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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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Answer:

156.0 ksi

Explanation:

The formula of compressive strength is  CS = F / A

where F is the force or load while A is the cross-sectional area

Given the inertia Ag = 35.1 in^4 , which is less than 40, the steel has a short column.

E = 29000 ksi

L = 12ft

r = 2.69 in

therefore;

CS = π^2 E / (kL^2/r)^2

   = π^2 29000 / (0.8 x 12^2 / 2.69)^2

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Air enters a compressor at 100 kPa and 25 ⁰C. It is compressed to 2 MPa and exits the compressor at 540 K. The compressor is at
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Answer:

(a) The reversible work is 207 kJ/kg

(b) The irreversibility rate is -38.39 kJ/kg

Explanation:

State1 : p1 = 100kpa, T1= 25+273 =298k

From air table, h1 =298.18 kJ/kg, s10= 1.69528 kJ/kgK

State 2a:p2=2mpa,t2=540k (actual condition 2a)

h2a= 544.35 kJ/kg,s2a0=2.29906

actual work input to the compressor =wout=h1-h2+Qin

=298.18-544.35+(-150)kJ/kg(- sign indicate heat loss)

=(-246.17)kJ/kg(-ve sign indicates the work is given into the system

a) Reversible work= Win actual - any irreversiblities present

                             =246.17 + irreversibilty

b) irreversibility = T0(Entopy generation Sgen) for air, Sgen

                         =s20-s10-Rln(p2/p1), T0=250C

                         =(25+273)(s2a0-s10-Rlnp2/p1+Qout/Tsurr)

    = 298x[(2.29906-1.69528-0.287kJ/kgK xln(2000kpa/100) + 150 /298]

  = -38.39 kJ/kg

a)Reversible work = Win actual -any irreversiblities present                  

                           =246.17 + irreversibilty

                           =246.17+-38.39

                          =207 kJ/kg

8 0
3 years ago
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A stream of vinyl chloride monomer (VCM, H2C=CHCl, molecular weight = 62.5lb/lb-mol) in air is being produced from a process to
Andreas93 [3]

Question:

The question is incomplete. The table of value was not added to your question. Below is the table of value and the plotted graph attached.

p          R= VCM/AC               p/R

0.0001          1                    0.0001

0.0005        1.7                   0.000294

0.001          3                   0.000333

0.005          6                   0.000833

0.01            8                   0.00125

0.05          10                   0.005

Answer:

(a) Yes

(b) W.C = 9.1428

(c) mass of activated carbon = 27.35 Ib-mole/activated carbon

Explanation:

(a) From the graph attached, it is evident that Langmuir isotherm fits the data well. So, the answer is yes.

 

(b)

Since the input stream has a VCM partial pressure of 0.02 psia, this would be used to find working capacity of activated carbon.

From the graph it can be found by interpolation that p/R value corresponding to 0.02 psia is 0.0021876. Thus R = 9.1428 = WC

It should be noted that in adsorption process, activated carbon is in equilibrium with 0.001 psia of VCM. Thus, from the data R= 1 i.e. this amount of VCM is not adsorbed and lost. This value may be deducted based on the definition of working capacity (WC).

(c)

The molar flow rate of VCM = 1000*62.3*0.02/379*60*14.7

                                               = 0.003739 Ib-mole/hr

Mass of activated carbon required = 0.003739 *8*9.1428*100

                                   = 27.35 Ib-mole/AC

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3 years ago
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