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

Question #201590

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


1
Expert's answer
2021-06-11T22:58:11-0400

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"


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