ENGINEERING COLLEGE

A 5000-ft long X-65 pipeline is laid down on seabed with two PLETS (One at each end). The pipe OD=7-in with 0.5-in wall thickness. The pipeline was laid at environmental temperature of 40 °F (As- laid temperature). When pipeline is put into operation, the oil flow was produced at 140 °F. If the thermal expansion coefficient of the pipe material is 6.5*10-/°F and its modulus of elasticity is 30,000 ksi, determine the compressive load applied by the pipeline on a PLET due to its thermal expansion. Assume no temperature change and no seabed friction along the pipeline span.

Answers

Answer 1
Answer:

Answer: 199.1 kip

Explanation:

Given that

Outer diameter is Do = 7 in

Inner diameter Di = ( Do - ( 2×0.5)) = 6 in

Length = 5000 ft = 60000 in

Now change in length of the pipe due to temperature difference

SL = L∝ΔT

= 60000 × 6.5×10^-6(140-40)

SL = 39 in

Also

sL = PL/AE

A = cross sectional area of pipe = π/4(Do^2 - Di^2)

so

P = SL×A×E / L

= (39 × π/4(7^2 - 6^2)×30000) / 60000

= 199.1 kip

compressive load applied by the pipeline on a PLET due to its thermal expansion is 199.1 kip

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(a) For a given material, would you expect the surface energy to be greater than, the same as, or less than the grain boundary energy? Why? (b) The grain boundary energy of a small-angle grain boundary is less than for a high-angle one. Why is this so?

Answers

Answer:

(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.

(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.

Explanation:

(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.

(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.

A cylindrical specimen of cold-worked steel has a Brinell hardness of 250.(a) Estimate its ductility in percent elongation.(b) If the specimen remained cylindrical during deformation and its original radius was 5 mm (0.20 in.), determine its radius after deformation.

Answers

Answer:

A) Ductility = 11% EL

B) Radius after deformation = 4.27 mm

Explanation:

A) From equations in steel test,

Tensile Strength (Ts) = 3.45 x HB

Where HB is brinell hardness;

Thus, Ts = 3.45 x 250 = 862MPa

From image 1 attached below, for steel at Tensile strength of 862 MPa, %CW = 27%.

Also, from image 2,at CW of 27%,

Ductility is approximately, 11% EL

B) Now we know that formula for %CW is;

%CW = (Ao - Ad)/(Ao)

Where Ao is area with initial radius and Ad is area deformation.

Thus;

%CW = [[π(ro)² - π(rd)²] /π(ro)²] x 100

%CW = [1 - (rd)²/(ro)²]

1 - (%CW/100) = (rd)²/(ro)²

So;

(rd)²[1 - (%CW/100)] = (ro)²

So putting the values as gotten initially ;

(ro)² = 5²([1 - (27/100)]

(ro)² = 25 - 6.75

(ro) ² = 18.25

ro = √18.25

So ro = 4.27 mm

Imagine you have been asked to find the following object pictured on the left in the accompanying array on the right.RED AND WHITE SQUARES ALL LOOK SIMILAR
What type of search do you think this would be?

Answers

This type of search would be a linear search. A linear search involves looking at each item in the array one by one until the desired item is found.

In this case, you would look at each square in the array until you find the one that matches the object pictured on the left.  This is a common search method for small arrays or when the array is unsorted.

The algorithm continues until the target element is found, or until it reaches the end of the array and fails to find the target element. Linear searches are useful in scenarios where the array is small and unsorted, as the algorithm does not need to compare every element in the array to find the target element.

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brainly.com/question/30026271

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In same Fig. A-3, a force F, having a slope of 2 vertical 3 horizontal, produces a clockwise moment of 330 ft-lb about A and a counterclockwise moment of 420 ft-lb about B. Compute the moment of F about C.​

Answers

The moment of force at the given slope about point C is 210 ft-lb.

The given parameters:

  • Vertical slope, = 2
  • Horizontal slope = 3
  • Clockwise moment = 330 ft-lb
  • Counterclockwise moment = 420 ft-lb

The magnitude of the two moments are in the following simple ratio;

330:420 = 11:14 (divide through by 30)

  • the A coordinate = (0, 5)
  • the B-coordinate = (5,0)

The line of action of the force passes line AB at the final following coordinates;

total ratio of 11:14 = 11 + 14 = 25

= (11 )/(25) * 5, \ \ (14)/(25) * 5\n\n= (2.2, \ 2.8)

The position of C = (3, 1)

The resultant position of point C = (3 - 2.2,  2.8-1) = (0.8, 1.8)

The moment of force at the given slope about point C is calculated as;

3(1.8) + 2(0.8) = 7

Recall that this is the simplest form of the moment produced by the force.

Moment about C = 7 x 30 = 210 ft-lb

Thus, the moment of force at the given slope about point C is 210 ft-lb.

Learn more about moment of force here: brainly.com/question/6278006

Find the minimum sum of products expression using Quine-McCluskey method of the function. F(A,B,C,D)= Σ m(1,5,7,8,9,13,15)+ Σ d(4,6,11)

Answers

Answer:

Digital electronics deals with the discrete-valued digital signals. In general, any electronic system based on the digital logic uses binary notation (zeros and ones) to represent the states of the variables involved in it. Thus, Boolean algebraic simplification is an integral part of the design and analysis of a digital electronic system.

Explanation:

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A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic field of the stator is rotating

Answers

Answer:

The answer is below

Explanation:

A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic field of the stator is rotating. 2- the speed of the rotor when the slip is 0.05. 3- the frequency of the rotor currents when the slip is 0.04. 4- the frequency of the rotor currents at standstill.

Given that:

number of poles (p) = 4, frequency (f) = 60 Hz

1) The synchronous speed of the motor is the speed at which the magnetic field of the stator is rotating. It is given as:

n_s=(120f)/(p)=(120*60)/(4)=1800\ rpm

2) The slip (s) = 0.05

The speed of the motor (n) is the speed of the rotor, it is given as:

n=n_s-sn_s\n\nn=1800-0.05(1800)\n\nn=1800-90\n\nn=1710\ rpm

3) s = 0.04

The rotor frequency is the product of the supply frequency and slip it is given as:

f_r=sf\n\nf_r=0.04*60\n\nf_r=2.4\ Hz

4) At standstill, the motor speed is zero hence the slip = 1:

s=(n_s-n)/(n_s)\n \nn=0\n\ns=(n_s-0)/(n_s)\n\ns=1

The rotor frequency is the product of the supply frequency and slip it is given as:

f_r=sf\n\nf_r=1*60\n\nf_r=60\ Hz
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