BEE Whole-Testpaper Contributed by Not Specified updated on Jan 2021

Section - II: SHORT DESCRIPTIVE QUESTIONS Marks: 8 x 5 = 40

(i) Answer all Eight questions
(ii) Each question carries Five marks

S-1 What is a Sankey diagram and what are its uses ? Explain with an example.

ANSThe Sankey diagram is a very useful tool to proportionally represent an entire input and output energy flow in any energy equipment or system such as boilers, fired heaters, furnaces etc. after carrying out an energy balance calculation, This diagram represents visually various outputs and losses so that energy managers can focus on finding improvements in prioritized manner.


ANS NPV = -100,000 + (75000-50000)/(1+0.15) + 75000/(1+0.15)2 + 75000/(1+0.15)3

= -100,000 + 21,739 + 56,710 + 49, 313
= Rs. 27,762 

S-3 In a compressed air Dryer, electrical heater is used for regeneration of silica gel. The present Electrical energy consumption is 100 kWh/day. The management intends to replace the electrical heater by steam coil. 

a)How much steam is need per day? 

b) Calculate cost savings/year. Cost of power is Rs.4/kWh and cost of steam is Rs. 500/ton (Assuming only latent heat of steam used. Latent heat of steam is 540 kCal/kg. Efficiency of steam heating is 70%, operating days = 300)

ANSLatent heat of steam = 540 kCal/kg
a)Amount of steam required = 100 * 860/(540*0.7 = 227.5 kg/day
b)Cost of power = 100 * 4 = Rs.400/day Cost of steam = 0.2275 * 500= Rs. 113.75/day Annual cost savings = (400 – 113.75) * 300 = 286.25*300 = Rs. 85,875/-

S-4Renovation and Modernization (R&M) of a 210 MW coal fired thermal power plant was carried out to enhance the operating efficiency from 28% to 32%. The specific coal consumption was 0.7 kg/kWh before R&M. For 8000 hours of operation per year, and assuming the coal quality remains the same, calculate 

a)the coal savings per year and 

b)the expected avoidance of CO2 into the atmosphere in Tons/year if the emission factor is 1.53 kg CO2/kg coal

ANSSpecific coal consumption of conventional power plant = 0.7 kg/kWh Specific coal consumption of power plant after R&M= 0.7 x (0.28/0.32) = 0.6125 kg/kWh
Coal savings per annum = (0.7-0.6125) x 210 x 1000 x 8000 
= 1,47,000 Tonnes per annum CO2 avoidance = 1,47,000 x 1.53 = 224910 tonnes per annum

S-5 Explain Time of Day (TOD) Tariff and how it is beneficial for the power system and consumers?

i. In Time of the Day Tariff (TOD) structure incentives for power drawl during off-peak hours and disincentives for power drawl during peak hours are built in. 
ii. Many electrical utilities like to have flat demand curve to achieve high plant efficiency. 
iii. ToD tariff encourage user to draw more power during off-peak hours (say during 11pm to 5 am, night time) and less power during peak hours. Energy meter will record peak and off-peak consumption and norma period separately. 
iv. TOD tariff gives opportunity for the user to reduce their billing, as off peak hour tariff is quite low in comparison to peak hour tariff. 
v. This also helps the power system to minimize in line congestion, in turn higher line losses and peak load incident and utilities power procurement charges by reduced demand

S-6 An energy meter connected to a 3 phase, 18.75 kW pump shows 108 units consumption for six hours of operation. The load on the motor was steady. The consumer doubted the energy meter reading and electrical parameter such as current, voltage and power factor were measured. The measured values were 430 V line volts, 25 amps line current and 0.80 Power Factor. Find out if the energy meter reading is correct.

ANS Energy consumption = √ 3 x 0.430 (kV) x 25(A) x 0.80(PF) x 6(hours) = 89. 37 kWh The consumption shown by energy meter is very high; higher by 21%

S-7 What are the three flexible mechanisms available under Kyoto protocol for achieving GHG reduction targets. Explain briefly the mechanism applicable to India.

ANS 1. Joint implementation
2. Emissions trading
3. Clean Development Mechanism (CDM)
CDM is applicable to India. It is a mechanism by which the developed countries finance/fund projects intended to reduce GHG emissions in return for emission reduction credits. This will help them to achieve the emission reduction targets. Alternatively the emission reductions achieved by developing countries can also be purchased,

S-8 What is meant by the following terms ?

a) Normalising of data

b) Benchmarking

ANS a) Normalising of data
The energy use of facilities varies greatly, partly due to factors beyond the energy efficiency of the equipment and operations. These factors may include weather or certain operating characteristics. Normalizing is the process of removing the impact of various factors on energy use so that energy performance of facilities and operations can be compared.
a) Benchmarking
Comparison of energy performance to peers and competitors to establish a relative understanding of where our performance ranks.

Section – III: LONG DESCRIPTIVE QUESTIONS Marks: 6 x 10 = 60 

(i) Answer all Five questions
(ii) Each question carries Ten marks


L-3 Production rate from a paper machine is 340 tonnes per day (TPD). Inlet and outlet dryness to paper machine is 50% and 95% respectively. Evaporated moisture temperature is 80 °C. To evaporate moisture, the steam is supplied at 3.5 kg /cm2. Latent heat of steam at 3.5 kg /cm2 is 513 kCal/kg. Assume 24 hours/day operation. 

i) Estimate the quantity of moisture to be evaporated.

ii) Input steam quantity required for evaporation (per hour).

Note: Consider enthalpy of evaporated moisture as 632 kCal/kg.
ANS Production rate from a paper machine = 340 TPD
= 14.16 TPH (tonnes per hour)

Inlet dryness to paper machine = 50%
Outlet dryness from paper machine = 95%

i) Estimation of moisture to be evaporated:
Paper weight in final product = 14.16 x 0.95 
= 13.45 TPH

Weight of moisture after dryer = 0.708 TPH

Paper weight before dryer on dry basis = 13.45 TPH

Weight of moisture before dryer = ((13.45 x 100)/50) – 13.45 = 13.45 TPH

Evaporated moisture quantity : 13.45 - 0.708 = 12.742 TPH

ii) Input steam quantity required for evaporation
Evaporated moisture temperature = 80 °C
Enthalpy of evaporated moisture = 632 kCal/kg
Heat available in moisture (sensible & latent) = 632 x 12742
= 8052944 kCal /h

For evaporation minimum equivalent heat available should be supplied from steam

Latent Heat available in supply steam (at 3.5 kg/cm2 (g)) = 513 kCal/kg

Quantity of steam required = 8052944/ 513 
= 15697.75 Kg/hr
= 15.698 MT/hour

L-6a)Explain the functioning of an ESCO in performance contracting.

b) Name two macro and micro factors considered in the sensitivity analysis of major energy saving projects.
Ans a)
ESCOs are usually companies that provide a complete energy project service, from assessment to design to construction or installation, along with engineering and project management services, and financing. 
In performance contracting, an end-user (such as an industry, institution, or utility),seeking to improve its energy efficiency, contracts with ESCO for energy efficiency services and financing. An agreed portion of the savings is shared with the ESCO. The ESCO gets back the invested money plus profit over a period of time

b)Micro factors 
• Operating expenses (various expenses items) 
• Capital structure 
• Costs of debt, equity 
• Changing of the forms of finance e.g. leasing 
• Changing the project duration

Macro factors Macro economic variables are the variable that affects the operation of the industry of which the firm operates. They cannot be changed by the firm’s management. Macro economic variables, which affect projects, include among others: 
• Changes in interest rates 
• Changes in the tax rates 
• Changes in the accounting standards e.g. methods of calculating depreciation
• Changes in depreciation rates 
• Extension of various government subsidized projects e.g. rural electrification
• General employment trends e.g. if the government changes the salary scales 
• Imposition of regulations on environmental and safety issues in the industry
• Energy Price change
• Technology changes

…….……. End of Section – III ………..….