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vredina [299]
3 years ago
13

1. A pipeline constructed of carbon steel failed after 3 years of operation. On examination it was found that the wall thickness

had been reduced by corrosion to about half the original value. The pipeline was constructed of nominal 100 mm (4 in) schedule 40 pipe, inside diameter 102.3 mm (4.026 in), outside diameter 114.3 mm (4.5 in). Estimate the rate of corrosion in ipy and mm per year.
2. The pipeline described in question 1 above was used to carry wastewater to a hold-up tank. The effluent is not hazardous. A decision has to be made on what material to use to replace the pipe. Three suggestions have been made: a. Replace with the same schedule carbon steel pipe and accept renewal at 3-year intervals. b. Replace with a thicker pipe, schedule 80, outside diameter 114.3 mm (4.5 in), inside diameter 97.2 mm (3.826 in). c. Use stainless steel pipe, which will not corrode.

The estimated cost of the pipes, per unit length is: schedule 40 carbon steel $5, schedule 80 carbon steel $8.30, stainless steel (304) schedule 40 $24.80. Installation and fittings for all the materials adds $16.5 per unit length. The downtime required to replace the pipe does not result in a loss of production. If the expected future life of the plant is 7 years, recommend which pipe to use.
Engineering
1 answer:
jek_recluse [69]3 years ago
3 0

Answer:

check the explanation

Explanation:

1.

Thickness Loss = t =\frac{t_{o}-t_{i}}{2} = \frac{114.3-102.3}{2} = 2mm

t_{f} = \frac{1}{2}*6 = 3mm

Hence Rate of Corrosion = 6*\frac{1-0.5}{3} = 1mm/year = 0.03 inches per year

2.

As the expected future life is 7 years,

40 carbon steel pipe has to be replaced every 3 years as given in the question,

Cost per unit length is the sum of material cost and installation cost.

Cost of 40 carbon steel = (5 dollars + 16.5 dollars) * 3 = 64.5 dollars

For 80 carbon steel pipe, first calculate the thickness loss,

\frac{114.3-97.2}{2} = 8.55mm

The critical thickness is given to be 3mm, Hence change in thickness is 8.55-3 = 5.5mm

This 80 carbon steel pipe has to be replaced one more time

Hence, Cost per unit length is the sum of material cost and installation cost.

Cost of 80 carbon steel = (8.3 dollars + 16.5 dollars) * 2 = 49.6 dollars

The best is of stainless steel which does not undergo corrosion at all and thus it needs to be replaced only once throughout the plant operation. Its cost = 24.8 dollars + 16.5 dollars = 41.3 dollars

Hence, stainless steel is the recommended pipe to be used.

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postnew [5]

Answer:

Ts = 413.66 K

Explanation:

given data

temperature = 20°C

velocity = 10 m/s

diameter = 5 mm

surface emissivity = 0.95

surrounding temperature = 20°C

heat flux dissipated = 17000 W/m²

to find out

surface temperature

solution

we know that here properties of air at 70°C

k = 0.02881 W/m.K

v = 1.995 ×10^{-5} m²/s

Pr = 0.7177

we find here reynolds no for air flow that is

Re = \frac{\rho V D }{\mu } = \frac{VD}{v}    

Re = \frac{10*0.005}{1.99*10^{-5}}

Re = 2506

now we use churchill and bernstein relation for nusselt no

Nu = \frac{hD}{k} = 0.3 + \frac{0.62 Re6{0.5}Pr^{0.33}}{[1+(0.4/Pr)^{2/3}]^{1/4}} [1+ (\frac{2506}{282000})^{5/8}]^{4/5}

h = \frac{0.02881}{0.005}0.3 + \frac{0.62*2506{0.5}0.7177^{0.33}}{[1+(0.4/0.7177)^{2/3}]^{1/4}} [1+ (\frac{2506}{282000})^{5/8}]^{4/5}

h = 148.3 W/m².K

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q conv = h∈(Ts- T∞ )

17000 = 148.3 ( 0.95) ( Ts - (20 + 273 ))

Ts = 413.66 K

5 0
4 years ago
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A sinusoidal voltage source produces the waveform, v t = 1 + cos 2πft. Design a system with v t as its input such that an LED wi
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Answer:

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Consider a piston-cylinder device with a piston surface area of 0.1 m^2 initially filled with 0.05 m^3 of saturated water vapor
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The friction force f = 10000 N

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Given data:

Surface area of Piston = 1 m^{2}

Volume of saturated water vapor = 100 K Pa

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= pressure inside the cylinder × piston area

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b)  To calculate heat transfer.

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