API 1110 Additional Procedure Report

8/8/2019 API 1110 Additional Procedure Report

1/36

Effects of Temperature Change on Pressure

Testing and Maintaining Environmental

Compliance

Arthur D. Bosshart II

TECO Energy

07/09/08


2/36

Arthur D Bosshart IITECO Energy

Effects of Temperature Change on Pressure Testing and Maintaining Environmental Compliance 2

Table of Contents

Abstract ... 3

Introduction . 3

Field Tests and Analysis ... 4

Proposed Procedure .. 9

Ideal Testing Conditions ... 11

Conclusions . 13

Acknowledgments .. 14

Bibliography 15

Appendix .. A) Calculations B) Example spread sheet walkthrough ...

C) Assumptions ...D) Nomenclature ..... E) Simplified Field Method ........

F) APT Lookup Table ........ G) Compliance Curve ...........

H) Simplified Field Method Sample Calculation ..... I)Field Pressure Testing Packet ........ J)

Completed Line Pressure Test Worksheet ........ K)Calculation of the P Coefficient ........ L)Calculation of a Specific APT Value ...........

16161821212224252527

333436


3/36



Abstract

The API 1110 pressure test method is used for determining leaks within piping segments in

contact with soil or above water. A criticism of this current method is that it does not account for

temperature effects on the contained substance in the piping. These temperature variances

affect the measured pressure during a test and depending on testing conditions can pass a failed

test or fail a passed test. The proposed alternate method and associated calculations account for

a temperature change that can occur during a pressure test and correlates an unaccounted

volume of substance to the Table BPP located in Rule 62-762.601, Florida Administrative Code

(F.A.C.) to determine pass or fail.

Introduction

The purpose of this paper is to define the allowable pressure tolerances of the API 1110

pressure test method between the theoretical pressure change and the actual pressure change

associated with a temperature change of a contained substance within piping during a pressure

test. A volumetric difference, over a testing duration, can be associated with the difference

between the theoretically calculated and actual measured pressures and then correlated to the

Table BPP found in Rule 62-762.601, F.A.C. to identify a passed or failed test in accordance with

API 1110 pressure test method. Ideal testing conditions will then be identified to ensure accuracy

and confidence in a given pressure test. There are a number of influences that affect the final

results of pressure tests such as the change in substance temperature within the piping, ambient

temperature changes, piping temperature change, piping material, substance within the pipe,

thermal expansion rates, compressibility, percentage of pipe directly in contact with sunlight, and

volume of piping isolated for testing. Some of the factors noted above have a greater effect than

others.

The most important operational condition to monitor, other than pressure, is the

temperature of the substance within the pipe. Depending upon the type of substance, a change

in temperature can have dramatic results. Using #2 diesel fuel as an example, if a temperature

change of 1 oF occurs within an isolated volume, the pressure will increase or decrease by 70 psi


4/36



to 100 psi. This is because liquids are not easily compressible like gases. The pressure will vary

depending on the volumetric thermal expansion rate of #2 diesel fuel and the piping material.

Another important factor to account for is the energy transfer between the outside

environment and the pipe and between the pipe and the substance contained therein. Given that

the piping section being tested is isolated, energy transfer is the reason why a substances

temperature will increase during a test. The temperature change is determined by the thermal

conductivity of the piping material, the energy flux, schedule of the piping, the heat capacity of the

liquid, initial ambient temperature, final ambient temperature, initial exterior pipe surface

temperature, final exterior pipe surface temperature, initial interior pipe surface temperature and

final interior pipe surface temperature. These relationships are described in Fourier's law.

Fourier's law describes linear heat flow rate between two surfaces. If a portion of the

piping is exposed to the sun the exterior surface temperature of the piping will increase. The

magnitude of the increase is determined by the solar radiation intensity and if the pipe is coated

with an energy absorbing color like black then a greater temperature change will occur.

Currently, there is no protocol in the current API 1110 pressure test method that requires

that the liquid temperature be monitored throughout the duration of a test. This is a potential

deficiency of this test. Without monitoring the temperature of the substance within the pipe,

pressure fluctuations cannot be accounted for other than reasoning that a product loss had

occurred meaning that there is a leak in the pipe that was pressure tested. This reasoning is not

accurate. If the heat capacity of the contained substance and thermal conductivity of the

substances surroundings are accounted for, there will be a substantial delay between the

ambient or environmental temperature change and the temperature change of the substance.

Field Tests and Analysis

For purposes of this paper, two examples of API 1110 pressure tests that were recently

conducted in the field at TECO Energy facilities will be examined to illustrate the temperature

changes, pressure changes, and energy transfers between materials.


5/36



Example #1:

This API 1110 pressure test was conducted at the TECO Big Bend Plant on February 12,

2007 located in Apollo Beach, FL from 2:00 p.m. to 3:19 p.m. The test was performed on a


6/36



section of piping nearly 900 feet in length constructed of carbon steel. Contained within the pipe

is #2 diesel fuel. This portion of piping is entirely aboveground and transports fuel from a pump

house, over a canal, along the side of the canal, and into an aboveground storage tank system.

The weather conditions during this test were partly cloudy with a light breeze.

When analyzing the data associated with Example #1, it is important to note the pressure

increase and the temperature increase. The initial pressure and temperature readings were

recorded at 107psi and 65 oF, respectively. Readings were recorded every 15 minutes throughout

the testing duration and before depressurization, the final reading was recorded at 234psi and

70 oF. The temperature recordings were of the ambient environmental temperature. Although the

ambient temperature increased by 5 oF, the temperature of the #2 diesel fuel only increased by

1.285 oF. This 1.285 oF temperature rise occurred mainly due to solar radiation incident upon the

pipes exterior surface. The change in ambient temperature appears to have had a smaller

impact. When using Fourier's law to describe the heat transfer occurring within this test, one

needs to look at the piping schedule, the temperature of the piping exterior surface, the energy

transfer through the pipe, and the heat capacity of the #2 diesel. Although the ambient

temperature was only around 70 oF, the temperature of the pipes exterior surface was most likely

around the range of 80 oF to 85 oF. This temperature difference is attributed to the energy

absorbing ability of the pipes exterior coating and the solar radiation directly incident upon the

pipe, radiation diffused through the surrounding environment, and radiation reflected upon the

pipe s surface from the water below. The #2 diesel fuel did not reach a thermal steady state with

the pipes exterior surface because of the length of the test, the rate at which heat is conducted

through the carbon steel piping, and the heat capacity of the contained liquid.


7/36



Example #2

This API 1110 pressure test was conducted at the TECO Big Bend Plant on February 12,

2007 located in Apollo Beach, FL from 3:20 p.m. to 4:40 p.m. The test was performed on a


8/36



section of piping between 400 and 600 ft. in length and constructed of carbon steel. Contained

within the pipe was #2 diesel fuel. This portion of piping is underground and transports fuel from

another portion of piping to an aboveground day storage tank system. The weather conditions

during this test were partly cloudy.

When analyzing the data associated with Example #2, it is important to note the relatively

stable pressures and temperatures recorded. The initial pressure and temperature readings were

recorded at 160psi and 72 oF, respectively. Readings were recorded every 15 minutes throughout

the testing duration and before depressurization, the final reading was recorded at 160psi and

70 oF. The temperature recordings were of the ambient environmental temperature. When

looking at the first four readings of the test it is important to note that the temperature decreases

and the pressure increases. This occurs because underground temperature remains relatively

stable throughout the day because the ground acts as an insulator and limits the impact of the

ambient temperature fluctuation. Also, the pipe is not absorbing energy from solar radiation

incident upon the exterior surface. Although the ground acts as an insulator, heat flow will still

occur from the ground surface to the soil in contact with the pipe, from the soil in contact with the

pipe to the pipes exterior surface, from the pipes exterior surface to the pipes interior surface,

and from the pipes interior surface to the #2 diesel fuel. Temperature under the soil surface

remains stable because heat transfer occurs through multiple layers with varying thermal

conductivities. Given the above information, one can see this delay in energy transfer when one

analyzes the fourth and fifth readings. During the time these last readings were taken, heat is

flowing in the opposite direction from the warmer liquid to the cooler interior surface of the pipe, to

the cooler exterior surface of the pipe, and all the way to the ground surface. Note that only a

0.15 oF change occurred to cause a 14.1 psi pressure drop.

The analysis of this pressure tests results includes a comparison of the initial recorded

pressure to other recorded pressures during the test and evaluation for a substantial pressure

drop to confirm a failed test. If there is no substantial pressure drop, then the test passes. In the

two examples given, both tests passed because neither fell substantially below the initial pressure


9/36



reading. A failed test can be overlooked because of increased pressure from an increase in the

con tained substances temperature and failed tests may have actually passed due to a

temperature decrease. The contained substances temperature must be measured during testing

and an expected pressure change must be calculated from the temperature data and compared

to the measured pressure change. The difference between the calculated and the measured

pressure values is what must be analyzed.

Hoop stress is defined in ASME B31.8 as the stress in a pipe wall, acting

circumferentially in a plain perpendicular to the longitudinal axis of the pipe and produced by the

pressure of the fluid in the pipe. Barlow's Formula is the common method used to determine

hoop stress in the wall of pipe. Pressure levels observed during API 1110 pressure testing will

not cause piping deformation because the observed pressure levels used during testing are well

below the design pressure limits of the piping system. Minimal volume change occurs because a

minimal piping deformation occurs therefore; the effects of hoop stress are negligible.

By monitoring the temperature of the contained substance in the piping during a pressure

test, the complexity of Fourier's heat transfer calculations and the need to monitor the ambient

and the exterior pipe surface temperature can be eliminated. If the temperature change of the

contained substance is known, then the energy change of the liquid can be used to determine the

temperature change of the pipe eventually determining the thermal expansion of the pipe. By

eliminating the need to monitor the external surface temperature of the pipe, the potential for

inaccuracies is decreased substantially and facilitates the development of an additional procedure

that can be used in conjunction with the API 1110 pressure test method to meet the standards in

Table BPP of Rule 62-762.601, F.A.C. See Appendix A for an example calculation.

Proposed Procedure

The following procedure is proposed to be conducted in association with the current API

1110 pressure testing. This proposed procedure will account for temperature change during a

pressure test to ensure compliance with the requirements of Table BPP table of Rule 62-762.601,


10/36



F.A.C. The first step is to identify known conditions and factors. These include the material of

which the pipe being pressure tested is constructed, the contained substance within the pipe, and

the physical state of that contained substance (i.e., solid, liquid, gas).

The next step is to gather referenced data for the known conditions and factors. The

thermal expansion coefficients of the pipe material and the contained substance must be

referenced, along with the compressibility of the contained substance. Piping as-built drawing or

a process flow diagram (PFD) should be referenced to find the length of pipe that is to be isolated

for testing. Once the length is determined, then the volume of the contained substance can be

calculated by using the inside diameter of the pipe.

The pressure test should be performed in accordance with API 1110 monitoring the

pressure and ambient temperature, but a temperature probe should also be used to take readings

of the substances temperature contained within the isolated pipe. Temperature of the contained

substance should be monitored with the same regularity of the pressure readings. It is important

to be aware of the integrity of the temperature probes seal to assure that the temperature reading

point is not a source of pressure loss.

Once the testing is completed and all data is gathered, the theoretical calculated

pressure can be determined. This calculation may be broken into two separate theoretical

pressure calculations if circumstances such as those seen in Example #2 occur. The test may

also be broken into separate calculations if the need to bleed off or increase pressure occurs.

This will provide an explanation for such abnormal occurrences while maintaining the accuracy of

the test. Please see Appendix E for a field simplified field method for determining compliance.

When the theoretical calculation(s) is complete, the deviations, if present, between the

theoretical calculated pressure(s) and the actual measured pressure(s) should be analyzed and

an unaccounted volumetric flow determined. This unaccounted volumetric flow can be compared

to Table BPP found in Rule 62-762.601, F.A.C. and set forth below, to determine a pass or a fail.


11/36



Frequency of Testing Line Segment CapacityLess than 50,000 gallons

Line Segment Capacity greaterthan 50000 gallons but less than100000 gallons

Line segment capacitygreater than 100000 gallons

Leak Rates for MonthlyTesting

3 gallons/hr 4 gallons/hr 5 gallons/hr

Leak Rates for

Quarterly Testing

1 gallon/hr 2 gallons/hr 3 gallons/hr

Leak Rates for AnnualTesting

0.2 gallons/hr 0.5 gallons/hr 1 gallons/hr

Using a spreadsheet developed for this purpose and using data from Example #1, it can be

determined that the pressure test in Example #1 was a passing test assuming a 1 oF temperature

change occurred. It was determined that 0.161469 gallons/hr were unaccounted for during this

test. The pipe segment tested has a capacity less than 50,000 gallons, therefore by referencing

the Table BPP above it is found that an unaccounted volumetric flow of 0.2 gallons/hr or less is

required to be in compliance. A full set of calculations of the theoretical pressure for Example #1

is shown in Appendix A. Calculations were also conducted for Example #2 which was also

determined to be a passing pressure test. Neither of the two examples actually monitored the

temperature of the contained substance during testing and therefore these calculated examples

are demonstrative only. Please see Appendix E for a field simplified field method for determining

compliance.

Ideal Testing Conditions

The proposed additional procedure to API 1110 accounts for temperature change and

accurately calculates a volumetric loss from a piping segment that is pressure tested. However,

there are a few scenarios that could potentially produce an inaccurate result. Therefore, it is of

the upmost importance to identify ideal testing conditions to optimize the accuracy of the results

produced to ensure compliance with Table BPP in Rule 62-762.601, F.A.C.

The greatest potential threat to the accuracy of the proposed addition to the API 1110

pressure test method is from the temperature gradients of the contained substance from which

temperature readings are being recorded. An example of one scenario is if a pressure test is

performed on an aboveground segment of piping where half is exposed to solar radiation and half


12/36



is covered by shade. It is a safe assumption to assume the portion of pipe exposed to solar

radiation will absorb more energy than the portion of piping covered by the shade. Consequently,

the energy flow through the pipe to the contained substance on one end will be greater than on

the other causing a temperature gradient across the pipe. If the temperature readings of the

contained substance are being measured on the warmer half of the pipe, then the value of the

theoretically calculated pressure will be higher because of a larger temperature change during the

pressure test on the half exposed to solar radiation than on the half covered by shade. This

would cause the calculations to represent that a greater volumetric loss had occurred and could

possibly fail a passing test. The reverse is true if the temperature measurements were recorded

from the cooler half of the pipe. A failed test could pass because the theoretically calculated

pressure would be lowered and would not account for the half of the pipe exposed to the solar

radiation. These temperature gradients are illustrated below. Calculations can be conducted

from weather data, solar engineering equations, and heat transfer equations but, the proposed

procedure would become too complex and could lead to a greater potential for error.

To assure accurate results, where portions of piping are exposed to intense day time

solar radiation, such piping should be tested during early morning or twilight hours to minimize

temperature gradients. If a pipe segment is entirely exposed to direct solar radiation or only

exposed to diffused and reflected solar radiation, the chance for temperature gradients is

minimized and testing can be conducted at anytime. Piping segments, where a portion of pipe

greater than 10% is aboveground and the remaining portion is below ground, should be pressure

tested separately so potential errors from temperature gradients can be avoided. Temperature

T = 1 oFT = 1.5 oF T = 2.5 oF

T = 4 oF

No Direct Exposure to Solar Radiation Direct Exposure to Solar Radiation


13/36



measurements should be recorded from the middle 20% of the piping segment isolated for

testing. If the above-mentioned actions are not possible, pressure testing should be conducted

during early morning or twilight hours and during months where there is not a rapid temperature

increase or decrease throughout the day.

Outside of the temperature change of the contained substance, the volumetric change is

what must be noted when measuring pressure. Changing pipe diameters in underground piping

is prevalent in older industrial facilities. If changes in diameter occur within a piping segment

isolated for pressure testing, the volume of a contained substance can still be calculated. If the

isolated pipe segment is underground, the temperature change of the piping will be very stable, at

most a 2F to 4F temperature rise. If the pressure test is conducted on above ground piping then

the temperature change might be more unstable. Volumetric thermal expansion of the piping, at

a varying range of temperature change of the pipe material is small to begin with, so if a

comparison between the volumetric thermal expansion differences between a 4in pipe and a 6in

pipe, the difference would be negligible, though linear thermal expansion will have an observable

variation. If a large temperature change of the contained substance occurs (more than 5F),

though highly unlikely, it might create a larger radial temperature gradient within the 6in pipe and

could provide a different end result than the 4in pipe. This is an unlikely scenario though during a

one hour test.

Conclusions

The existing API 1110 pressure test method is accurate unless a temperature change in

the contained substance occurs. This proposed method is to be used in conjunction with the API

1110 pressure test method which allows for highly accurate results that account for a temperaturechange during a test. This proposed method will allow the Department of Environmental

Protection (DEP) to have full confidence that, if this additional procedure is used during API 1110

pressure testing, an accurate grade of pass or fail will be determined in accordance with Table

BPP of Rule 62-762.601, F.A.C.


14/36



Acknowledgements

The cooperation and assistance of the members of the Solid Waste Subcommittee of the

Florida Electric Power Coordinating Group, Inc. is greatly appreciated. Special thanks to Mr.

Randy Melton and Mr. Stan Kroh of TECO Energy, representatives of Marathon Oil, Mr. Michael

Petrovich of Hopping Green & Sams, P.A., and the Florida Petroleum Council. All queries

regarding the proposed method and technical aspects should be directed to Mr. Arthur Bosshart,

TECO Energy, [email protected]


15/36



Bibliography

1) API Recommended Practice 1110, Fourth Edition, March 1997

2) Chapter 62-762, Florida Administrative Code Draft Rule September 28, 2007

3) American Society of Mechanical Engineers (ASME) B31.8 - 2003 Gas Transmission

and Distribution Piping Systems


16/36



Appendix

Appendix A

A calculation walkthrough step-by-step for Example 1:

1) Volumetric Thermal Expansion of Liquid (#2 Diesel Fuel)

V L = L*V Lo*T L = 0.00046 * 325581.3953 * 1 = 149.7674419

-volume is converted to in 3

2) Volume of pipe material is calculated

= 3.14 ro = 3.3125 in. (Referenced Value, The outside radius of the pipe in inches.)LTo = 11275.30538 (The length of pipe in inches at initial temperature)VLo = 325581.3953 in

3 (Value must be converted to in 3)

Vp = (*r o2*LTo ) - VLo (Initial Volume @ Initial Temperature)

Vp = (3.14*(3.3125^2)* 11275.30538)*.0043) 1400 = 270.4681151 gal.

3) Mass of the pipe is calculated

MP = V P*D P = 270.4681151 * 65.36954953 = 17680. 37884

- Volume is in gallons, density is in pounds per gallon

4) Mass of contained substance

ML = D L*V L = 6.7 * 1400 = 9380

- Volume is in gallons, density is in pounds per gallon

5) Energy change of the substance from the measured temperature change is calculated

Q = M L*Cp L*TL = 9380 * 0.43 * 1 = 4033.4

- Energy is given in BTU

6) Rearranging the equation from step 5, the temperature change of the pipe is determined from thecontained substances energy change by its temperature change

T P = Q/(M P*Cp P) = 4033.4 / (17680.37884 * 0.12) = 3.583333333

- Temperature is given in degrees Fahrenheit

7) Linear Thermal Expansion of Piping (Carbon Steel)


17/36


18/36


19/36



VLo = Initial Liquid VolumeT L = Change in Temperature

L 0.00046 oF -1

VLo 325581.3953 in 3

T L 1 oF

V L 149.7674419 in 3 232.5581395

Pipe Surface Area

Ape = 2r o2 + 2r oLTo

Ape = Surface Area of Pipe Exterior

ro = Radius of Pipe Exterior

*assume that the minimal increase in piping surface due to temperature rise has no effect on heat transfer from the pipe exterior surface tothe liquid contained within the pipe. Therefore, piping surface area at initial temperature is used for calculation.

ro 3.3125 in

Ape 234623.4484 in 2

Volumetric Thermal Expansion of Piping (Carbon Steel)

L p = p* L To *T p

p = Thermal Expansion Coefficient of Piping MaterialLTo = Length of Pipe @ Initial Temperature

T p = Change in Piping Temperature

p 0.0000078 oF -1

LTo 11275.30538 in 939.60878 ftT p 3.583333333 oF Use heat transfer Equation

Lp 0.315144785 in

VpTo = r i2LTo *assuming no residual air pockets in pipe

VpTo = Volume of Pipe @ Initial Temperature

ri = radius of inside piping diameter

ri 3.0325 inVpTo 325581.3953 in 3

VpT2 = r i2LT2

VpT2 = Volume of Pipe @ Final Temperature

LT2 = Length of Pipe @ Final Temperature

LT2 11275.62052 in

VpT2 325590.4953 in 3

V p = V pT2 VpTo

V p = Change in Piping Volume

V P 9.1 in 3


20/36



Overall Change in Volume

V = V L - V P V = Overall Change in Volume

V 140.6674419 in 3

Mass of PipingMP= D P*VP

ML = Mass of Isolated Piping

DL = Density of Piping Material

VP 270.4681151 gallons 1 in^3 = .0043 gallonsDP 65.36954953 lbs/gallon 1 kg/m^3 = .008345404 lbs/gallon

MP 17680.37884 lbs.Mass of Substance in Pipe

ML = D L*V L

MP = Mass of Substance contained in Pipe

DP = Density of Substance contained in PipeVL 1400 gallons

DL 6.7 lbs/gallon

ML 9380 lbs.Energy in Substance

Q=M L*Cp L*TL Q = Heat

Cp L = Heat capacity of Substance

Cp L 0.43BTU/lboF

Q 4033.4 BTU

Temperature Change of Pipe

T P = Q/(M P*Cp P)

Cp P = Heat Capacity of Pipe Material

Cp P 0.12BTU/lboF

T P 3.583333333oF

This is your results portion of the spread sheet:

UnaccountedVolume

V La = (P TC - P AM )*V Lo/ V La = Calculated Volume Difference associated with pressure change

PTC = Theoretical Calculated Pressure

PAM = Actually Measured Pressure


21/36



PTC 200.97 psi

V La -0.21260126 gallons V La /hr 0.1614693 gallons/hr

Appendix C

Assumptions

1) A minimal increase in piping surface due to temperature rise has no effect on heat

transfer from the pipe exterior surface to the liquid contained within the pipe.

Therefore, piping surface area at initial temperature is used for calculation.

2) The system to be tested is assumed to be comprised entirely of a homogeneous

liquid.

3) Hoop Stress on piping does not cause piping deformation because pressure levels

used during testing are well below the design pressure of the piping system. No

volume change occurs because no deformation of the piping material occurs.

4) The radial temperature gradient of the pipe material is negligible because pipe walls

are less than .6 in. and during normal testing conditions an extreme temperature

change will not occur over single test duration.

Appendix D

Nomenclature

= Compressibility (#2 Diesel)

P = Change in Pressure

V L = Change in Volume of Liquid

L = Thermal Expansion Coefficient of Liquid

VLo = Initial Liquid Volume

T L = Change in Temperature

Ape = Surface Area of Pipe Exterior


22/36



ro = Radius of Pipe Exterior

p = Thermal Expansion Coefficient of Piping Material

LTo = Length of Pipe @ Initial Temperature

T p = Change in Piping Temperature

VpTo = Volume of Pipe @ Initial Temperature

ri = radius of inside piping diameter

VpT2 = Volume of Pipe @ Final Temperature

LT2 = Length of Pipe @ Final Temperature

V p = Change in Piping Volume

V = Overall Change in Volume

ML = Mass of Isolated Piping

DL = Density of Piping Material

MP = Mass of Substance contained in Pipe

DP = Density of Substance contained in Pipe

Q = Heat

CpL = Heat capacity of Substance

CpP = Heat Capacity of Pipe Material

Appendix E

Simplified Field Method

For #2 Diesel it is found that a linear pressure change of 9.4 PSI occurs for every .1 oF change in

temperature within a perfectly isolated volume. This 9.4 PSI change per unit of temperature will

be known as the P coefficient in the pressure testing packet in appendix I. The P Coefficient is

found by relating compressibility to a change in volume and defining a change in temperature

between the initial temperature (T i) and the final temperature (T f) to be .1oF. Using equations 1

through 12 found in Appendix A of Effects of Temperature Change on Pressure Testing and

Maintaining Environmental Compliance and defining referenced coefficients for liquid


23/36



compressibility, piping thermal expansion, liquid thermal expansion, pipe density, liquid density,

pipe heat capacity, and liquid heat capacity the P coefficient for any substance may be found.

The Allowable Pressure Tolerance (APT) is calculated by defining an acceptable volumetric loss

which for purposes of compliance in Flori da is defined to be .2 gallons/hour (V La /hr) by Table

BPP within 62-762 F.A.C. Equation 1 is the formula used for calculating ATP where

compressibility () and piping volume at the initial temperature are shown (V Lo). APT is the

allowable pressure difference between the theoretically calculated pressure (P Tc ) and the actually

measured pressure (P am ).

APT = ((V La/hr)())/V Lo (1)

APT = P Tc - P am (2)

Procedure Simplified

Find as-built drawings of piping system to be pressure tested and note the lengths of each

section that is to be isolated. Determine the size of piping and the schedule of piping then find

the inside radius of the pipe using reference material. Calculate the volume within the section of

pipe that is to be isolated for pressure testing using the following equation:

VTo = r 2

L (3)

Given that 9.4 PSI occurs for every .1 oF change in temperature, the noted temperature change

during the testing period should be divided by .1 oF to determine the magnitude factor to be used

later. When the pressure testing is complete the magnitude factor (MF) should be multiplied by

9.4 PSI and then added to the initially measured pressure (P i) from the beginning of the test to

determine the theoretically calculated pressure (P Tc ). See Equation 4.

P Tc = P i + 9.4(MF) (4)During the test, the APT lookup table must be present. Using the piping volume that was

calculated prior to testing, the APT can be found using the APT lookup table. By subtracting the

APT defined for the given piping volume from the P Tc , the lower limit of the actually measured


24/36



pressure can be defined. If the measured pressure during the duration of a pressure test falls

below this defined lower limit then the test fails.

Appendix F

APT Lookup Table9.4 PSI / .1 oF Test Duration .75 hr Test Duration 1 hr Test Duration 1.25 hr Test Duration 1.5Piping Volume

(Gal.) APT (PSI) APT (PSI) APT (PSI) APT (PSI) V La /hr500 65.25 87.00 108.75 130.50 0.2600 54.38 72.50 90.63 108.75 0.2700 46.61 62.14 77.68 93.21 0.2800 40.78 54.38 67.97 81.56 0.2900 36.25 48.33 60.42 72.50 0.2

1000 32.63 43.50 54.38 65.25 0.2

1100 29.66 39.55 49.43 59.32 0.21200 27.19 36.25 45.31 54.38 0.21300 25.10 33.46 41.83 50.19 0.21400 23.30 31.07 38.84 46.61 0.21500 21.75 29.00 36.25 43.50 0.21600 20.39 27.19 33.98 40.78 0.21700 19.19 25.59 31.99 38.38 0.21800 18.13 24.17 30.21 36.25 0.2

1900 17.17 22.89 28.62 34.34 0.22000 16.31 21.75 27.19 32.63 0.22100 15.54 20.71 25.89 31.07 0.22200 14.83 19.77 24.72 29.66 0.2

2300 14.18 18.91 23.64 28.37 0.22400 13.59 18.13 22.66 27.19 0.22500 13.05 17.40 21.75 26.10 0.22600 12.55 16.73 20.91 25.10 0.2

2700 12.08 16.11 20.14 24.17 0.22800 11.65 15.54 19.42 23.30 0.22900 11.25 15.00 18.75 22.50 0.23000 10.88 14.50 18.13 21.75 0.2

3100 10.52 14.03 17.54 21.05 0.23200 10.20 13.59 16.99 20.39 0.23300 9.89 13.18 16.48 19.77 0.23400 9.60 12.79 15.99 19.19 0.2

3500 9.32 12.43 15.54 18.64 0.23600 9.06 12.08 15.10 18.13 0.23700 8.82 11.76 14.70 17.64 0.23800 8.59 11.45 14.31 17.17 0.2

3900 8.37 11.15 13.94 16.73 0.24000 8.16 10.88 13.59 16.31 0.24100 7.96 10.61 13.26 15.91 0.24200 7.77 10.36 12.95 15.54 0.24300 7.59 10.12 12.65 15.17 0.2


25/36



4400 7.41 9.89 12.36 14.83 0.24500 7.25 9.67 12.08 14.50 0.24600 7.09 9.46 11.82 14.18 0.24700 6.94 9.26 11.57 13.88 0.2

4800 6.80 9.06 11.33 13.59 0.24900 6.66 8.88 11.10 13.32 0.2

5000 6.53 8.70 10.88 13.05 0.2

Appendix G

Compliance Curve

Compliance Curve #2 DieselAPT = (( VLa /hr)*) / V Lo

0.00

20.00

40.00

60.00

80.00

100.00

120.00

0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000

Piping Volume (Gallons)

A l l o w a

b l e P r e s s u r e

T o

l e r a n c e

" A P T " ( P S I )

.75 hrs. Test Duration

1 hr. Test Duration

1.25 hrs. Test Duration


Appendix H

Simplified Field Method Sample Calculation

For example, an underground pipe that contains #2 diesel fuel is pressure tested in the state of

Florida. Before the test it was determined that the pipe was 6in. in diameter and had a schedule

40 wall thickness. The volume of the pipe that is to be isolated was calculated to have a capacity

of 1400 gallons. The line was properly packed and the initial temperature and pressure readings


26/36



were 70 oF and 100PSI respectively. The test lasted 1.5 hours and the final temperature and

pressure readings were 71.1 oF and 171 PSI respectively. Is the piping system in compliance?

Using the simplified procedure we know that a linear pressure change of 9.4 PSI occurs for every

.1 oF change in temperature within a perfectly isolated volume therefore the magnitude factor (MF)

is first calculated. The magnitude factor has no units.

MF = (71.1 oF - 70 oF)/.1 oF = 11

Next, the theoretically calculated pressure (P Tc ) is determined.

P Tc = P i + 9.4(MF) = 100 PSI + 9.4 PSI(11) = 203.4 PSI

Using the APT lookup table, we find that for a piping system of 1400 gallons, the allowable

pressure tolerance (APT) is 46.61 PSI. Next, the lower limit for compliance is determined.

Lower Limit = 203.4 PSI 46.61 PSI = 156.79 PSI

The final temperature reading from the test was 171 PSI. The final reading is greater than the

lower limit therefore the pressure test passes. See Appendix J to view the completed Line

Pressure Testing form using this example.


27/36



Appendix ILine Pressure Testing Packet Instructions

PretestPretest data should be acquired prior to pressure testing.Line 1 -- The facility representative should identify an accurate pipe capacity through calculation or bestengineering judgment. Information regarding the pipe capacity and the contained substance should begiven to the entity or individuals conducting the pressure test prior to testing.Line P Values for the P coefficient may be found on page 2 of the DEP Line Pressure Testing Packet inthe P Coefficient Lookup Table. Enter this value in to line P.

Test DataBefore the first temperature and pressure readings are recorded, the piping segment being pressure testedmust be properly packed. Meaning the piping segment must be completely filled with the containedsubstance with no vapor or air pockets. Please reference for proper packing procedures section 3.7 of APIRecommended Practice 1110 Fourth Edition, March 1997.

Line 2 After the line is properly packed and pressurized the desired test pressure, enter the initial pressurereading once testing begins into line 2.Line 3 Enter the initial temperature reading once testing begins into line 3.Line 4 Enter the time which testing begins into line 4.Line 5 When testing is complete enter the final pressure reading into line 5.Line 6 Enter the final temperature reading into Line 6.Line 7 Enter the time which the test ended into line 7.

Post Test CalculationLine 8 Subtract line 7 from line 4 and enter the length of the test, in hours, into line 8. If a test, forexample, lasts 90 minutes, divide 90 by 60 to convert the test duration to hours, which in this example is1.5hrs.Line 9 Subtract line 6 from line 3 and enter the change in temperature into line 9. If this value is anegative number, then simply multiple the value by -1 and enter the positive number into line 9.Line 10 Divide the value from line 9 by 0.1 and enter it into Line 10.Line 11 Determine the theoretically calculated pressure by multiplying the value from line P by line 10and adding the product of the multiplication to the initial pressure reading on line 2. Enter this value intoline 11.Line 12 -- Values for the allowable pressure toler ance APT may be found on page 3 of the DEP LinePressure Testing Packet in the APT Lookup Table. Enter this value into line 12. Use the Facility IdentifiedIsolated Piping Segment Capacity found on line 1 and the test duration found on line 8 to determine theAPT value.Line 13 Determine the lower limit value by subtracting line 11 from line 12. Enter this value into line 13.The lower limit is the limit that that the pressure cannot drop below during testing to be compliant.


28/36



DEP Line Pressure Testing WorksheetFacility Owner _______________________ Date _________________Facility Name _______________________ Facility ID _________________Line Segment Tested _______________________Pretest Data

Data Description Data Entry LineNumber

Contained Substance ____________________ P=_________________________ PThe P coefficients can be found using a referenced value of the contained substance found on page 2 of the pressure testing packet.Facility Identified Isolated Piping Segment Capacity (V)Capacity must be given in gallons

___________________________ (1)

Test DataIs the test segment properly packed? For proper line packing procedures refer to Section 3.7 of API

Recommended Practice 1110 Fourth Edition, March 1997

Y N

Initial Test DataInitial Pressure Reading (P i)units of pressure must be given in PSI

____________________________ (2)

Initial Temperature Reading (T i)units of Temperature must be given in o F

____________________________ (3)

Time Began ____________________________ (4)Final Test Data

Final Pressure Reading (P F)units of pressure must be given in PSI

____________________________ (5)

Final Temperature Reading (T F)units of Temperature must be given in o F

____________________________ (6)

Time Ended ____________________________ (7)Post Test CalculationTest Duration (t)units of time must be given in hours

t = line 7 line 4 = __________________________ Hours (8)

Temperature Change (T) units of Temperature must be given in o F

T = line 6 line 3 = __________________________ oF (9)

Magnitude Factor (MF)The Magnitude Factor is unit less

MF = line 9 / 0.1 oF = ____________________________ (10)

Theoretically Calculated Pressure (P Tc)units of Temperature must be given in o F PTc = line 2 + line P x line 10 = _________________

o

F (11)Allowable Pressure Tolerance (APT)units of Pressure must be given in PSI

APT = _____________________________________ PSI (12)The APT is found using the table on page 3 of the pressure testing packet. Using the Facility Identified Isolated Piping Se gment Capacity on line (1), find the associated APT value.Lower Limitunits of Pressure must be given in PSI

Lower Limit = line 11 line 12 = __________________ PSI (13)

Is the lower limit pressure (Lower Limit), on line 13, less than the final pressure (P F) reading, online 5? Y N

(14)

Test ResultIf the lower limit pressure (Lower Limit), on line 13, is greater than the final pressure (P F) reading, on line5 then circle the compliance determination of FAIL, otherwise circle the compliance determination of PASS.

PASS FAILPressure Test Conducted by _____________________________________Pressure Test Technician _____________________________________Technician Signature _____________________________________Facility Representative _____________________________________Facility Representative Signature _____________________________________


29/36



P Coefficient Lookup Table Enter the correct P coefficient into the DEP Line Pressure Testing Worksheet into line P

Gasoline#2 Diesel 9.4Biodiesel

#6 Fuel OilBenzeneMethanolEthanol


30/36



APT Lookup Table #2 Diesel Fuel Oil9.4 PSI / .1 oF Test Duration .75 hr Test Duration 1 hr Test Duration 1.25 hr Test Duration 1.5Piping Volume

(Gal.) APT (PSI) APT (PSI) APT (PSI) APT (PSI) V La /hr100 326.25 435.00 543.75 652.50 0.2200 163.13 217.50 271.88 326.25 0.2

300 108.75 145.00 181.25 217.50 0.2400 81.56 108.75 135.94 163.13 0.2500 65.25 87.00 108.75 130.50 0.2600 54.38 72.50 90.63 108.75 0.2700 46.61 62.14 77.68 93.21 0.2800 40.78 54.38 67.97 81.56 0.2900 36.25 48.33 60.42 72.50 0.2

1000 32.63 43.50 54.38 65.25 0.21100 29.66 39.55 49.43 59.32 0.2

1200 27.19 36.25 45.31 54.38 0.21300 25.10 33.46 41.83 50.19 0.21400 23.30 31.07 38.84 46.61 0.2

1500 21.75 29.00 36.25 43.50 0.21600 20.39 27.19 33.98 40.78 0.21700 19.19 25.59 31.99 38.38 0.21800 18.13 24.17 30.21 36.25 0.21900 17.17 22.89 28.62 34.34 0.2

2000 16.31 21.75 27.19 32.63 0.22100 15.54 20.71 25.89 31.07 0.22200 14.83 19.77 24.72 29.66 0.22300 14.18 18.91 23.64 28.37 0.2

2400 13.59 18.13 22.66 27.19 0.22500 13.05 17.40 21.75 26.10 0.22600 12.55 16.73 20.91 25.10 0.2

2700 12.08 16.11 20.14 24.17 0.22800 11.65 15.54 19.42 23.30 0.22900 11.25 15.00 18.75 22.50 0.23000 10.88 14.50 18.13 21.75 0.23100 10.52 14.03 17.54 21.05 0.2

3200 10.20 13.59 16.99 20.39 0.23300 9.89 13.18 16.48 19.77 0.23400 9.60 12.79 15.99 19.19 0.23500 9.32 12.43 15.54 18.64 0.23600 9.06 12.08 15.10 18.13 0.23700 8.82 11.76 14.70 17.64 0.23800 8.59 11.45 14.31 17.17 0.2

3900 8.37 11.15 13.94 16.73 0.24000 8.16 10.88 13.59 16.31 0.24100 7.96 10.61 13.26 15.91 0.24200 7.77 10.36 12.95 15.54 0.24300 7.59 10.12 12.65 15.17 0.24400 7.41 9.89 12.36 14.83 0.24500 7.25 9.67 12.08 14.50 0.24600 7.09 9.46 11.82 14.18 0.2


31/36



4700 6.94 9.26 11.57 13.88 0.24800 6.80 9.06 11.33 13.59 0.24900 6.66 8.88 11.10 13.32 0.2

5000 6.53 8.70 10.88 13.05 0.2

5100 6.40 8.53 10.66 12.79 0.25200 6.27 8.37 10.46 12.55 0.25300 6.16 8.21 10.26 12.31 0.25400 6.04 8.06 10.07 12.08 0.25500 5.93 7.91 9.89 11.86 0.25600 5.83 7.77 9.71 11.65 0.2

5700 5.72 7.63 9.54 11.45 0.25800 5.63 7.50 9.38 11.25 0.25900 5.53 7.37 9.22 11.06 0.26000 5.44 7.25 9.06 10.88 0.2

6100 5.35 7.13 8.91 10.70 0.26200 5.26 7.02 8.77 10.52 0.26300 5.18 6.90 8.63 10.36 0.26400 5.10 6.80 8.50 10.20 0.2

6500 5.02 6.69 8.37 10.04 0.26600 4.94 6.59 8.24 9.89 0.26700 4.87 6.49 8.12 9.74 0.26800 4.80 6.40 8.00 9.60 0.2

6900 4.73 6.30 7.88 9.46 0.27000 4.66 6.21 7.77 9.32 0.27100 4.60 6.13 7.66 9.19 0.27200 4.53 6.04 7.55 9.06 0.2

7300 4.47 5.96 7.45 8.94 0.27400 4.41 5.88 7.35 8.82 0.27500 4.35 5.80 7.25 8.70 0.27600 4.29 5.72 7.15 8.59 0.2

7700 4.24 5.65 7.06 8.47 0.27800 4.18 5.58 6.97 8.37 0.27900 4.13 5.51 6.88 8.26 0.28000 4.08 5.44 6.80 8.16 0.2

8100 4.03 5.37 6.71 8.06 0.28200 3.98 5.30 6.63 7.96 0.28300 3.93 5.24 6.55 7.86 0.28400 3.88 5.18 6.47 7.77 0.28500 3.84 5.12 6.40 7.68 0.28600 3.79 5.06 6.32 7.59 0.28700 3.75 5.00 6.25 7.50 0.28800 3.71 4.94 6.18 7.41 0.28900 3.67 4.89 6.11 7.33 0.29000 3.63 4.83 6.04 7.25 0.29100 3.59 4.78 5.98 7.17 0.29200 3.55 4.73 5.91 7.09 0.29300 3.51 4.68 5.85 7.02 0.29400 3.47 4.63 5.78 6.94 0.29500 3.43 4.58 5.72 6.87 0.29600 3.40 4.53 5.66 6.80 0.2


32/36



9700 3.36 4.48 5.61 6.73 0.29800 3.33 4.44 5.55 6.66 0.29900 3.30 4.39 5.49 6.59 0.2

10000 3.26 4.35 5.44 6.53 0.2

Compliance Curve #2 DieselAPT = (( VLa /hr)*) / V Lo

0.00

20.00

40.00

60.00

80.00

100.00

120.00

0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000

Piping Volume (Gallons)

A l l o w a

b l e P r e s s u r e

T o

l e r a n c e

" A P T " ( P S I )

.75 hrs. Test Duration

1 hr. Test Duration




33/36



Appendix JCompleted Line Pressure Testing Worksheet


34/36



Appendix KStep-by- Step Calculation of the P Coefficient

The P Coefficient is found by relating compressibility to a change in volume and defining achange in temperature between the initial temperature (T i) and the final temperature (T f) to be.1 oF. For #2 Diesel it is found that a linear pressure change of 9.4 PSI occurs for every .1 oFchange in temperature within a perfectly isolated volume. This calculation is independent of theamount of volume but requires an initial value for volume to begin the calculation of the Pcoefficient. In the example below, we will use the capacity of one of our actual piping systems.

1) Volumetric Thermal Expansion of Liquid (#2 Diesel Fuel)

This is the change in volume due to the change in temperature. L = .00046

oF -1 (Referenced coefficient, must be in oF -1)VLo = 325581.3953 in

3 (Value must be converted to in 3)TL = T f Ti = .1

oF (Define your temperature change as .1 oF)

VL = L*VLo*TL = 0.00046 * 325581.3953 * .1 = 14.97674419 in3

2) Volume of pipe material is calculated

This is not the capacity of the pipe. This is the amount of piping material in in 3. Subtractsthe outside volume if pipe was a solid cylinder from the inside capacity of the pipe.Answer must be given in gallons. We are using a 6 in. pipe.

= 3.14 ro = 3.3125 in. (Referenced Value, The outside radius of the pipe in inches.)LTo = 11275.30538 (The length of pipe in inches at initial temperature)VLo = 325581.3953 in

3 (Value must be converted to in 3)

Vp = (*r o2*LTo ) - VLo (Initial Volume @ Initial Temperature)

Vp

= (3.14*(3.3125^2)* 11275.30538)*.0043) 1400 = 270.4681151 gal.

3) Mass of the pipe is calculated

The mass must be calculated to determine energy transfer.DP = 65.36954953 lbs./gal (Referenced Coefficient, must be in pounds per gallon)VP = 270.4681151 gal. (From Equation 2)

MP = VP*DP = 270.4681151 * 65.36954953 = 17680. 37884 lbs.

4) Mass of contained substance

This is calculated to determine energy transfer.DL = 6.7 lbs./gal (Referenced Coefficient, must be in pounds per gallon)VLo = 1400 gal. (Initial volume @ the initial temperature. Must be in gallons)

ML = D L*VL = 6.7 * 1400 = 9380 lbs.

5) Energy change of the substance from the measured temperature change iscalculated.


35/36



This is calculated to determine energy transfer from the pipe material to determine thetemperature change of the pipe material and eventually determine the thermal expansionof the pipe.ML = 9380 lbs. (From Equation 4)Cp L = .043 BTU/lb.

oF (Referenced Coefficient, Must be in BTU/lb. oF) TL = T f Ti = .1

oF (Define your temperature change as .1 oF)

Q = M L*CpL*TL = 9380* 0.43 * .1 = 403.34 BTU

6) Temperature Change of Piping Material

Rearranging the equation from step 5, the temperature change of the pipe is determinedfrom the contained substances energy change by its temperature change. Q = 403.34 BTU (From Equation 6)Cp P = .12 BTU/lb.

oF (Referenced Coefficient, Must be in BTU/lb. oF) MP = 17680.37884 lbs. (From Equation 3)

TP = Q/(M P *CpP) = 403.34 / (17680.37884 * 0.12) = .3583333333oF

7) Thermal Expansion of Piping (Carbon Steel)

The Thermal Expansion will not substantially vary from material to material and thereforewill not affect the determined value of the P Coefficient. The differences from materialto material are negligible.p = .0000078 oF -1 (Referenced coefficient, must be in oF -1)LTo = 11275.30538 in. (The length of pipe in inches at initial temperature)Tp = .3583333333

oF (From Equation 6)

Lp = p* L To *Tp = 0.0000078 * 11275.30538 * .3583333333 = 0.0315144785 in.

8) Volume of the pipe at the final temperature is calculated from its thermalexpansion

This is used to determine the overall change in volume with respect to all thermalexpansions affecting the isolated system. = 3.14 ri = 3.0325 in. (Referenced Value, The inside radius of the pipe in inches.)LTo = 11275.30538 (The length of pipe in inches at initial temperature)Lp = 0.0315144785 in. (From Equation 7)

VpT2 = *r i2*(LTo

+ Lp) = 3.14 * 3.0325 2 * (11275.30538 + .315144785) = 325582.3053 in 3

9) The volume change of the pipe is calculated

This is the volume change of the pipe due to thermal expansion.VpT2 = 325582.3053 in

3 (From Equation 8)

VLo = 325581.3953 in3

(Value must be converted to in3

)Vp = VpT2 VLo = 325582.3053 - 325581.3953 = 0.91 in

3

10) The overall change in volume accounts for the expansion rates of the substanceand the pipe.

VL = 14.97674419 in3 (From Equation 1)

VP = .91 in3 (From Equation 9)


36/36


V = V L - VP = 14.97674419 - 0.91 = 14.06674419 in3

12) Determination of the P Coefficient.

The theoretical pressure change is calculated using the bulk modulus of elasticity of thesubstance. The bulk modulus is the inverse of the substances compressibility. Theequation below can be arranged to use compressibility or the bulk modulus of elasticity.V = 14.06674419 in 3 (From Equation 11)VL0 = 325581.3953 in

3 (Value must be converted to in 3) = 217500 psi (Referenced Coefficient, Compressibility must be in psi)

P coefficient = P = (V/V L0) / = (14.06674419 / 325581.3953) * 217500) = 9.397 psi

Appendix LStep-by-Step Calculation of a Specific APT Value

The Allowable Pressure Tolerance (APT) is calculated by defining an acceptable volumetric loss

which for purposes of compliance in Florida is defined to be .2 gallons/hour (V La /hr) by Table

BPP within 62-762 F.A.C. Equation 1 is the formula used for calculating ATP where

compressibility () and piping volume at the initi al temperature (V Lo) are accounted used in

determination. APT is the allowable pressure difference, defined by an environmental authority,

between the theoretically calculated pressure (P Tc ) and the actually measured pressure (P am ).

We will use the referenced and calculated values from Appendix K.

1) APT = (((V La/hr)())/V Lo)*t

VLa /hr = 0.2 gal/hr (This is the minimal accepted leak rate defined by environmental authority) = 217500 psi (Compressibility of #2 Diesel)VLo = 1400 gal. (Value must be in Gallons)t = 1.5 hrs. (Duration of test, given in hours)

APT = (((V La/hr)())/V Lo)*t = ((0.2 * 217500)/1400)*1.5 = 46.61 psi

Documents

API 1110 Additional Procedure Report