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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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Flauer [41]

Answer:

CARBON

Explanation:

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5 0
3 years ago
In the circuit given below, R1 = 17 kΩ, R2 = 74 kΩ, and R3 = 5 MΩ. Calculate the gain 1formula58.mml when the switch is in posit
Elenna [48]

Answer:a

a) Vo/Vi = - 3.4

b) Vo/Vi = - 14.8

c) Vo/Vi = - 1000

Explanation:

a)

R1 = 17kΩ

for ideal op-amp

Va≈Vb=0 so Va=0

(Va - Vi)/5kΩ + (Va -Vo)/17kΩ = 0

sin we know Va≈Vb=0

so

-Vi/5kΩ + -Vo/17kΩ = 0

Vo/Vi = - 17k/5k

Vo/Vi = -3.4

║Vo/Vi ║ = 3.4    ( negative sign phase inversion)

b)

R2 = 74kΩ

for ideal op-amp

Va≈Vb=0 so Va=0

so

(Va-Vi)/5kΩ + (Va-Vo)74kΩ = 0

-Vi/5kΩ + -Vo/74kΩ = 0

Vo/Vi = - 74kΩ/5kΩ

Vo/Vi = - 14.8

║Vo/Vi ║ = 14.8  ( negative sign phase inversion)

c)

Also for ideal op-amp

Va≈Vb=0 so Va=0

Now for position 3 we apply nodal analysis we got at position 1

(Va - Vi)/5kΩ + (Va - Vo)/5000kΩ = 0           ( 5MΩ = 5000kΩ )

so

-Vi/5kΩ + -Vo/5000kΩ = 0

Vo/Vi = - 5000kΩ/5kΩ

Vo/Vi = - 1000

║Vo/Vi ║ = 1000  ( negative sign phase inversion)

3 0
3 years ago
What are the maximum weights for single, tandem and 5-axle combinations?
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7 0
3 years ago
The 1000-lb elevator is hoisted by the pulley system and motor M. The motor exerts a constant force of 500 lb on the cable. The
klemol [59]

The power that must be supplied to the motor is 136 hp

<u>Explanation:</u>

Given-

weight of the elevator, m = 1000 lb

Force on the table, F = 500 lb

Distance, s = 27 ft

Efficiency, ε = 0.65

Power  = ?

According to the equation of motion:

F = ma

3(500) - 1000 = \frac{1000}{32.2} * a

a = 16.1 ft/s²

We know,

v^2 - u^2 = 2a (S - So)\\\\v^2 - (0)^2 = 2 * 16.1 (27-0)\\\\v = 29.48m/s

To calculate the output power:

Pout = F. v

Pout = 3 (500) * 29.48

Pout = 44220 lb.ft/s

As efficiency is given and output power is known, we can calculate the input power.

ε = Pout / Pin

0.65 = 44220 / Pin

Pin = 68030.8 lb.ft/s

Pin = 68030.8 / 500 hp

     = 136 hp

Therefore, the power that must be supplied to the motor is 136 hp

5 0
3 years ago
A hydraulic jump is induced in an 80 ft wide channel.The water depths on either side of the jump are 1 ft and 10 ft.Please calcu
krek1111 [17]

Answer:

a) 42.08 ft/sec

b) 3366.33 ft³/sec

c) 0.235

d) 18.225 ft

e) 3.80 ft

Explanation:

Given:

b = 80ft

y1 = 1 ft

y2 = 10ft

a) Let's take the formula:

\frac{y2}{y1} = \frac{1}{5} * \sqrt{1 + 8f^2 - 1}

10*2 = \sqrt{1 + 8f^2 - 1

1 + 8f² = (20+1)²

= 8f² = 440

f² = 55

f = 7.416

For velocity of the faster moving flow, we have :

\frac{V_1}{\sqrt{g*y_1}} = 7.416

V_1 = 7.416 *\sqrt{32.2*1}

V1 = 42.08 ft/sec

b) the flow rate will be calculated as

Q = VA

VA = V1 * b *y1

= 42.08 * 80 * 1

= 3366.66 ft³/sec

c) The Froude number of the sub-critical flow.

V2.A2 = 3366.66

Where A2 = 80ft * 10ft

Solving for V2, we have:

V_2 = \frac{3666.66}{80*10}

= 4.208 ft/sec

Froude number, F2 =

\frac{V_2}{g*y_2} = \frac{4.208}{32.2*10}

F2 = 0.235

d) El = \frac{(y_2 - y_1)^3}{4*y_1*y_2}

El = \frac{(10-1)^3}{4*1*10}

= \frac{9^3}{40}

= 18.225ft

e) for critical depth, we use :

y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3

= 3.80 ft

7 0
3 years ago
Read 2 more answers
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