DOC

NATIONAL CERTIFICATION EXAMINATION

By Jerome Hunt,2014-12-28 11:19
21 views 0
NATIONAL CERTIFICATION EXAMINATION

     1

    Model Question Paper - 2006

    NATIONAL CERTIFICATION EXAMINATION

     FOR

    ENERGY MANAGERS AND ENERGY AUDITORS

    PAPER 3 : ENERGY EFFICIENCY IN ELECTRICAL UTILITIES

Section 1 : OBJECTIVE TYPE

1. Power generating capacity in India, from coal based thermal power plants is

    about

    b) 70%

    2. Power loss in a line is the product of -

    a) Resistance and Square of the current

    3. Power factor is called as of the angle between kW and kVA

     a) Cosine

    th4. If the fundamental frequency of the power system is 50 Hz, 5 harmonic will

    have a frequency of -- Hz

    b) 250

    5. MD recorder is recording ---

    b) time-integrated demand over a predefined recording cycle 6. For a synchronous speed of 1500 rpm, at a given mains frequency of 50 Hz, the

    motor will have the number of poles as ---

    b) 4

    7. Squirrel cage motors are normally --- efficient than Slip ring motors

    a) more

    8. Core losses in a motor accounts for --- of its total losses

    b) 20-25%

    9. For efficient overall operation in a Plant, it is desirable to have

    a) High value of efficiency and high power factor

    10. Fixed losses in a motor consist of ---

    a) Magnetic core loss, friction and Windage losses

    11. For every 250 mm Wc pressure drop increase across at the suction path due to

    choked filter etc., the compressor power consumption increases by about ----

    percent for the same output

    b) 2

    12. The likely estimate on equivalent power wastage for a leakage from 7 bar

    compressed air system through 1.6 mm orifice size is ---kW

    c) 0.8

    13. Vertical type reciprocating compressors are used in the capacity range of ---

    cfm

    a) 50-150

     1

     2 14. The compression ratios for axial flow compressors are ---

    b) higher

    15. The temperature at which moisture condenses is called as ---

    b) Dew point

    16. The driving force for refrigeration in vapour absorption refrigeration plants is

    b) mechanical energy

    0 17. In general, designed chill water temperature drop across the chillers is ---C

    a) 5

    18. The compressor that has recently become practical in the Market is

    compressor

    b) Scroll

    19. Screw compressor is also called as compressor

    d) helical rotary

    20. The output of a reciprocating compressor is ---

    a) nearly constant

    21. Specific ratio is used to define fans, blowers and compressors as per ---

    c) ASME

    22. System curve of a fan depicts the relation between ---

    a) air flow and pressure

    23. Axial flow fans are equipped with ---blades

    a) variable pitch

    24. Peak efficiency of a radial fan is ---

    c) 69-75%

    25. The most efficient axial fan is of type

    c) vane axial

    26. Frictional losses in a pumping system is proportional to

    -2 b) Q

    27. The intersection point of the pump curve and the system curve is called ---

    b) best efficiency point

    28. Increased suction lift from open wells, the delivery flow rate ---

    b) decreases

    29. For large capacity of centrifugal pumps, design efficiencies are in the range of

    around ---

    a) 85%

    30. Throttling the delivery valve of a pump results in increased ---

    c) both (a) and (b)

    31. Natural draft cooling towers are mainly used in

    b) Power stations

    32. Cooling capacity of cooling tower is ---

    a) the heat rejected

    33. Which one of the following has the maximum effect on cooling tower

    performance?

    c) fill media

    34. This fill material is more energy efficient

    b) film fill

    35. Some of the components of cooling tower are:

     2

     3

    c) both (a) and (b)

    36. Colour rending index is measured in the scale of ---

    a) 1-100

    37. Which of the following lighting source has the least life?

    a) incandescent

    38. The average rated life of CFL is hr

    b) 10,000

    39. This device aids initial voltage build-up for a Tube light

    a) Choke

    40. For better street lighting, ---type of lamp is recommended

    a) HPSV

    41. Power requirement of DG set is determined by:

    b) minimum load

    42. Designed power factor of a DG set is

    b) 0.8

    043. The radiator cooling temperature for DG sets should be in the range of C

    d) 80-90

    44. High pressure gas injection is resorted to in the case of ---

    b) slow speed dual-fuel engines

    45. Water-cooled DG sets are ---air- cooled DG sets in performance

    c) more efficient

    46. Hydrodynamic principle of speed control is used in ---

    b) Fluid coupling

    47. MD controller is used to switch off ---loads, in logical sequence

    b) non-essential

    48. Identify the ―odd one‖ out from the following:

    d) capacitor based sensor

    49. Eddy current Drive is used to ---

    b) vary the output speed

    50. Amorphous core transformers are available at present up to KVA capacity

     d) 1600

    Section ii : SHORT DESCRIPTIVE QUESTIONS 10 X 5 =50 MARKS

    S-1 “One Unit saved in Industry is equal to Two Units generated in the Power Station”-

     Justify this Statement.

The generated power from the Power station is transmitted and distributed to consumers.

    Though the T & D losses are said to be around 17% in India and hence efficiency being 87%, all

    these losses may not constitute technical losses as, un-metered consumption and pilferage are

    also accounted in this loss.

    Before the generated power reaches Industry, it passes thro‘ transformers having 95%

    efficiency and in Industry the motor efficiency is about 90%. Another 30% is lost in mechanical

    system that includes coupling/driving system and driven equipments like pumps, valves etc. Thus

    the overall efficiency becomes 0.83 x 0.95 x 0.9 x 0.70 = 0.50 i.e. 50% efficiency.

     3

     4

    Hence, one unit saved in the Industry is equal to two units generated in the Power station.

S-2 What is an energy-efficient motor?

    An energy-efficient motor is the one in which, design improvements are incorporated to increase operating efficiency over motors of standard design. These improvements are in reducing motor losses, use of lower-loss silicon-steel, a longer core, thicker wires to reduce resistance, thinner laminations, smaller air gap between stator and rotor, copper instead of aluminium bars in the rotor, superior bearings and a smaller fan. As per the stipulations of BIS, energy-efficient motors are designed to operate with out loss in efficiency at loads between 75% and 100% of rated capacity.

S-3 What are the major compressed air system components?

Compressed air system consists of the major components as stated below:

     Intake Air Filters, preventing dust entering compressor

     Inter-stage Coolers, reducing the temperature of air before it enters the next stage

    to reduce the work of compression and increase efficiency

     After Coolers, removing moisture in the air by reducing the temperature in a water-

    cooled heat exchanger

     Air-dryers, removing the remaining traces of moisture

     Moisture Drain Traps, used for the removal of moisture in the compressed air

     Receivers, provided to store and smoothen the pulsating air output.

S-4 What are the several heat transfer loops in a refrigeration system?

The several heat transfer loops in a refrigeration system are:

     Indoor air loop: indoor air is driven in the leftmost loop by the supply air fan through a

    cooling coil, where it transfers its heat to chilled water. The cool air then cools the

    building

     Chilled water loop: It is driven by the chilled water pump and water returns from the

    cooling coil to the chiller‘s evaporator to be re-cooled

     Refrigerant loop: Using a phase-change refrigerant, the chiller‘s compressor pumps heat

    from the chilled water to the condenser water

     Condenser water loop: Water absorbs heat from the chiller‘s condenser and the

    condenser water pump sends it to the cooling tower

     Cooling tower loop: The cooling tower‘s fan drives air across an open flow of the hot

    condenser water, transferring the heat to the outdoors.

    S-5 What are the design and selection criteria in respect of fan?

    Accurate determination of air-flow and required outlet pressure are most important

    in the proper selection of fan type and size. The required air flow depends on the process

    requirements, normally determined from heat transfer rates or combustion air or flue gas

    quantity to be handled. System pressure requirement is to be computed, though it is

    difficult. Pressure drop across the length, bends, contractions and expansions in the ducting

    system, pressure drop across filters, drop in branch lines etc. A very conservative approach

    is adopted allocating large safety margins, resulting in over-sizing of fans, operating at flow

    rates much below their design values and consequently of poor efficiency.

     4

     5

    S-6 What are the affinity laws relating to a rotodynamic pump performance?

    The equations relating to rotodynamic pump performance of flow, head, and power absorbed, to speed are known as ―Affinity Laws‖. They are:

    1. Q is proportional to N 22. H is proportional to N 33. P is proportional to N

    Where,

    Q = Flow rate

    H = Head

    P = Power absorbed

    N = Rotating speed

In other words,

    Flow is proportional to speed

    Head is proportional to the square of speed

    Power is proportional to the cube of speed

    S-7 What is meant by Cycles of concentration and how is it related to cooling tower Blow down?

    Cycles of Concentration (COC) is the ratio of dissolved solids in circulating water to the dissolved solids in make up water.

    Blow down losses depend up on Cycles of concentration and evaporation losses and is related as:

    Blow down = Evaporation loss / (COC 1)

    S-9 What is „lamp efficacy‟? How savings in Industrial sector can be achieved by using high efficiency lamps?

    Lamp efficiency is the ratio of light output in lumens to power input to lamps in watts. High efficacy gas discharge lamps which are suitable for different types of applications, offer appreciable scope for energy conservation.

    In Industrial sector, instead of HPMV lamps of 250 W low efficacy, if changed to HPSV lamps of 150 W, there will be a saving of 100W in power saving will be 37%. For GLS lamps of 13 W existing, 9W CFL lamps, if used saves 31% power.

S-9 Briefly explain two important factors, to decide the type of DG set.

    The two most important factors to decide the type of DG set are: power and speed of the engine.

    The power requirement is determined by the maximum load. The engine power rating should be 10-20% more than the power demand by the end use.

    Speed of the engine should be such of a value that the fuel efficiency at that speed is the greatest. The DG sets should be run at this rated speed to avoid poor efficiency and to prevent build up of engine deposits due to incomplete combustion, leading to higher maintenance costs.

    S-10 Why Amorphous core transformers are preferred over conventional transformers?

     5

     6

    Amorphous core transformers provide excellent opportunities to conserve energy, due to

    low core loss. The reduction in energy loss over the conventional transformers is roughly

    around 70%. Amorphous core transformers are of less weight and hence easier to handle.

    Section-iii: LONG DESCRIPTIVE QUESTIONS Marks: 5 x 10 =50

    L-1. The maximum demand approved by a utility is 5500 kVA and tariff provides for minimum billing demand of 80% of approved. Review of past 12 months records of bills reveals that the monthly maximum demand recorded is around 4200 kVA.

    Will there be any benefits in surrendering part of contract demand? If so what is the kVA that you recommend for surrendering?

    Give the costs savings by surrendering demand, if unit rate for kVA is Rs. 200. ANS: The maximum demand approved by a utility is 5500 kVA and tariff provides for minimum billing demand of 80% of approved. Review of past 12 months records of bills reveals that the monthly maximum demand recorded is around 4200 kVA.

    Will there be any benefits in surrendering part of contract demand? If so what is the kVA that you recommend for surrendering?

    Give the costs savings by surrendering demand, if unit rate for kVA demand is Rs 200.

Approved maximum demand : 5500 kVA

    Minimum billing demand (MBD) : 4400 kVA

    Maximum demand recorded : 4200 kVA

    The approved maximum demand is selected such a way that the gap between MD recorded and minimum billing demand are narrowed down.

     A different scenario is to be created.

    Contract demand 5500 5400 5300 5200

    (Minimum Billing Demand) 80% of contract demand 4400 4320 4240 4116

    Monthly demand recorded 4200 4200 4200 4200

    From above, it would be advantage to surrender ‗300 kVA‘ demand and set new approved demand

    at 5200 kVA.

    The actual savings possible is calculated below:

    New contract (approved) demand :5200 kVA

    New minimum billing demand :4160kVA)

    Existing MD recorded :4200kVA

    Present billing demand :4200 (since recorded kVA is more than MBD)

     6

     7

    Original minimum billing demand : 4400kVA (Before surrendering demand)

    Reduction demand value taken for billing: 200 kVA

    Savings in demand charges/month Rs 200 * 200kVA

     Rs40,000

    L-2. What are the factors to be considered while selecting a motor? Explain in detail

ANS:

     Torque Requirement A.

    The primary consideration defining the motor choice for any particular application is the torque required by the load. The relationship between the maximum torque generated by the motor (break-down torque) and the torque requirements for start-up (locked rotor torque) and during acceleration periods is very important. The thermal loading on the motor is determined by the duty/load cycle. One important consideration with totally enclosed fan cooled (TEFC) motors is that the cooling may be insufficient when the motor is operated at speeds lower than its rated speed.

    B. Sizing to Variable Load

    Industrial motors frequently operate under varying load conditions due to process requirements. A common practice in cases where such variable loads are found is to select a motor based on the highest anticipated load. In many instances, an alternative approach is typically less costly, more efficient and provides equally satisfactory operation. With this approach, the optimum rating for the motor is selected on the basis of the load duration curve for the particular application. Thus, rather than selecting a motor of high rather than selecting a motor of high rating that would operate at full capacity for only a short period, a motor would be selected with a rating slightly lower than the peak anticipated load and would operate at overload for a short period of time. Since operating within the thermal capacity of the motor insulation is of greatest concern in a motor operating at higher than its rated load, the motor rating is selected as that which would result in the same temperature rise under continuous full-load operation as the weighted average temperature rise over the actual operating cycle.

    L-3. List out different basic capacity control methods for the fans and blowers

    Different basic capacity (volume) control methods adopted in fans and blowers are as follows:

    1. Changing the rotational speed is the most efficient. If the volume requirement is

    constant, it can be achieved by selecting appropriate pulley sizes. If the volume

    varies with the process, adjustable-speed drives can be used.

    2. Changing the blade angle is a method used with some vane-axial fans.

     7

     8

    3. Restricting the air flow is accomplished with dampers or valves which close off the

    air flow at the inlet or outlet. Inlet vanes, which swirl the air entering the

    centrifugal fan or blower, are more efficient than dampers or butterfly valves.

    4. Venting the high-pressure air, or recirculating it to the inlet, is often used with

    positive-displacement blowers. It is sometimes used with fan systems, but is the

    least efficient method as there is no reduction in the air being moved.

L-4. Explain the methodology of refrigeration plant energy audit.

ANS: The cooling effect produced is quantified as tons of refrigeration.

    1 ton of refrigeration = 3024 kCal/hr heat rejected.

    The specific power consumption kW/TR is a useful indicator of the performance of refrigeration system. By measuring refrigeration duty performed in TR and the Kilo Watt inputs measured, kW/TR is used as a reference energy performance indicator. The refrigeration TR is assessed as TR = Q (C ( (T T) / 3024 pio

    Where Q is mass flow rate of coolant in kg/hr

    C is coolant specific heat in kCal /kg deg C p

    0T is inlet, temperature of coolant to evaporator (chiller) in C i

    0T is outlet temperature of coolant from evaporator (chiller) in C. o

    The above TR is also called as chiller tonnage. In a centralized chilled water system, apart from the compressor unit, power is also consumed by the chilled water (secondary) coolant pump as well condenser water (for heat rejection to cooling tower) pump and cooling tower fan in the cooling tower fan. Effectively, the overall energy consumption would be towards ;

    ; Compressor kW

    ; Chilled water pump kW

    ; Condenser water pump kW

    ; Cooling tower fan kW, for induced / forced draft towers

    The specific power consumption for certain TR output would therefore have to include :

    ; Compressor kW/TR

    ; Chilled water pump kW/TR

    ; Condenser water pump kW/TR

     8

     9

    ; Cooling tower fan kW/TR

    and overall kW/TR as a sum of the above.

    In case of air conditioning units, the air flow at the Fan Coil Units (FCU) or the Air Handling Units (AHU) can be measured with an anemometer. Dry bulb and wet bulb temperatures are measured at the inlet and outlet of AHU or the FCU and the refrigeration load in TR is assessed as:

    Q ρ h h;;inoutTR 3024

    Where Q is the air flow in CMH

    3 is density of air kg/m

    h is enthalpy of inlet air kCal/kg in

    h is enthalpy of outlet air kCal/kg out

    Use of handy psychometric charts can help to calculate h and h from dry bulb, wet bulb inout

    temperature values which are, in-turn measured, during trials,by a whirling psychrometer.

    L-5. Describe the step by step methodology of lighting system audit in a plant?

ANS:

    Improvement option in lighting at any facility would involve following step by step approach: a. Step-1: Inventorise the lighting system elements with respect to device rating, population &

    use profile.

    b. Step-2: Measure and document lux levels at various plan locations at working place, at day

    time and night times w.r.t lamps ON or OFF during the said period.

    c. Step-3: Use a portable load analyzer to measure and document the voltage and power

    consumption profile at various lighting load distribution panels

    d. Step-4: Compare the measured lux values with standard values and identify locations of

    under-lit and over-lit areas

    e. Step-5: Analysis of failure rates of lamps, ballasts and actual life expectancy levels from

    the past data.

    f. Step-6: Based on above careful assessment has to be carried out along with energy saving

    potential, investment required and payback calculations.

     9

Report this document

For any questions or suggestions please email
cust-service@docsford.com