Question

A consultant has proposed that a pulse-jet baghouse with bags that are 15 cm in diameter and 5 m in length. Estimate the net number of bags required if the manufacturer’s recommended air-to-cloth ratio for aggregate plants in 0.05 m/s, and a requirement that 1/8 of the bags are off-line for cleaning

Answer 1

0,5 bags

Total number of bags = 0.5 + 4 = 4.5 bags or 5 bags

Step-by-step explanation:

Convert 15cm to meters = 15/100 = 0.15m

Flow rate Q = Area x velocity

= π d x v

= π (0.15) x 1

= 0.4712

A = Q/v = 0.4712/ 0.05

= 9.435 m^2

The net number of bags = A/π dh

= 9.425/π (0.15)(5)

= 4 bags

Requirement that 1/8 of the bags are off-line for cleaning

= Net number of bags / 8

= 4/8

= 0.5 bag or 1bag

Total number of bags

= 4bags + 1bag

= 5bags