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**2nd NATIONAL CERTIFICATION EXAMINATION 2005
FOR
**

**ENERGY AUDITORS
**

**MODEL TEST SERIES-1
**

**Paper – 4: Energy Performance Assessment for Equipment and Utility Systems
**

**Duration: 2 HRS
Max. Marks: 100
**

**General instructions:
**o Please check that this question paper contains **16 **questions
o The question paper is divided into three sections
o All questions in all three sections are compulsory
o All parts of a question should be answered at one place
o **Open Book Examination
**

**Section – I: Short Questions
Marks: 10 x 1 = 10
**(i) Answer all **Ten **questions
(ii) Each question carries **One **mark

1 How boiler rated capacity is specified?

Conventionally, boilers are specified by their capacity to hold water and the steam generation rate. Often, the capacity to generate steam is specified in terms of equivalent evaporation (kg of steam / hour at 100oC). Equivalent evaporation- “from and at” 100oC. The equivalent of the evaporation of 1 kg of water at 100oC to steam at 100oC.

2 What are the components of surface heat loss in the furnaces and its dependent factors affecting loss?

Radiation loss: The low components of Surface heat loss are Surface temperature and emissivity and Convection loss: Radiation loss is defined as Surface temperature and wind Velocity

3 How Efficiency and Power Factor varies in motors with VSD Drives.

The variable frequency drive should have an efficiency rating of **95% or better at full
load**.

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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).

4 **What are the three reasons for poor fan performance in the field?
**

**a) Leakage, recirculation or other defects in the system
b) Excessive loss in system component close to the fan
**c) Disturbance to fan performance due to bend or other system components

5 **How do you determine system resistance for a pump?
**

**System resistance = frictional head + total static head
**

**Frictional head in a system of pipes roughly varies as the square of the capacity
flow. If system resistance for full flow is known, resistance for other flow can be
calculated as % of full flow.
**

6 What are the measuring instruments required for a compressed air delivery test by nozzle method ?

• Thermometers or Thermocouple • Pressure gauges or Manometers • Differential pressure gauges or Manometers • Standard Nozzle

7 A refrigeration plant operating at 600 T is consuming 564 kW of compressor power. Find out the kW/ton, COP and EER

KW /ton = 0.94

COP = 3.72

EER = 12.8

8 What is Lamp Circuit Efficacy?

Lamp Circuit Efficacy is the amount of light (lumens) emitted by a lamp for each watt of
*power consumed by the lamp circuit, i.e. including control gear losses. *This is a more
meaningful measure for those lamps that require control gear. Unit: lumens per circuit watt
(lm/W

9 What is the capacity factor of wind turbine

The Capacity Factor (CF) is simply the wind turbine's actual energy output for the year divided by the energy output if the machine operated at its rated power output for

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the entire year. A reasonable capacity factor would be 0.25 to 0.30 and a very good capacity factor would be around 0.40. It is important to select a site with good capacity factor, as economic viability of wind power projects is extremely sensitive to the capacity factor

10 **Name four type of wastage due to poorhouse keeping?
**

Wastage of water oil steam due to the following

**• Leaking taps / valves / flanges
**• Spillages
• Overflowing tanks
• Worn out material transfer belts

**Section - II: Long Questions
Marks: 2 x 5 = 10
**(i) Answer all **Two **questions
(ii) Each question carries **Five **marks

1

Compare the advantage of biogas generation over the direct burning of 25 kg biomass? Bio gas generation has higher efficiency and additional manure generation compared to burning process as shown below

2 What are the Factors Affecting Furnace Performance ?

The important factors, which affect the efficiency, are listed below for critical analysis. • Under loading due to poor hearth loading and improper production scheduling • Improper Design • Use of inefficient burner • Insufficient draft/chimney • Absence of Waste heat recovery • Absence of Instruments/Controls • Improper operation/Maintenance • High stack loss • Improper insulation /Refractories

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**Section - III: Numerical Questions
**

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**Marks: 4 x 20 = 80
**(i) Answer all **Four **questions
(ii) Each question carries **Twenty **marks

1 A Boiler Efficiency trial was conducted in an Furnace Oil fired process boiler during Energy Audit Study and the following data were collected.

**Boiler Data :
**

Boiler Rated Capacity = 10 TPH (F&A 100oC) Rated Boiler Efficiency = 84% Actual steam generation pressure = 7 kg/cm2 (g) Saturated Feed water Temperature = 45oC

**Fuel Oil Data :
**

Furnace Oil Consumption = 600 litre per hour GCV of Oil = 10200 Kcal/kg Specific gravity of oil = 0.92 % Carbon = 84% % Hydrogen = 12% % Sulphur = 3% % Oxygen = Nil % Nitrogen = 1%

**Flue Gas Data :
**

% O2 in Flue Gas = 5.5% by volume CO = Nil Flue Gas Temperature = 240oC Specific Heat of Flue Gas = 0.24 Moisture at Ambient air = 0.03 kg/kg of air Ambient air temperature = 40oC

Determine the boiler efficiency by indirect method by assuming 2% boiler surface heat loss. Also find out the fuel oil saving in litre per hour, if efficiency of boiler was improved to Rated efficiency.

**Solution :
**

**Theoritical air Requirement for Furnace Oil
**

Theoritical air required based on ultimate analysis.

11.6 x 84 + ((34.8 x (12-0 )) + 4.35 x 3 100

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= 11.6 x 84 + 34.8 x 12 + 4.35 x 3 = 14.05 100

Excess air Supplied (EA) = 5.5 x 100 21-5.5

Excess air for 5.5 % O2 in flue gas

= 35.48%

Actual air supplied (ASS) = 1 + EA x Theoretical air 100

1 + 35.48 x 14.05 = 19.03 kg of air/kg of oil 100

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 = 19.94 kg of air / kg of oil

**Calculation of All Losses :
**

1. Dry flue gas loss = 19.94 x 0.24 (240-40) x 100 10200

= 9.38%

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 (given) = 2%

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( i )Boiler Efficiency = 100 – (9.38 + 7.14 + 0.5 +2%)

= 100-19.02 = 81%

Efficiency Improvement from Existing to Rated = 3%

Fuel Input after improvement in boiler efficiency = 600 x .81 = 578.57 lit./hr 0.84

( ii ) Oil Saving per hour = 600-578.57 = 21.42 litre/hr.

2 A process plant requires 28 tonnes of steam per hour. The power requirement is 2250 kW. The plant operates for 8000 hours per annum.

Steam is generated at 2 bar 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/tonne.

Power is drawn from the grid at Rs. 4/kWh. The contract demand is 3000 kVA and the company is charged for 100 % of the contract demand at Rs. 300/kVA.

The company is planning for a back pressure cogeneration plant using the same coal with the following parameters. The power and steam demand were fully met by the cogeneration plant.

Boiler generation pressure - 18 bar, 310OC Boiler efficiency - 81 % A contract demand of 1000 kVA with the grid was kept for emergency purpose. Investment required - Rs. 20 crores Generated power = 2250 kW Find out the IRR over a project life cycle of 6 years

Coal consumption 28,000 x [(2706.23/4.18) – 80] 0.75 x 4000

5296 kg/hr

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Fuel cost per annum 5.296 x 2000 x 8000

Rs. 8.5 crores

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

Total electricity bill per annum Rs. 8.3 crores

Total energy bill per annum Rs. 16.8 crores

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**With cogeneration plant
**

Enthalpy of steam at turbine inlet 3052.56 kJ/kg

Coal consumption 28,000 x [(3052.56/4.18) – 80] 0.81 x 4000

5780 kg/hr

Incremental coal consumption 484 kg/hr

Incremental fuel cost per annum 0.484 x 8000 x Rs.4000

Rs. 1.5 crores

Maximum demand charges per annum 1000 x 12 x Rs.300

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Rs. 0.36 crores

Total cost per year 8.5 + 1.5 + 0.36 = 10.36 crores

Savings = 16.8 – 10.36 = Rs. 6.44 crores

Investment Rs. 20 crores

IRR 27 %

3 **A centrifugal pump at base of cooling tower pumps 120 m3/hr at pressure 2.1 kg/
cm2. The cooling tower range was measured to be 4oC. What is the power input at
the motor? (Efficiency of pump is 65% and motor efficiency is 82%).
**

**Pump is throttled such that cooling tower range was 6oC. What is the new flow rate
under throttled conditions? (Pressure drop across the throttle valve is 0.5 kg/cm2).
**

**If instead of throttling, the existing impeller is replaced with a new impeller at a
cost of Rs.25000/- find the operating point and differential savings vis-à-vis
throttling case considering 8760 operating hours and unit cost of Rs.4/- (assume
pump efficiency of 63% and motor efficiency of 82%)
**

**Head load in cooling tower with range of 40C = 120 x 1000 x 4
= 480 x 103 kCal/hour
**

**Total pump head = 2.1
= 2.1 kg/cm2
Motor input power = 120 x 21 x 1000 x 9.81
3600 x 0.65 x 0.82 x 1000
**

** = 12.88 kW
**

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**Under throttled conditions
Flow rate in cooling tower with range of 60C = 480 x 103 / (6 x 1000)
= 80 m3/hour
**

**Flow rate = 80 m3/hr
Head = 21+ 5
= 26 m
**

**Motor input power = 80 x 26 x 1000 x 9.81
3600 x 0.65 x 0.82 x 1000
= 10.62 kW
**

**After replacing the existing impeller with a new impeller to give the required flow
rate of 80 m3/hr, the new head will be
**

**Considering the fan law
**

**Under impeller replaced conditions
**

**Q1 = 120 m3/hour, H1 = 21 m, Q2 = 80 m3/hr, H2=?
**

** (Q1/Q2)2 = (H1/H2)
**

** H2 = H1 x (Q2/Q1)2
**

** = 21 x (80/120)2
= 9.33 m
**

**Motor input power = 80 x 9.33x 1000 x 9.81
3600 x 0.63 x 0.82 x 1000
**

** = 3.94 kW
**

**Annual saving = ( 10.62-3.94) x 8760 x 4
= Rs.96360/-
**

**Investment = Rs. 234067/-
**

**Simple payback period = 25000 / 96360
= 0.26 year
~ 3.2 months
**

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4 In an air conditioning system, the air flow rate is 80,000 m3/hr at a density of 1.2 kg/m3.. The inlet and outlet enthalpies of air at Air Handling Unit is 105 kJ/kg and 79 kJ/kg. The COP of the system is 3.82. It is proposed to replace it with a vapour absorption system at a cost of Rs. 50 lacs with an annual maintenance cost of Rs.4 lacs.. The steam consumption will be 4.5 kg/hr/TR. The steam is to be generated by a coal fired boiler with an evaporation ratio of 4.5. The energy electrical energy cost is Rs.4 per kWh and the system operates for 8000 hrs per annum. The cost of coal is Rs.2000/tonne. Find out the payback period for the investment.

Refrigeration load
**= m x (hin – hout)
4.18 x 3024
**

M = mass of airflow = 80000 x 1.2 = 96000 kg/hr

Refrigeration load (TR) = [96000 x (105 – 79)] / (4.18 x 3024)

=198 tons

KW/TR = 3.516 / COP

COP = 3.82

KW/TR = 3.516 / 3.82

= 0.92

Power drawn (kW) = 198 x 0.92

= 182 kW

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Annual energy cost = 182 x Rs.4 x 8000

= Rs.58.24 lacs

Steam required for VAR = 198 x 4.5

= 891 kg/hr

Cost of steam = 2000 / 4.5

= Rs.444/ton

= Rs 0.44 /kg

Steam cost for VAR = 0.44 x 891

= Rs.392/hr

Annual cost = Rs. 392 x 8000

= Rs.31.36 lacs

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Annual maintenance cost = 4 lakhs

Payback period = 50 / 31.36 –4)

= 22 months

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