A pulsejet baghouse is desired for a finished cement plant C

A pulse-jet baghouse is desired for a finished cement plant. Calculate the number of bags required to filter 500 m^3/min of air with a dust loading of 3.0 g/m^3. Each bag is 3.0 m long with a 0.13 m diameter. If the average pressure drop is 1.0 kPa and the main fan is 60% efficient, calculate the fan power in kW. If the pulse air volumetric flow rate is 0.5% of the filter airflow rate and the pulse air pressure is 6.0 atm, calculate the power drawn by a 50% efficient compressor (in kW).

Solution

Given: Q = 500 m³/min, L = 3g/m³, l=3m, d = 0.13m,P = 1kPa = 1000Pa, P_pulse-air = 6 atm = 607950 Pa

Total filtering area required = 0.26*Q*L^0.18*1.1 = 174.2680868 m²

Area of 1 bag = (3.14*d*l)+(3.14*0.25*d²) = 1.2385m

Number of bags required = 174.2680868/1.2385 = 141

Output power = P*Q = 1000*50/60 = 833.333 W

Fan power = 833.33/0.6 = 1.388889 kW

Pulse volumetric air flow rate = 0.5*10^-2*500/60 = 0.041666667

Power drawn by compressor = 0.041666667*607950/0.5 = 50.662 kW