Answer to Question #201590 in Mechanical Engineering for Christian Jay T. G

Question

in a steam power plant, 2.5 kg of coal is consumed per second and air required for combustion is 16 kg per kg coal fired. Air enters at 27°C and the barometric pressure is 98.5 kPa. The flue gases enter the chimney at a temperature of 300°C and leave at a temperature of 220°C. The ash loss is expected to be 12 % of the consumed fuel. A theoretical dra

Answer 1

If we were to calculate the height of the chimney

The specific volume of 1 kg of flue gas at NTP

PV=RT$\implies V_a=\frac{8.314}{30}*273=0.74668 m^3/kg$

The specific volume of m kg of air per kg fuel at temperature T

$V_a=0.74668m*\frac{T_a}{273}$

Density of air $\rho_a=\frac{1}{0.74668}*\frac{273}{T_a}=1.33926(\frac{273}{T_a})$

Mass of the gas will be m+1 per kg of fuel burnt and the temperature T.

Density of flue gas $\rho _g= \frac{m+1}{0.74668*m}*\frac{273}{T_g}=\frac{365.62}{T_g}*\frac{m+1}{m}$

So, $\triangle P =1.33926 (\frac{1}{T_a}-\frac{m+1}{m}*\frac{1}{T_g})$

$\rho gh=365.62gH (\frac{1}{T_a}-\frac{m+1}{m}*\frac{1}{T_g})$

$1000*h*10^{-3}=365.62 H(\frac{1}{T_a}-\frac{m+1}{m}*\frac{1}{T_g})$

$h=365.62H (\frac{1}{T_a}-\frac{m+1}{m}*\frac{1}{T_g})$

$50=365.62H (\frac{1}{300}-\frac{16}{15}*\frac{1}{575}) \implies H= 92.51m$