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const2013
4 days ago
6

A piece of corroded metal alloy plate was found in a submerged ocean vessel. It was estimated that the original area of the plat

e was 800 cm2 and that approximately 7.6 kg had corroded away during the submersion. Assuming a corrosion penetration rate of 4 mm/yr for this alloy in seawater, estimate the time of submersion in years. The density of the alloy is 4.5 g/cm3.
Engineering
1 answer:
Mrrafil [253]4 days ago
8 0

Answer:

t = 5.27 years

Explanation:

Firstly, the corrosion penetration rate is defined by the formula;

CPR = (KW)/(ρAt)

Where;

K = constant based on exposed area A.

W - mass lost over time

t- duration

ρ - density

A - area exposed

From the problem, we have;

W = 7.6kg or 7.6 x 10^(6) mg

CPR = 4 mm/yr

ρ = 4.5 g/cm³

Area = 800 cm²

K is a constant valued at 87.6cm

Rearranging the CPR formula to isolate t, we derive;

t = KW/(ρA(CPR))

t = (87.6 x 7.6 x 10^(6))/(4.5 x 800 x 4) = 46233.3 hours

The duration in question needs to be expressed in years.

Thus, converting hours to years;

There are 8760 hours in a year.

Therefore;

t = 46233.3/8760 = 5.27 years.

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Consider a process carried out on 1.00 mol of a monatomic ideal gas by the following two different pathways.
grin007 [219]

Answer:

90 L.atm

Explanation:

According to the provided details:

First pathway:

A( 3 atm, 20 L) → C ( 1 atm, 20 L) → D (1 atm, 50 L)

Second pathway:

A(3 atm, 20 L) → B( 3 atm, 50 L) → D ( 1 atm, 50 L)

As the number of moles is 1.00 moles

To calculate wAB;

A → B signifies the transformation is happening at a steady pressure;

Thus,

wAB = pressure multiplied by the change in volume

wAB = P(V₂ - V₁)

wAB = 3 atm (50 L - 20 L)

wAB = 3 atm (30 L)

wAB = 90 L.atm

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17 days ago
A bar of 75 mm diameter is reduced to 73mm by a cutting tool while cutting orthogonally. If the mean length of the cut chip is 7
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Answer:

r=0.31

Ф=18.03°

Explanation:

Provided:

Original diameter of bar = 75 mm

Diameter post-cutting = 73 mm

Average diameter of the bar d= (75+73)/2=74 mm

Average length of uncut chip = πd

Average length of uncut chip = π x 74 =232.45 mm

Thus, cutting ratio r

Cutting\ ratio=\dfrac{Mean\ length\ of cut\ chip}{Mean\ length\ of uncut\ chip}

r=\dfrac{73.5}{232.45}   r=0.31

Therefore, the cutting ratio equals 0.31.

Now, the shearing angle is given as

tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }

Next by substituting the values

tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }

tan\phi =\dfrac{0.31cos15 }{1-0.31sin15 }\

Ф=18.03°

Concluding, the shearing angle is 18.03°.

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A well-insulated rigid tank contains 1.5 kg of a saturated liquid–vapor mixture of water at 200 kPa. Initially, three-quarters o
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Answer:

The change in entropy of the steam is 2.673 kJ/K

Explanation:

The mass of the liquid-vapor mixture is 1.5 kg

The mass in the liquid phase is calculated as 3/4 × 1.5 kg = 1.125 kg

The mass in the vapor phase is calculated as 1.5 - 1.125 = 0.375 kg

According to the steam tables

At a pressure of 200 kPa (200/100 = 2 bar), the specific entropy of steam is found to be 7.127 kJ/kgK

The entropy of steam can be calculated as specific entropy multiplied by mass = 7.127 × 0.375 = 2.673 kJ/K

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