An average pulverized-coal power plant has anefficiency of 33%. Suppose a new ultra-supercritical coal plant increases that to 42%. Assume coalburning emits 24.5 kgC/10° kJ. (1 kWh = 3600 kJ ajd 1 metric ton = 10 kg) If CO2 emissions are eventualy taxed at $50 per metric ton, what would the tax savings be for the supercritical plant ($/kW). 24 K9 soal 2t大* 50 1'27K24.589 k - 11.9 KJ 278*10 Krh kwh with n=033 11- 36 05 11-9 0-3 with2-o42 = 28.33 & Ewh 0:42 Saving = 36.05-28.33=7.-72 KWh lf coal that delivers 24x106 kJ of heat per metric ton costs $40/metric ton, what would be the fuel savings for the ultra-supercritical plant ($/kWh)?

Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
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An average pulverized-coal power plant has anefficiency of 33%. Suppose a new ultra-supercritical
coal plant increases that to 42%. Assume coalburning emits 24.5 kgC/10° kJ.
(1 kWh = 3600 kJ and 1 metric ton = 10 kg)
If CO2 emissions are eventualy taxed at $50 per metric ton, what would the tax
savings be for the supercritical plant ($kW).
np:033
24 K9 Soal
11.9
l6 KJ 278*10 Kwh
korh
with n=033 >
0-3
11-1
: 36 .05
=28.33 &
0:42
with 2=o42
Saving = 36.05-28.33 =772
Ewh
KWh
If coal that delivers 24x106 kJ of heat per metric ton costs $40/metric ton, what
would be the fuel savings for the ultra-supercritical plant ($/kWh)?
24x16 FJ
Transcribed Image Text:An average pulverized-coal power plant has anefficiency of 33%. Suppose a new ultra-supercritical coal plant increases that to 42%. Assume coalburning emits 24.5 kgC/10° kJ. (1 kWh = 3600 kJ and 1 metric ton = 10 kg) If CO2 emissions are eventualy taxed at $50 per metric ton, what would the tax savings be for the supercritical plant ($kW). np:033 24 K9 Soal 11.9 l6 KJ 278*10 Kwh korh with n=033 > 0-3 11-1 : 36 .05 =28.33 & 0:42 with 2=o42 Saving = 36.05-28.33 =772 Ewh KWh If coal that delivers 24x106 kJ of heat per metric ton costs $40/metric ton, what would be the fuel savings for the ultra-supercritical plant ($/kWh)? 24x16 FJ
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