Improvement in the efficiency of thermal power plant

1. BY ANSHU AGRAWAL
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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. 3
WORKING

4. 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 %

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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.

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THEROMOELECTRIC EFFECT
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 = 훼(Th-Tc)
훼 = Seebeck's coefficient.

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THERMO ELECTRIC MATERIAL PROPERTIES
Material having low thermal conductivity
and a high electrical conductivity are
required for this kind of generator
Performance Equation
ZT=훼^2휎T/K
훼=Seebek coefficient , 휎 =conductivity of material
K=thermal conductivity. ,ZT =figure of merit

8. 8
Efficiency v/s figure of merit curve

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PRINCIPLE OF THERMO ELECTRIC GENERATOR
The temperature difference between the hot and cold side
of the TE geneartor is :
ΔT=Th-Tc
Th = hot side temperature(K)
Tc = cold side temperature(K)
The open circuit output voltage is:
Voc= 훼. ΔT
output current
I= Voc / (R +RL )
R = TE generator internal resistance
For optimal efficiency RL =1.32R

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THERMOELECTRIC GENERATOR

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A OVERVIEW of this method
Consider power plant of 2000MW & efficiency=33%
Input power =6060.6MW
LET 40 % of the input energy is wasted during the condensation
process.
INPUT energy to condenser=2424.25
LET Nanowires 6×6×1 mm bismuth telluride (Bi2Te3) TE pellet
is selected for the design.
seebek coefficient (훼) =287 V/K at 327 Kelvin.
(conductivity) (휎) =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 Kelvin
ZT= [(287 × 10-6)2. (1.1×105). (327)] / (1.20) = 2.47

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T(HOT) =400K. ; T(COLD) = 300 K,
ΔT=400 -300 = 100 K
Voc =(287 × 10-6). 100 = 28.7 mV
the electrical resistivity is:
휌= 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 load
P= (48.98)2.(0.334×10-3) = 801.28 mW.

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The total heat input to the TE couple is represented by
Qin= 훼IΔT +1/2(I 2R) +K 2 Δ T
the TE Couple Input heat power is =9.640W
The wasted heat is
QC = 9.640 – 0.801 = 8.839 W
So the efficiency is :
ŋ= (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

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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%.

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