Anderson, Loren Runar et al "ECONOMICS OF BURIED PIPES AND TANKS "

Structural Mechanics of Buried Pipes

Boca Raton: CRC Press LLC,2000
CHAPTER 30 ECONOMICS OF BURIED PIPES AND TANKS

The engineer is responsible for cost effectiveness. Derivation of Equations

Cost effectiveness of any project is the extent to
which the returns on investment outweigh the costs. The interrelationships of money values, P, S, and R,
Costs are money expenditures; but costs also include for n periods at interest rate i compounded at the
time, effort, overhead, insurance, warranty, public end of each period, are as follows.
relations, etc. Returns include money, but also
reputation, etc. All costs and returns must be S = P + Pi + i(P+Pi) + i[P+Pi+i(P+Pi)] + ....
reduced to monetary equivalents. And for analysis,
all monetary equivalents must be reduced to the S = P(1+i) + Pi(1+i) + Pi(1+i) 2 + Pi(1+i) 3 + ...
same basis — a present worth, P; or a periodic + Pi(1+i) n-1 . . . . . (30.1)
payment (annually), R; or a future lump sum, S. For
planning and design, it is customary to compare the Rewriting,
monetary equivalents of costs for all possible S = P[(1+i) 2 + i(1+i) 2 + i(1+i) 3 + i(1+i) 4 + ...
alternatives after reducing them to the same basis of + i(1+i) n-1] . . . . . (30.2)
monetary equivalence — usually equal periodic
payment, R, (or sometimes present worth, P). For Multiplying both sides of Equation 30.1 by (1+i)
sale of a project, or settlement of claims, present
worth, P, is usually the best monetary equivalent. S(1+i) = P[(1+i) 2 + i(1+i) 2 + i(1+i) 3 + i(1+i) 4 + ...
The present worth of a project that is to continue + i(1+i) n] . . . . . (30.3)
forever by repeated replacements is called
capitalized cost, P . The interest on capitalized cost Subtracting Equation 30.2 from Equation 30.3,
provides the periodic payments. Monetary
equivalence depends upon time and interest rates S = P(1+i) n . . . . . (30.4)
because the value of money is not constant.
Consider each R of a series of periodic payments to
VALUE OF MONEY be a separate P. For this analysis, let R be due at
the beginning of each pay period. Then, evaluating
Nomenclature and adding up the sums of all of the R's in the series,
while working backward from the last R-payment to
P = present worth, an amount of money now, the first from one year past the last R,
S = sum, the value after n periods of time at
interest rate i, S = R(1+i) + R(1+i) 2 + R(1+i) 3 + ...+ R(1+i) n
R = equal periodic payments over n periods at . . . . . (30.5)
interest rate i, that accumulates sum S or
repays present worth P, Multiplying both sides by (1+i),
P = capitalized cost (interest = R),
i = interest rate, compounded at the end of S(1+i) = R(1+i) 2 + R(1+i) 3 + ...+ R(1+i) n+1
each pay period, . . . . . (30.6)
n = number of pay periods of time (usually in
years). Subtracting Equation 30.5 from Equation 30.6 yields
Equation 30.7. Eliminating S between Equations
30.4 and 30.7 yields Equation 30.8.

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If R is at the beginning of each pay period, year repayment of purchase cost, P = $1,000,000
from Equation 30.12 in n = 4 years at i = 7%
Si = R[(1+i) n+1 - (1+i)] . . . . . (30.7) interest.

Pi(1+i) n = R[(1+i) n+1 - (1+i)] . . . . . (30.8) R = $100,000

+ $1,000,000(0.07)(1.07)4/[(1.07)4 - 1]
S = P(1+i) n . . . . . (30.9)
R = $100,000 + $295,228 = $395,228
P = R(1+1/i) . . . . . (30.10)
Find the present worth, P, if R = $395,228. From
The three variables P, R, and S, can be interrelated Equation 30.12,
by the three Equations, 30.7, 30.8, and 30.9, for any
given interest rate, i, and any number of pay periods, P = R[(1+i) n - 1]/i(1+i) n
n. Equation 30.10 is capitalized cost, P , which is = $395,228[(1.07)4 - 1]/(0.07)(1.07)4
the present worth of all series payments, R, if the
project is to continue on forever by replacements; P = $1,338,721
i.e., for n = oo .
If the life of the tanks is 50 years, and payments are
If R is at the end of each pay period, made over the 50-year life rather than 4 years, what
is the annual end-of-year payment? From Equation
Si = R[(1+i) n - 1] . . . . . (30.11) 30.12, if P = $1,338,721, i = 7%, and n = 50 pay
periods,
Pi(1+i) n = R[(1+i) n - 1] . . . . . (30.12)
R = $1,338,721(0.07)(1.07)50/[(1.07)50 - 1]
n
S = P(1+i) . . . . . (30.13)
R = $97,003.50
P = R/i . . . . . (30.14)
If the tank farm must be replaced every 50 years,
The three variables P, R, and S, can be interrelated find the capitalized cost. From Equation 30.14, P =
by the three Equations, 30.11, 30.12, and 30.13 for R/i, where R = $97,003.50 and i = 7 % . P =
any given interest rate, i, and any number of pay $97,003.50/0.07 = $1,385,764.32.
periods, n. Equation 30.14 is capitalized cost, P ,
which is the present worth of all periodic payments, P = $1,385,764
R, if the projec t is to continue on forever; i.e. for
infinite time, n = oo . MONETARY EQUIVALENCE

In general, R is used as a basis for comparing Monetary equivalence includes overhead and all
alternatives for least cost. P is used as a basis for direct costs such as purchases and installation. But
sale or for settlement of a claim. monetary equivalence also includes maintenance and
the cost of risk. Risk includes public relations,
Example 1 insurance, and legal counsel, in addition to the cost
of replacement.
The purchase cost of tanks in a buried tank farm is Example 2
$1,000,000. Installation costs are $100,000 per year
at year end for 4 years. What is the annual end-of- Purchase price of a buried tank is $10,000.
period payment R to pay off the project in four Installation cost is $40,000. Maintenance is $500 per
years? R is annual installation cost plus the end-of- year. The cost of a leak (replacement, damage, and

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liability) is estimated to be $100,000. For this records of previous events with magnitudes and
analysis, assume that insurance covering leaks can times noted. The best monetary equivalent of the
be obtained for $2000 per year. The tank is time-indexed magnitude of an event is an insurance
designed for 50 years of service. If the interest rate premium — a series payment, Rc r, which is equal to
is 8%, what is the equivalent annual R at beginning- the cost of the event times the probability that the
of-pay periods for the project? This R can be event will occur in any one pay period. A
compared with the R's for alternatives such as more reasonable profit for the insurance carrier is added.
expensive (dual-containment) tanks with less For an insurance carrier, with many portfolios, the
probability of a leak. For beginning-of-pay periods, probability that a 100-year event will occur in any
utilizing Equation 30.8, one year is 1/100. The unpredictable timing of each
is balanced out in the totality of portfolios. The
R = $500 + $2,000 monetary equivalent of the magnitude of an event is
+ $50,000(0.08)(1.08)50/[(1.08)51 - 1.08] Pc r. The probability is 1/nc r. The periodic payment
to cover the event i s Rcr = Pc r/nc r , where nc r is the
R = $500 + $4,000 + $3784 = $6284. time index (100 years, etc.).

R = $6284 Example 3
The tank of Example 2 will float out of its
What is the present worth, P, of the project in the embedment if it is empty when a 50 year flood
event of sale or legal action? Already known from occurs. The cost of flotation is tank replacement at
Example 2 is R = $6284. From Equation 30.8, $50,000 plus public liability estimated to be $100,000.
What insurance premium would be justified if
P(0.08)(1.08)50 = $6284[(1.08)51 - 1.08] reasonable profit for the carrier is 30%? Assuming
flotation of an empty tank, from Example 2, the cost
P(3.7521) = $6284[49.5737] = $311,521.39 of the disaster is the purchase price plus installation,
plus liability; i.e., Pcr = $10,000 + $40,000 + $100,000
P = $83,025 = $150,000. But the probability of a flood that could
cause flotation in any one time period is only one in
COST OF FAILURE 50 years; i.e., 1/ncr = 0.02. R = ($150,000)(0.02) =
$3000. Including a 30% profit for the insurance
The cost of failure could be staggering, depending on carrier, the insurance premium is $3900. This is
damage and liabilities, but the probability of failure based on the assumption that the tank is always
may be remote. The monetary equivalence of empty. But what is the probability that the tank is
failure in any pay period is the approximate cost of empty if it is used for gasoline storage at a service
failure times the probability of failure. Two station?
techniques for finding the monetary equivalence of
failure follow. Example 4
Find the probability, Pe, that a tank will float if it
Probability of a 100-Year Event: is not always empty. The subscript, e,
A 100-year event (or any time-indexed event) is distinguishes probability from present worth. If the
defined as the average period of time between tank is used to store gasoline at a service station, it
occurrences of an event of magnitude greater than will be refilled as soon as it is empty. Assume that
some given magnitude. If there is no periodicity, an gasoline is pumped out at a constant rate between
event can occur at any time. For example, the refillings. The first problem is that the tank may
periodicity of storms, floods, and earthquakes has float with some gasoline in it. What is the least
not yet been predicted with confidence, despite amount of gasoline in the tank that acts as ballast
promising new techniques. Available only are and

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Figure 30-1 Buried tank (top) showing the level of the contents (cross-hatched) lower than which the tank
will float; and (bottom) the volume of contents as a function of time between refills, showing (cross-hatched)
the time, between refills, during which flatation can occur.

Figure 30-2 Systems diagram of

a buried tank under worst-case
conditions of external loads.

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keeps the tank from floating? See the cross-hatched System Diagram
areas of Figure 30-1. Assume saturated soil with The first step is a diagram of the system. See
flood level at or above ground surface. The tank Figure 30-2. The soil cover, H, is minimum for an
volume is 12,000 gallons. HS-20 dual-wheel load. It is understood that the
D = 7 ft, tank meets specifications for quality.
L = 42 ft,
WT = 8 kips = weight of the tank, Fault Tree Construction
w T = 190 lb/ft = tank wt/unit length, A fault tree is a trunk and root system. The trunk is
H = 2 ft = height of soil cover, the event (failure), and the roots are the faults
g s = 57.6 pcf = unit weight of soil, (causes of failure). Figure 30-3 is a table of
w s = 1980 lb/ft = buoyant soil wt/unit length, symbols. In this example, the failure is a fuel
w = 2170 lb/ft = total wt/ft acting down, LEAK. It is shown at the top of the fault tree of
w = 2400 lb/ft = buoyant force acting up, Figure 30-4 in a rectangle indicating that it is an
Dw = 230 lb/ft = wt/ft of gasoline at a level lower event. The two causes (faults) considered in this
than which the tank floats, example, are shown in a horizontal row below the
w = 1616 lb/ft = wt/ft of gasoline in the tank event. They are connected to LEAK through an
when the tank is full (unit weight of gasoline and gate (bullet) indicating that both POOR
= 42 pcf), EMBEDMENT and EXCESSIVE LOAD are
T = time between refillings, necessary to cause the leak. The probabilities, Pe,
DT = time during which the tank can float (with of the faults must be multiplied together to get
less than 230 lb/ft of gasoline ballast). through the and gate. POOR EMBEDMENT is an
event (rectangle) which could be the result of a
What is the probability, Pe, that the tank will float HARD SPOT or LOOSE SOIL in the embedment.
considering gasoline is in the tank between Because either loose soil or a hard spot is sufficient
refillings? P e = DT/T , where DT = (230/1616)T = to cause a leak, they are connected to POOR
0.1423T. Pe = 0.1423. EMBEDMENT through an or gate (arrowhead).
The probabilities, Pe , of the faults must be added
together to get through the or gate.
Probability of Failure by Fault Tree
HARD SPOT is a diamond, a basic fault
A justifiable R, either for insurance coverage or undeveloped. If the constructor had a history of
elimination of the probability of failure, is the cost of problems with hard spots, a probability might be
failure times the probability that it will occur in any developed. Without such a history, probability can
pay period. A model for estimating probability of only be assumed and varied to determine the effect
failure is the fault tree — a logic diagram that starts of hard spots on the probability of a leak. This could
with an undesired event (failure) and traces back serve to evaluate encroachment of the constructor’s
through all of the causes (faults) assessing and hard spots into the safety zone (safety factor zone).
relating the probability of each fault to the probability Or from the history (track record) of the
of failure. An example illustrates the fault tree. constructor, suppose that repair of a hard spot
shows up in the bedding of one out of every 20
Example 5 tanks. Probability is 0.05.

A buried fuel tank costs $50,000 installed. A leak LOOSE SOIL is a circle because it is a basic fault,
could result in liabilities up to $100,000. What is the but developed. From geotechnical tests, the soil was
probability of a leak? found to be loose in one out of every 40 tanks,
(0.025).

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Figure 30-4 Example of a Fault Tree for finding the probability of a leak in a tank that is buried in poor
embedment, and is subjected to excessive loads.

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With loose soil in one out of every 40 tanks (0.025), Example 6
and hard spots in one out of every 20 tanks (0.050),
the Pe 's are shown in Figure 30-4 for HARD SPOT What beginning-of-period payment, R, can be
and LOOSE SOIL. They are added together justified to cover the cost of a leak in the tank of
through the or gate yielding a value of Pe = 0.075 = Example 5 if the cost of the leak is $150,000. The
probability of a POOR EMBEDMENT. probability of a leak is 1/296 = 0.003375 in any one
year. R = $150,000(3.375)10-3.
EXCESSIVE LOAD is an event (rectangle) that R = $506.25 per year.
might be caused by e ither a 50-YEAR FLOOD or
LIVE LOAD. If either one or the other is
sufficient, the two are connected to EXCESSIVE SAFETY FACTORS
LOAD through an or gate.
Simply defined, a safety factor is the ratio of
The 50-YEAR FLOOD is a triangle because it is performance limit to performance. Not so simple is
developed elsewhere. See Example 3. The a number for that safety factor. For above-ground
probability is shown on 50-YEAR FLOOD; Pe = structures, classical performance is measured by
0.02 in any payment period (year). The LIVE stress in the material. Performance limit is strength.
LOAD is a diamond (undeveloped). Why not For buried structures, performance is, usually,
assume that a ready-mix concrete truck could pass deformation — including leaks and excessive
over the tank inadvertently — say once every 40 movement of the soil or structure. Failure is usually
years? Then the probability in any payment period reduced to a monetary equivalent. What is the
(year) is Pe = 0.025 as shown on LIVE LOAD. monetary equivalent of failure? Are there any
mitigations for failure? What are the encroachments
The probability of EXCESSIVE LOAD is the sum into the safety zone (demilitarized zone)? And, in
of the Pe's for the 50-YEAR FLOOD and LIVE the event of legal action, who is responsible for the
LOAD. For EXCESSIVE LOAD, P e = 0.02 + encroachments? What are the monetary equivalents
0.025 = 0.045. of encroachments? Whose are the responsibilities?
Legal counsel usually becomes involved in these
The probability of a LEAK in any pay period is the questions. However, information upon which
product of the Pe's for POOR EMBEDMENT and counsel directs legal proceedings must come from
EXCESSIVE LOAD. Pe = 0.075(0.045). The the engineer. Reduction to monetary equivalence
probability of a leak is, often comes from the engineer. Comparative or
Pe = 0.003375 = 1/296 or approximately one in every contributory encroachment into the safety factor
300 tanks per year. zone requires engineering knowledge.

This 1-in-300 probability of a leak per year might Example 7

justify either a periodic payment to reduce the
possibility of a leak, or insurance premiums that A 12,000-gallon tank for storing gasoline, leaked just
cover the leak. after installation. The leak was a circumferential
crack on the bottom at midspan. The tank was
It is noteworthy that the above simple example does buried in dry soil (no water table). What are the
not include such effects as contents of the tank, time relative responsibilities for the leak (in percent) of
related consolidation of sidefill soil, vacuum in the the engineer, manufacturer, and installer? Neglect
tank, and liquefaction of the embedment. A more contributory negligence of other parties in this
complex fault tree might be justified. hypothetical case. It is known that an HS-20 axle
load of 32 kips passed over the tank at midspan.
See

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Figure 30-2. The dual wheels were separated by 6 Percent Encroachment into the Safety Zone
ft. (Safety zone = 0.53+0.40+0.66=1.59):
D = 84 inches,
L = 42 ft, ENGINEER = 33%
t = 0.6 inch, MANUFACTURER = 25%
H = 2 ft, INSTALLER = 42%
g = 120 pcf = unit weight of dry soil,
g G = 42 pcf = unit weight of gasoline,
sf = 5 ksi = yield strength of the tank wall, PROBLEMS
w T = 200 lb/ft = weight per unit length.
30-1 What is the probability of a leak in a farm of
The safety factor is two based on yield strength of 10 buried tanks if the probability of a leak in any one
5 ksi. Design stress is 2.5 ksi. Therefore the safety tank is 1/300 during any pay period?
zone is from 2.5 to 5 ksi. (1/30)

Weight per unit length of tank is the tank full of 30-2 Two tanks are under consideration for a
gasoline, plus prismatic soil load on top. If the tank project. Tank A costs $10,000 plus installation cost
is a simply supported beam, including live load, the of $5,000. Service life is 30 years if an insurance
longitudinal stress in the bottom at midspan is 3.83 premium of $400 is paid at the beginning of each
ksi. year. Tank B costs $15,000 plus installation cost of
$7,000 and service life of 50 years. What are the
ENGINEER specified compacted soil with no equal series payments, R, at year end for the two
organics or large rocks. The tank was designed for alternatives? Interest rate is 8%.
40% of simply supported beam stresses — assuming (A, $1764; B, $1798)
that the bedding would provide some support. But
the beam was simply supported. Therefore, design 30-3 If the tanks of Problem 30-2 are to be
encroachment is (3.83-2.50)/(5-2.5) = 0.53. replaced by identical tanks at the end of each
service life, what are the capitalized costs, P , of
MANUFACTURER supplied a tank with wall tanks A and B?
strength of 4 ksi according to post-leak tests — not (A, $22,055; B, $22,479)
5 ksi as advertised. The encroachment is (5-4)/(5-
2.5) = 0.4. 30-4 The head of a tank leaks near the bottom. H
= 2 ft, D = 84 inches, t = 0.187 inch. Find
INSTALLER leveled the tank on timbers at the contributory negligence of:
ends thereby forcing the tank to act as a simply Manufacturer — insertion is only 0.3 inch.
supported beam with no bedding. There was no Constructor — a timber is left under the end of the
compaction under the haunches. Had a bedding tank for vertical alignment.
reduced stresses to 40% of the simply Owner — a 16-kip wheel load passed over the tank
supported beam stresses, the maximum stress in the the evening before the leak was detected.
tank would be (0.4)(3.83) = 1.53 ksi. The
encroachment is (3.83-1.53)/(5-1.53) = 0.66.

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