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Coal-steam electricity generation

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hdr
#1
Mar9-12, 11:31 AM
P: 4
1. The problem statement, all variables and given/known data
A coal-fired steam power plant operates on the ideal Rankine cycle. Feedwater from the
condenser at P1
= 10 kPa is pumped to the boiler at P2
= P3
= 20 MPa, where it is
superheated and exits the boiler at T3
= 550 C.
The moisture content in the turbine is not to exceed 12% (i.e. steam quality is to be
maintained above x = 0.88 throughout the turbine). Up to two reheat stages can be added to
the steam cycle to satisfy this requirement.
If needed, one stage can be added to reheat the steam to T5
= 525 C at P4
= P5
= 10 MPa. If
more reheating is required, a second stage can be added to reheat the steam to T7
= 500 C at
P6
= P7
= 2.5 MPa.

(a). How many reheat stages are required for this power plant (no reheat, single reheat, or
double reheat)? Provide calculations in support of your answer.
(b). What is the thermal efficiency ηth
of the power cycle?
(c). Determine the annual emission of sulfur dioxide, mSO2
, in metric tons, from this power
plant, if the net generating capacity of the plant is 600 MW, and the boiler is 88%
efficient in transferring combustion heat into the feedwater superheat and steam reheat
stages. The boiler burns, to completion, Power River Basin coal that has a heating value
of 22,800 kJ/kg and contains 0.46 wt% sulfur.
Assume that all of the sulfur content of the coal is emitted as SO2
in the combustion flue
gas, and that the coal-burning plant has a capacity factor of 0.85; i.e. the boiler is
operational for 85% of the hours in one calendar year. (1 metric ton = 1000 kg)


2. Relevant equations
Thermodynamics energy balance equations


3. The attempt at a solution
I completed the first 2 parts of the problem and I found out that one reheating stage is enough to active a steam quality more than 88.2%. I also calculated the thermal efficiency of 38/6% of this power plant.
My question is about part 3. How can I approach?
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Joffan
#2
Mar9-12, 01:10 PM
P: 361
Seems like that's easier than what you've done already - just calculate the total amount of electricity generated, work back through the efficiencies for the heat required, get the mass of coal and multiply out the fraction of sulfur into SO2.
sawhai
#3
Mar9-12, 01:58 PM
P: 28
I am sorry this comes from my bad background in chemistry, I get scared whenever feel that chemistry is involved! So the total electricity generation in a year is 600MW*365*24*3600= 18.9216*10^9 MW/year
I am not sure how to calculate the mass of the coal and as used the efficiencies for the heat required? Can you explain more please?


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