A particular filter was tested for compliance by passing a test dust through the

Question

A particular filter was tested for compliance by passing a test dust through the filter medium at 32 L/min for 90 minutes. Before the test, the pressure drop through the filter was 17 mm H2O at a flow rate of 85 L/min and after the test was 43 mm H2O at a flow rate of 85 L/min. The test dust was at a concentration of 54 mg/m3 . If this filter element were to be used by a worker for an extended time, breathing at a rate of 40 L/min (average), how often will the filter element have to be changed if the worker’s pressure drop tolerance is 35 mm H2O and if he is breathing a dust which is at an average concentration of 16 mg/m3 .

Note that if the worker’s respiration rate is 40 L/min, remember that he spends only about half of the time inhaling, and the other half exhaling. Thus the flow rate through the filter during inhalation is more than 40 L/min.

Explanation / Answer

The mass of the dry soil particles is given by (m2-m1) = 20.00g

The mass of water displaced by the soil particles given by

                                                                                     (m4-m1) - (m3-m2)

                                                                                   = (50.03) -(42.48) = 7.55g

Gs = (m2-m1)/[(m4-m1)-(m3-m2)]

= (20.00g)÷(7.55g)

= 2.65

For the sample of natural soil, the unit weight is equal to the actual weight divided by the

total volume,

= (1kg × 9.81N/kg × 0.001kN/N) ÷ (0.52×10-3m3)

= 18.865 kN/m3

The water content w = mw/ms = 0.27. For the 1kg sample, we know that mw+ms = 1kg,

hence

1.27 × ms = 1kg

ms = 0.7874kg and mw = 0.2126kg

The volume of water vw = mw/w

= 0.2126kg ÷ 1kg/litre

= 0.2126litre

The volume of solids

vs = ms/s

= 0.7874kg÷2.65kg/litre

= 0.2971litre

The specific volume v is defined as the ratio vt/vs = 0.52litre/0.297litre

v = 1.75

The saturation ratio is given by the volume of water divided by the total void volume,

=0.2126litre ÷(0.52litre - 0.297litre) = 0.9534

Sr = 95.34%

If the soil were fully saturated, the volume of water would be (0.52litre - 0.297litre) =

0.223litre. The mass of water would be 0.223kg, and the water content would be

0.223kg ÷0.7874kg

W saturates = 28.32%

The overall mass of the 0.52litre sample would be 0.223kg + 0.7874kg = 1.0104kg, and its

unit weight (1.0101kg × 9.81N/kg × 10-3kN/N) ÷ (0.52×10-3m3)

sat = 19.06kN/m3

If the soil were dry but had the same specific (and overall) volume, the mass would be equal

to the mass of solids alone, and the unit weight would be

(0.7874kg × 9.81N/kg ×10-3kN/N) ÷(0.52×10-3m3)

dry = 14.86 kN/m3

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