Answered

A simple Rankine cycle coal-fired power plant has given states identified in the following table. The power plant produces 2.1 billion kW-hr/year of electricity. Ignore losses in the pump and turbine. If the generator is 98% efficient and boiler is 84% efficient, and coal costs $30/tonne, determine:

a. The thermal efficiency of the cycle.
b. The total coal energy content required per year.
c. The annual coal expenditures.
d. The mass of CO2 emitted per year from the plant.
e. The mass of CO2 emitted per million kW-hr produced.

State Location h(KJ/kg)
1 boiler exit, turbine entrance 2784.3
2 turbine exit, condenser entrance 2041.6
3 condenser exit, pump entrance 340.5
4 pump exit, boiler entrance 346.6

Answer :

mirianmoses

Answer:

Explanation:

Thermal Efficiency=0.98*0.84=0.8232

Total energy produced=2784.3*0.98+2041.6*0.84+340.5+346.6=5150.658 kJ/kg

Toal coal consumed=2.1*10^9*3600/5130.658=1473495.213 tonnes

Total cost=4.42*10^7 $

Mass of CO2 produced=Total Coal consumed*(Mass of CO2/Mass Of C)=54028158.11 tonnes

Mass of CO2/million kWH=25727.694

OR

Work done by turbine(Wt)

=(h1-h2) = 2784.3 - 2041.6 = 742.7 KJ/Kg

Work done on Pump(Wp) = (h4-h3) = 6.1 KJ/kg

Work Done by Boiler(Wb) = (1/efficiency)*(h1-h4) =(1/0.84) *(2784.3-346.6) = 2902.02 KJ/Kg

Thermal Efficiency = (Wt-Wp)/Wb = 25.38 Percent

Total Work by Turbine = (1/0.98)*2.1 Billion KW-hr/year

Total Coal Energy = (1/0.2538)*(1/0.98)*2.1 Billion KW-hr/year = 8.443 Billion KW-hr/year

Toatal coal required for an year

m*742.7 = (1/0.98)*2.1 Billion KW-hr/year = (1/0.98)*2.1*10^9*60 KJ/year

m = 1731.14 Tonnes

annual coal cost =1731.14*30 = 51934.2 dollars

2 metric tons of CO2 per one Kilowatt-hour (from graph)

Total CO2 emission = 2.1*10^9*2 = 4.2 Billion Metric tonnes

Total CO2 emitted per million KW-hr Produced = 2 million Metric tons

Other Questions