1.Design and Selection

.Design data

.Design of the two steel material

.Fracture Assessment

.NDT methods

2.Sensitivity Test

.Lifetime assessment

.Revised assessment

3.Fatigue Assessmen

## Design data

**DESIGN AND SELECTION**

*(a).Design Data Provided*

Table 1: Design Parameters

Crack length (mm) |
Pipe diameter (m) |
Static head (m) |
Water hummer (m) |

3.0 |
2.5 |
501 |
765 |

Table 2: Steel Material Properties

PROPERTIES |
STEEL A QT 445 |
STEEL B BS150490LT50 |

C |
0.15-0.21 |
<0.20 |

Si |
<0.90 |
0.10-0.50 |

S |
<0.04 |
<0.03 |

P |
<0.04 |
<0.03 |

Mn |
0.80-1.10 |
0.9-1.60 |

Cr |
0.50-0.80 |
<0.25 |

Mo |
0.25-0.60 |
<0.10 |

Ni |
- |
<0.30 |

Zr |
0.05-0.15 |
- |

Cu |
- |
<0.30 |

B |
0.0005 |
- |

Nb |
- |
0.01-0.06 |

Yield strength (MPa |
700 |
350 |

UTS (MPa) |
800 |
500 |

Fracture toughness Kic |
100 |
130 |

Elongation (%) |
18 |
20 |

Price of rolled plate (per 1000kg |
965 |
525 |

Density (kg/m |
9750 |
9750 |

(b) *Design of the two steel Material*

It is noted that the maximum allowable stress (for the purpose of determining the thickness of both steel materials) is fixed at 0.6 times the material yield strength. The penstock is normally built in a similar fashion as the pressure vessels. Consideration is given for the thin walled cylinders hence theories of hoop and circumferential stresses come in handy in this case:

Firstly, the wall thickness is determined by considering the loading conditions and the yield stress

Consider steel A:

The allowable design stress= Yield stress/factor of safety=0.6x 700= 420MPa

The effective head, he= hw+hwh= 501 + 765= 1266m

Next, internal pressure P in determined:

P= ρgh= 1000x 9.81x 1266= 12.419MN/m^{2}

Radius of pipe= d/2= 2.5/2= 1.25m

Now, we assume the pipe is designed in a fashion similar to pressure vessel hence thickness is given as:

Hoop stress (maximum principal stress) = Pr/t

Implying: tA= (1.25x12.419)/420 = 0.03696m=36.96mm

Now, considering Steel B:

The allowable design stress= Yield stress/factor of safety=0.6x 350= 210MPa

The effective head, he= hw+hwh= 501 + 765= 1266m

Next, internal pressure P is determined:

P= ρgh= 1000x 9.81x 1266= 12.419MN/m^{2}

Radius of pipe= d/2= 2.5/2= 1.25m

Hoop stress (maximum principal stress) = Pr/t

Hence tB= (1.25x12.419)/210 = 0.07392m= 73.92mm

Steel B wall pipe will have to be thicker than that of steel A in order to withstand the same loading conditions. Although, per a thousand kg, steel B is cheaper than steel A, however in the long run more capital would be needed to purchase steel B material given the doubling of material quantity due to increased thickness (Bannister, 1998).

In fact considering the cost implications:

Cost A and Cost B can be compared:

Cost A/Cost B= tA$A/tB$B

36.96x 96.5/52.5x73.92= 0.9190

This implies that Material A will cost about 92%relative to material B

(c) **Fracture**** assessment**

(i) **The effects of cracks on the integrity of the structure using FAD is considered.**

** **

** **

From the graphs, Steel A has a fracture toughness of between 113.93MNm^{-3/2} and 66.74MNm^{-3/2} while B has a fracture toughness of between 127.49 and 72.62. From here, the workable values of fracture toughness can be deduced hence : at 0oC, steel A and B fracture toughness of 102MNm^{-3/2} and 128MNm-3/2 respectively. Therefore, it can be said that steel A is less tough given the design values. It should be noted that in arriving at the fracture toughness values, te following equation was used: Kc= 16(Cv)^{0.5}

## Design of the two steel material

(ii) *Criticality of material cracking under repetitive loading*

It is understandable that the penstock will undergo a number of repetitive loadings. It is essential to determine the critical cracking size from which material failure is inevitable. This is done by using the iterative approach such as Newton-Raphson method. In this case, the following equation is considered for iteration: the calibration function:

Since steel B gives a much bigger value of critical crack length, it can be stated that steel B is structurally more resilient than A in terms of fracture toughness. The critical crack size for Steel A is **?****.7398** mm, which is greater than 6mm, which is the assigned detectable flaw size. With a total crack propagation length of only 0.7398mm, steel A may not withstand repetitive loading as the critical crack size is nearer the design value of 6mm.

(iii) *NDT methods and its accuracy*

Accuracy of NDT Methods:

NDT methods like ultrasonic inspection often are used to detect flaws on the surface pipe in a real-time fashion (Bannister, 1998). At minimum, a crack size of 6mm would be perceived and this will facilitate repairs before cracking failure supersedes. Notably, with time, the geometry calibration factor, Y would go down as the stress intensity ratio, Kr and load ratio, Lr both increase with crack size. Thus assessment would focus on slightly shifting along the failure curve so that crack size can be detected earlier than that of 6mm. Normally crack propagation is rapid and would bear catastrophic results if not checked in real time. The NDT therefore provide tools of assessment of its status. However, the accuracy of these methods would have to be increased so that chance of failure can be fixed at the minimum level. For instance, as mentioned, one can reduce the detectable flaw size by half which will then significantly thwart material failure (Hudson & Rich,1986).

(iv) Failure assessment diagram (FAD) Analysis

The geometry calibration function factor for both steel materials A and B are

The resulting curve has an equation given by:

Y= 1.12-0.231t+10.67t^{2}+ 30.39t^{3}

The type of failure in each case can then be determined by using two points on the curve as assessments

Figure 3 shows a line of best fit is drawn between the two assessment points P1 (0,0.1633) and P2 (0.6193, 1.1431). The line crosses the boundary of the R6 Kr failure curve at approximately (0.5,0.9582). The relatively steep gradient of the line indicates that Steel A is very likely to undergo brittle fracture, i.e. the crack will spread rapidly with a very limited extent of plastic deformation in the structure that is ‘contained’ or ‘small scale’ plasticity. The assessment point for Steel A, under maximum allowable hoop stress lies outside the predicted R6 failure curve given by Equation 3.3.1, and thus failure will occur under the current conditions. The critical crack length calculated for Steel A (6.1046mm) is also extremely close to the detectable flaw size, (6mm) and this is a reason for why the material fails in a brittle manner. Hence, Steel A is not viable.

## Fracture assessment

Steel B

**2.SENSITIVITY TEST**

As per Figure 3, the line of best-fit is plotted in Figure 4 between the two assessment points, P1 (0,0.1022) and P2(0.6095,0.4088). For further investigation, the best-fit line is extrapolated until it intersects the failure curve. It crosses the failure curve at approximately (0.98,0.6003), indicating a plastic-elastic collapse. The relatively gradual gradient of this line typifies ductile failure, i.e. the crack will propagate slowly, and is accompanied by a large amount of plastic deformation. The higher critical crack length for Steel B (17.4415mm) as compared to Steel A (6.1045mm) is a contributing factor to this. However, in this case, both assessment lines lie within the safe/acceptable threshold under the R6 Kr failure curve, signifying that Steel B will not fail under the current conditions. Based on the FAD Assessment alone, it can be concluded that Steel B is the better, safer option. Catastrophic failure is very unlikely to occur without signs of warning first, and under the given conditions, Steel B will not undergo failure

Figure 5 shows that the second assessment point on the FAD for Steel A, which was previously seen to lie outside the R6 specified failure curve, has now re-located to within the safe threshold. Hence, Steel A will not fail under the current conditions. It is also observed that Steel A will now eventually undergo ductile failure, instead of brittle failure. Therefore, a higher accuracy of NDT methods would have allowed a more thorough and precise comparison of the two steels (McStraw, 1996).

(b) *Lifetime assessment*

It is also crucial to check on the durability of the penstock. This is done by estimating the fatigue lifetime using the technique as discussed below:

Following the previous analysis, Steel B has been determined to be a better-suited material for the design of the penstocks. Steel A had been shown to have less toughness and be more likely to fracture due to unstable crack propagation.

The crack propagation life can be estimated using the relation between the stress intensity range, Δ? and the crack growth rate (Paris’ Law)

The stress cycle arises from the pressure change from 0m to 525m head of water within the pressure vessel. Using this information, the range of hoop stresses acting on the vessel can be calculated.

## NDT methods

By utilizing a small step size, (Δ?), an iterative numerical integration can be carried out to determine the solution. The initial crack size, ?? is equivalent to the given detectable crack/flaw size of 6mm. The final crack size, , will be greater than the critical crack size, ?? of the steel, as calculated previously:

3.**FATIGUE ASSESSMENT **

The graph below can be used to estimate the service life of both materials. From the obtained crack size values of steels A and B are respectively. The corresponding service life are: 0.5 and 2.35 respectively

The crack length propagation plot in Figure 10 shows that the time taken for the initial crack to propagate to critical size within Steel B is 2.35 years, well below the required service life of 50 years. Design improvements are to be made accordingly. The fatigue life plot for Steel A predictably shows that failure due to fatigue will occur very quickly. However, it is noted that the toughness data provided is for the parent material yet the point of propagation of cracks are likely to be in the heat affected zones (Molzen & Hornbach, 2000).

**CONCLUSION **

Notably, the internal pressure P acting on the walls on the pipe greatly contributes to cracking. However, the penstock is treated as a thin walled cylinder such that variations of stresses are not considered. There are points along the pipe when stress would be maximum than the design value. In order to accommodate these changes, the design engineer would design the pipe with some factor of safety so as to cushion against catastrophic failure.

Now, there are techniques that are used to calculate the crack size. Matlab can be useful especially for iteration. However, LEFM can be used as well although it is not strictly applicable to the section thicknesses of Steels A and B, for values of K approaching KIC, as the plastic zone at the crack tip is significant. By integrating the plastic zone correction factor, smaller values of the critical crack length would be obtained. The size of the crack tip can either be calculated using Irwin’s model, which estimates the elastic-plastic boundary using elastic stress analysis, and the strip-yield model (O'Brien, 2015)

In the section ‘Sensitivity Analysis’, it was shown that the risk of failure could be substantially reduced by improving the accuracy of the Non-Destructive Testing (NDT) methods. The minimum detectable crack length should be reduced to 3mm, as this ensures that both Steels A and B will not undergo failure under given conditions. Steel A can therefore be compared with Steel B in a more critical and thorough manner and a more informed choice can be made. The reduction of minimum detectable crack length also allows a more accurate estimation of crack propagation, so that risks of failure can be identified in advance and dealt with accordingly (Saadat, 2015).

The section on Fatigue Assessment does not take into account the damage caused by any variation in the number of cycles, Nf. Variation in size, number and order of stress cycles will lead to cumulative fatigue damage of the penstock. Therefore, fatigue damage must be evaluated by adding the detrimental effects of each individual cycle, to ensure that any chance of failure is minimized.

Cracks are to be repaired when they are approximately equal to 15mm, thus providing a safety window of about 12 years to accommodate for the risk of failure. Maintenance of cracks of smaller lengths than 15mm will increase the operational cost of the station, without any substantial increase in safety.

The inspection frequency for the intermediate penstock must be between 1 and 5 years, but no more than 5 years . Therefore, the inspection interval should be co-ordinate with the maintenance interval, i.e. stripping of the walls and repainting.

Anderson T. (2005). Fracture Mechanics. Hoboken: CRC Press.

Bannister A. (1998). Structural integrity assessment procedures for European Industry. Swindon: British Steel plc Deuflhard P. (2011). Newton Methods for Nonlinear Problems: Affine Invariance and Adaptive Algorithms.] Hudson, C & Rich,T. (1986). Case histories involving fatigue and fracture mechanics. Philadelphia, PA: ASTM.

Jivkov A. 2015. The Design of a Penstock Pipe for a Hydro-electric Pumped Storage Station. 1st edKaechele L. (1963). Review and Analysis of Cumulative Fatigue-Damage Theories.McStraw, B. (1996). Inspection of Steel Penstocks and Pressure Conduits. United States Department of the interior bureau of reclamation denver, colorado.Molzen, M & Hornbach, D. (2000). Evaluation of Welding Residual Stress Levels Through Shot Peening and Heat Treating. 1st ed.

O'Brien R. Welding handbook. (1991). Miami, Fla.: American Welding Society.Saadat H. 2015. Power System Analysis [Internet]. Psapublishing.com. 2015 [cited 20 November. Available from: __https://www.psapublishing.com/__

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