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A bag house is to be designed to handle 1000 m3/min of air. The filtration takes place at constant pressure so that the air velocity through each bag decreases during the time between cleaning according to the relation \(u = \frac{1}{{0.267 + 0.08t}}\) where, u is in m/min of cloth and t is the time in min. The bags are shaken in sequence row by row on a 30 min cycle. Each bag is 20 cm in diameter and 3 m height. The bag house is to be square in cross-section with 30 cm spacing between bags and 30 cm clearance from the walls. Calculate the minimum number of bags required.
1. 663
2. 753
3. 553
4. 853

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Correct Answer - Option 3 : 553

Concept:

The average velocity is given by

\({{\rm{V}}_{{\rm{avg}}}} = \frac{1}{{\rm{t}}}\mathop \smallint \limits_0^{\rm{t}} {\rm{udt}}\)

Let, At = Total area , Ab = Area of each bag

\({{\rm{A}}_{\rm{t}}} = \frac{{{{\rm{Q}}_{\rm{g}}}}}{{{{\rm{V}}_{{\rm{avg}}}}}}\)

Ab = πdl 

Where, d = diameter of the bag, l = length of the bag

N = Number of bags = \({\rm{N}} = \frac{{{{\rm{A}}_{\rm{t}}}}}{{{{\rm{A}}_{\rm{b}}}}}\)

Calculation:

Given:

The ratio of flow rate air to cloth is (u) = \(\frac{1}{{0.267 + 0.08{\rm{t}}}}\)(m3/m2 min of cloth)

Time, t = 30 min, diameter of bag, d = 0.2, Length of bag, l = 3 m

Flow rate, Qg = 1000 m3/min.

Put the values in the equation, we get the average velocity

\({{\rm{V}}_{{\rm{avg}}}} = \frac{1}{{\rm{t}}}\mathop \smallint \limits_0^{\rm{t}} {\rm{udt}} = \frac{1}{{\rm{t}}}\mathop \smallint \limits_0^{\rm{t}} \frac{{{\rm{dt}}}}{{0.267 + 0.08{\rm{t}}}}\)

\({{\rm{V}}_{{\rm{avg}}}} = \frac{1}{{30}}\mathop \smallint \limits_0^{30} \frac{{{\rm{dt}}}}{{0.267 + 0.08{\rm{t}}}} = \frac{{28.78}}{{30}} = 0.959\;{\rm{m}}/{\rm{min}}\)

\({{\rm{A}}_{\rm{t}}} = \frac{{1000}}{{0.959}} = 1042.390\;{{\rm{m}}^2}\)

Ab = π × 0.2 × 3 = 1.8849 m2

\({\rm{N}} = \frac{{1042.390}}{{1.884}} = 553.005\) ≈ 553

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