Question

A pulse-jet baghouse is desired for a finished cement plant. Calculate the number of bags required to filter 500 m3/min of air with a dust loading of 3.0 g/m3. 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).

Answer 1

Answer:

1) 4.41 * 10^-4 kw

2) 2.20 * 10^-4 kw

Explanation:

Given data:

Filter = 500 m^3/min

dust velocity = 3g/m^3

bag ; length = 3 m , diameter = 0.13 m

change in pressure = 1 kPa

efficiency = 60%

1) Calculate the Fan power

First :

Calculate the total dust loading = 3 * 500 = 1500 g

To determine the Fan power we will apply the relation

$n_{o} = \frac{\frac{p}{eg*Q*h} }{1000}$    = $\frac{\frac{p}{(3*10^{-3})* 981*( 500/60) *3 } }{1000}$

fan power ( $n_{0}$ ) = 4.41 * 10^-4 kw

2) calculate power drawn

change in P = 6 atm = 6 * 10^5 pa

efficiency compressor = 50%

hence power drawn = 4.41 * 10^-4 kw  * 50% = 2.20 * 10^-4 kw