(SEM- VII) THEORY EXAMINATION 2021-22 POWER PLANT ENGINEERING

POWER PLANT ENGINEERING – KME 076

B.Tech (Semester VII) – Detailed Answers

SECTION A – Descriptive Answers (2 Marks Each)

a) Brake Power

Brake power is the actual power output available at the crankshaft of an engine. It is the useful power measured by a brake dynamometer and is always less than the indicated power due to mechanical losses such as friction.

b) Components of Fixed Cost

Fixed costs are independent of power generation. They include capital cost, interest on capital, depreciation, insurance charges, taxes, and salaries of permanent staff.

c) Conventional Power Plants

Conventional power plants include:                     Steam (thermal) power plant

Hydroelectric power plant                                    Nuclear power plant

Diesel power plant                                                Gas turbine power plant

d) Boiler Efficiency

Boiler efficiency is the ratio of heat actually utilized in producing steam to the heat supplied by fuel.

Boiler efficiency=Heat output in steamHeat input from fuel\text{Boiler efficiency} = \frac{\text{Heat output in steam}}{\text{Heat input from fuel}}Boiler efficiency=Heat input from fuelHeat output in steam​ 

e) Applications of Diesel Engine Power Plant

Diesel power plants are used for standby power supply, peak load plants, small power stations, emergency power supply in hospitals, and remote area electrification.

f) Turbo Charging

Turbo charging is a method of increasing engine power by forcing more air into the combustion chamber using a turbine driven by exhaust gases. This improves volumetric efficiency and engine output.

g) Types of Fuel Cells

Different types of fuel cells are:                              Alkaline fuel cell

Proton Exchange Membrane (PEM) fuel cell          Solid Oxide Fuel Cell (SOFC)

Molten Carbonate Fuel Cell (MCFC)                       Phosphoric Acid Fuel Cell (PAFC)

h) Breeding

Breeding is the process in which fertile nuclear material such as Uranium-238 is converted into fissile material like Plutonium-239 inside a nuclear reactor.

i) Transformer Protection

Transformer protection involves protecting the transformer from faults such as overloads, short circuits, overheating, and insulation failure using relays, fuses, Buchholz relay, and differential protection.

j) Fossil Fuel Pollution

Fossil fuel pollution results from burning coal, oil, and gas, releasing harmful gases like CO₂, SO₂, NOx, and particulate matter. This leads to global warming, acid rain, and air pollution.

SECTION B – Long Answers (10 Marks Each)

2(a) Depreciation of Equipment

Given:                                                                        Initial cost = ₹5,00,000
Salvage value = ₹1,00,000                                         Life = 15 years

(i) Straight Line Method

Annual Depreciation=5,00,000−1,00,00015=₹26,667\text{Annual Depreciation} = \frac{5,00,000 - 1,00,000}{15} = ₹26,667Annual Depreciation=155,00,000−1,00,000​=₹26,667

Value after 5 years:

=5,00,000−(5×26,667)=₹3,66,665= 5,00,000 - (5 × 26,667) = ₹3,66,665=5,00,000−(5×26,667)=₹3,66,665

(ii) Sinking Fund Method

A=(P−S)i(1+i)n−1A = \frac{(P - S)i}{(1+i)^n - 1}A=(1+i)n−1(P−S)i​

Substituting values and compounding annually at 10%, the value of equipment after 5 years is calculated by deducting accumulated depreciation from original cost.

2(b) Supercritical and Supercharged Boilers

Supercritical Boilers:
These boilers operate above the critical pressure (221 bar) of water, eliminating the distinction between liquid and vapor phases, resulting in higher efficiency.

Supercharged Boilers:
These boilers use pressurized air supplied by compressors to improve combustion and increase boiler output.

2(c) Reheating in Gas Turbine Plant

Reheating improves thermal efficiency by expanding gases in stages. After partial expansion in the first turbine, gases are reheated and expanded again, increasing work output and reducing exhaust losses.

2(d) Fast Breeder Nuclear Reactor

A fast breeder reactor uses fast neutrons and liquid metal coolant like sodium. It produces more fissile material than it consumes, improving fuel utilization.
(Neat labeled diagram to be drawn in exam.)

2(e) Properties of Conductor Materials

Good conductor materials should have high electrical conductivity, low resistivity, high tensile strength, corrosion resistance, and low cost.

Materials used:
Copper, aluminum, silver, steel-reinforced aluminum conductors.

SECTION C – Very Long Answers (10 Marks Each)

3(a) Load Factor Calculation

Connected Load:                                                               Lamps = 10 × 60 = 600 W

Heaters = 2 × 1000 = 2000 W                                          Total = 2600 W

Daily energy consumption:                                                Lamps: 8 × 60 × 5 = 2400 Wh

Heaters: 2 × 1000 × 3 = 6000 Wh                                     Total = 8400 Wh/day

Average Demand:

=840024=350W= \frac{8400}{24} = 350 W=248400​=350W

Load Factor:

=3501500=0.233= \frac{350}{1500} = 0.233=1500350​=0.233

Monthly Energy Consumption:

=8400×30=252 kWh= 8400 × 30 = 252 \text{ kWh}=8400×30=252 kWh 

3(b) Cost of Electrical Generation

Cost of electrical generation includes fixed cost, semi-fixed cost, and running cost. It determines the tariff and economic feasibility of a power plant.

4(a) Layout of Steam Power Plant

A steam power plant consists of boiler, turbine, condenser, cooling tower, feed water pump, and generator. Steam produced in the boiler drives the turbine which generates electricity.

4(b) Fluidized Bed Combustion (FBC)

In FBC, fuel particles are suspended in air, resulting in better combustion, higher efficiency, and lower emissions.
(Neat sketch to be drawn.)

5(a) Diesel Engine Performance

Given:                                                                                   Fuel consumption = 20 kg/hr
Brake power = 80 kW                                                          Mechanical efficiency = 80%

Indicated power:

IP=BPηm=100 kWIP = \frac{BP}{η_m} = 100 \text{ kW}IP=ηm​BP​=100 kW

Brake SFC:

=2080=0.25 kg/kWh= \frac{20}{80} = 0.25 \text{ kg/kWh}=8020​=0.25 kg/kWh

Indicated SFC:

=20100=0.20 kg/kWh= \frac{20}{100} = 0.20 \text{ kg/kWh}=10020​=0.20 kg/kWh

Thermal efficiencies are calculated using CV = 45000 kJ/kg.

5(b) Pressure Ratio in Brayton Cycle

Increasing pressure ratio increases efficiency up to an optimum value. Beyond this, turbine work decreases due to higher compressor work.

6(a) Controlled vs Uncontrolled Nuclear Reaction

6(b) Solar Collectors

Types include flat plate collectors, parabolic trough collectors, parabolic dish collectors, and heliostat systems.

7(a) Circuit Breaker

A circuit breaker is a protective device that interrupts current during faults.
Types include oil, air, SF₆, and vacuum circuit breakers.

7(b) Bus Bar

A bus bar distributes electrical power to multiple circuits.
Types include single bus bar, double bus bar, ring main, and sectionalized bus bar arrangements with their respective merits and demerits.