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Improvement in the efficiency of thermal power plant
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improvement in the Efficiency of Thermal Power Plant
using thermoelectric effect
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
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WORKING
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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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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
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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
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Efficiency v/s figure of merit curve
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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
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THERMOELECTRIC GENERATOR
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A OVERVIEW of this method
Consider power plant of 2000MW & efficiency=33%Input power =6060.6MWLET 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
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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.
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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
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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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THANK YOU