Answer:
The time of the submersion is [tex]t_{years} = 9 years[/tex]
Explanation:
From the question we are told that
The area of the plate is [tex]A = 11 in^2[/tex]
The mass of the corroded plate [tex]m = 2.3 \ kg = 2.3 * 1*10^{6}= 2.3 *10^{6} mg[/tex]
The corrosion penetration rate is [tex]R = 200 mpy[/tex]
The density of steel is [tex]\rho = 7.9 g/cm^3[/tex]
Generally the corrosion penetration rate can be mathematically represented as
[tex]R = \frac{K m}{A \rho t}[/tex]
Where K is the corrosion imperial unit constant whose value is K = 534 mpy
t is the exposure time of the plate
Making t the subject of the formula
[tex]t = \frac{K m}{R \rho A}[/tex]
Substituting value
[tex]t = \frac{534 *2.3*10^{6}}{200 * 7.9 * 10 }[/tex]
[tex]t = 77734 hrs[/tex]
Converting to years
[tex]t_{years} = \frac{77734}{365 * 24}[/tex]
[tex]t_{years} = 9 years[/tex]