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3E Plus Technical Addendum

Introduction

This addendum summarizes some of the methodologies and equations used in 3E Plus. Theaddendum provides background information for users who are interested in the technical foundations of3E Plus. For a user to use 3E Plus, its not essential to understand the information in this document,

but it may be helpful. The addendum contains six sections, which are indexed below.

Section 1 Mathematical Model for Economic Thickness Determination 2

Section 2 Series Present Worth Factor for Fuel Inflation 12

Section 3 Heat Transfer Equations 14

Section 4 Dew Point Calculation 18

Section 5 Apparent Thermal Conductivity Equations 19

Section 6 Reduction of CO2, NOx, and Carbon Equivalent (CE) 21

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Section 1 Mathematical Model for Economic Thickness Determination

The list of factors that influence the economic thickness of industrial insulation is extremely long. Inorder to develop a realistic model that contains a tractable number of input parameters, a limitednumber of factors will be considered. The factors considered in the model of this report can be dividedinto three general categories:

Insulation Related CostsInsulation CostInsulation Maintenance

Heat Loss or Gain Related CostsEquipment CostEnergy CostEquipment MaintenanceCondenser Make-Up Water Cost

Tax SavingsInsulation DepreciationInsulation MaintenanceEquipment DepreciationEquipment Maintenance

Energy CostCondenser Make-Up Water Cost

Other factors such as property taxes on the increased value of insulation and plant equipment,investment tax credits, and tax savings on interest on borrowed money are specifically ignored. Theremay also be other factors that certain companies or analysts would add to this list. However, it is feltthat the factors given above certainly represent the major contributions to the total cost of owning apiece of insulation, and that the computer economic thickness from this model is well within theuncertainty associated with a more sophisticated model.

The factors that influence the economic thickness are similar for either the hot or cold surface models.While heating plant equipment costs are important for hot surfaces, the chiller cost is the correspondingcost for cold surfaces. The only cold surface factor that does not have an analogous hot surface term is

the condenser make-up water cost. For a water-driven condenser, the water evaporated into theatmosphere must be replenished.

The total cost of owning a piece of insulation can be written as the sum of heat loss and insulationrelated costs less any tax savings. Symbolically, this can be written as:

Insulation Related CostTotal Cost = + Heat Flow Costs

Tax Savings(1-1)

The economic thickness is defined as the insulation thickness that yields the minimum total cost ofowning insulation. Since money obviously has a time value, one must choose a particular time at whichcosts are evaluated. The procedure adopted on this model is the Equivalent Uniform Annual Cost

(EUAC) method in which all costs are converted to an equivalent uniform annual cost by use of theappropriate interest discount factors. This procedure is mathematically equivalent to the Present Valueor Future Value methods provided the time period chosen is the least multiple of the insulation life andplant equipment life. Hereafter, the EUAC method will be referred to simply as the annual cost. Theremainder of this section will be devoted to a discussion of the equations used to calculate the annualcost of the various factors that influence the economic thickness.

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Insulation Related Costs

Insulation Cost: The unit cost of insulation CI($/ft or $/ft2) is already expressed as a present value. This

cost can be uniformly distributed over the life of the insulation by use of the series present worth factor,(P/A,i,nI).

=

I

I

niA

PCA

,, (1-2)

Throughout this practice, the symbol A will represent any equivalent uniform annual cost.

For those readers unfamiliar with the concept of the series present worth factor, it can be thought ofconceptually as an "effective" life that considers the time value of money. For example, if money doesnot have a time value, the (P/A,0,nI) = nI: it is obvious that the insulation cost has been divided into nIequal parts. In general, (P/A,i,nI) < nIwith the difference between the two terms increasing as theinterest rate increases.

Insulation Maintenance: First year's insulation maintenance is assumed to be proportional to the initialinsulation cost. In each succeeding year the maintenance is allowed to inflate at the same rate as fuel.It is probable that maintenance charges will not inflate at the same rate as fuel. However, it should becloser to the fuel inflation rate than to zero. Rather than introduce a general inflation rate as anadditional input parameter, the fuel inflation rate is chosen as representative of the rate at whichinsulation maintenance inflates. The resulting annual cost of insulation maintenance is

=

I

FI

MII

niA

P

iniA

P

fCA

,,

,,,0

(1-3)

The term (P/A0,i,nI,iF) is the series present worth factor that converts the annual (inflating) cost to apresent value and can also be thought of as an effective life. If the fuel inflation rate is zero, then(P/A,i,nI,0) = (P/A,i,nI). If both interest rates are zero, then both series present worth factors are equalto nI, and they simply cancel out. Additional information on this factor can be found in section 2.

Heat Related Costs

For any specified thickness, the heat loss, Q, (Btu/hr-ft or Btu/hr-ft2) can be calculated. The heatproducing equipment can have a life nEthat is different from the insulation life n I.

Incremental Equipment Cost (Heating): The cost of adding an increment in heating capacity is specifiedby the parameter mE, which represents the dollar investment in adding a 1,000,000 Btu/hr increase in

capacity. Since this quantity is already in present value terms, the series present worth factor must beapplied to convert this to an annual cost.

=

E

E

niA

P

QmA

,,

(1-4)

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Note that the equipment life nEis used instead of the insulation life. Every effort should be made toobtain a value that is representative for the particular project being analyzed. Since we have used anincremental equipment cost equation, the total annual cost cannot be used in an absolute sense.

Energy Cost: Fuel prices are assumed to be inflating at a fixed annual rate of i F. If the present day fuel

price is CFo($/1,000,000 Btu), the system operates H hours each year, and Eis the thermal efficiency

at which fuel is converted to heat, then the annual cost of the lost energy is

=

I

FI

E

F

niA

P

iniA

P

QHCA

,,

,,,0

0

(1-5)

Equipment Maintenance: The annual equipment maintenance is assumed to be proportional to the costof the annual heat loss or gain. Since the fuel cost is inflating at an annual rate of i F, the equipmentmaintenance also inflates at the same rate.

=

I

FI

E

FME

niA

P

iniAP

QHCfA

,,

,,,00

(1-6)

Tax Savings for Hot Surfaces

Maintenance, fuel costs, and depreciation all produce savings from the point of view of federal incometaxes. Operating expenses for a business are subtracted from the gross income in order to determinethe taxable income.

TAXABLE INCOME = GROSS INCOME - EXPENSES

The reduction in federal income tax due to expenses is simply the effective tax rate times the expenses.

INCOME TAXES = (TR)(GROSS INCOME - EXPENSES)

Insulation Depreciation: Straight-line depreciation will be assumed. This procedure allows an annualdeduction of CI/nIfrom gross income for a tax savings of

( )I

I

n

CTRA= (1-7)

Insulation Maintenance: The annual insulation maintenance cost is simply multiplied by the effective taxrate.

( )

=

I

FI

MII

niA

P

iniA

P

fCTRA

,,

,,,0

(1-8)

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Equipment Depreciation: Assuming straight-line depreciation, this becomes

( )E

E

QmTRA

= (1-9)

Equipment Maintenance: The annual equipment costs are multiplied by the effective tax rate.

( )

=

I

FI

E

FME

niA

P

iniA

P

QHCfTRA

,,

,,,00

(1-10)

Energy Cost: The annual energy costs are multiplied by the effective tax rate

( )

=

I

FI

E

F

niAP

iniA

P

QHCTRA

,,

,,,0

0

(1-11)

As can be readily seen, the tax rate has a very strong effect. For example, with a tax rate of 48%,natural gas at $2.00/MMBtu effectively costs only (1 - 0.48) X $2.00 or $1.04/MMBtu. The federalincome tax structure discourages the application of insulation. If the tax savings are subtracted fromthe insulation and heat loss related costs, the net annual cost can be written as

( )

( )( )

+

+

+

+

=

MEIFI

E

f

E

EE

I

I

FI

MIII

fTRniP

Aini

A

PHC

n

TRni

P

AmQ

n

TR

niA

P

iniA

P

fTRniP

ACA

11,,,,,,,

,,

,,,

1,,

0

0

0

(1-12)

The above equation can be written more compactly as

bQaCA I+= (1-13)

Where the factors a and b are independent of insulation thickness.

The procedure for determining the economic thickness is to evaluate the annual cost for each discrete

insulation thickness (including zero thickness) and to choose the thickness that gives the minimumannual cost. Since most commercial and industrial insulations are only available in nominal thicknessincrements of one-half inch, there appears to be little value in determining the precise insulationthickness that minimizes Equation 1-12 for two reasons. One, the minimizing thickness is probably notcommercially available. Second, the cost versus thickness relationship is not known analytically. Themost important terms that contribute the greatest amount to the total annual cost, are (a) insulation costand depreciation, (b) energy cost and its tax savings, (c) insulation maintenance, and (d) equipmentmaintenance, with the latter two terms being an order of magnitude smaller than the first two terms.

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Discounted Payback Period

The concept of payback period is often used in judging the merits of an economic venture. In itssimplest form, payback period is the time required for the savings resulting from an investment to repaythe cost of the investment. For the case of thermal insulation, the savings is the economic value of thereduction in heat loss or gain due to increased insulation, while the cost of the investment is theinsulation cost. If one ignores the time value of money and any inflation, the resulting payback period istermed simple payback period. The discounted payback period considers the time value of money andany anticipated inflation costs. Some care is necessary in precisely defining the energy savings andinsulation cost. For the model presented in this report, all costs and savings will be expressed inpresent value terms. The present value of the savings is the difference between the present value ofthe cost for two different alternatives, A and B.

PV Savings = (PV Cost)A (PV Cost)B (1-14)

Alternative A will generally be the "bare pipe" condition for a new job. However, there may be instanceswhere this is not true. Alternative B will be whatever insulation thickness is under consideration.Therefore, one can calculate savings for each thickness of additional insulation that is greater than thereference thickness. The annual cost equations used to develop the economic thickness model can beused as a guide for the payback period model. To convert a uniform annual cost to a present value,

multiply by the series present worth factor (P/A,i,np) where npis the payback period in years. Realizingthat the insulation cost increment is CI= CI- CRand that Q is the heat savings due to insulation, thenthe present value of the savings can be written as

( )( )

( )

+

+

=

I

P

FPMII

FP

E

fME

E

P

E

n

niA

P

TRiniA

PfTRC

iniA

PHCTRf

n

niA

P

TRmQSavingsPV

,,

,,,11

,,,11

,,

1

0

0

0

(1-15)

The payback period will be the value of np that produces the Present Value of the Savings (PVSAVINGS) equal to zero. In other words, the present value of the heat loss related savings equals thepresent value of the insulation-related costs. Note that all of the tax savings appear as reductions inboth the heat loss related savings and insulation related savings. The equipment and insulation lifeappear only in the depreciation terms. The unknown discounted payback period npappears in thevarious series present-worth factors. Since Equation 1-15 is non-linear in the payback period, sometype of iterative procedure is necessary in order to solve for np. A bisection method for solving non-linear algebraic equations was chosen for this procedure. The method requires an initial guess of thepayback period and is obtained from the simple payback period. For I = i F= 0, both of the seriespresent worth factors in Equation 1-15 are equal to np. Solving for the value of npthat yieldsPV Savings = 0, one obtains

( )( ) ( )

++

=

I

MII

E

fME

E

E

EIP

n

TRfTRC

HCTRf

n

TRmQ

QmCn

111 0

(1-16)

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If income tax, maintenance, and incremental plant equipment are ignored, the simplepayback period reduces to the familiar

HQC

Cn

f

IP

0

=

A problem occurs when the operating temperature approaches the ambient temperature. In somecases, the iteration procedure will not converge. In this case, the program substitutes the simplepayback period. Typically this number is longer than the project life.

It must be realized that equation 1-15, in general, plots as a quadratic. Since a negative payback periodhas no physical meaning, the program sets it equal to zero if it is found to be negative. The discountedpayback period and the simple payback period are very close for situations in which the payback periodis relatively short or the time value of money is relatively unimportant for short time periods.

The payback period generally increases with increasing insulation thickness. This might suggest tryingto minimize the payback period. However, this is an erroneous concept. The payback period conceptignores any benefits that occur after the initial investment has been repaid. Although payback period

has some uses, it should never be used as the sole indicator of how much insulation to apply.

The present value of the fuel savings for the life n Iis often of interest. This quantity can be calculatedfrom Equation 1-15 if replaced by nIand set CI= 0.

Heat Gain Related Costs

Cold surfaces are different from hot surfaces in that condensation control must be considered inaddition to the economic thickness. If the outer surface temperature of the insulation system is belowthe design dew point temperature, then condensation can occur. Since water degrades the thermalperformance of insulation and reduces its expected life, this condition must be avoided. If thecondensation control thickness exceeds the economic thickness, then the condensation control

thickness should be applied. If the condensation control thickness is smaller than the economicthickness, then the economic thickness should be applied. See Section 3 for a discussion about dewpoint calculations.

Incremental Equipment Cost (Cooling): In the design of new systems, the addition of insulation cancause a reduction in the required chiller capacity. The cost model of this practice assumes that thechiller incremental cost varies linearly with the heat gain of the cold line.

Since we are always comparing the total annual cost of two different insulation thicknesses, any coststhat are independent of insulation thickness will simply cancel out in the analysis. Therefore, the chillercost will be the cost of adding an increment in chiller capacity

=

C

c

niAP

QmA

,,

(1-17)

where mcis the incremental cost of increasing the chiller capacity. The computer input will be in theform of $/1,000,000 Btu/hr. For example, suppose that a 100-ton unit costs $70,000 and a 150-ton unitcosts $100,000. The incremental chiller cost will be

( )( ) hrMMBtumc

/

50$

12100150

000,70000,100=

=

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If an absorption type chiller is used instead of an electric driven chiller, then mcshould also include theincremental heating plant cost if applicable.

Energy Cost: For the most general case of an absorption chiller, the efficiency of the heat source (E)and the chiller coefficient of performance (COP) influence the energy consumption. If today's fuel price(Cf0) is inflating at an annual rate of i F, the annual cost is

=

I

FI

C

F

niA

P

iniA

P

COP

QHCA

,,

,,,0

0

(1-18)

Chiller Maintenance: The annual chiller maintenance is assumed to be proportional to the cost of theannual heat gain. Since fuel costs are inflating at an annual rate of iF, the chiller maintenance alsoinflates at the same rate.

=

I

FI

C

FMC

niA

P

iniA

P

COP

QHCfA

,,

,,,0

0

(1-19)

Condenser Make-Up Water (Incremental): The condenser make-up water requirements are assumed tobe proportional to the chiller capacity. Since we are looking at incremental costs, the incremental make-up water costs can be written as

=

I

WI

W

ni

A

P

iniA

P

BQHCA

,,

,,,0

0 (1-20)

The 2000 ASHRAE Systems Handbook recommends B = 3.2 gal/ton-hr for electric driven systems and6.2 gal/ton-hr for absorption systems (6.2 is the default). Note that the water costs are allowed to inflateat an annual rate of iw.

Tax Savings for Cold Surfaces

All maintenance, fuel cost, water cost, and depreciation factors produce tax savings of the effective taxrate times the appropriate costs.

Insulation Depreciation:

( )I

I

n

CTRA= (1-21)

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Insulation Maintenance:

( )

=

I

FI

MII

ni

A

P

iniA

P

fCTRA

,,

,,,0

(1-22)

Chiller Depreciation:

( )C

Cn

QmTRA= (1-23)

Chiller Maintenance:

( )

=

I

FI

E

fMC

niA

P

iniA

P

COP

QHCfTRA

,,

,,,0

0

(1-24)

Energy Cost:

( )

=

I

FI

E

f

niA

P

iniA

P

COP

QHCTRA

,,

,,,0

0

(1-25)

Condenser Water Make-Up Cost:

( )

=

I

WI

W

niA

P

iniAP

BQHCTRA

,,

,,,0

0 (1-26)

The net annual cost is obtained by subtracting the tax savings from the sum of the insulation and heatgain related cost:

( )

( )( )

( )

+

+

+

+

+

=

IWIW

MCIFI

E

f

C

CC

II

FI

MIII

niP

Aini

A

PBHCTR

fTRniP

Aini

A

P

COP

HC

n

TRni

P

AmQ

n

TR

niA

P

iniA

P

fTRniP

ACA

,,,,,1

11,,,,,,,

,,

,,,

1,,

0

0

0

0

0

(1-27)

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The previous equation can be written more compactly as

bQaCA I+= (1-28)

where the factors a and b are independent of insulation thickness.

Discounted Payback Period for Cold Surfaces

The discounted payback period is significant only for the economic thickness condition; if thecondensation control thickness exceeds the economic thickness, then payback period has no meaning.If the reference condition is bare pipe for a new job or heat gain through existing insulation for a retrofitjob, the present value of the savings can be written as:

( )( )

( )

( )

+

+

+

=

I

P

FPMII

WPW

FP

E

fMC

C

P

C

n

niA

P

TRiniA

PfTRC

iniA

PBHCTR

iniA

P

COP

HCTRf

n

niA

P

TRmQSavingsPV

,,

,,,11

,,,1

,,,11

,,

1

0

0

0

0

0

(1-29)

Zero additional insulation thickness may be an acceptable engineering solution for hot surfaceconditions, provided personnel protection is not important. However, for cold surfaces the minimumallowable thickness is the condensation control thickness. If the condensation control thicknessexceeds the economic thickness, then the payback period is not defined. Using this concept, the

present value of the savings can be written as:

( )( )

( )

( )

+

+

+

=

I

P

FPMII

WPW

FP

E

fMC

C

P

C

n

niA

P

TRiniA

PfTRC

iniA

PBHCTR

iniA

P

COP

HCTRf

n

niA

P

TRmQSavingsPV

,,

,,,11

,,,1

,,,11

,,

1

0

0

0

0

0

(1-30)

where CI= CI-CIR. If the reference condition is bare pipe then, CIR= 0 and Equation 1-30 reduces toEquation 1-29.

In some cases, some thickness may have zero payback periods! At first glance this may seem toindicate that all of the thicknesses will pay for themselves immediately. In reality, by defining the

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discounted payback period as we have, we have allowed a mathematical quirk to occur. There may becases where the correct value for nPin equation 1-28 is actually negative. This quantity then has nophysical meaning, but is mathematically correct. If this occurs, the computer program automatically setsthe payback period to zero. A negative value for nPcan be caused when the costs associated toenergy are much greater than the costs associated with insulation. It must be realized that equation 1-30 plots as a quadratic. The relative costs of energy and insulation will determine where the curvecrosses the time axis.

Section 1 List of Symbols

nE Life (or remaining life if retrofit) of heating plant, yearsmI Life of insulation, yearsQ Heat loss/gain, Btu/hr-ft, or Btu/hr-ft2mE Dollar investment in adding a 1,000,000 Btu/hr increase in physical capacity, $/1,000,000 Btu/hrCf0 Present day fuel price, $/1,000,000 Btuif Fuel inflation rate, (% per year)nE Heat conversion efficiency in plant, %H Annual hours of operation, hoursfME Percentage of fuel cost that is spent each year for physical plant maintenance, %

I After tax rate-of-return, %TR Effective income tax rate, %a,b Constants independent of insulation thicknessQ Heat savings due to additional insulation, Btu/unit area

CI Increment in insulation cost, $/unit areanp Payback period, yearsA Annual CostB Water operated condenser factor, gal/ton-hrCI Initial installed cost of insulation, $/ft or $/ft

2COP Coefficient of performance of the refrigeration systemCw0 Present day cost of condenser make-up water, $/MgalfMI Fraction of initial insulation cost spent each year on insulation maintenance, %iw Inflation rate of the cost of condenser make-up water, %

mc Dollar investment in adding a 1,000,000 Btu/hr increase in capacity of the chiller, $/MMBtu/hr(A/P,i,nI) Capital recovery factor(P/A,i,nI,iF) Series present worth factor accounting for inflation(P/A,i,nI) Series present worth factor not accounting for inflationPV Savings Present Value of Savings, $

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For those readers unfamiliar with the series present worth factor, it can be thought of conceptually asan "effective life" that considers the time value of money. If there is no inflation or time value of money

nnA

Pn

A

P=

=

,0,0,,0,

0

(2-7)

In general,

nIniA

PF

,,,

0

(2-8)

for all values of I. For the inflating series present worth factor,

iiforniniA

PFF

,,,

0

(2-9)

or

iiforniniA

PFF

,,,

0

(2-10)

Thus, the present value of fuel costs is converted to a uniform annual amount by dividing by the capitalrecovery factor or equivalently dividing by the series worth factor.

=ni

A

P

iniA

P

CA

F

,,

,,,0

0 (2-11)

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Section 3 Heat Transfer Equations

The equations used for calculating heat transfer and various temperatures in 3E Plus are presented inthis section. The equations are based on ASTM C 680 with modifications to the calculation of surfaceconvection. These modifications agree with practices commonly used, and its important to note thatthe modifications are currently under review by ASTM Task Group TG-5.2.

Background Information

Heat transfer through insulation on flat systems can be calculated using

( )( )s

AP

RR

tt

A

Q

+

= (3-1)

where Q/A (also known as q) is the heat flux or heat transfer per unit area, tPis the processtemperature, tAis the ambient temperature, R is the thermal resistance of the insulation(s), and Rsis theouter-surface resistance. R and Rscan be defined as

o

S

n

j j

j

hR

k

xR

1;

1=== (3-2)

where xjis the insulation thickness of layer j, kjis the apparent thermal conductivity of insulation j, andhois the outer-surface conductance that combines the effects of radiant and convective heat transfer.NOTE: it is assumed that the surface resistance on the process-temperature side is sufficiently small incomparison to the other resistances, so it is excluded. In addition, the program includes the conductiveresistance of the equipment/pipe, and that resistance is used in the calculation of both bare andinsulated heat transfer.

For further discussion of the conductive heat transfer through the insulation, please consult ASTM C680, since the program is based on ASTM C 680. Where the program deviates from ASTM C 680, is inthe calculation of ho. As stated earlier horepresents the outer-surface conductance, which is a

combination of radiant and convective heat transfer, so

cro hhh += (3-3)

The radiant component, hr, is calculated using the Stefan-Bolzmann equation:

os

osr

TT

TTh

=)( 44

(3-4)

where

= effective surface emittance between outside surface and the ambient surroundings,dimensionless; = Stefan-Bolzmann constant, 0.1714 10-8Btu/(hft2R4) (5.6697 10-8W/(m2K4))

Ts= absolute surface temperature, R (K); and,To= absolute surroundings temperature, R (K). The program assumes that the ambient air is thesame temperature, so Ta= To.

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The convective component, hc, is calculated based on dimensionless groups defined for fluids incontact with solid surfaces.

The dimensionless numbers are:

Nusselt,f

cL

k

LhNu = or

f

cD

k

DhNu = ; (3-5)

Rayleigh,f

p

Lk

LTcgRa

=

3)(or

f

p

Dk

DTcgRa

=

3)(; (3-6)

Reynolds,

VLL=Re or

VDD =Re ; (3-7)

Prandtl,f

p

k

c=

Pr ; (3-8)

where,L = characteristic dimension for horizontal flat surfaces (hydraulic length), vertical

walls (height) and vertical cylinders (height).D = characteristic dimension for horizontal cylinders (diameter) and spheres (diameter).cp = specific heat of ambient fluid, Btu/(lb.R) (J/(kgK))

hc

= average convection conductance, Btu/(hft2F) (W/(m2K))kf = thermal conductivity of ambient fluid, Btu/(hftF) (W/(mK))V = free stream velocity of ambient fluid, ft/h (m/s)

= kinematic viscosity of ambient fluid, ft2/h (m2/s)g = acceleration due to gravity, ft/h2(m/s2)

= volumetric thermal expansion coefficient of ambient fluid, R-1(K-1)

= density of ambient fluid, lb/ft

3(kg/m

3)

T = absolute value of temperature difference between surface and ambient fluid, R (K).

It needs to be noted that the above fluid properties must be calculated at the film temperature, T f, whichis the average of surface and ambient fluid temperatures. For this practice, it is assumed that theambient fluid is dry air at atmospheric pressure. The properties of air can be found in references suchas Tables of Thermodynamic and Transport Properties of Air... inNBS Circular 564 U.S. Dept. ofCommerce by Hilsenrath, et al.

Also, for each geometric shape and surface orientation the overall average Nusselt number is to becomputed from the average Nusselt number for forced convection and the average Nusselt number fornatural convection. The relationship is

( ) ( ) ( )j

n

j

f

j

NuNuNu += (3-9)

where the exponent, j, and the constant, , are defined based on the geometry and orientation.

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Flat Surface Convection Calculations

Forced Convection From Heat Transferby Churchill and Ozoe as cited in Fundamentals of Heat andMass Transfer(p. 354) by Incropera and Dewitt the relation for forced convection by laminar flow overan isothermal flat surface is

4/13/2

3/12/1

,

]Pr)/0468.0(1[PrRe6774.0

+= LLfNu 5105Re

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Section 4 Dew Point Calculation

The dew point for a mixture of air and water vapor is defined as the temperature where the water vaporcondenses when cooled at constant pressure. The dew point can easily be determined from therelative humidity and the design ambient temperature. The relative humidity () is the ratio of the partialpressure of the water vapor as is exists in the mixture (P v) to the saturation pressure at the sametemperature (Pg).

g

v

P

P= (4-1)

The variation of the saturation pressure with temperature must be determined experimentally. Aconvenient empirical relationship is the Goff equation

( ) ( )

( )( ) 2195983.21101042873.0

101100574.102802.5179586.10

176955.43

1

129692.8

4

+

++=

LOGPLOG (4-2)

where P is the vapor pressure in atmospheres and = 273.16/T with T in K. This relationship is validover the temperature range of 58F to 212F. In terms of temperature (F):

( )

+

=16.27332

9

5

16.273

T

(4-3)

The calculation procedure is as follows:

1. Use equation 4-2 to determine Pgcorresponding to ta

2. Use equation 4-1 to determine Pv

3. Solve equation 4-2 for dpusing an iterative solution for non-linear equations.4. Solve equation 4-3 for Tdp

When solving step 3 a variation of Newtons method is used to accelerate convergence. Convergenceoccurs rapidly, because log P is nearly a linear function of temperature of the practical range of interest.

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Page 19 of 21

Section 5 Apparent Thermal Conductivity Equations

The thermal conductivity file supplied with the 3E Plus program includes the apparent thermalconductivity equations of several generic types of thermal insulation. These equations are derived fromASTM material specifications.

Material and

Equation

Maximum Use

Temp. (F)

450F M F BOARD ASTM C612-00a T1B274 10404277.710426568.42208473. mm ttk ++=

450

850F M F BOARD ASTM C612-00a T2264 10064601.110292883.4215392. mm ttk ++=

850

1000F M F BOARD ASTM C612-00a T3264 10408288.110572084.22301582. mm ttk ++=

1000

1200F MF BOARD ASTM C612-00a T4B274 109.361318101.073707.2247394 mm ttk ++=

1200

1800F M F BLOCK ASTM C612-00a T5

274 102.618991107.166724.1928559 mm ttk ++= 1800

450F M F BLANKET ASTM C553-00 T2264 102.487349103.556493.2694291 mm ttk ++=

450

850F M F BLANKET ASTM C553-00 T4264 101.276007103.1274.2230754 mm ttk ++=

850

1000F MF BLANKET ASTM C553-00 T5264 102.952804102.522329.2739618 mm ttk ++=

1000

1200F MF BLANKET ASTM C553-00 T7264 101.611744101.312481.2396443 mm ttk ++=

1200

MF Metal Mesh BLANKET C592-00 T2

274 107.023157104.691053.2053742 mm ttk ++= 1200

650F Min. Fiber P&Tank C1393-00a264 101.6666110822843.2.2448986 mm ttk ++=

650

850F Min. Fiber P&Tank C1393-00a274 106.694144105.229674.2290639 mm ttk ++=

850

1000F Min.Fiber P&Tank C1393-00a274 103.028062105.297851.2414944 mm ttk ++=

1000

850F Mineral Fiber PIPE C547-95264 101.142861102.94283.2080004 mm ttk ++=

850

1200F Mineral Fiber PIPE C547-95274

106.785715103.17857.2142858 mm ttk ++= 1200

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Page 20 of 21

Material andEquation

Maximum UseTemp. (F)

CalciumSilicate BLK+PIPE C533-99284 107.142491104.585749.3554278 mm ttk ++=

1200

Perlite BLOCK+PIPE ASTM C610-99 274 102.678513104.246484.451856 mm ttk ++=

1200

Melamine PIPE+FLAT ASTM C1410-98264 101.591361102.921287.2691389 mm ttk ++=

350

Phenolic SHEET+TUBE C1126-98 Gr2264 101.28389103.870224.192792 mm ttk ++=

257

Polyolefin SHT+TUBE C1427-99a G1284 108.750668103.254081.3281455 mm ttk ++=

200

PIR(-50 to 50F MEAN)ASTMC591-00212108.583068.0002-.18 mm ttk =

50

PIR (> 50F MEAN) ASTM C 591-00 274 104.293448105.858669.1405089 mm ttk += 300

1.6#Polystyrene SHT ASTM C578-99274 101.390192104.060314.1686797 mm ttk ++=

165

Elastomeric SH+TUBE ASTM C534-99274 103.599066102.990193.277474 mm ttk ++=

220

MF Metal Mesh BLANKET C592-00 T2274 107.023157104.691053.2053742 mm ttk ++=

1200

Cellular Glass PIPE C552-00 Gr 2264 101.000015104.399919.3550009 mm ttk ++=

800

Cellular Glass BOARD C552-00 Gr2274 107.905024104.738999.3117792 mm ttk ++=

800

MF Insulat'g CEMENT ASTM C195-00274 107.252833101.060789.6976953 mm ttk +=

1900

Insul+Finish CEMENT ASTM C449-00263 103.749988103.499989.3500023 mm ttk +=

1200

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Section 6 Reduction of CO2, NOx, and Carbon Equivalent (CE)

3E Plus calculates heat transfer for a given system. From this heat transfer, the program calculatesgreenhouse gas emissions by multiplying the heat transfer and an EPA factor for a given fuel. NOTE: Ifthe user selects OTHER as the fuel type, the emissions are not calculated because the EPA factor isunknown. Also, when the fuel choice is electricity, the program adjusts for delivered electricity ratio bymultiplying Q times 2.95. The equation used to calculate emissions is

( )Efficiency

ValueEPAQEmission

*000,000,1

*= (6-1)

where Q is in Btu/hr/unit, and EPA Value is in lbs./Btu. The table below presents the EPA Values forCO2and NOx. Carbon Equivalent is calculated based on its molecular weight ratio to CO2. The ratio is12/44, so

EmissionsEmissions COCE 2*44

12= (6-2)

Energy Source CO2 Emissions NOx Emissions

Natural Gas 109 lbs. per MM Btu 0.23369 lbs. per MM Btu

Fuel Oil 164 lbs. per MM Btu 0.266759 lbs. per MM Btu

LPG 137 lbs. per MM Btu 0.21 lbs. per MM Btu

Coal 207 lbs. per MM Btu 0.570997 lbs. per MM Btu

Electricity 155 lbs. per MM Btu 0.338189 lbs. per MM Btu

Year 2000 Data outsourced from GPRA data call by ADL for DOE and EPA