A piece of corroded steel plate was found in a submerged ocean vessel. It was estimated that the original area of the plate was
17 in.2 and that approximately 2.1 kg had corroded away during the submersion. Assuming a corrosion penetration rate of 200 mpy for this alloy in seawater, estimate the time of submersion in years. The density of steel is 7.9 g/cm^3.
1 answer:
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Answer:
The correct answers are:
a. % w = 33.3%
b. mass of water = 45g
Explanation:
First, let us define the parameters in the question:
void ratio e = =
Specific gravity =
% Saturation S = ×
=
×
water content w = =
a) To calculate the lower and upper limits of water content:
when S = 100%, it means that the soil is fully saturated and this will give the upper limit of water content.
when S < 100%, the soil is partially saturated, and this will give the lower limit of water content.
Note; S = 0% means that the soil is perfectly dry. Hence, when s = 1 will give the lowest limit of water content.
To get the relationship between water content and saturation, we will manipulate the equations above;
w =
Recall; mass = Density × volume
w =
From eqn. (2) =
∴
putting eqn. (6) into (5)
w =
Again, from eqn (1)
substituting into eqn. (7)
∴
With eqn. (7), we can calculate
upper limit of water content
when S = 100% = 1
Given,
∴
∴ %w = 33.3%
Lower limit of water content
when S = 1% = 0.01
∴ % w = 0.33%
b) Calculating mass of water in 100 cm³ sample of soil ( )
Given, , S = 50% = 0.5
%S = ×
=
×
0.50 =
mass of water =