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vredina [299]
2 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]2 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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Small droplets of carbon tetrachloride at 68 °F are formed with a spray nozzle. If the average diameter of the droplets is 200 u
Licemer1 [7]

Answer:

the difference in pressure between the inside and outside of the droplets is 538 Pa

Explanation:

given data

temperature = 68 °F

average diameter = 200 µm

to find out

what is the difference in pressure between the inside and outside of the droplets

solution

we know here surface tension of carbon tetra chloride at 68 °F is get from table 1.6 physical properties of liquid that is

σ = 2.69 × 10^{-2} N/m

so average radius = \frac{diameter}{2} =  100 µm = 100 ×10^{-6} m

now here we know relation between pressure difference and surface tension

so we can derive difference pressure as

2π×σ×r = Δp×π×r²    .....................1

here r is radius and  Δp pressure difference and σ surface tension

Δp = \frac{2 \sigma }{r}    

put here value

Δp = \frac{2*2.69*10^{-2}}{100*10^{-6}}  

Δp = 538

so the difference in pressure between the inside and outside of the droplets is 538 Pa

7 0
2 years ago
A curve in a speed track has a radius of 1000 ft and a rated speed of 120 mi/h. (From Sample Prob. 12.7 is the definition of rat
forsale [732]

Answer:

tan \theta = \frac{(176ft/s)^2}{1000 ft 32.2 ft/s^2}= 0.962

\theta = tan^{-1} (0.962) = 43.89

Explanation:

If the question is: Determine the banking angle θ

We have the forces involved on the figure attached.

For this case we know that the weight is given by:

W = mg

And for this case the centripetal acceleration would be given by:

a=\frac{v^2}{r}

If we analyze the sum of forces on x and y we have:

\sum F_x = m a_x

F + W sin \theta = ma cos theta

And if we solve for the force we got:

F = ma cos \theta - mg sin \theta = \frac{mv^2}{r} cos \theta - mg sin \theta

\sum F_y = m a_y

N - W cos \theta = ma sin \theta

If we solve for the normal force we got:

N =W cos \theta + ma sin \theta = \frac{mv^2}{r} sin \theta + mg cos \theta

In order to find the banking angle we use the fact that F =0

0 = \frac{mv^2}{r} cos \theta - mg sin \theta

tan \theta= \frac{v^2}{rg}

The velocity on this case is 120 mi/h if we convert this into ft/ s we got:

120 mi/h * \frac{5280 ft}{1mi} *\frac{1hr}{3600 s}= 176 ft/s

And then we have this:

tan \theta = \frac{(176ft/s)^2}{1000 ft 32.2 ft/s^2}= 0.962

\theta = tan^{-1} (0.962) = 43.89

5 0
2 years ago
A group of statisticians at a local college has asked you to create a set of functions that compute the median and mode of a set
iVinArrow [24]

Answer:

Functions to create a median and mode of a set of numbers

Explanation:

def median(list):

   if len(list) == 0:

       return 0

   list.sort()

   midIndex = len(list) / 2

   if len(list) % 2 == 1:

       return list[midIndex]

   else:

       return (list[midIndex] + list[midIndex - 1]) / 2

def mean(list):

   if len(list) == 0:

       return 0

   list.sort()

   total = 0

   for number in list:

       total += number

   return total / len(list)

def mode(list):

   numberDictionary = {}

   for digit in list:

       number = numberDictionary.get(digit, None)

       if number == None:

           numberDictionary[digit] = 1

       else:

           numberDictionary[digit] = number + 1

   maxValue = max(numberDictionary.values())

   modeList = []

   for key in numberDictionary:

       if numberDictionary[key] == maxValue:

           modeList.append(key)

   return modeList

def main():

   print "Mean of [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]: ", mean(range(1, 11))

   print "Mode of [1, 1, 1, 1, 4, 4]:", mode([1, 1, 1, 1, 4, 4])

   print "Median of [1, 2, 3, 4]:", median([1, 2, 3, 4])

main()

3 0
2 years ago
Read 2 more answers
A railcar with an overall mass of 78,000 kg traveling with a speed vi is approaching a barrier equipped with a bumper consisting
sergij07 [2.7K]

Answer:

v₀ = 2,562 m / s  = 9.2 km/h

Explanation:

To solve this problem let's use Newton's second law

              F = m a = m dv / dt = m dv / dx dx / dt = m dv / dx v

              F dx = m v dv

We replace and integrate

            -β ∫ x³ dx = m ∫ v dv

            β x⁴/ 4 = m v² / 2

We evaluate between the lower (initial) integration limits v = v₀, x = 0 and upper limit v = 0 x = x_max

        -β (0- x_max⁴) / 4 = ½ m (v₀²2 - 0)

         x_max⁴ = 2 m /β   v₀²

         

Let's look for the speed that the train can have for maximum compression

         x_max = 20 cm = 0.20 m

         

         v₀ =√(β/2m)   x_max²

Let's calculate

          v₀ = √(640 106/2 7.8 104)    0.20²

          v₀ = 64.05  0.04

          v₀ = 2,562 m / s

          v₀ = 2,562 m / s (1lm / 1000m) (3600s / 1h)

          v₀ = 9.2 km / h

5 0
2 years ago
A shipment of rebar that weighs 745 kg would weigh roughly how much in pounds​
Andre45 [30]

Answer:

Dont no but will check

Explanation:

6 0
2 years ago
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