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enpe 563 quesiton 1

6) In this scenario, we are to examine materials that have to withstand high temperatures (650 degrees Celsius) without enduring much stress in terms of pressure, as well as the pipeline to maintain its diameter. The temperature limitation eliminates materials like PVC Polyvinylchloride (material used often in piping in chemical plants) from consideration because it deforms at 85 degrees Celsius. The primary property that should be considered is creep resistance. Since the pipeline is in a high temperature environment, it is of great importance that it has a good creep resistance. Creep resistance is represents the material’s resistance to high temperatures. The secondary property that should be considered is yield stress. It is important that piping does not deform, because if the diameter of the piping is distorted, it could pose serious problems. Also, since this is a high temperature pipeline a deformation could cause a dangerous and volatile situation. For example, if the piping diameter shrunk it would increase the pressure inside the pipeline which therein will increase the temperature making the high temperatures even higher. This pressure temperature relationship can be seen with the ideal formula PV=nRT, where P is directly proportional with T. The three best materials for this scenario are: - Ferralium 255 - Stainless Steel 410 - Grey Class 20 (ASTM A48) Ferralium has a high ductility, strength, and hardness which will help the material keep a consistent diameter which is of crucial importance for a pipe (σy = 683MPa). 410 is a martensinic stainless steel that has a high yield stress and creep resistance (σy = 1005MPa, σcr = 28 MPa at 650 degrees Celsius for 10 000 hours). Grey Class 20 is used for piping in a sewer environments and has a yield stress of σy = 80MPa. Since Grey Class 20 has a lower yield stress, it can be eliminated from consideration. Between Ferralium and 410, it is observed that 410 has a relatively higher yield stress and an excellent creep resistance when compared to Ferralium. Therefore, I would conclude that 410 would be the best material for the job.

3) In this question we are to examine materials that must withstand high pressure. A BOP is used in a well as a safety precaution. If the wellbore pressure increased there is a chance that a blowout may occur. If this pressure starts to occur they can close off the well, and increase the mud density to counteract this problem. In this scenario, we must look at the yield strength, ultimate strength, cost and safety. Safety is a major component of this analysis, a BOP can potentially save lives, if it’s designed with the wrong material, it may fail under the extreme pressures that it’s subjected to. If the material gives out people may die. With this reasoning, we can assume this item to be a performance item, meaning that cost is not a major factor. The Primary property that should be considered is the Ultimate stress of the material. Ultimate stress is the maximum stress that the material can withstand without failing. Ultimate stress can be calculated using the formula:  σ=P'/A Where: P’ is a force and A is the area of the BOP. Given 200 000 psi we can calculate the ultimate stress, by using the formula:   P'=F/A σ = PA/A  =P Where: P is pressure in psi and F is a force in N.

The secondary property that should be considered is the yield stress of the material. The yield stress is stress until deformation. It is important that the material does not deform. If the BOP was to deform (the diameter became smaller) the maximum pressure exerted on the system will increase, making the possible material chosen ineffective. The two best materials for this case are: · 18Ni(350) ( σu – 1140MPa, σy – 827MPa, εf - 18%) · Stainless steel 410 ( σu – 1475MPa, σy – 1005MPa, εf - 11%)

In order to pick one of these materials we must compare the values given. The ultimate stress that this system will be under is 137MPa. Both of these materials have a much higher ultimate and yield stress than the stress that is measured. The yield stress and ultimate stress are higher for SS410 but the strain at fracture is a lot lower. Both of these materials have an Rn higher than 1.2. The factor that set these two materials apart is the plane strain fracture toughness of 18Ni(350) is much higher than that of SS410. Given that the Kic and ε<span style="font-family: 'Cambria Math','serif';">f of 18Ni(350) are higher this is the better choice of material.

Q7

A surge tank is a storage reservoir placed in a flowline through which liquids and gases are flowed to neutralize sudden pressure surges. The design criterion for the tank will be mainly focused on how well it can withstand pressure since there is pressurized nitrogen being charged into it. Because this is a performance item, cost is not a major factor. Safety is of paramount importance.

Since this is a pressurized vessel, it can blow in case of a small fracture if the right material is not used. High strength and good ductility are the controlling factors for material selection. Because strength is important in material selection, materials with high yield strength and high ultimate strength are selected. Ductile materials are selected based upon strain at fracture. The higher the strain percent the more ‘forgiving’ the material is. For critical applications like these, material with strain higher than 20% is selected. Fracture toughness also plays an important part in material selection for the tank because even a smallest fracture could eventually lead to a catastrophic explosion.


 * Pressurized Vessel Material Selection:**

My first choice of material for the vessel is steel alloy AISI 4340. AISI 4340 has a very high hardenability, high yield and ultimate strength and a relatively high plastic strain at fracture. Following are its properties: σu – 745 MPa σy – 470 MPa εf – 22% (indicates high ductility i.e. material is more forgiving)

My second choice is another steel alloy AISI 4140. AISI 4140 has a high hardenability, good fatigue resistance and good impact resistance with a high ductility. Following are its properties: σu – 655 MPa σy – 415 MPa εf – 26%

AISI 4340 is my top pick because it has a higher yield and ultimate strength compared to AISI 4140 but both are very good materials to build pressure vessels out of.

There is a rubber bladder secured at the mid-point in the interior of a pressurized vessel indicating that the rubber selected in the manufacture of the bladder should have a high elasticity modulus and a high tear strength.


 * Rubber Selection:**

My first choice for rubber would be butyl rubber. Products used to prevent gases from passing through the material are based on butyl rubber. The polymer consists of isobutene with a minor part of isoprene. The gas permeability increases with increased temperature for all rubber types, but for butyl rubber it is very low. Butly rubber also has a high tear strength and a high elastic modulus.

My second choice is chlorobutyl rubber. Chlorobutyl rubber has properties similar to butyl rubber but with improved ozone and environmental resistance and greater stability at high temperatures. Since high temperatures are not involoved in this scenario, chlorobutyl rubber is not necessary,

Q8

A 4 metre tall 2 metre diameter pressure vessel is to be designed with the following factors in mind:

- Tank has an acidic environment of pH=4 - There is oxygen and brine containing sulphate and bicarbonate ions in the tank - Tank operated under a constant pressure and temperature of 45°C

Dissolved oxygen is a major factor in natural corrosion of steel. Corrosion is further accelerated when dissolved oxygen is reduced at metal surfaces at a low pH. The tank contains brine which comprises of sulphate ions and bicarbonate ions both of which contribute to corrosion. Hence corrosion and strength are controlling factors in selecting materials for this pressurized vessel.

A FCC material has excellent corrosion resistivity, excellent ability to withstand high or low temperatures, good strength and ductility, therefore a material with FCC crystal structure is selected.

My first choice for material is Alloy 400 consisting mostly of nickel and copper. Alloy 400 has high strength and excellent corrosion resistance to sea water (brine), sulphuric acid and alkalines. It also has a high yield and ultimate sress indicating that they can be used in design of pressurized vessels. Following are some of its characteristics: σu – 550 MPa σy – 240 MPa εf – 40% HB – 140 E – 180 GPa

My second choice would be Ferralium 255. It is a duplex ferritic-sutenitic stainless steel. It has high strength, ductility, hardness and corrosion and erosion resistance. It is corrosion resistant to most acids and has good stress corrosion cracking (SCC). Following are some of its properties: σu – 855 MPa σy – 683 MPa εf – 26% E – 210 GPa

Alloy 400 is my first choice because it has a better corrosion resistance to sea water (brine) than Ferralium 255.


 * Question 9: thermal conductivity predictions: I`ll ask someone for it tomorrow (sunday)***

Question 10: I`ll send through email, I got it from a friend, but then I think we should take a look at the handbook ourselves so that we dont have the same answers as him.
 * Question 11:**
 * You work for a pressure relief manufacturer and he has assigned you the task of selecting a material for the helical spring to be used on a new series of pressure relief units. The units are to be attached to the top of small pressure vessels that operate at 340 degrees.**

To answer this question, let's walk through the main steps of material selection: > Therefore the things to consider: modulus of resilience UR, fatigue, yield strength σy, strain εy, corrosion resistance, σcr (Creep resistance - resistance to heat), C (cost) > > > Because the units are being used at the top of small pressure vessels that operate at high temperatures (340°C), an assumption will be made that a hot-coiled spring would be required. From research, it is known that most hot-coiled coil springs are made from bars of carbon or alloy steel (large portion of the cost is in the material and therefore will increase the cost of hot-coiled coil springs.
 * //Major function and application of the material, and properties to consider//: Pressure relief units are designed to withstand a certain pressure. The pressure relief valve is installed at the top of a pressure vessel with a spring in it. The valve is designed to relieve the excess pressure and opens up upon reaching this maximum point (the pressure inside the vessel is high enough to work against the spring in the valve) and vents out to prevent overpressure in the vessel. Springs need to be able to store elastic energy (be able to convert kinetic energy as it gets compressed to potential energy and then reproduce kinetic energy....while being able to return to original shape)
 * //Design Considerations:// Since the spring will just be used within the PSV (Pressure Safety Vessel), the appearance is not a key concern, however the longevity of the spring should be considered. It would be dangerous to make a spring that is not durable and thus unable to perform its task of preventing overpressure in the vessel once the yield has been reached.
 * //Fabrication ability and availability://
 * Cost: Springs are used for performance, its durability is important in triggering the pressure safety valve to open when the max pressure has been reached, therefore, one cannot be stingy about the material used to produce these springs. With that being said, however, if these springs need to be mass-produced for many new pressure relief units, then cost will become an issue and we will wish to minimize it.

Performing the CCR: __Ur*fatigue*εy__ __C


 * Question 11 NOT COMPLETE :( sorry, ill finish it soon***

Question 4 :  4) The main property of the material selected is its magnetic ability (magnetic permeability-μ). A non-magnetic sub is required; therefore the material must be non-magnetic. Since the sub is attached to drill string it must meet the requirements of the drill string, therefore it must have high yield and ultimate stress. The material that best meets this requirement is nickel alloy K500. This material is aged hardened and is non- magnetic. The magnetic permeability of nickel is 125E-6, the smaller the μ the less likely the material will become magnetic. Being age hardened it has a higher ultimate and yield stress (σu – 1100MPA, σy – 790MPa). Being attached to a drill string safety is also important, this material has a εf = 20%, which puts it in the extremely critical region (this isn’t an important property but it helps make our choice easier).__

__ Hey guys, Here are the materials I ended up choosing for each of the questions. Since I wrote all of my stuff on paper, I can scan/upload them if you'd like. Just let me know. For some of them I had more than 1 material listed, but the ones below are my "top choice".

1) AISI 4140 2) AISI 9255 (primary) and AISI 4140 (secondary) 3) 440C 4) Alloy K500 5) Stainless Steel 316 6) Stainless Steel 316 7) Butyl Bladder and AISI 4340 8) Alloy 400 9) Not done yet.. 10) Too long to type 11) AISI 9255

<span style="font-family: 'Times New Roman','serif'; font-size: 12pt;"> 5) In this scenario we are trying to build a heat exchanger. The main purpose of a heat exchanger is like its name suggest – heat transfer. In the heat exchanger there is 2 streams that are exchanging the heat. Residual hydrogen sulphide and carbon dioxide are in the pipes, adding to the risk of corrosion.  <span style="font-family: 'Times New Roman','serif'; font-size: 12pt;">  <span style="font-family: 'Times New Roman','serif'; font-size: 12pt;"> The primary property of this material must b thermal conductivity, the second property will be its corrosion resistance and a third property that may be examined is its creep resistance, this is the materials resistance to high temperatures. In theory would assume that k (thermal conductivity) and Cr would cancel each other, but we must find the material that maximizes these values. <span style="font-family: 'Times New Roman','serif'; font-size: 12pt;"> <span style="font-family: 'Times New Roman','serif'; font-size: 12pt;"> Some materials which could be used are : 1) Stainless steel 316 (Cr – 97MPA @650 C for 100000 hrs, K= 16.3, great corrosion resistance) <span style="font-family: 'Times New Roman','serif'; font-size: 12pt;">  2)No8366 ( good SSC resistance, higher Cr when exposed to sea water) <span style="font-family: 'Times New Roman','serif'; font-size: 12pt;"> 3)S2910 (great corrosion resistance, reistance to sour SCC)  <span style="font-family: 'Times New Roman','serif'; font-size: 12pt;">    <span style="font-family: 'Times New Roman','serif'; font-size: 12pt;"> The best choice out of these materials is stainless steel 316, this material has a very good creep resistance while maintaining a high k value. No8366 would be good to use for parts of the heat exchanger due to its excellent corrosion resistance. S2910 would be ideal if the heat exchanger was subjected to sea water.