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Kitty [74]
2 years ago
6

A sample of sand weighs 490 g in stock and 475 in Oven Dry (OD) condition, respectively. If absorption capability of the sand is

1.1%, calculate the weight of the specimen in SSD condition.
Engineering
1 answer:
Ivahew [28]2 years ago
5 0

The weight of the specimen in SSD condition is 373.3 cc

<u>Explanation</u>:

a) Apparent specific gravity = \frac{A}{A-C}

Where,

A = mass of oven dried test sample in air = 1034 g

B = saturated surface test sample in air = 1048.9 g

C = apparent mass of saturated test sample in water = 975.6 g

apparent specific gravity = \frac{A}{A-C}

                                         = \frac{1034}{1034-675 \cdot 6}

Apparent specific gravity = 2.88

b) Bulk specific gravity G_{B}^{O D}=\frac{A}{B-C}

G_{B}^{O D}=\frac{1034}{1048.9-675 \cdot 6}

       =  2.76

c) Bulk specific gravity (SSD):

G_{B}^{S S D}=\frac{B}{B-C}

=\frac{1048 \cdot 9}{1048 \cdot 9-675 \cdot 6}

G_{B}^{S S D} = 2.80

d) Absorption% :

=\frac{B-A}{A} \times 100 \%

=\frac{1048 \cdot 9-1034}{1034} \times 100

Absorption = 1.44 %

e) Bulk Volume :

v_{b}=\frac{\text { weight of dispaced water }}{P \omega t}

=\frac{1048 \cdot 9-675 \cdot 6}{1}

= 373.3 cc

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P= 1.5 kW, cos θ= 0.707 and θ=45 degree

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θ= 25.84 °

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