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fomenos
3 years ago
11

A corroded metal gusset plate was found on a bridge. It was estimated that the original area of the plate was 750 cm2 and that a

pproximately 2.5 kg had corroded away during use. Assuming a corrosion penetration rate of 0.5 mm/year for this metal, estimate the time since construction of the bridge. The density of the alloy is 10 g/cm3 .
Engineering
1 answer:
fomenos3 years ago
6 0

Answer:

estimate the time since construction of the bridge is 6.66 year

Explanation:

given data

area of the plate =  750 cm²

corroded away during use = 2.5 kg  

corrosion penetration rate = 0.5 mm/year

density of the alloy = 10 g/cm³

solution

we know here area is express as

CPR = \frac{kw}{\rho A t}       ................1

put here value ans we get t

t = \frac{87.6\times 2.5 \times 10^6}{10\times 750 \times 0.5}                  

t = 58400  hour

time = 6.66 year

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We need to define the variables,

So,

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Therefore, the probability that the repair time is more than 4 horus can be calculate as,

P(x>4)=1-P(x4)= 1-F_x(4)\\P(x>4) = 1-e^{-0.5*4}\\P(x>4) = 1-0.98\\P(x>4) = 0.018

The probability that the repair time is more than 4 hours is 0.136

b) The probability that repair time is at least 12 hours given that the repair time is more than 7 hoirs is calculated as,

P(x\geq 12|x>7)=P(X\geq7+5|x>7)\\P(x\geq12|x>7)=P(X\geq5)\\P(x\geq12|x>7)=1-P(x\leq 5)\\P(x\geq12|x>7)=1-e^{-0.5(2)}

P(x\geq 12|x>7)=0.6321

The probability that repair time is at least 12 hours given that the repair time is more than 7 hours is 0.63

3 0
3 years ago
g The parameters of a certain transmission line operating at 휔휔=6 ×108 [rad/s] are 퐿퐿=0.35 [휇휇H/m], 퐶퐶=75 [pF/m], 퐺퐺=75 [휇휇S/m],
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Explanation:

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Therefore,

-\alpha-0.094 \mathrm{Np} / \mathrm{m} . \quad 3-2.25 \mathrm{rad} / \mathrm{m}, \text { and } \lambda-2 \pi / \beta-\underline{2.79} \mathrm{m}

Z_{0}-\sqrt{\frac{Z}{Y}}-\sqrt{\frac{R+j \omega L}{G+j \omega C}}-\sqrt{\frac{17+j 2.1 \times 10^{2}}{75 \times 10^{-6}+j 2.4 \times 10^{-2}}}-\frac{93.6-j 3.64 \Omega}{4}

5 0
3 years ago
A liquid refrigerant (sg=1,2) is flowing at a weight flow rate of 20,9 N/h. Refrigerant flashes into a vapor and its specific we
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Answer:

Explanation:

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= 20.9 / 11.5 m³

= 1.8174 m³

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5 0
3 years ago
Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 10 MPa, 450°C, and 80 m/s, and the exit
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Answer:

a) The change in Kinetic energy, KE = -1.95 kJ

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Explanation:

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For an adiabatic steady state flow of steam:

KE = \frac{V_2^2 - V_1^2}{2} \\.........(1)

Where Inlet velocity,  V₁ = 80 m/s

Outlet velocity, V₂ = 50 m/s

Substitute these values into equation (1)

KE = \frac{50^2 - 80^2}{2} \\

KE = -1950 m²/s²

To convert this to kJ/kg, divide by 1000

KE = -1950/1000

KE = -1.95 kJ/kg

b) The power output, w

The equation below is used to represent a  steady state flow.

q - w = h_2 - h_1 + KE + g(z_2 - z_1)

For an adiabatic process, the rate of heat transfer, q = 0

z₂ = z₁

The equation thus reduces to :

w = h₁ - h₂ - KE...........(2)

Where Power output, W = \dot{m}w..........(3)

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To get the specific enthalpy at the inlet, h₁

At P₁ = 10 MPa, T₁ = 450°C,

h₁ = 3242.4 kJ/kg,

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h₂ = 3242.4 + 0.92(2392.1)

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w = 3242.4 - 2392.54 - (-1.95)

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W = 12 * 851.81

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c) The turbine inlet area

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3 0
4 years ago
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harina [27]

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5 0
3 years ago
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