Respuesta :
Answer:
1. [tex]\eta_{p} = 33.34%[/tex]
2. [tex]m_{coal} = 0.396 kg/kWh[/tex]
3. carbon emission = 0.297 kg/kWh[/tex]
4. carbon-dioxide emission = 1.089 kg/kWh
5. Min flow of cooling water = = 1457.14 kg/kWh
Given:
Heat Rate = 10,800 kJ/kWh
Heating value = 27,300 kJ/kg
Solution:
In order to calculate the plant efficiency, [rex]\eta_{p}[/tex]
Heat rate in coal fired steam power plant = [tex]\frac{3600}{\eta_{p}}[/tex]
Therefore,
[tex]\eta_{p} = \frac{3600}{Heat rate}[/tex]
[tex]\eta_{p} = \frac{3600}{10800}\times 100 = 33.34%[/tex]
Now,
To calculate mass of coal per kWh, [tex]m_{coal}[/tex]:
[tex]m_{coal} = \frac{Required heat rate}{heating value per kg}[/tex]
[tex]m_{coal} = \frac{10800}{27300} = 0.396 kg/kWh[/tex]
Now,
Rate of emission carbon and carbon-dioxide from the plant:
In accordance to the question, 75% of the bituminous coal being burned is carbon, thus carbon emission is given by:
carbon emission = [tex]0.75\times m_{coal}[/tex]
carbon emission = [tex]0.75\times 0.396 = 0.297 kg/kWh = 29.7%[/tex]
Now, for carbon-dioxide emission:
Since, molecular weight of carbon-dioxide = 44 kg
Thus carbon-dioxide emission:
[tex]0.297\times \frac{44}{12} = 1.089 kg/kWh[/tex]
The minimum flow of cooling water per kWh if the allowance in temperature is [tex]10^{\circ}[/tex]:
[tex]\frac{0.85\times \frac{2}{3}\times 10800}{4.2} = 1457.14 kg/kWh[/tex]
