Question 3.10: A sedimentation analysis by the hydrometer method (152 H) wa...

A sedimentation analysis by the hydrometer method (152 H) was conducted with 50 g \left(=M_{s}\right)of oven dried soil. The volume of soil suspension is V=10^{3} cm ^{3}. The hydrometer reading R_{a}=19.50 after a lapse of 60 minutes after the commencement of the test.

Given: C_{m} \text { (meniscus) }=0.52, L \text { (effective) }=14.0 cm , C_{o}(\text { zero correction })=+2.50, G_{s}=2.70 and \mu=0.011 poise.

Calculate the smallest particle size, which would have settled a depth of 14.0 cm and the percentage finer than this size. Temperature of test = 25° C.

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From Eq. (3.24)

 

D=\sqrt{\frac{30 \mu}{\left(G_{s}-1\right) \gamma_{w}}} \sqrt{\frac{L}{t}}=K \sqrt{\frac{L}{t}} (3.24)

 

where \mu=0.01 \times 10^{-3}( gm – sec ) / cm ^{2}.

 

Substituting

 

D=\sqrt{\frac{30 \times 0.01 \times 10^{-3}}{(2.7-1)}} \times \sqrt{\frac{14}{60}}=0.0064 mm.

 

From Eq. (3.31)

 

C_{s g}=\frac{1.65 G_{s}}{2.65\left(G_{s}-1\right)} (3.31)

 

R_{c}=R_{a}-C_{o}+C_{t}

 

From Table 3.6 for T=25{ }^{\circ} C , C_{T}=+1.3. Therefore,

 

R_{c}=19.5-2.5+1.3=18.3

 

From Eqs (3.32) and (3.31), we have

 

P^{\prime}=\frac{C_{s g} R_{c}}{M_{s}} \times 100 (3.32)

 

P^{\prime} \%=\frac{C_{s g} R_{c}}{M_{s}} \times 100, \quad C_{s g}=\frac{1.65 G_{s}}{2.65\left(G_{s}-1\right)}

 

C_{s g}=\frac{1.65 \times 2.7}{2.65(2.7-1)}=0.99, \quad P^{\prime} \%=\frac{0.99 \times 18.3}{50} \times 100=36.23 \%

 

Table 3.6 Temperature correction factors C_{T}
Temp °C C_{T} Temp °C C_{T}
15 -1.10 23 +0.70
16 -0.90 24 +1.00
17 -0.70 25 +1.30
18 -0.50 26 +1.65
19 -0.30 27 +2.00
20 0.00 28 +2.50
21 +0.20 29 +3.05
22 +0.40 30 +3.80

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