Regn No: _________________
Name: ___________________
(To be written by the
candidates)
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:
- 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
o
All
parts of a question should be answered at one place
Section – I: OBJECTIVE
TYPE Marks: 50 x 1 = 50
(i) Answer all 50 questions
(ii) Each question carries one mark
(iii) Put a (Ö) tick mark
in the appropriate box in the answer book
1.
|
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 b) 294 rpm c)
1176 rpm d) none of the above
|
2.
|
For centrifugal fans, the relation between
Power (kW) and Speed (N) is given by
a)
 = b) = c) = d) none of the above |
3.
|
The pressure flow
characteristic curve of a centrifugal fan changes with the following flow
control method
a) inlet damper b) outlet damper c)
inlet guide vane d) none of
the above
|
4.
|
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%
b) 56% c) 43.4% d) none of the above
|
5.
|
Which of the following is not true of air
receivers?
a)
increases the pressure of air b)
smoothen pulsating air output
c)
storage of large volumes of air d)
a source for draining of moisture
|
6.
|
If inlet and outlet water temperatures of a cooling tower are 42oC
and 36oC respectively and atmospheric DBT and WBT are 39 oC
and 32 oC respectively, then the effectiveness of cooling tower is
a)
60% b) 75% c) 85.7% d) none of the above
|
7.
|
The
efficiency of forward curved fans compared to backward-inclined fans is__
a)
lower b) same
c) higher d) none of the above
|
8.
|
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
a) 16.8
kW b) 25.2 kW c)
28.0 kW d) none of the above
|
9.
|
What is the
impact on flow and head when the impeller of a pump is trimmed?
a) both head and flow
decreases b) flow decreases and pump
head increases
c) both flow and pump head increases
d) none of the above
|
10.
|
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 b)
40 m c) 50 m
d) none of the above
|
11.
|
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 b) 200 to 230 deg C
c) 340
to 370 deg C d) none of the above
|
12.
|
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
b)
1000 kVA
c) 300 kVA d) none of the above
|
13.
|
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 d) all of the above
|
14.
|
The jacket
cooling water in a diesel engine flows at 12.9 m3/hr 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
b) 387 kW c) 430 kW d)
none of the above
|
15.
|
The maximum back pressure drop (mmWC)
allowed in a DG set is
a) 100-150 b) 150-200 c)
250-300 d) none
of the above
|
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 b) lower c) higher d)
none of the above
|
18.
|
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 b) large production
shops/hangars
c) conference halls d) street
lighting
|
20.
|
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 % b) 50% c) 60 % d) none of the above
|
22.
|
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 b) full load kVAr of
motor
c) No-load kVAr of motor d) none of the above
|
23.
|
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
b) 1000 kVAr c) 500 kVAr d) none of the above
|
24.
|
Power factor is the ratio of
a) kVAr/ (kW2+kVAr2)1/2 b) kW/ (kW2+kVAr2)1/2
c) (kW2+kVAr2)1/2/kW d)
kVAr/kW
|
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 b) 2943 RPM c) 2955 RPM d) none of the above
|
26.
|
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
b)
8 kVAr c) 10 kVAr d)
none of the above
|
27.
|
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
b) 1/3 c) 1/4 d)
none of the above
|
28.
|
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 b)
leading reactive power
c) lagging
reactive power d) none of the above
|
30.
|
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 b) Illuminance
c) Lux d) Luminous Efficacy
|
33.
|
Which is the most energy efficient lamp?
a) GLS b) LPSV c)
HPMV d) FTL
|
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 b) 1000 lumens per square
feet
c) 10 lumen per square metre
d)
one lumen per square metre
|
36.
|
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
b) range
c) enthalpy of inlet air d) outlet wet bulb
temperature
|
38.
|
Which of the
following ambient conditions will evaporate maximum amount of water in a
cooling tower
a) 35oC DBT and 30oC
WBT b) 40oC
DBT and 37oC WBT
c) 38oC DBT and 37oC
WBT d) 35oC
DBT and 29oC WBT
|
39.
|
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
b) 140 m3/hr c) 180 m3/hr d) none of the
above
|
40.
|
Lowering the
Cycles of Concentration (C.O.C) in circulating water in a cooling tower, the
blow down quantity will
a) decrease
b) not change c) increase d) none of the above
|
41.
|
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 b)
flow rate
c) suction
pressure d) system air leakages
|
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 b) 85 cfm c) 170 cfm
d) 256 cfm
|
44.
|
Dynamic
air compressors are mainly of the following type
a) centrifugal compressors b) two stage screw
compressors c) two stage
reciprocating compressors d)
none of the above
|
45.
|
The flow output of which of the
following changes with the discharge pressure
a) centrifugal compressor b)
reciprocating compressor
c) screw compressor d) none of the
above
|
46.
|
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 b) dew-point temperature
c)
ambient dry bulb temperature d) none of the above
|
48.
|
The refrigeration load in TR when 20 m3/hr
of water is cooled from a 13 oC to 8 oC is about
a) 80.3 b) 39.6 c) 33 d)
none of the above
|
49.
|
One
ton of refrigeration (TR) is equal to
a)
3516 W b) 200 BTU/min c)
50.4 kcal/min d) all of the above
|
50.
|
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 …….
Section – II:
SHORT DESCRIPTIVE QUESTIONS Marks: 10 x 5 = 50
(i) Answer all Ten questions
(ii) Each question carries Five marks
S-1 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 r
(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)
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 = 140 m3/hr
Specific
heat of water = 1 kcal/kg °C
Inlet
water temperature = 40 °C
Outlet
water temperature = 36 °C
Ambient
WBT = 31 °C
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 = 36-
31 = 5oC
S-7 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
22.71 kW x 860 kcal/kwh
= 19531 kcals/hr
Refrigeration tones = 19531/3024
= 6.46 Tonnes
S-8 Calculate
the free air delivery (FAD) capacity of a compressor in m3/hr for
the following observed data:
Receiver capacity: 0.4 m3
Initial pressure (with empty
receiver): 0 kg/cm2
(g)
Final pressure: 7
kg/cm2 (g)
Initial air temperature: 35oC
Final air temperature: 55 oC
Additional holdup volume: 0.05 m3
Compressor pump up time: 4.5 minutes
Atmospheric pressure: 1.026 kg/sq.
cm absolute
Ans:
= 
=
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 …….
Section – III:
LONG DESCRIPTIVE QUESTIONS Marks: 5 x 10 = 50
(i) Answer all Five questions
(ii) Each question carries Ten
marks
L-1
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.
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) Comments
on housekeeping issues like drip leakages, insulation, vibrations, etc.
L-2 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.
§ 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.
§ 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.
, DPst = 60
mmWC, hSt = ?
, P = 6.9 kW
= 20,000/3600 = 5.56 m3/sec
Fan static hSt =
Volume in m3/sec x DPst 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.
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) Supply
voltage = 410 V
(b) Current
dram = 9.7 A
(c) PF = 0.8
(d) Supply
frequency = 50.1 Hz
(e) RPM = 1480
Find out the following at the motor operating conditions:
1. Power
input in kW
2. % motor
loading
3. % 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.
Ans:
Minimum
payable demand = 5000 x 0.75 = 3750 kVA
Margin
available for reduction of MD = 4500 – 3750 = 750 kVA
Present
MD in kW = 4500 x 0.82 = 3690 kW
Desired
power factor = 3690/ 3750 = 0.984
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…….
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