oC WBT
c) 38oC DBT and 37oC WBT
39.
b) 140 m3/hr
c) 180 m3/hr
d) none of the above
Lowering the Cycles of Concentration (C.O.C) in circulating water in a cooling tower, the
blow down quantity will
a) decrease
41.
b) 40oC DBT and 37oC WBT
d) 35oC DBT and 29oC WBT
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
by
a) 160 m3/hr
40.
b) range
d) outlet wet bulb temperature
b) not change
c) increase
d) none of the above
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
c) suction pressure
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Bureau of Energy Efficiency
b) flow rate
d) system air leakages
5
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
44.
b) 85 cfm
c) 170 cfm
Dynamic air compressors are mainly of the following type
a) centrifugal compressors
c) two stage reciprocating compressors
45.
b) two stage screw compressors
d) none of the above
The flow output of which of the following changes with the discharge pressure
a) centrifugal compressor
c) screw compressor
46.
d) 256 cfm
b) reciprocating compressor
d) none of the above
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
c) ambient dry bulb temperature
48.
The refrigeration load in TR when 20 m 3/hr of water is cooled from a 13 oC to 8 oC is
about
a) 80.3
49.
b) 39.6
c) 33
d) none of the above
One ton of refrigeration (TR) is equal to
a) 3516 W
50.
b) dew-point temperature
d) none of the above
b) 200 BTU/min
c) 50.4 kcal/min
d) all of the above
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 …….
_______________________
Bureau of Energy Efficiency
6
Paper 3 –Set B Solutions
Section – II: SHORT DESCRIPTIVE QUESTIONS
(i)
(ii)
S-1
Marks: 10 x 5 = 50
Answer all Ten questions
Each question carries Five marks
What are the technical aspects of energy efficient motors with respect to insulation
life, slip and starting torque?
Ans:
Lower temperatures in energy efficient motors translate to long lasting
insulation. Generally, motor life doubles for each 10°C reduction in operating
temperature.
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
efficiency.
Ans:
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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Bureau of Energy Efficiency
7
Paper 3 –Set B Solutions
Ans:
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
Ans:
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?
Ans:
Velocity V, m/s = CP x (2 x 9.81 Δp x y)1/2
y
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
Specific heat of water
Inlet water temperature
Outlet water temperature
Ambient WBT
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= 140 m3/hr
= 1 kcal/kg °C
= 40 °C
= 36 °C
= 31 °C
8
Paper 3 –Set B Solutions
Ans:
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
S-7
= 36- 31 = 5oC
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.
Ans:
Power input to compressor
=
=
0.89 x 8.5
7.57 kW
Cooling effect
=
=
7.57 x 3.0
22.71 kW
=
19531 kcals/hr
=
=
19531/3024
6.46 Tonnes
22.71 kW x 860 kcal/kwh
Refrigeration tones
S-8
Calculate the free air delivery (FAD) capacity of a compressor in m3/hr for the
following observed data:
Receiver capacity:
Initial pressure (with empty receiver):
Final pressure:
Initial air temperature:
Final air temperature:
Additional holdup volume:
Compressor pump up time:
Atmospheric pressure:
0.4 m3
0 kg/cm2 (g)
7 kg/cm2 (g)
35oC
55 oC
0.05 m3
4.5 minutes
1.026 kg/sq. cm absolute
Ans:
=
P2  P1 V  273  t1 

  
P0
t  273  t 2 
=
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Bureau of Energy Efficiency
9
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
motors
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
respectively.
Ans:
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 …….
_______________________
Bureau of Energy Efficiency
10
Paper 3 –Set B Solutions
Section – III: LONG DESCRIPTIVE QUESTIONS
(i)
(ii)
L-1
Marks: 5 x 10 = 50
Answer all Five questions
Each question carries Ten marks
Briefly explain the step-by-step approach for conducting Energy Performance
Assessment of DG set on shopfloor.
Ans:
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
difference.
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Bureau of Energy Efficiency
11
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)
L-2
Comments on housekeeping issues like drip leakages, insulation, vibrations,
etc.
List down any 10 energy conservation opportunities in compressed air system.
Ans:

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
percent.

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
load.
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Bureau of Energy Efficiency
12
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
compressor.

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
settings.

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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Bureau of Energy Efficiency
13
Paper 3 –Set B Solutions

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
wastage.

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.
Ans:
(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.
_______________________
Bureau of Energy Efficiency
14
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)
(b)
(c)
(d)
(e)
Supply voltage
Current dram
PF
Supply frequency
RPM
=
=
=
=
=
410 V
9.7 A
0.8
50.1 Hz
1480
Find out the following at the motor operating conditions:
1.
2.
3.
Power input in kW
% motor loading
% slip
Ans:
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
structure.
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.
_______________________
Bureau of Energy Efficiency
15
Paper 3 –Set B Solutions
Ans:
Minimum payable demand = 5000 x 0.75 = 3750 kVA
Margin available for reduction of MD
=
4500 – 3750 = 750 kVA
Present MD in kW
=
3690 kW
=
0.984
=
4500 x 0.82
Desired power factor = 3690/ 3750
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
months
……. End of Section - III …….
_______________________
Bureau of Energy Efficiency
16