1

improvement in the Efficiency of Thermal Power Plant

using thermoelectric effect

BY ANSHU AGRAWAL

2

INTRODUCTION

• WHY IMPROVEMENT IN EFFICIENCY

REQUIRED

• HOW A THERMAL POWER PLANT

WORKS

• WHAT ARE THE LOSSES

• WHAT CAN BE DONE TO IMPROVE

THE EFFICIENCY

3

WORKING

4

LOSSES

• Practical limitations in heat transfer, all the heat

produced by combustion is not transferred to the water.

some heat is lost to the atmosphere as hot gases.

• The coal contains moisture

• The steam is condensed for re-use. During this process

the latent heat of condensation is lost to the cooling

water. This is the major loss and is almost 40 % of the

energy input

• Losses in the turbine blades and exit losses at turbine

end

• 5 % loss in the Generator. Another 3 % is lost in the

step-up transformer

• This brings the overall efficiency of the power plant to

around 33.5 %

5

TO IMPROVE THE EFFICIENCY OF THERMAL POWER PLANT

• Using the heat of flew gases( Economiser & Air preheater)

• Increasing efficiency of Generator.

• Using the dry coal .• convert some of the condenser

wasted energy to electricity using thermoelectric material.

6

According to Thermoelectric effect if Two junctions connected back to back are held at two different temperatures Th and Tc then EMF E appears between their free contacts:E = (Th-Tc) Seebeck's coefficient.

THEROMOELECTRIC EFFECT

7

THERMO ELECTRIC MATERIAL PROPERTIES

Material having low thermal conductivity and a high electrical conductivity are required for this kind of generatorPerformance Equation

ZT=T/K

=Seebek coefficient , conductivity of material K=thermal conductivity. ,ZT =figure of merit

8

Efficiency v/s figure of merit curve

9

PRINCIPLE OF THERMO ELECTRIC GENERATOR

The temperature difference between the hot and cold sideof the TE geneartor is :T=Th-Tc Th = hot side temperature(K) Tc = cold side temperature(K)

The open circuit output voltage is:Voc= output current I= Voc / (R +RL ) R = TE generator internal resistanceFor optimal efficiency RL =1.32R

10

THERMOELECTRIC GENERATOR

11

A OVERVIEW of this method

Consider power plant of 2000MW & efficiency=33%Input power =6060.6MWLET 40 % of the input energy is wasted during the condensation process.INPUT energy to condenser=2424.25LET Nanowires 6×6×1 mm bismuth telluride (Bi2Te3) TE pellet is selected for the design. seebek coefficient () =287 V/K at 327 Kelvin. (conductivity) (1.1×105 S/m thermal conductivity (K)=1.20 W.m-1.K-1. melting point is about 858 Kelvin and it's useful in temperature between 300 to 400 KelvinZT= [(287 × 10-6)2. (1.1×105). (327)] / (1.20) = 2.47

12

T(HOT) =400K. ; T(COLD) = 300 K, T=400 -300 = 100 K  Voc =(287 × 10-6). 100 = 28.7 mV the electrical resistivity is:= 9.09 × 10-6 the internal resistance is:R = (9.09 × 10-6). (1×10-3) / (6×6×10-6) = 0.252 m So RL is:RL =1.323393R = 0.334 Ohm  I= (28.7) / (0.252 + 0.334) =48.98 A Heat supplied to loadP= (48.98)2.(0.334×10-3) = 801.28 mW.

13

The total heat input to the TE couple is represented by Qin= T +1/2(I 2R) +K 2 Tthe TE Couple Input heat power is =9.640W

The wasted heat is QC = 9.640 – 0.801 = 8.839 W

So the efficiency is := (0.801/ 9.640) ×100 = 8.31%

As PC = 2424.25 MW, so the output power of the TE generator is:

PC out = 2424.25 × (8.31 / 100) =201.505 MW

14

TOTAL OTPUT OF THE POWER PLANT=2000+201.505=2201.505MW

TOTAL EFFICIENCY=100 × 2201.505 / 6060.6 =36.33 %

INCREASE IN EFFICIENCY =3.3%

USING NEW AND EFFICIENT MATERIAL EFFICIENCY CAN BE INCREASED MORE THAN 3.3%.

15

THANK YOU