Penstock Design for a Hydro-electric Pumped Storage Station Report_Ayman_Siddique

1. 2015
Design of Penstock Pipe
for a Hydro-electric
Pumped Storage Station
MACE 61057: STRUCTURAL INTEGRITY (DR. ANDREY ZIVKOV)
AYMAN SIDDIQUE (ID: 7669311)

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TABLE OF CONTENTS

TABLE OF FIGURES ........................................................................................................................................... 1
LIST OF TABLES ................................................................................................................................................ 1
1. INTRODUCTION ....................................................................................................................................... 2
2. STATIC DESIGN/ OVERLOAD ASSESSMENT ............................................................................................... 3
3. FRACTURE ASSESSMENT .............................................................................................................................. 4
3.1 Determining KIC for Steel A and Steel B .................................................................................................. 4
3.2 Critical Crack Sizes in Steel A and Steel B................................................................................................ 5
3.3 Failure Assessment Diagram (FAD) Analysis ........................................................................................... 5
3.4 Sensitivity Analysis ................................................................................................................................. 7
4. FATIGUE ASSESSMENT ............................................................................................................................... 10
5. DESIGN IMPROVEMENTS ........................................................................................................................... 11
6. CRITICAL DISCUSSION AND FINAL RECOMMENDATIONS ............................................................................ 12
7. REFERENCES............................................................................................................................................... 14
8. APPENDIX ................................................................................................................................................... 15

TABLE OF FIGURES

Figure 1: Schematic diagram of a hydro-electric power station [3] .......................................................... 2
Figure 2: Fracture toughness data for Steel A .......................................................................................... 4
Figure 3: FAD for Steel A .......................................................................................................................... 6
Figure 4: FAD for Steel B .......................................................................................................................... 6
Figure 5: FAD for Steel A (Detectable crack length = 3mm) ...................................................................... 7
Figure 6: Pressurized cylinder with semi-elliptical crack ........................................................................... 8
Figure 7: Variation of Y with crack ratio a/c for Steel A and Steel B .......................................................... 9
Figure 8: FAD for Steel A using different values of a/c .............................................................................. 9
Figure 9: FAD for Steel B using different values of a/c ............................................................................ 10
Figure 10: Fatigue growth for Steel B, thickness = 38.5mm .................................................................... 11
Figure 11: FAD of Steel B with increased thickness of 85mm ................................................................. 11
Figure 12: Fatigue growth of Steel B, thickness = 85mm ........................................................................ 12

LIST OF TABLES

Table 1: Material Properties of Steel A and Steel B …........................................................................................2
Table 2: Individual Parameters.…………………………………………………………………………………………………………………… 2
Table 3: Charpy Data for Steel B…………………………………………………………………………………………………………… ……..4

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1. INTRODUCTION
Hydro-electric power stations utilise water from a source at a higher altitude, in order to convert
potential/kinetic energy into electrical energy. [1,2] The momentum of the water intake drives the turbines,
which in turn power the generators that produce electricity. Pumped storage stations have two reservoirs to
recirculate the water, and hence store electrical energy. Once the turbines are driven, the water is pumped
from the lower reservoir to the higher reservoir to create an immediate reserve of water. The implementation
of both the hydro-electric and pumped storage stations is quicker and more reliable than any other type of
station [1], and therefore, an informed design of the specified combined station and its components is of
utmost importance.

This design task is concerned with the
structural integrity of the intermediate
penstocks. Penstocks are the steel pipes
that convey water from the intake
reservoir to the turbine of the hydro-
electric power station. Their function can
be visually observed from Fig 1. The
intermediate penstocks are unsupported
(no presence of surrounding rock or
concrete) and are to be capable of
withstanding the full water pressure
(hydraulic load) throughout the entirety of
its design life. [2]
Two steels, A and B, are to be evaluated in this report as a suitable material to be used in the fabrication of the
intermediate penstock, using their respective given material properties in Table 1, and the assigned individual
parameters shown in Table 2:







Table 1: Material Properties of Steel A and Steel B

This report will assess and compare the two Steels A and B, via fracture and fatigue assessments, based on the
assigned individual parameters. Design improvements required to meet the assigned criteria for the penstocks
will be made, and evaluated, in order to provide a final recommendation, and suggest a suitable inspection
interval. All MATLAB codes used for analysis in the subsequent sections are included in the Appendix.
Detectable
flaw
size, mm

Pipe
diameter,
m
Static
head of
water, m
Max head of
water
during
hammer, m

6 2.2 525 750
Table 2: Individual Parameters
Figure 1: Schematic diagram of a hydro-electric power station [3]
The penstocks are subjected to ultra-sonic non-
destructive testing (NDT), and a minimum detectable
crack length of 6mm has been determined. The
assigned maximum head of water or pressure transient
is 750m. The intermediate penstock is exposed to a
pressure change during the station’s change in modes,
i.e. changing from pumping to generating and vice
versa. [2] There is an additional residual stress of
70MPa, which is created by localised areas of cooling
and heating during welding.

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2. STATIC DESIGN/ OVERLOAD ASSESSMENT

The maximum allowable design stress is stated to be equal to 0.6 times the yield stress, and this value has
been determined by using the prescribed factor of safety. It is therefore necessary to estimate what wall
thickness will be needed in either steel in order to allow for safe operation under static conditions. The
maximum possible hoop stress occurs due to the 'water hammer' transient pressure. The penstocks are
fabricated like pressure vessels by longitudinally and circumferentially welding shaped steel plates. They are
modelled as thin-walled cylinders, and hence the hoop stress can be calculated:

??????
ℎ=
??????
�??????� .�
�
⁄=0.6??????
� ….(2.1)
where ??????
ℎ is the hoop stress, ??????
�??????� represents the maximum pressure due to the water hammer, ??????
� is the
tensile yield stress, whereas � and � represent the radius and thickness of the specified penstock, respectively.

??????
�??????� is calculated using the maximum head of water, ℎ
�??????�, which corresponds to:
??????
�??????� = ��ℎ
�??????� ….(2.2)

where � is the density of water (1000 ��/�
3
) and � represents acceleration due to gravity(9.81 �/�
−2
)

Substituting Eq 2.2 into Eq 2.1 and rearranging for t gives:

�=
��ℎ
�??????�.�
0.6??????
�
⁄ …..(2.3)
The Eq 2.3 allows the thickness, t to be calculated for both Steel A and Steel B. From Figure 1.1 the yield
stresses for Steel A and Steel B are 750 MPa and 300 MPa, respectively. The radius, � of the pipe is 1.1m.

�
�=
1000∗ 9.81∗750∗1.1
0.6∗700∗10
6
= 19.3 ��, �
�=
1000∗ 9.81∗750∗1.1
0.6∗350∗10
6
=38.5 ��

It can therefore be determined that the minimum thicknesses for Steel A and Steel B, are 19.3mm and
38.5mm, respectively. Additionally, the hoop stresses for the Steels A and B are calculated to be 420MPa and
210MPa, respectively.

The material cost depends of the weight of steel used, which is a function of the wall thickness, t. Since the
diameter and length of the vessel are fixed, the weight depends only on the wall thickness used.
In the following equation, �
�,�
�,£
�,£
� are the thicknesses and prices of rolled plate/1000kg of Steels A and B,
respectively .
Thus, the ratio of the costs of Steels A and B, or the relative cost is:

����
�
����
�
=
�
�£
�
�
�£
�
=
0.0193∗965
0.0385∗525
=0.92

Therefore, the price of steel A is 92%, relative to that of steel B. One of the key objectives of during the design
process is the minimization of costs, and the ratio given in Eq 2.4 will be used in the final evaluation.

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3. FRACTURE ASSESSMENT
3.1 Determining KIC for Steel A and Steel B

Steel A:
The design brief includes fracture toughness data for Steel A, measured on a 40mm thick plate. This is
shown in Figure 2:
At 0℃ a fracture toughness of
approximately between 100 ���
−3
2
⁄

& 120 ���
−3
2
⁄
would be a
conservative choice. Figure 3.1 shows
that the steel would need to be at
roughly −50℃ for 100 ���
−3
2
⁄
to be
the correct value. The operating
temperature is not known and the
test-piece thickness of 40mm is greater
than the calculated thickness for Steel
A (19.3mm). The lower bound fracture
toughness of 100 ���
−3
2
⁄
will
therefore be used for analysis, as it is
assumed to be the worst case scenario.


Steel B:
Charpy test data is given for steel B, in Table 3 However, this data is deemed inaccurate/unreliable and could
not lead to a very accurate value of fracture toughness.







Using the current SINTAP recommended equation �
??????�=12 √�
?????? , the corresponding calculated fracture
toughness value is 89 ���
−3
2
⁄
, at 0℃. It was expected of Steel B to have approximately the same or greater
fracture toughness as Steel A.

It is therefore suitable to use an earlier and less convervative correlation, �
??????�=16 √�
?????? , which results in a
fracture toughness of approximately 120 ���
−3
2
⁄
at 0℃ for Steel B.

Fracture toughness for both Steels A and B at 0℃ have been determined. Hence:
�
??????�
�
= ��� ��??????
−�
�
⁄
….(3.1.1) ; �
??????�
�
= ��� ��??????
−�
�
⁄
……. (3.1.2)


Figure 2: Fracture toughness data for Steel A
Table 3: Charpy data for Steel B

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3.2 Critical Crack Sizes in Steel A and Steel B

For simplicity, the growth of a long surface edge crack, lying parallel to the axis of the pipe is considered. This
crack is influenced by the hoop stress generated when the vessel is filled with water.
The calibration function for a similar geometry is given as an edge crack in a finite width plate in tension. A widely
used equation to determine the calibration function is:

??????=1.12−0.231(
??????
�
⁄)+10.55(
??????
�
⁄)
2
− 21.72(
??????
�
⁄)
3
+30.39(
??????
�
⁄)
4
……….(3.2.1)

Since the geometry calibration function varies with crack size, ?????? , the critical crack size, ??????
� is the solution to the
non-linear equation: �
??????�=???????????? √�??????
� ……….(3.2.2)
The non-linear equation can be solved by using a simple iterative method like the Newton-Raphson technique.
Initially, the equation 3.2.1 is substituted into equation 3.2 and rearranged in the form f(ac) = 0. The next
iteration is calculated by the equation:

??????
�
??????+1
=
�(??????
�
??????
)
�′(??????
�
??????
)
⁄ − ??????
�
??????
(3.2.3)

The benefit of the Newton-Raphson method is that is quick to converge, and is stable. An initial estimation of
the critical crack size is required for the first iteration, and in this case, a size of 1mm was chosen. The MATLAB
program was used for the purpose of this calculation, and the relevant code is attached in the appendix. [11]

The calculated critical crack lengths, ??????
� for the two steels are:

Steel A: ??????
�
�
=�.����???????????? ; Steel B: ??????
�
�
=��.����????????????

It can be stated that Steel B is much more durable than Steel A, since the critical crack length in order to
initiate catastrophic crack growth for Steel B is approximately 3.33 greater than that of Steel A. The critical
crack size for Steel A is 6.1046mm, which is greater than 6mm, which is the assigned detectable flaw size. With
a total crack propagation length of only 0.1046mm, it can be predicted that failure will occur very quickly and
hence Steel A is deemed unsuitable.

3.3 Failure Assessment Diagram (FAD) Analysis

The geometry calibration function, Y has been calculated for both steel using Equation 3.2.1:
??????
�=�.�????????????� (??????���� �) ??????
�=�.���?????? (??????���� �)
The �
� curve used is the currently accepted curve in the R6 specification, given by:
�
�=(1+0.5�
�
2
)
−
1
2
⁄
∗[0.3+0.7exp(−0.6�
�
6
)] …… (3.3.1)
The type of failure occurring in each steel can be estimated by plotting two assessment points on the same
plot as the R6 �
� curve mentioned previously.
The first point considers the secondary stresses, or stresses which do not contribute to plastic collapse. The
secondary stress for this vessel is the residual stress due to welding, which is present even in the absence of an
applied primary (hoop) stress. Therefore it is plotted at �
�=0. Let this point be P1, and in the following Failure
Assessment Diagrams(FADs) for Steels A and B, is represented by a red marker.
The second point utilises both the primary and secondary stresses, where the maximum primary stress is due
to the pressure of the water hammer, as mentioned previously. Let this point be P2, represented by a green
marker in the following diagrams.
Values of �
�
??????
,�
�
�
,�
�
??????
,??????�� �
�
??????+�
were calculated using the following equations:

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�
�
??????
=
??????
??????
??????
�(1−
??????
�
)
⁄
…(3.3.2) �
�
??????,�
=??????
??????
??????,�
√�??????
�
??????�
⁄…(3.3.3) �
�
??????+�
=(??????
??????
+??????
�
)*?????? (
√�??????
�
??????�
⁄)…(3.3.4)
??????�∶(0,�
�
�
) , where for Steel A and Steel B, are (0, 0.1633) and (0, 0.1022), respectively. This point is signified
in the subsequent FAD plots as Lr(s), Kr(s). �
�
�
is calculated using Equation 3.3.3.
??????�:(�
�
??????
,�
�
�
+�
�
??????
), where for Steel A and Steel B, are (0,1.1431) and (0,0.4088), respectively. This point is
denoted in the subsequent FAD plots as Lr(p), Kr(p+s).
�
�
�
+�
�
??????
is calculated using Equation 3.3.4.
STEEL A:

Figure 3: FAD for Steel A

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.
STEEL B:









Figure 4: FAD for Steel B

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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.

3.4 Sensitivity Analysis
Accuracy of NDT Methods:
The minimum detectable crack depth is stated to be 6mm, i.e. the smallest detectable crack which could
be treated during the lifetime of the vessel. The geometry calibration factor, Y will decrease, whereas the
stress intensity ratio, Kr and load ratio, Lr both increase with crack size, as observed from their respective
equations. As a result, the assessment point will shift closer, within the failure curve of the material with
increasing crack size. Increasing the accuracy of the NDT methods will decrease chance of failure. For
example, reducing the detectable flaw size by half (i.e. 3mm) will significantly delay failure, as shown in
the following FAD:
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.

Figure 5: FAD for Steel A (Detectable crack length = 3mm)

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Accounting for Residual Stress:
The welding process used for the steels induce residual tensile stress, which substantially decrease fatigue life.
Cracks are initiated in metals which are exposed to high tensile stresses over a significant number of load
cycles. []
The residual stress value is used explicitly to calculate the stress intensity factor, K, by the equation
�
??????�=???????????? √�??????
�. The assessment points are displaced vertically along the y-axis of the FAD, by changing the
magnitude of the residual stress.
The residual stress is constantly applied to the penstock, and does not contribute to the yielding of the vessels
material. It is therefore concluded that an under-estimate of the stress could indicate that the vessel is indeed
safe from a brittle and sudden fracture due to crack propagation, where in practice it is more likely to do so.
More than 90% of failures occurring in mechanical components occur due to crack nucleation/propagation
arising from high tensile loading conditions, and as a result, adversely affects the load-bearing capacity of the
component. In this case, one of the key criteria for the intermediate penstock is to maximize its load-bearing
capabilities, and therefore reduce residual stresses [8]. This will be outlined in section ‘6. Critical Discussion
and Final Recommendations’


Shape of the defect:
The crack has been assumed to be a simple
edge crack until now, where its width was not
taken into consideration. In order to provide a
more accurate model, the crack is assumed to
be a shallow semi-elliptical crack which is
parallel to the axis of the pipe. Subsequently,
various crack shapes and their influence on
fracture analysis can be analyzed.The shape
and dimensions of the crack both have an
impact on both the internal collapse pressure,
??????
� and the geometry calibration function, ??????.
Figure 3.5 shows the semi-elliptical crack and
its dimensions.
The dependent ratios or the geometry calibration function are:

??????
�
⁄ (Crack length/Thickness) ;
??????
�
⁄ (Crack length/crack width)

The vessel thickness, t, and the crack depth, a, are known. The only unknown is the crack width, c. Using
the following ratios:
??????
�
�
=
6
19.3
=�.��� ,
??????
�
�
=
6
38.5
=�.��

Figure 6: Pressurized cylinder with semi-elliptical crack

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The corresponding a /t ratios
for Steel A and Steel B are used
in conjunction with the graph
shown in Fig, which is in the
appendix. It displays the
variation of the geometric
calibration factor Y with the
crack width, c, or more
specifically, the a/c ratios. [5]
The plot of the variation of Y
against ratio a/c for both Steels
A and B is shown in Figure 7:


The plot shows that as c tends to infinity, the geometry reverts to the initial, simplified model of a single edge-
crack under tension. Comparing the newly obtained values of Y to the original values in 3.3 FAD Analysis, it can
be seen that they are very similar. Current values of Y are 1.7 and 1.4 for Steels A and B, respectively.
The internal collapse hoop stress/collapse pressure for various crack widths are to be calculated. The internal
collapse pressure can be calculated using the following equation:
??????
�=
????????????
��
(1−
1−??????
√1+1.05??????
2
)
⁄
∗�….(3.4.1), where ??????=
�−??????
�
⁄ ….(3.4.2) and �= √�
2
��
⁄ ……(3.4.3)
The collapse hoop stress can therefore be calculated using the following expression: ??????
�=
??????
��
�
⁄….(3.4.5)
Graphs showing the co-relation between collapse hoop stress and crack width for both Steel A and Steel B can
be found in the Appendix. They both show that as c tends to infinity, ??????
� tends towards ????????????
��. These subsequent
values of ??????
� are 420 MPa and 210 MPa for Steels A and B, respectively.
In order to assess the extent to which crack widths affect the probability of failure of the two candidate steels,
new FADs are plotted, using different values of Y (which had been calculated from corresponding values of
a/c).
STEEL A:
Figure 7: Variation of Y with crack ratio a/c for Steel A and Steel B
Figure 8: FAD for Steel A using different values of a/c

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STEEL B:
The assessment points for a/c = 0 are the same as those calculated when assuming a simplified edge-crack
geometry. Predictably, the crack is not expected to be seen, and the main method of plastic deformation will
be of the yielding of steel. The ratio Kr is found to decrease with decreasing value of c, and the lower stress
concentrations can be attributed to the smaller crack size.
From Figures 8 it is apparent that, for a straight edge crack, Steel A would fail under maximum stress. Shifting
to a more realistic value (that of a semi-elliptical crack) ensures that failure does not occur. Steel B does not
undergo failure This only provides further validation that Steel B is the more favourable candidate.
4. FATIGUE ASSESSMENT
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.
Fatigue crack growth rate using LEFM assumptions will follow. The crack propagation life can be estimated
using the relation between the stress intensity range, ∆� 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.
∆??????= ??????
�??????�− ??????
�??????�=
??????
525.�
�
�
⁄−0=
��ℎ.�
�
�
⁄=
1000∗9.81∗525∗1.1
0.385
⁄ =��.� �???????????? …(4.1)
Likewise, for Steel A, ∆?????? is calculated to be 29.4 MPa.
The stress intensity range, ∆� can be calculated using the value of ∆??????, using:
∆�=??????∆??????√�??????…………………………………………. (4.2)
The change in crack sizes, ‘a’ can be calculated with respect to the number of cycles, ‘N’ by use of Paris’ Law. �,
� are material constants of the given steels, and the stated values are 10
-11
, 3 respectively.


�??????
�??????
=�(∆�)
�
…………………………………….. (4.3)
By utilising a small step size, (∆??????), 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:
Figure 9: FAD for Steel B using different values of a/c

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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 2.3 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. This is included in the appendix.
5. DESIGN IMPROVEMENTS

As concluded in the section ‘4. Fatigue Assessment’, changes are required in the design of the penstock. An
appropriate design parameter to change would be the thickness of the pressure vessel. By increasing the
thickness to approximately 85mm, the critical crack length is increased to 27.6 mm, using the same procedure
in section ‘3.2 Critical Crack Sizes’.
Figure 11 shows the FAD for Steel B, incorporating the new specified thickness of 85mm. The probability of
fracture and yielding is reduced by increasing the thickness, which subsequently lowers the hoop stress. It can
be seen that Steel B also undergoes plastic collapse.











Figure 10: Fatigue growth for Steel B, thickness = 38.5mm
Figure 11: FAD of Steel B with increased thickness of 85mm

13. 12 | P a g e


The fatigue life plot illustrated in Figure 11 shows that the time taken for the initial crack to grow to the critical
crack length size of 27.6mm is approximately 56 years. This is an acceptable value for the design life of the
penstock, with an appropriate 6 years of safety to accommodate for any load discrepancies.


6. CRITICAL DISCUSSION AND FINAL RECOMMENDATIONS

Generally, due to a hydraulic pressure P acting on the faces of the penstock pipe, the crack located at the
inside surface is additionally loaded. [1] The stress intensity due to this loading is given by
�
??????=??????√�?????? . However, since the penstock pipe is treated as a thin-walled cylinder in ‘2. Overload Assessment’,
??????≪??????, and therefore, the additional loading effect has been not been considered.

It must be noted that only the upper estimates for the critical crack lengths have been calculated using the
MATLAB code. LEFM 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. Smaller values of the critical crack length
would have been obtained if a plastic zone correction factor would have been used. [1] 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 [4]

In the section ‘3.4 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.
Figure 12: Fatigue growth of Steel B, thickness = 85mm

14. 13 | P a g e


As mentioned in sub-section ‘Accounting for Residual Stress’, the tensile residual stress problems can be
resolved by processes which induce compressive tensile stresses such as post weld heat treatment (PWHT),
shot peening and spot heating. [9] It must also be mentioned that the weld metal used in the welding
procedure decreases in toughness (COD toughness, or Crack Opening Displacement toughness). This
associated with the coarsening of the microstructure due to increases in temperature. This occurs in the sub-
critical Heat-Affected Zones (HAZs) [12]. The fracture toughness values for only the parent material (Steels A
and B) can be derived from the given information. This value is KIC = Kmat at fracture. Though the properties of
the weld material are not stated or given, it must have approximately the same fracture toughness values as
that of the parent material (steel). This would ensure safety, and a penstock fabrication of good quality, i.e.
lower risk of failure occurring at the weld joints.

The section ‘4. 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. [10]

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 [5]. Therefore, the inspection interval should be co-ordinated with the maintenance interval, i.e.
stripping of the walls and repainting. [2]

15. 14 | P a g e

7. REFERENCES
1. Hudson C, Rich T. Case histories involving fatigue and fracture mechanics. Philadelphia, PA: ASTM; 1986.
2. Jivkov A. The Design of a Penstock Pipe for a Hydro-electric Pumped Storage Station. 1st ed. 2015.
3. Saadat H. Power System Analysis [Internet]. Psapublishing.com. 2015 [cited 20 November 2015]. Available from:
http://www.psapublishing.com/
4. Anderson T, Anderson T. Fracture Mechanics. Hoboken: CRC Press; 2005.
5. Plastic Collapse Handbook. 1st ed. 2002.
6. Bannister A. Structural integrity assessment procedures for European Industry. Swindon: British Steel plc; 1998.
7. McStraw B. Inspection of Steel Penstocks and Pressure Conduits. UNITED STATES DEPARTMENT OF THE
INTERIOR BUREAU OF RECLAMATION DENVER, COLORADO; 1996.
8. Molzen M, Hornbach D. Evaluation of Welding Residual Stress Levels Through Shot Peening and Heat Treating.
1st ed. 2000.
9. O'Brien R. Welding handbook. Miami, Fla.: American Welding Society; 1991.
10. Kaechele L. Review and Analysis of Cumulative Fatigue-Damage Theories. 1963;
11. Deuflhard P. Newton Methods for Nonlinear Problems: Affine Invariance and Adaptive Algorithms. 2011.
12. Neves J, Loureiro A. Fracture toughness of welds—effect of brittle zones and strength mismatch. Journal of
Materials Processing Technology. 2004;153-154:537-543.

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8. APPENDIX


A. Ratio of Y with a/t

B. Plastic Limit Stresses for Steels A and B

17. 16 | P a g e





C. Crack Length Propagation for Steel A

D. MATLAB Code: Critical Crack Length
% This script finds the solution to the non -linear crack size equation,
using the Newton-Raphson Method.
%Number of iterations to be specified. In this case, 10 iterations will
suffice:

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n=10;

%Critical Fracture Toughness for Steel A and Steel B, respectively
K1c=[100*10^6;120*10^6];

%Maximum allowable hoop st ress (60% of yield stress);
%Steel A - 420MPa
%Steel B - 210MPa
sigmah =[420*10^6;210*10^6];

%Minimum thickness calculated for both Steel A(18mm) and Steel B(45mm);
t=[0.0193;0.0385];
%An initial value of a=1mm for both Steels A and B
a=[0.001;0.001];

%Newton-Raphson method iterations carried out using a for loop:

for i=1:n
f=((1.12.*a.^0.5-(0.231.*(a.^1.5).*(t.^( -1)))+(10.55.*(a.^2.5).*(t.^( -2)))-
(21.72.*(a.^3.5).*(t.^( -3)))+(30.39.*(a.^4.5).*(t.^( -
4)))).*(pi^0.5).*sigmah) -K1c;

f1 =((0.56.*a.^-0.5-(0.3465.*(a.^0.5).*(t.^( -1)))+(26.375.*(a.^1.5).*(t.^( -
2)))-(76.02.*(a.^2.5).*(t.^( -3)))+(136.755.*(a.^3.5).*(t.^( -
4)))).*(pi^0.5).*sigmah);

a1=a-(f./f1);
a=a1;
end
%The final output a is the critical crack length in m for both Steel A and
Steel B

%Critical crack length a is converted from m to mm:
a=a*1000

a =

6.1046
17.4415


E. MATLAB Code: Crack Length Propagation
clear
clc
%Input known variables
%Assigned parameter: Maximum head of water
h=525;
%Assigned parameter: Radius of the pip e
r=1.1;
%Assigned parameter: Material Constant
C=10^-11;
%Assigned parameter: Material Constant
m=3;
%Calculated parameter:Final crack size

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af=0.085;
%Initial Crack size
ai=0.006;
%Pipe thickness
t=0.085;
%Calculate stress range
dsigma=(h*r*1000*9.81)/(t*10^6);
%Input number of iterations
Na=5000;
%Calculate crack resolution (step -size)
da=(af-ai)/Na;
%Initialize the solution
a=ai;
N=0;
Yr(1)=0;
%Calculation Loop - Algorithm Shown in Figure 4.1
for i=1:Na-1
G=(a/t);
Y=1.12-(0.231*G)+(10.55*G^2) -(21.72*G^3)+(30.39*G^4);
dk=Y*dsigma*((pi*a)^0.5);
dadn=C*(dk^m);
dN=da/dadn;
N=N+dN;
Yr(i+1)=N/7500;
a=a+da;
end
%Change a into millimeters
a=1000*linspace(ai,af,Na);
%Plot
plot(Yr,a)
grid on
xlabel('Service Life (years)' )
ylabel('Crack Length (mm)')
title('Crack Length Propagation vs Service Life (Steel A/B) ' )

F. MATLAB Code: Plastic Collapse

clear;

%Enter your own input variables:
a=0.006;%Detectable flaw size
t=0.385;%Thickness
c=(0:0.2:2);%Range of values of c
r=1.1; %Radius

sigmay=350*10^6;
n=(t-a)/t;
rho=((c.^2)/(r*t)).^0.5;

%Plastic collapse hoop stress
Pc=(n*sigmay*t)./((1 -((1-n)./(1+(1.05.*(rho.^2))).^0.5)).*r);

sigmac=Pc.*r/(t*1000*(10^5));

plot(c,sigmac,'-bo','LineWidth', 2);
xlabel('c (m)');
ylabel('Sigma_c (MPa)');

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title('Plastic Limit Stress vs Crack Length (Steel B) ' )
grid on;
hold on;

G. MATLAB Code: FAD

clear;
Lr=[0:0.02:1.4];
kr=(1-0.14.*Lr.^2).*(0.3 + 0.7.*exp( -0.65*Lr.^6));

%SteelB
Lr1=0;
Lr2=0.6095;
Kr1=0.092;
Kr2=0.276;

plot(Lr,kr,'LineWidth',3);
legend('Limit');
hold on;
title('Level 2 Failure Assessment Diagram (FAD) for Steel B (Increased
Thickness to 85mm )' );
xlabel('Lr');
ylabel('Kr');
hold on;
scatter(Lr1,Kr1,'r','o','markerfacecolor','r' )
hold on;
scatter(Lr2,Kr2,'g','o','markerfacecolor','g' )
hold on;
grid on;
legend('R6 Kr curve','Lr(s)=0 ,Kr(s)(Steel B)' , 'Lr(p),Kr(p+s)(Steel B)' )