In one month a person uses $852$ kWh of electricity. Using a fossil-fueled vapor power plant, find how much coal is needed for one person. The assumptions are:
Boiler efficiency of the Rankine cycle is $80$%
Efficiency of the electrical generator is $80$%
Coal burns completely with air and all incoming and outgoing streams are $300$K, $2$atm in the boiler
Ideal Rankine cycle with steam as working fluid
Sat. vapor enters turbine at $8$MPa and sat. liquid exits the condenser at $0.008$MPa
First, I found the total how much kW the generator would need to generate at any given moment $$852 \frac{kWh}{month}*\frac{1 month}{30 days}*\frac{1day}{24hours}=1.183kW$$ Then applying the effiency of the generator I get $1.47875$kW that needs to be the net output of the system.
Going through the system analysis with steam tables I found that $h_1=2758 \frac{kJ}{kg}$,$h_2=1794.77\frac{kJ}{kg}$,$h_3=173.88\frac{kJ}{kg}$,$h_4=181.94 \frac{kJ}{kg}$ $$\frac{\dot{W}_t}{\dot{m}}=963.23\frac{kJ}{kg}$$ $$\frac{\dot{W}_p}{\dot{m}}=8.06\frac{kJ}{kg}$$
Knowing that the $\dot{W}_{cycle}= 1.47875$kW I can find the mass flow rate $$\dot{m}=\frac{1.47875}{963.23-8.06}=0.00155\frac{kg}{s}$$
Now I analyzed the combustion reaction to be $$C+O_2\rightarrow CO_2$$ where 1 kmole of oxygen is needed to burn 1 kmole of carbon, then applying this to the energy balance equation I get $$(\bar{h}^o_f+\Delta\bar{h})_{CO_2}-(\bar{h}^o_f+\Delta\bar{h})_{C}-(\bar{h}^o_f+\Delta\bar{h})_{O_2}=\frac{\dot{Q}}{\dot{n}_C}$$ the enthalpy of formation of both $O_2$ and carbon is zero and the $\Delta \bar{h}$ of carbon is also zero. Then plugging in for the values at $300$K $$(-393520+(9431-9364))-(8736-8682)=\dot{Q}$$ $$\frac{\dot{Q}}{\dot{n}_C}=-393507\frac{kJ}{kmol}$$ Now to change the $kmol$ to $kg$ I multiplied by the molecular weight of carbon $$\dot{Q}=-4726019.07\frac{kJ}{kg}$$ Now applying the boiler effiency which I believe to mean that $80$% of the heat added from combustion is transfered into the working fluid. $$0.8\frac{\dot{Q}_{in}}{\dot{m}_C}=\frac{\dot{Q}_{out}}{\dot{m}_{water}}$$ And from the Rankine cycle analyse, we can find the heat added from the boiler which in the equation above is labeled as $\dot{Q}_{out}$ $$\dot{Q}_{in}=\dot{m}(h_1-h_4)$$ So rearranging for $\dot{m}_C$ we get $$\dot{m}_C=\frac{\dot{m}(h_1-h_4)}{(\dot{m}_{water})(.8\dot{Q}_{in})}$$
So is my work correct or have I made so error in my math and/or assumptions?
