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Brums [2.3K]
2 years ago
12

3. An underground copper pipe is used to transport oil under a factory. It is very important that this pipe does not oxidize, so

engineers use impressed current cathodic protection. a. Would zinc be an appropriate metal to use for cathodic protection? Why or why not? (4 points) b. Engineers attach a sacrificial anode to the copper pipe. To use impressed current cathodic protection, they must attach a battery between the cathode and anode. Which material should the positive side of the battery be attached to? Explain how you know. (4 points)
Engineering
1 answer:
Akimi4 [234]2 years ago
5 0

Answer:

a) Copper corrosion is prevented by an external current source that provides a condition around the copper that halts the rust through current cathodic protection

Due to the location of the pipeline, an material such as graphite, titanium, or niobium can be used which lasts longer that zinc can be used for longevity of the protection

b) The positive side of the output DC terminals are attached to the anodes such as graphite, while the negative terminals are connected to the copper pipe to be protected

Explanation:

Making the protected metal the cathode will make the metal be the source of electrons into the reaction taking place such that the release of the positive metal ions that results in disintegration takes place at the anode and as such the protected metal made the cathode, to which the negative terminal of the battery is attached is protected.

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3/63 A 2‐kg sphere S is being moved in a vertical plane by a robotic arm. When the angle θ is 30°, the angular velocity of the a
miss Akunina [59]

Answer:

Ps=19.62N

Explanation:

The detailed explanation of answer is given in attached files.

5 0
2 years ago
A steel rotating-beam test specimen has an ultimate strength Sut of 1600 MPa. Estimate the life (N) of the specimen if it is tes
ziro4ka [17]

Answer:

the life (N) of the specimen is 46400 cycles

Explanation:

given data

ultimate strength Su = 1600 MPa

stress amplitude σa = 900 MPa

to find out

life (N) of the specimen

solution

we first calculate the endurance limit of specimen Se i.e

Se = 0.5× Su   .............1

Se = 0.5 × 1600

Se = 800 Mpa

and we know

Se for steel is 700 Mpa for Su ≥ 1400 Mpa

so we take endurance limit Se is = 700 Mpa

and strength of friction f  = 0.77 for 232 ksi

because for Se 0.5 Su at 10^{6} cycle = (1600 × 0.145 ksi ) = 232

so here coefficient value (a) will be

a = \frac{(f*Su)^2}{Se}    

a = \frac{(0.77*1600)^2}{700}  

a = 2168.3 Mpa

so

coefficient value (b) will be

a = -\frac{1}{3}log\frac{(f*Su)}{Se}

b =  -\frac{1}{3}log\frac{(0.77*1600)}{700}

b = -0.0818

so no of cycle N is

N =  (\frac{ \sigma a}{a})^{1/b}

put here value

N =  (\frac{ 900}{2168.3})^{1/-0.0818}

N = 46400

the life (N) of the specimen is 46400 cycles

5 0
2 years ago
If you see rough patches, loose gravel, or potholes on the road, you should ______.l
Black_prince [1.1K]
Avoid them because they can cause you to lose traction.
4 0
2 years ago
What kind of datatype is being input for Number1, Number2, Number3?
Contact [7]

Answer: integer

Explanation:

Its python code.

The datatype for number1, number2 and number3 it will be integer and even the output will be integer.

the programmer will decide whether to convert integer to float or not.

The program want a user to inputting a value number1, number2 and number3, then the interpreter will add the numbers and get the total and then the average will be total of numbers / 3.

so it will print number1, number3, number3, the sum and average.

6 0
2 years ago
A 10-mm drill rod was heat-treated and ground. The measured hardness He was found to be 300 Brinell. Estimate the endurance stre
lesya [120]

Answer:

The endurance strength for the rod is 434.6 MPa

Explanation:

Since the rod is used in rotating bending, we need to use Marin equation given by

S=k_ak_bS'_e

Here S stands for the endurance strength for rotating  bending, S'_e is the endurance strength, and k_a \text{ and } k_b are the parameters for Marin surface modification factor.

Endurance strength.

We can start finding the endurance strength, from the directions we know that the hardness H_e was found to be 300 Brinell, thus for such value we can find the ultimate tensile strength using

S_{ut}=3.41H_e

Replacing the hardness we get

S_{ut}=3.41(300) MPa \\ S_{ut}=1023 MPa

Now since the ultimate tensile strength has a value less than 1400 MPa, we can find the endurance strength using

S'_e =0.5S_{ut}

Replacing the tensile strength we get

S'_e=0.5(1023) MPa \\ S'_e = 511.5 MPa

Parameters for Marin surface modification factor.

From the directions we know that the drill rod has a ground surface finish, so then from tables we get

a=1.58 \text{ and } b = -0.085

Thus the surface factor will be

k_a=a(S_{ut})^b

Replacing values and the ultimate tensile strength

k_a=(1.58)1023^{-0.085}\\k_a=0.8766

Then we can find the rotating shaft factor, for a diameter of 10 mm, we can use the equation

k_b=1.24d^{-0.107}

Replacing the diameter we get

k_b=1.24(10)^{-0.107}\\k_b=0.9692

Estimating endurance strength for rotating shaft.

We can replace now all values we have found in Marin equation.

S=k_ak_bS'_e

S=(0.8766)(0.9692)(511.5) MPa

S=434.6 MPa

Thus the endurance strength for the rod is 434.6 MPa

3 0
2 years ago
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