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Considerations on Real Gas Expansion: Otis P. Armstrong P.E., -Dec/1. 2011 r4

Abstract:This topic is on effects of friction in real gas expansion, called non-reversible conditions.

The subtlety of this effect can be masked by the accompanying temperature decrease of

gas expansion. Use of Clausius Equality (Entropy) can properly account for friction in gas

flow. Many methods overlook the correct accounting for friction.

It is irrefutable: friction produces heat, as shown on the following H-S diagram. A decrease

in gas expander mechanical efficiency is accompanied by a subsequent increase in outlet

temperature. A result of friction between gas and machinery surfaces. This is noticed on a

H-S diagram by a counter-clockwise rotation away from the minimum isentropic

temperature towards isothermal operation. Also for compression, friction will increase

outlet temperature, a result of reduced efficiency, the same is true for nozzle flow.

Likewise for valves: gas friction increases the outlet temperature. This is noticed on the

H-S diagram by a counter-clockwise rotation away from the minimum isentropic

temperature towards the Free Expansion temperature. The Joule or Free Expansion

temperature is lower than isothermal expansion temperature and proportional to the

Joule Coefficient, J.

Many thermodynamic concepts of gas expansions are just idealization concepts. Actual

equipment shall always have friction effect, which will produce heat. For any real gas

expansion, generation of heat by friction produces a temperature increase.

Temperature and pressure equations for Isothermal, Isenthalpic, Constant Internal Energy,

and Isentropic gas processes are presented. The basis for these processes are the Joule-

Thompson , JT, and Joule coefficient , J, of real gases. Calculations for, and uses of, JT

& J are detailed in this discussion. HTML calculation widgets are embedded in the .ppt

document for the readers utilization.

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Summary, Recommendations & Calculation MethodsA General Energy Balance is suitable for prediction of Real Flow Phenomena thru valves, pipes, Turbines,

Compressors, pumps, and heaters. This general energy balance greatly simplifies the explination of

thermodynamics for either compressible or incompressible fluids into a unified concept.{Ws + dP/ +VdV/g + [Pd(1/) + U] - (Q)h ≡ 0} or (U+(V2/2g)/J)1+(Q-W)=(U+(V2/2g)/J)2+(PV)+ F.

Friction, F, may be determined by difference, by losses definition, or by (f L/D)(G/)2/2gJ, the details are

outlined on pages 14 and 15, plus other discussions. The term fL/D is also the Crane K factor for flow

apertures. Also Crane K, (TP410) is related to valve coefficient as detailed on page 14. Details for skin

friction are: K f L/D = (29.84d2/CV)2 = (1/Kd)2 = 1/ = 1/(n)

The Clausius Equality {F=(TS-Qh)} is correct accounting of frictional heat, for either gas or liquid flow with

friction. Dissipation of energy by Friction to heat always increases temperature relative to a reversible

process. A correct heat balance must always calculate zero heat addition for Both a Joule Expansion {dU=0}

and a Joule Thompson Expansion, {dH=0}.

The determination of dH & dU to include pressure correction is easily implemented by the temperature

change from Joule or JT coefficients. Methods to evaluate these coefficients and associated temperature

change are presented in detail. U =Cv(T - TJX) & H = Cp (T - TJTX) : The C‟s are low pressure heat

capacity. The temperature sign convention is easily remembered. For at either U or H = 0, the

temperature change will be negative for all but 3 quantum gasses, H2 for example. This is detailed in the

section “Inversion Point”. These rules offer simple EOS thermodynamic consistency checks for simulation

work.

The Schultz method, combined with the JT coefficient greatly simplifies evaluation of gas compression and

expansion : T2 = T1{P2/P1)(m)} m=(Z/Cp)() & =(+X)expansion =(1/ + X)compress : The term X is related to the

JT Coefficient, in Rankin/atmosphere, Cp in BTU/#/F, and density in #/CF as: X=(0.37Cp) . The details are

on page 40. Compressor head and Power are determined by use of real gas enthalpy change, as explained

above for total energy balance. A complete section is detailed in the appendix.

The ratio of Joule Expansion temperature change to Joule Thompson Expansion temperature change is the

ratio of specific heats: (TJTX/TJX) = (Cv/Cp). The temperature drop associated with a Joule Free expansion

is greater than a JT, the difference may be attributed to friction effect.

Black Powder is found in raw gas pipelines and sales gas pipelines. Installation of high friction gas valve

requires immediate upstream gas filter to prevent black powder plugging, unless vendor warranty otherwise.O P Armstrong PE/Dec.2011 printed Feb 2013

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IntroductionDiscussion with some engineers indicated a lack of grasp

on the heating effect of friction in gas flow. Additional

review of literature showed a lack of presentation, that

dissipation of energy via friction produces heat,

irrespective of the medium in which friction is dissipated.

The TS diagram for gas on right depicts friction effect in

either compression or expansion. It shows that friction

increases outlet temperature in both cases.

Even-so, some engineering routines perpetuate the cooling myth by misleading statements like: “in

liquids friction produces heat but for gasses friction is converted to “internal energy”‟, hello?? Just

what is heat? Followed by such statements as “was any heat added? Or was any work done by the gas”.

This thought is routinely & incorrectly expressed by: (H+V2/2g/778)1 = (H + V2/2g/778)2 ” .

Some process simulator programs do not take into account details of mechanical design. Thus the myth

that friction pressure drop in gases does not heat a gas is routinely perpetuated. True: a normal gas

expansion cools, but friction in flow apertures reduces the amount of cooling. Why?, because friction

acts to reduce the output of useful work by energy dissipation to heat production. Heat generation

always acts to increase temperature.

Here is a typical quote from a web page: “flow in oil pipelines acts to heat oil, where-as flow in gas

pipelines acts to cool the gas.” It is NOT flow friction which acts to cool the gas but rather the

expansion resulting from pressure reduction. In BOTH cases, flow friction acts to produce heat. It is

just that the heating effect of flow friction in gas is masked by the normal cooling effect of gas

expansion.

A field test with a high friction valve demonstrated that friction in gas flow acts to counter the

expansion cooling effect. The field trial IR temperature measurements were personally supervised.

Results of this trial showed the valve outlet temperature exceeded the temperature determined by

simple JT expansion. Methods to quantify this effect are presented in this discussion.

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Introduction, c‟td:Dissipation of work or momentum via friction will produce

heat, irrespective of the medium in which friction is

dissipated. The TS diagram on right shows friction effect in

either compression or expansion. Friction increases outlet

temperature, irrespective of expansion or compression.

The result of friction or “irreversible” losses causes the

outlet temperature to INCREASE to point 2‟. This is above

the “reversible” expansion temperature, point 2.

There are an accumulation of routines to perpetuate the cooling myth of gas friction. Looking at friction

on a molecular level, a gas molecule which losses momentum to a stationary object, gives up translation

energy. By conservation of energy, this organized momentum energy is translated to un-organized thermal

energy, so dS must increase. As correctly depicted in the above diagrams. The Clausius inequality for any

process involving friction requires: d(losses)>(Tds-d(Q)h). Likewise for flowing gas, Momentum lost to

intermolecular collision, turbulence, stationary walls of either a containment pipe, a valve flow orifice,

or a thermometer yields a temperature increase.

The thermometer correction factor based on the fluid velocity is: [T.bulk =T.tw-0.95V2/(778Cp2g)]. The

reading of a thermometer inserted into a pipe with flowing fluid is high by the above correction. Why?

Because momentum or irreversible energy loss happened, which locally heats the thermowell. This is

termed stagnation temperature. Friction effect of stagnation temperature requires high velocity aircraft

to need special leading edge materials. Concord jet nose cone was built at the limit of aluminum

materials for the achieved velocity. For a gas flowing inside a tube, the boundary layer volume to total

expanded volume is a small fraction, (momentum transfer efficiency). The small ratio of boundary layer

volume to a much greater bulk volume gives a net decreases in bulk of temperature due to gas volume

expansion. The effect of friction in gas is not a special case but just the nature of friction. Friction always

produces heat, irrespective of the involved media. Some practical aspects are: increase flare line

momentum loss to maximum limits to minimize cooling, improve SRV‟s by use of high turbulence valves to

gain maximum possible temperature in flare flow lines.

5

H-S C3=

Unzip compressed

folder to storage.

JT calculator is the

HTML file & executes

in mobile or other

browser with java

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Field Trial: High Friction Gas Valve Temperature drop compared to Isenthalpic Performance

The percent efficiency is T actual, divided by T isenthalpic, times 100. A valve with

zero efficiency is a constant internal energy valve.

Note: the BWR JT temp is 465R or u.jt of 1.08, R/atm.

A pipeline rule of thumb is 7F/100psid, by this rule the high friction valve out

performed the conventional rule. By the 7/100 rule the outlet T calculates to about 8F.

Small ports are required to obtain

high skin friction losses. These ports

can result in valve trim plugging or

erosion for dirty gases. Port plugging

during this trial indicated gas filters

were required due to dirty gas.

Looking at long term service, filter

costs may prove economical, if

outlet gas heat is of value or if

hydrate inhibitors are used.

Test Results show a 70% increase in

outlet temperature compared to JT

isenthalpic Performance. This is a

result of friction heating. One

method to create friction heating is

to increase turbulence by square

edge port trim.

prop unit viral R K actualW/m w ac entric 0.0720 20.3000 Hi

C p btu/mol/R 11.6 11.6 F ric tion

Tc K elvin 213.9 213.9 Valve

P c Pc ,atm 45.2 45.2 70% eff

T in T1Rankine 520.0 520.0 30% dH

P in P1atm 62.3 62.3 to heat

P out P2atm 11.2 11.2

ujt.c lc Rankin/atm0.823 0.769 na

T2 jt.c lc Rankin 478.0 480.7 492

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Nozzle Efficiency & Gas HeatingMany engineering routines do not acknowledge friction

heating in gasses. Instead placing emphasis on idealized

thermodynamic conditions. Such ideal conditions are not

meet in practice.*

As shown to right, effect of friction is to heat gas. Step 1 to

2 is the idealized thermodynamic temperature drop of a

nozzle. Step 1 to 2‟ is actual temperature drop. Nozzle

efficiency is given by Faires, p406:

n = (T1 – T2‟)/(T1-T2) =(kd)2

For the gas valve discussed above these calculate as:

T1=60, T2=19F, T2‟=32F which is n =(60-32)/(60-19)=0.68 & kd=0.68=0.83

Where T2 is the ideal isentropic temperature.. T2‟ is actual outlet temperature. Kd is coefficient of

discharge. Since kd is tabulated for many physical configurations it is possible to calculate actual outlet

temperatures based on equipment type and basic thermodynamic properties. The following calculations

detail this concept. The term, kd , may be correlated to either a Cv or to Crane K, fL/D.

•Here is typical idealization from an engineering text (in courtesy, no reference to source): discussing flow thru a metering

nozzle situated inside a constant diameter pipe as an idealized throttling process. „the discharge from the nozzle, with it‟s

high kinetic energy, swirls about… and dissipates it‟s KE, thus KE is reconverted to internal energy and Ws=-H=0.‟ This is only

true in pedagogical circles. For practicing engineers, permanent pressure losses are the reality of actual operating

equipment. Because not all velocity head is converted back to pressure head, some velocity head is dissipated to what are

termed „permanent pressure losses‟ to be lost as low grade thermal energy. The engineering equation that correctly describes

a real process is: [H + V2/(2gJ)]1= [H + V2/(2gJ)]2 + F. Where F are friction losses. The pedagogical concept focus on JT cooling

neglects details such as, friction heating, operational losses, and a realistic thermodynamic frame work.

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Determination Fluid Temperature ChangeJT or Isenthalpic Expansion : T =(CpJT)P, H =0 & if P2<P1 then T2<T1 & S>0 no work adiabatic

Joule Free Expansion: T = (CpJ)V , U =0 & if V2>V1 then T2<T1 no work adiabatic

Isothermal Temperature: T = 0 , PV=C & if dQ >0 T2=T1 & S>>0 & dQ by Clausius Equality

Polythermal Temperature: T 0 , PV=ZRT & dQ either >0 or <0 T2=?T1 & S varies & no work

Isentropic Temperature: T = T1((P2/P1)(1-1/k) –1), S =0 & if P2<P1 then T2=T1((P2/P1)(1-1/k) & <T1

Polytropic Temperature: T = T1((P2/P1)m –1), S >0 & T2 = T1{P2/P1)(m)} m=(Z/Cp)() & X=(0.37Cp)

(1.986Z/Cp) Cp in molar units, Cp is Btu/CF/Rankin , is JT in Rankin/atm, =(+X)expansion =(1/ + X)compress

Each of the above process have specific properties and energy balance requirements. One must be careful

to apply process conditions that conform to assumptions stated. In other words, compare apples to apples.

Conditions for the Polytropic process are properly determined by the Schultz Method. A review of Schultz

method showed for limiting cases of: a) efficiency equals zero, it reduces to the Isenthalpic Expansion Joule

Thompson expansion and for b) efficiency of 100% it is an ideal gas Isentropic expansion. Both the

Isenthalpic and Isentropic Processes require adiabatic processes.

Adiabatic flow equation derivation uses premise of H0 ={H+ V2/2gJ} which is only true for frictionless flow.

This article advocates that where irreversibility is present H0 ={H + V2/2gJ – F}. Derivations omitting

frictional effects then proceed to mix apples (frictionless energy balance) against energy balance with

friction, oranges. With advent of numerical computation such assumptions of isentropic energy balances

should be dropped. The replacement is to break pipe segments into smaller parts and numerically integrate

the correct energy balance of {H + (G/)2/2gJ}(I -1) = {H + (G/)2/2gJ – dL(f/D)(G/LM)2/2gJ}(I) and

momentum balance: {P/ + (G/)2/2g}(I -1) = {P/ + (G/)2/2g – dL(f/D)(G/LM)2/2g}(I).These equations are

correct only for no heat addition and no shaft work. A classic example would be initial flow inside a flare

relief line where transit time minimizes heat gain from pipe metal sensible heat.

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Pressure loss of valves may also be defined by CV, which at a given valve size may be defined in terms of K,

as given in Table at column 2. This slide demonstrates that frictional resistance, whether expressed in

form of Crane K, CV, or Co, or efficiency are all identical forms of irreversible friction process which

produce heat. Refer to Perry 5th Ed. Page 5-18 “No friction term occurs in this expression since friction

represents a conversion of mechanical energy into HEAT”.

Caution: Knowledgeable texts on friction place emphasis on manufacturer rating of equipment. One reason

for such caution is different types of energy dissipation. The above deal only with skin friction, other

forms are vortex &/or centrifugal. For example the Twister process which gives more cooling than JT by

mitigating friction heating.

Valve Flow Coefficient, CV

CV,gpm(SG/psi), & Crane pressure loss is 144(psi)/ =KV2/2g, Equate psi for CV, & Crane K, using

water density of 62.4#/CF & SG= /62.4 & gpm=2.45Vd2. When making these substitutions & eliminate the

density, SG, Velocity, & flow terms one arrives at CV,=29.84d2/K or K= (29.84d2/CV)2. The constant for the

Crane formula is 29.9, which the reader may back calculate the density is nearly the 62.4#/CF used for

this coefficient calculation. In summary Crane CV, relation is not an empirical equation. If skin friction is

accurately represented by fL/D= Crane K, so also is relation between CV, and Crane K.

Gas Valve Flow Coefficient

The CV, of gas is represented thusly: CV,=gpm(SG/psi) & #/hr =gpm/(500SG), (#/hr)/(500SG)*(SG/psi)

& SG=/62.4 & =1/, place SG inside the root radical leads to CV=(#/HR)(1/500){(/psi)62.4}

=#/HR(7.9/500)(/psi) = (pph/63.3) (/psi), where is average CF/# for gas. Valve books have

extended formula for choked flow. For friction, the above is proposed to best rep‟s total energy concept.

Summary: K= fL/D = (29.84d2/CV)2 = (1/Co)2 = 1/ = 1/(n)

Revised slide

Friction losses for valves, throttling processes and many flow apertures are well documented, Crane

TP410, and easily calculated. The Crane method defines friction loss as multiple of velocity head, the K

method. Where Crane‟s K is product of friction factor and equalivalent length divided by diameter, a

dimensionless number. The below table gives some examples.

Friction Loss of Flow Apertures Summary: K f L/D = (29.84d2/CV)2 = (1/Kd)

2 = 1/ = 1/(n)

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Appendix List

1. FRICTION2. What are Joule & Joule Thompson Coefficients?3. Joule Thompson Coefficient Inversion Point4. Summary of Equations for Joule Coefficient5. Joule Coefficient by VdW EOS –limited utility6. Joule Coefficient by RK EOS7. Joule Coefficient by Viral EOS8. Joule Coefficient by BWR EOS9 Summary of Equations for JT Coefficient, JT

10 JT EOS COMPARE RESULTS11 Van der Waal’s Simplified JT Coefficient12. Van der Waal’s Exact JT Coefficient13. JT by the Two Term Viral EOS14. Real gas JT coefficient as function of Z15. Forms & Solutions of Redlich Kwong EOS 16 Determine (Z/oR) by Redlich Kwong h

method17. Determine (Z/oR) by BWR Reduced Density

EOS18. Energy Balance Summary19. Clausius Equality & Energy Balance20. Clausius Equality & Friction Heat21. Energy Balance Corrected for Friction22 Demo of Energy Balance for Liquid Pump, a. b23. Nozzle Efficiency & Friction I & II24. Gas Expansion Energy Balance SvN EX10-4

Corrected for Frictional Heat, 3 pages

25. General Enthalpy equation of Compression, Expansion, & Flow Apertures26. Real Gas Enthalpy change by JT coefficient, JT

& pressure change:27. Pressure Effect on dH by Cp correction vs. use of Joule Coefficient.28. Review of Polytropic Head Calculation Methods 7pp The Schultz Method29. Express = Cp-Cv for real gas as z = Cp-Cv

30. Compare Gas X Calculation Methods 131. Compare Gas X Calculation Methods 232. Expansion of N2 Energy Balance for Turbine33. Cross Correlations Coefficients & Properties 134. Cross Correlations Coefficients & Properties 235 Corresponding States as related to EOS errors36 Kinetic Gas Theory Review37 Dimensionless Numbers for Kinetic Gas Theory38 Dimensionless Numbers of Gas Expansion39 Kinetic Gas Theory & Friction Energy Balance40 Properties of Common Gasses41 Review of C3= Gas Properties42. Review of differential forms & Math used here: 43. JX & JTX by Berthelot EOS44. dT(JX vs. JTX) is Cp/Cv

45. Field Trial Details: MultiStage Hi Friction Valve46. Black Powder in Raw Gas Pipelines47 Summary

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Friction is not fiction, nor is it imaginary. Friction is the law of the universe: you can‟t win, you can‟t

break even and you can‟t get out of the game. A frictionless device is parlance for a non-existent Rube

Goldberg perpetual motion device. Ditto for the words: ideal, reversible, and isentropic. Such equipment

is a thought only. All motion is associated with friction. Some friction is hardly noticeable, sonic wave.

Extreme friction is dangerous! Friction is degradation of organized energy to disorganized energy, aka low

grade thermal energy. Father Friction is beneficial in many ways: it keeps our feet & other transport

means from sliding out of control. It also reduces impacts when control is lost. But for all this, Father

Friction, extracts a price of additional exerted energy. Motion has resistance which is mostly friction. It

will be present until universal kinetic and potential energies are reduced to a uniform state of motion.

Friction is lost work, W. Where work is, there also will be friction. Friction is work lost to heat. These were

the observations of Joule and Rumford. About 1850, Mr. Joule quantified the equality of work and heat as

778ft#=1BTU. A few BTU = lots of work. The work of getting into the bath tub will not substantially change

water temperature but, a few degree of extra heat will quickly end a bath. Friction is associated with

increase in entropy, dS. A P-V process or cycle with zero Entropy, S, change is isentropic aka reversible

adiabatic, whilst the isenthalpic (H=0) process is between has positive S. The most probable state has

the highest positive S. The word adiabatic is another ideal concept, meaning zero heat transfer. All

isentropic changes (dS=0) must be adiabatic, but not all adiabatic changes need follow the dS=0 path. The

H-S chart for C3=, presented here, is constructed for adiabatic enthalpy changes but only the 100%

efficiency line is also isenthalpic. An adiabatic friction process has Q=0, as given by the C3= H-S chart.

Friction consumption is rapidly determined at various efficiency on such a chart as dH @e less dH

@isentropic. The math of friction is as follows: Entropy always increases:

TS>Q & TdS = (dU + PdV) = (dH – VdP) & Friction = (TS - Q)

eS=(T2/T1)Cv(1/2)

R = [(T2/T1)(1/2)(k-1) ]Cv = [(P2/P1)

k(1/2)(k) ]Cv = [(T2/T1)

k(P2/P1)(1-k) ]Cv

Summary: FFriction, Turbulence, & Work are identical energy forms (778ft#/BTU aka Joules’ Stirring

experiments) because Friction and Turbulence Always consumes Work.

Friction

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What are Joule & Joule Thompson Coefficients?This question is asked to better apply the (JT). It may appear to define gas flow processes. Internally, (JT)

is a property of the gas state. When the gas state is effectively modeled, it has both attraction and

repulsion terms. The cooling effect of gas pressure reduction arises when the gas molecules are dominated

by attractive force. As the gas expands, work is done to overcome the attractive forces and cooling is

effected from energy loss to overcome the attractive forces. The opposite is true if repulsive forces dominate.

The dimensionless Number, CpJT , is the ratio of gas expansion temperature change to friction

temperature change, for dP & JT 0, as TF = PF/(J)= TJT = (JTCp)P. Which is true only for

Joule Thompson Expansion: 0H=CpT+(CpJT)P. While for Joule Expansion: 0U=CvT+(CvJ)V’

A JTX is the isenthalpic case. Article 10 found Temperature relation for the general isentropic case as:

Ti@e=1 + TF = TJT, So if the expansion is without friction then, it is also isentropic. If TF is maximum, the

temperature of an isenthalpic expansion, identical to the JT Temperature. The Joule & Thompson

experiment was devised, circa 1850. Under conditions of an equally isolated and insulated Joule free

expansion friction dissipation caused heating on the B sphere just equal to the cooling of the A sphere. The

net result for the free expansion was temperature decrease too small for Joule to detect. Joule and Lord

Kelvin devised the JT experiment with a porous plug. The ratio of Joule Thompson expansion dT to the ratio

Joule Free expansion dT is proportional to Cv/Cp. On the C3= H-S diagram, the adiabatic JT Expansion from

an initial pressure of 125Psia to 62.5psia at JT Temperature has a zero dH with no heat added. A Joule

expansion follows a constant internal energy line. A Joule or Free Expansion is not an isothermal

expansion. Lord Kelvin & Joule both agreed Joule‟s device (& gas generally) was subject to frictional

heating. C W Smith 1977 Kent, Review of Lord Kelvin‟s Cambridge Papers

JT Expansion dT -5F 22/11atmDevised to eliminate friction heatingJoule Free Expansion dT -7F, if no Friction

B4: A @ 22atm: After A & B @

11atm. Inside copper spheres

was air, surrounded by an

insulated water jacket Va=Vb

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Joule Thompson Coefficient Inversion Point

Another phenomena with gas is the (JT) inversion point. This is the temperature at which a quantum

change happens so the dominate force between gas molecules is repulsive. Inversion temperature for

various gases at 1atm are as follows: He, 40K; Ne; 231K, air 659K, N2, 621K; and O2, 764K. When a gas is

expanded at P & T above the inversion point then, an expanded gas, is a heated gas. When a gas is

expanded below the inversion point, an expanded gas is a cooled gas. Operation below the IP is the basis of

the Linde air separation plant. Operation below the IP produces an important class of fuels, LPG and LNG.

Operation at cryogenic conditions requires very low levels of tertiary gases; H2O, CO2, H2S, to prevent gas

hydrate formation in the equipment.

The phenomena of expansion is unrelated to friction heating. Just the opposite, careful measurement is

needed to eliminate friction heating to accurately measure (JT). Mr. Joule‟s original experiment was in

reply to prior experiments which placed 2 spheres (1 of pressured gas other evacuated) connected by

stopcock into a water bath and found no change in water temperature upon opening stopcock between

spheres. Mr. Joule re-made the experiment by placing each sphere in individual water baths. After again

equalizing pressures found the water bath on pressure side had a decrease in temperature just equal to

temperature increase of the evacuated sphere‟s side. Joule & Thompson revised the experiment using

flowing gases and a highly compressed cotton plug between 2 thermometers. The porous plug reduced flow

enough to mitigate friction effect. Also the relative mass effect of Joule‟s method was eliminated. A

temperature decrease was measured for air at room temperature and an increase in temperature for

hydrogen was recorded at same conditions. The Joule expansion should have also measured a small net

decrease in temperature, but for the lack of accurate instruments, the relative gas mass to the combined

mass of copper spheres and water bath and friction effects. C W Smith 1977 Kent, Review of Lord Kelvin‟s Cambridge Papers

One empirical correlation for (JT) inversion pressure is Pri=an(Tr)n where n runs from 0 to 5 and a0 =-36.3,

a1=71.6, a2=41.6, a3=11.8, a4=1.67, & a5=0.091. Another empirical equation for approximation of HC gases

inversion point developed by Miller, presented on page 105 Walas ,1985 is: Pr = 24.21-18.54/Tr-0.825Tr2.

14

Summary of Equations for Joule Coefficient, J

J (T/V)U=c = -((U/V)T=c )/Cv = -{T[(P/T)V=c ]- P)}/Cv

1. The Van der Waals EOS, a & b are unique to the VdW EOSJ (T/V)U=c = {-2.72a/V2}/Cv & TJX = {-2.72a V-1}/Cv NOT Recommended

2. By any EOS for Z: [du u2-u1 & 1/V = P/ZRT] J = {-T2/V}(Z/T)/Cv

3. Viral 2 term form:Tjx (PR){1.09/TR

1.6 - 0.139 +0.89/TR4.2 - 0.083}(TC)/Cv for Pr>0.8 & TR>1

This Viral form is limited to Vr<2 and this limit is to be strictly observed to eliminate unrealistic answers.

4. RK EOS: a’ & b’ are unique to the RK EOSTjx [-1.5a’/(b’CvT)] ln{[V/(V+b’)]2/[V/(V+b’)]1} Joule X dT, Rankin deg.

5. BERTHELOT EOSTJX (PR)( T/CV ){1.266/(TR

3) - 1/(14.2TR)} & also Cv TJX = Cp TJTX

Use dT.jx = dT.jtx (Cp/Cv) is direct. The low pressure ratio is applied as both Cp and Cv are low pressure.

The VdW EOS dT’s are inconsistent for this data set. The VdW EOS is not regularly recommended for this work.

Joule X dT, Rankin degrees

EOS O2 CO2 C3= Air H2O

VdW -4.58 -0.49 -7.36 -0.51 -5.80

RK -6.44 -1.14 -21.70 -0.57 -32.80

Viral -6.16 -1.45 -23.50 -0.58 -41.72

BWR -6.24 -1.37 -28.60 -0.60 -44.92

B'lot -5.65 -1.47 -24.88 -0.50 -38.48

Avg -6.12 -1.36 -24.67 -0.56 -39.48

data BWRbest thisdata -28.40 -56.50

Conditions for dT.jx calcs

atm/R O2 CO2 C3= Air

P1 10.000 1.000 8.503 2.361

T1 76.4 60.0 100.0 70.0

P2 0.500 0.500 1.361 1.299

T2.jx 69.8 59.6 94.9 69.5

Z1 0.9696 0.9957 0.9332 0.9988

Z2.jx 0.9970 0.9971 0.9812 0.9991

dU 0.0000 0.0000 0.0000 0.0000

15

Isothermal Joule Coefficient of a Reversible Free Expansion (not to be confused with JT-X which has units of degree per pressure but Joule Coefficient units are degree per volume) or

J (T/V)U=c =(-1/Cv )(U/V)T=c =(-1/Cv)[ T(P/T)V=c – P] =R-mol/CF= 2.72(-a /Cv)/V2m ,

Tjx= (a/Cv)(1/V) van der Waals EOS &

CO2 from: 8.54CF/# to 17.1CF/# & Cv=6.71btu/mol/R, a=926atm/(CF/mol)2 oR = -0.5oR

free expansion, (1/17.1-1/8.54)(#/CF)(1mol/44#)(926atm(CF/mol)2)(2.72Btu/atm/CF)(mol-oR/6.71Btu)

The average J (0.5oR/(8.54) (#/CF)= -0.06 oR/(CF/#)

For O2 298K & 10Bar to vac. oR = - 4.9oR, V=10.72(536)/147=39CF/mol, & 390CF/m

oR =-(1/39-1/390)(mol/CF) (348atm(CF/mol)2)(2.72Btu/atm/CF)(mol-oR/5Btu) = -4.9 oR

The average J (4.9oR/(39) (mol/CF)= -0.13 oR/(CF/mol)

As with Enthalpy Correction for pressure, the Internal Energy may be density corrected:

U = CvT + (T(P/T)V=c -P )V = CvT - Cv J V = Cv(T - J V) = Cv(T - TJX)

Verification of sign convention when correcting dH & dU for pressure effect: only 3 gases increase in

temperature upon expansion at normal T’s, so, TJX & TJTX is normally negative and is identical to dT

for dH=0 & dU=0. So C*(-dT-(-dT)) = 0=C*(-dT+dT)= C*(0)=0.

For vdW gas: (Pr-3/Vr2)(Vr-1/3)=8/3Tr gives 2 unknowns & 2 equations to solve average Tr, But simple to look on P-H chart for CO2 and conditions are identified by internal energy line on which corresponding to the 2 identified volumes. Cp by Cp =6.075+0.00523R & 14.7psia & 60F is close approximation to initial conditions.

J, Joule Coefficient of a Free Expansion by VdW EOS

16

Isothermal Joule Coefficient of a Reversible Free Expansion (not to be confused with JT-X) or

J (T/V)U=c =(-1/Cv )(U/V)T=c =(-1/Cv)[ T(P/T)V=c – P]

Tjx= [-1.5a/(bCvT)] [ln{(V/(V+b))]2/[V/(V+b)]1} by RK EOS

RK EOS: P=RT/(V-b)-a/(T(V+b)V) & [ T(P/T)V=c – P] = 3a/(2T(V+b)V) (form kind of dV/V2)

dT=(-1.5a/(CvT))[dV/((V+b)V)] but wait is slight problem to get delta term, so decompose it

dV/{(V+b)V} =(1/b)[(dV)/V – dV/(V+b)] =(1/b)[lnV –ln(V+b)] |21 =(1/b)[ln{V/(V+b)}|2

1]

Tjx= [-1.5a/(bCvT)] {ln([V/(V+b)]2/[V/(V+b)]1)} by RK EOS integrated form

for CO2 from 8.54CF/# to 17.1CF/# & Cv=6.71btu/mol/R, a.vdw=926atm/(CF/mol)2, T60F, b.vdw=0.686cf/mol, 2.72Btu/atm-CF, a.rk =21677atm(CF/mol)2, b.rk =0.48CF/mol or 0.0109 CF/#, MW=44, Tc=548R, left apprentices is -1195.3R and right apprentices, which is 0.0064 unit-less

free expansion, (1.5)(21677R-atm/(CF/mol)2)(2.72BTU/atm/CF)(mol-R/6.8btu)(1.mol/0.48CF)/548R

dT=(-1157.5)(0.00064)= -0.75R slightly over vdW dT, such small dT is unresolved by P-H charts.

For the Oxygen example the oR= - 5.35 by RK EOS = 584.3LN(…

b.rk=0.696(b.vdw) & a.rk=Tc (a.vdw), as b.rk in cf/mol; divide by MW to put in CF/# b4 add w/V

J, Isothermal Joule Coefficient For Free Expansion by RK eos

17

J (T/V)U=c =(-1/Cv )(U/V)T=c =(-1/Cv)[ T(P/T)V=c – P]= (-1/Cv)[ T(P/T)V=c – P]

Viral P=(T/V)[1+(B)Pr/Tr]& T(P/T)=T{[1+(B)Pr/Tr](T/V)/T+(T/V)[1+(B)Pr/Tr]/T}

T(P/T)-P = (T/V)[1+(B)Pr/Tr]+(T2/V)[1+(B)Pr/Tr]/T} - P = {P+(T2/V)[1+(B)Pr/Tr]/T} - P

T(P/T)-P= (T2/V) {[1+(B)Pr/Tr]/T} = (T2/V){Pr/Tr(B)/T + (B)(Pr/Tr)/T}

T(P/T)-P=(T2/V) {Pr/Tr(B)/T - (PrTc)(B)/(T2)}=(Pr/V) {T Tc(B)/T - (B)Tc}

T(P/T)-P=(PrTc/V){TB/T-B} =(PrTc/V){(T/Tc)B/Tr-B} =(PrTc/V){(TR)B/Tr-B}T(P/T)-P=(PrTc/V){(TR)B/Tr-B} =(PrTc/V){(0.67/Tr

1.6 +0.72/Tr4.2)-B}

T(P/T)-P=(PrTc/V){(0.67/Tr1.6 +0.72/Tr

4.2)-(0.083-0.42/Tr1.6+ (0.139-0.172/Tr

4.2)}T(P/T)-P=(PrTc/V){(0.67/Tr

1.6 +0.72/Tr4.2)+(-0.083+0.42/Tr

1.6 - 0.139+0.172/Tr4.2)}

T(P/T)-P=(PrTc/V){1.09/Tr1.6 - 0.139 +0.89/Tr

4.2-0.083} cannot do integral dV/V as P & V related, use PV=Z T or P=(Z T/V), so (PrTc/V)=(Z T/V) Tc/(V Pc) The terms in brackets { } are dimensionless as & Tr , are dimensionless, as is Z. The integral of –dV/V2 is (1/V) & term RTc/Cv has units of oR ((btu/mol)(mol-R/btu)=R), the term (1/V) ZT/(Pc) when using atm, comes to V=ZRT/P & 1/V=P/(ZRT)avg so (1/V)=(P2-P1)/(ZRT)avg so (1/V) ZT/(Pc)=(Pr2 –Pr1)(ZRT)avg/(ZRT)avg.

Tjx {(Pr)}{1.09/Tr1.6 - 0.139 +0.89/Tr

4.2-0.083} (Tc)/Cv

Results for the prior 2 examples are CO2 oR= – 2.04 & Oxygen, O2 oR= - 4.54 oRB =(BPc/Tc) =(Bo+ B1) =f(Tr) use Bo=0.083-0.42/Tr

1.6. B1=0.139-0.172/Tr4.2

B =(BPc/Tc) =(0.083 - 0.42/Tr1.6 + (0.139-0.172/Tr

4.2))B/T=(B/Tr)(Tr /T)=(1/Tc)(dBo/dTr+dB1/dTr ) & B/T=(1/Tc)(0.67/Tr

2.6 +0.72/Tr5.2)

Terms for Bo & B1 , Perry & SvN 3rdEd ‘75 p87 simple correlation of Abbott (1975) used here: = 0.1745-0.0838Tr, Tr<0.65 P<10atm, otherwise use data tables for , & dV/V = ln(V2/V1)

J, Joule Coefficient For Free Expansion by Viral EOS

18

J (T/V)U=c =(-1/Cv )(U/V)T=c =(-1/Cv)[ T(P/T)V=c – P]= (-1/Cv)[ T(P/T)V=c – P]

For the BWR, the EOS is very complex. Thus is simpler to do the EOS using Numerical Z factor.

PV=ZRT & P=ZRT/V & (P/T)V=c = (ZR/V)(T/T)V=c + (RT/V)(Z/T)V=c = ZR/V +(RT/V)Z/T|V

& T(P/T)V=c = ZRT/V + (RT2/V)Z/T|V = P + (RT2/V)Z/T|V

[ T(P/T)V=c – P] = P + (RT2/V)Z/T|V – P = (RT2/V)Z/T|V

J (T/V)U=c = (-1/Cv)[ T(P/T)V=c – P]= (-1/Cv)[(RT2/V)Z/T|V]=

Tjx [(T2)]{ln(V2/V1)}{Z/T}/Cv (btu/mol)(mol-R/btu)=R

Best for small change in dT was determined as:

Tjx {Z}T/Cv units:(btu/mol)(mol-R/btu)=R, valid small dT & dP or T2 =T1/exp(ln(P1/P2)Z/Cv)

Alternative is to use reduced density method but was found to have higher error

Z/T= (Z/Tr)(Tr/ T) = -0.27[Pr /(Tr2ra)](1/ Tc ) which leads to:

T2 =T1/exp(ln(P1/P2)Z/Cv)

Most simple is to use dT.jx = k dT.jtx

J, Joule Coefficient For Free Expansion by BWR EOS

19

Summary of Equations for JT Coefficient, JT

JTX (T/P)h=c = -((h/p)T=c )/Cp = {T[(V/ T)p=c ]- V)}/Cp

1. The Van der Waals EOS, a & b are unique to the VdW EOS

JTX (T/P)h=c = {(2a/T)-b}/Cp Not recommended

2. By any EOS for Z: [ as du/u =d(ln[u]) ln(u2/u1) or du u2-u1 & 1/V = P/ZRT]

JTX = {T2/P}(Z/T)/Cp ={[ln(ZT2/ZT1)]/ln(T2/T1)}(ZRT/(PCp))average

ZT2 is numerically evaluated at average Pr and the average Tr estimated by JT for the

given EOS.

3. Viral 2 term form:

JTX = (T/Cp)((B)/T – B/T) & B =(Tc/Pc)(Bo+ B1)

TJTX =(Tc/Cp)(PR) {1.09/Tr1.6 - 0.139 +0.89/Tr

4.2 - 0.083}

(Bo+ B1)={(0.083-0.422/Tr1.6) – (0.139-0.172/Tr

4.2)}, Pr>0.8 & Tr>1This Viral form is limited to Vr<2 and this limit is to be strictly observed to eliminate

unrealistic answers.

4. Berthelot EOS:

TJTX = {[PR]}(T/CP){1.27/Tr3 - 1/(14.2Tr)}

The specific heat, CP is the low pressure specific heat for all equations.

20

For this data set, theBWR Method has lesserror for HC gases. Thisis especially true for theC3= data point.

However, for generalgasses; Air, H2 etc.,either the Viral, B’lot &RK method are better forestimate of JouleThompson coefficient.

This result is notsurprising. Because BWREOS is a tailored EOS andthe BWR form used hereis based on light HCgasses. Possibly, themore involved Lee-Kessler BWR EOS issuperior to all methods.

Even with all the advances in EOS development, the RK EOS remains a recommended method forthermodynamic properties. Modern VLE calculations typically use either the LK-BWR, S-RK or PR. Both thelatter 2 forms are adaptations of the basic RK EOS. All three modern methods of VLE calculations useaccentric factors and mixing rule interaction parameters to improve accuracy. As shown for Z calculations,limits for this BWR equation should be duly noted. None of these EOS were satisfactory for steam.

JT EOS Comparison

JT Result Compare Viral/BWR/RK/vdW/B'lot EOS v. Tabulated Data

term R/atm C3= C1 C1 air CO2 H2 CO2 H2O

u.jt.vir 2.79 0.71 0.78 0.49 1.76 -0.09 1.32 3.77

u.jt.bwr 3.49 0.71 1.01 -0.04 2.13 -0.03 2.19 3.03

u.jt.vdw approx 1.66 0.68 0.88 0.96 1.21 -0.04 1.11 1.24

u.jt.Ber'lot 3.18 0.69 0.82 0.56 1.92 -0.07 1.56 2.67

u.jt.rk.dlnZ 2.78 0.75 1.27 0.31 1.94 -0.06 2.05 2.19

u.jt.rk.h eqn 2.73 0.73 1.17 0.36 1.89 -0.13 1.91 2.17

u.jt.rk.Z2-Z1 2.87 0.75 1.30 0.30 1.94 -0.06 2.08 2.27

u ref HS or TS 3.50 0.62 0.92 0.31 2.12 -0.05 1.70 4.46

reference scheel Schams Faries FariesHS perry5th wiki jt perry 5th stmTabl

page p94.HS p214 p222 p223/153p p3-163 www p3-163 KernDQ

Pr.avg 0.11 0.27 1.20 4.44 0.23 0.23 0.49 0.08

Tr.avg 0.83 1.63 1.41 1.61 0.99 9.03 1.05 0.74

T.avg R 543 558 484 384 542 540 578 860

P.avg atm 4.95 12.50 55.00 165.00 17.00 3.00 36.05 16.50

21

The VDW method for JT is used to make a 1st estimate of dT. From this expansion temperature

the calculation proceeds with more exact expressions of JT to determine average Tr. Iterationconverges to an average JT by either R-K or Viral EOS. = universal gas constant,1.99BTU/mol/oR or 0.7302atm CF/oR /mol, oR, Rankin & conversion factor, (2.72BTU/atm/CF).

A JT approximation from van der Waals EOS is: (adapted from McMaster)

JT T/P = {(2a/T)-b}/Cp & b =Vc/3= 0.7302Tc/(8Pc), a = 27pc(b2)

JT: Van der Waal’s Simplified JT Coefficient

Units: b CF/mol, a atm(CF/mol)2 , T 0.73(atm-cf/mol/ oR)(oR) & Cp BTU/mol/oR

‘b/Cp (CF/mol)(oR mol/BTU)(1/ =oR mol/0.7302 atm CF)( =1.99BTU/ oR mol)= oR/atm

‘JT OR/atm =2.72{(2a/(0.73oR)-b}/Cp & b Tc/8Pc, a 27Pc(b)2

This review indicates that VdW is inferior to the Berthelot EOS for initial estimate of temperature change for either JX or JTX.

Count Rumford‟s observation that

work produced heat via friction.

Joules Jar which established ratio

between heat or friction and work

at about 778 BTU/Ft-#.

22

JT by from van der Waals EOS at high pressures is:

JT T/P = {-b/Cp}{[43 - 3(3-1)2]/[43 - (3-1)2]} & =Tr, =Pr

JT: Van der Waal’s Exact JT Coefficient

Units: b CF/mol & Cp BTU/R/mol so multiply above result by 2.72 for units of R/atm

‘b/Cp (CF/mol)(oR mol/BTU)(1/ =oR mol/0.7302 atm CF)( =1.99BTU/oR mol)= oR/atm

JT R/atm = 2.72{-b/Cp}{[43 - 3(3-1)2]/[43 - (3-1)2]}

Inversion point is where JT 0.00, this equation is one way to find I.P. pressure and temperature.However, the vdW EOS is not very accurate method to represent gas behavior. This is demonstrated by acomparison Zc.. The objective of any EOS is to mimic actual gas behavior. Most accepted EOS will do fairjob away from the Critical Point. But as conditions approach critical point larger deviations are expected.For the vdW EOS, estimation by this exact form was typically no better than the approximate method.This is likely due to the inherent limit of the vdW EOS gas having a Zc of 0.37. A Zc of 0.37 is not close tothe behavior of most real gases. The RK EOS is an improvement that fits a wider range of P & T than thegeneric BWR*.

For a vdW gas inversion point is defined curve defined by (2a-bT)V2-4abV+2ab2=0

Likewise, Boyle curve of a vdW gas is any point defined by (a-bT)V2-2abV+ab2=0,

Adapted from www. The VdW are generally poor estimations of either dT.jx or dT.jtx

23

JT:Two Term Viral EOS JT Coefficient

JT = (T/Cp)((B)/T – B/T) = (T(B)/T – B)/Cp & B =(Tc/Pc)(Bo + B1)

dB/dT=( /Pc)(dBo/dTr + dB1/dTr ) Where Bo & B1 = f(Tr) :Perry & SvN 3rdEd ‘75 p87

Method NOT Recommended as other EOS’s are more regular:# The viral is restricted beyond pure components by • complicated mixing rules• need to evaluate 3rd parameter, =f(Vapor Pressure @ Tr = 0.7)• need an Binary interaction parameter ki-j for mixtures. •Other limits of this viral form are: Vr>2, Best for non-polar or simular compounds

= 0.1745-0.0838Tr, Tr<0.65 P<10atm, otherwise use data tables for =(0.291-Zc)/0.08 Reid in Perry Eqn3-117, otherwise use data tables for Viral Terms for Bo & B1 , simple Abbott (1975): Bo=0.083-0.42/Tr

1.6. B1=0.139-0.172/Tr4.2.

Pitzer (1958) and Tsonopoulos (1974) Terms for Bo & B1 follow the form:Bo=A+B/Tr +C/Tr

2 +D/Tr3 +E/Tr

8. B1= =A’+B’/Tr +C’/Tr2 +D’/Tr

3 +E’/Tr8.. Viral Terms

A B C D E A' B ' C ' D' E 'P itz 1.4450E-01 -3.3000E-01 -1.3850E-01 -1.2100E-02 0.0000E+00 7.3000E-02 4.6000E-01 -5.0000E-01 -9.7000E-02 -7.3000E-03

T son 1.4450E-01 -3.3000E-01 -1.3850E-01 -1.2100E-02 -6.0700E-04 6.3700E-02 0.0000E+00 3.3100E-01 -4.2300E-01 -8.0000E-03

JT = ( Tc/Cp)(1/Pc) {1.09/Tr1.6 - 0.139 +0.89/Tr

4.2-0.083} & by Using Abbots formula & Compare to TJTX =(Tc/Cp)(PR) {1.09/Tr

1.6 - 0.139 +0.89/Tr4.2-0.083} & :

Tjx (Tc/Cv) (PR){1.09/Tr1.6 - 0.139 +0.89/Tr

4.2-0.083}

(Tjx /TJTX )=(Cp/Cv) {f(B.JX)/f(B.JTX)} =(Cp/Cv) {1, for this EOS} =(Cp/Cv)

24

JT: JT Coefficient as function of Z any EOS

For real gas PV=zT , & V = zT/P & JT = dT/dP |s=c

Expressed in terms of V, JT = [T(dV/dT)-V]/Cp

Td(zT/P)/dT–V=T[z/P +(RT/P)(dz/dT)]-zT/P ={T2/P}(dZ/dT)

JT =(T2/P)(dZ/dT)/Cp = 1.99(oR2/Patm)(Z/oR)/Cp {Schaum’s Thermo 1972 p163.eq2}

units: R=1.99btu/(oR) & Cp as btu//oR (oR/atm)[1.99btu/(oR)] (oR/btu)(oR/oR)=oR/atm

In above equation make substitution: PV=ZT so T/P = V/Z & d(lnX) = 1/X dX

JT = (VT/Z)(dZ/dT)/Cp =(V)(dlnZ/dlnT)/Cp =(ZT/P)(dlnZ/dlnT)/Cp

numerically evaluate the differential : dZ/dT = (ZT2-ZT1)/(T2-T1) at Pr average

It is often simpler to make the numerical evaluation than do complex algebra

JT=(Z T/P)(dlnZ/dlnT)/Cp = {[ln(ZT2/ZT1)]/ln(T2/T1)}(Z T/PCp)average

Recommendation: log evaluation gave best results of all dual term methods. For general HC work the RK & BWR (withimposed limits) do a reasonable job at dT.jtx estimation. The dT.jx is best estimated from (Cp/Cv)dT.jtx. The Joule and JTtemperature corrections are then used to pressure adjust Internal Energy and Total Energy: Elementary my dear Watson.Note: Methods disposed here are identical to H.L. Callendar’s seminal work: Properties of Steam & Thermodynamic Theory.of Turbines, London 1920. The Callendar EOS was later used to develop the ASME Steam Tables. Temperature and Pressureare the only necessary coordinates to define all thermo properties of a single component system. Setting dU in terms ofvolume and temperature and volume makes a perfect gas reduction to Cv=dU/dT. Because low pressure Cv and Cp areused to define H & U, it follows that low pressure ratio of Cp/Cv is adequate to estimate dT.jx from dT.jtx. For expandedpressure ranges , perhaps is better to use dT/T ={[ln... }}(1-1/k) dP/P , or ln(T2/T1)={[ln...]}(1-1/k)ln(P2/P1)?

25

Forms & Solutions of Redlich Kwong EOS

A. M-N Exact Cubic Solution (after LF Scheel GPC ‘72 p144/103 & WC Edminster GPC’62)

M = 0.427Pr/Tr2.5 - 0.00752(Pr/Tr)2 - 0.0867Pr/Tr - 1/3

N = M/3 - 0.0372(Pr/Tr1.75)2 + 0.037

X = (N/2)2 + (M/3)3 [Z(M-N) only valid X>0]

Z = (-N/2 - X 0.5)0.333 + (-N/2 + X0.5)0.333 + 1/3

Most valves, compressors, and expanders are near or at single phase region, X>0. It is then

reasonable to force a solution using ABS(X). Then verify correct solution with a general Z by hiterative method:

B. h method: with z = Z(M-N), from aboveh= 0.0867Pr/(zTr) = 0.0867 Tc/(VPc) (if V known)

Zrk by h iteration method: Zrk=1/(1-h)-h/(1+h), =4.934/Tr1.5

Excess Properties and Fugacity by the RK EOS (from Perry’s Eqn. 4-277/9

H’/RT = 1.5ln(1+h) + (1-Z) & S’/T = 0.5ln(1+h)-ln(Z-hZ) & ln=(S- H/T)/R

Mixing Rules: 1.Kay Pc=(yPc)I & Tc= (yTc)I &

2. Empirical a=(1/3)(yTc /Pc)I +(2/3)((y(Tc /Pc))I )2 b=(y (Tc /Pc))I & Pc =(b/a)2 & Tc=(b2/a)

The a & b terms of Redlich Kwong are not same as a, b of VDW. For RK, a=0.0371b(Tc)1.5 & b=.0867(Tc/Pc), atm, K. The RK EOS gives Vc=3.85b where-as vdW gives Vc=3b. All gases at any T and high pressure, the volume

approaches 0.26Vc, as calculated by RK EOS. b.rk=0.696b.vdw & a.rk=Tc a.vdw, as b.rk in cf/mol;

26

Zrk=[(1+h) - 4.93(h-h2)/(Tr1.5)]/(1-h2) & h = 0.0867Pr/(zTr) =K/(zTr)

(1-h2)Zrk - (1+h) + J(h-h2)/(Tr1.5)=0 =f(z,Tr) & sub for h w/ J=4.93 & K= 0.0867Pr

(1-(K/(zTr))2)Zrk - (1+ K/(zTr)) + J(K/(zTr) –(K/(zTr))2)/(Tr1.5) & by z2 Tr

3.5

z2 Tr3.5(1-(K/(zTr))2)Zrk- z2 Tr

3.5(1+ K/(zTr)) + z2 Tr3.5 J(K/(zTr) –(K/(zTr))2)/(Tr

1.5)

(Zrk3Tr

3.5 – ZrkTr1.5K2) - (Zrk

2 Tr3.5 + Zrk

1 Tr2.5 K) + Z2 Tr

3.5 J(K/(zTr) –(K/(zTr))2)/(Tr1.5)

(Zrk3 Tr

3.5 – ZrkTr1.5K2) - (Zrk

2 Tr3.5 + Zrk

1 Tr2.5 K) + Z2 Tr

2 J(K/(zTr) –(K/(zTr))2)(Zrk

3 Tr3.5 – ZrkTr

1.5K2) - (Zrk2 Tr

3.5 + Zrk1 Tr

2.5K) + J(Z Tr K – (K2))(Zrk

3 Tr3.5 – ZrkTr

1.5K2) - (Zrk2 Tr

3.5 + ZrkTr2.5K) + JK(Z Tr – K) = 0 =f(z,Tr)

Use implicit rule to get : Z/Tr = -(f/Tr)/(f/Z) & chain Z/T = (Z/Tr )/Tc

(f/Tr)= (3.5Zrk3 Tr

2.5 – 1.5K2 ZrkTr0.5) - (3.5z2 Tr

2.5 + 2.5K ZrkTr1.5) + Zrk JK

(f/Tr)= (3.5Zrk3 Tr

2.5 + KZrk (J– 1.5KTr0.5 - 2.5Tr

1.5) - (3.5z2 Tr2.5)

(f/Z)= (3z2 Tr3.5 –Tr

1.5K2) - (2z1 Tr3.5 + Tr

2.5K) + JK(Tr)

(f/Z)= Tr3.5(3z2 –2z1 ) + K(J Tr - Tr

2.5 – K Tr1.5)

If V is known it may be helpful to use the reduced form of the RK EOS with Zrkc=1/3 :

Pr =3Tr/(Vr-0.26) – 3.85/[(Tr0.5(Vr+0.26)Vr] (the RK EOS in reduced format)

Determine (Z/oR) by Redlich Kwong h method

27

Z =0.27Pr /(Tr r) & Z/T= (Z/Tr)(Tr/ T) = -0.27Pr /(Tr 2ra)(1/ Tc )

Solve reduced density at average Z by Benedict-Webb-Rubin EOS

f(r) = Ar6 + Br

3 +Cr2 +Dr + (Er

3)(1+Fr2)exp(-Fr

2) -G

f’(r) = 6Ar5 + 3Br

2 +2Cr1 +D + (Er

2)[3+Fr2 (3-2Fr

2)] exp(-Fr2)

(r)i+1 = (r)i - f(r)/ f’(r) use Z at average conditions for (r)0 = 0.27Pr /(Tr Za)

A=0.06423, B=(0.5353Tr -0.6123), C=(0.3151Tr -1.0467-0.5783/Tr2), & D=Tr

E=0.6816/Tr2 , F= 0.6845, & G=0.27Pr Ref: HP41 Petro Fluids Pac, Meehan & Ramey. Pc=(yPc)I Tc=(yTc)I

This Benedict Webb Rubin EOS method is attributed to Dranchuk, Purvis, & Robinson who fitted the BWR EOS to the Standing Katz Z factor chart for light hydrocarbons. G. Takacs (1976) found this method to have the lowest error among 8 common methods to estimate Z. Valid for 1.05<Tr<3.0 and 0<Pr<3. Estimate Z1 by RK method & apply critical property corrections for

N2, H2S, & CO2 via Wichert-Aziz method, if needed. http://www.oiljetpump.com/welltesting_oil_L.xls

Determine (Z/oR) by BWR Reduced Density EOS

comparis on of Z by Various R eferences

# P r Tr Zchrt pitz'r Zrk Zbwr.x ls Zref R eferenc e

1 9.5 1.99 nd 1+ 1.020 1.247 1.248 HP BWR clc

2 5.0 1.50 0.810 0.801 0.811 0.811 HP BWR clc

3 0.4 0.80 nd 0.74? 0.650 0.535 0.630 nC5 LeeK bwr

4 0.4 0.90 0.778 0.780 0.761 0.756 nC5 LeeK bwr

5 0.4 1.00 0.850 0.850 0.842 0.844 nC5 LeeK bwr

6 0.4 1.10 0.880 0.891 0.888 0.891 nC5 LeeK bwr

7 0.4 1.20 0.910 0.918 0.917 0.923 nC5 LeeK bwr

8 1.2 1.05 0.470 0.460 0.404 na limit test HP bwr

A comparison of Z calculation by Benedict Webb Rubinand RK EOS by excel is given in table on RHS tosundry reference sources. LI 1 & 2 are to testconvergence routine used in.xls to that of the HP41routine. LI 3 to 7 is comparison of BWR DPR to that ofBWR by Lee Kestler for nC5. A favorable comparisonwas found for the imposed limits of the DPR-BWR vs.the LK-BWR. The Z by Pitzer chart is also included.Pitzer chart has limits not imposed on EOS Eqn’s. Thehardy RK EOS is also included. At high Pr and also atcryogenic conditions the BWR is reported to havebetter Z factors as compared to the RK EOS.

28

Energy Balance SummaryThis section is written because energy balance for gasses was found to be misrepresented by a prolific and

widely used author of Chemical Engineering texts. The same misrepresentation is oft found by internet

search of same topic. A correct energy balance will work for any state, gas, liquid, solid, flow or non flow.

Energy does not respect the state of matter. A correct energy balance equation must predict zero heat and

correct expansion temperature for both adiabatic Joule Expansion (U)=0 and for adiabatic Joule Thompson

Expansion, (H)=0. These 2 expansions are litmus tests for any correct energy equation. The temperature

change and heat for both expansions are well documented by any EOS and by experimental data.

(U+(V2/2g)/J)1+(Q-W)=(U+(V2/2g)/J)2+ (PV)+FL RB Bird 1957 in DM Himmelblau & HF Rase p100 1990

(H + V2/2g)1 + EG = (H + V2/2g)2 + FL (var..Perry 5th Ed., Real Gas-EB, Kinetic Gas Theory-Energy Bal.)

{Ws + dP/ +VdV/g + [Pd(1/) + U] - Qh =0} & Qh = (TS - FL) Clausius/VL Streeter 5th Ed. p130

The reason to present these energy balance equations is pointing out that (H + V2/2g) =C is a very limited

case. Equation, (H + V2/2g) =C does not correctly convey energy balance for applications with friction.

Case 1. IDEAL NOZZLE/VALVE: No Work, No Friction Losses: (H + V2/2g)1 = (H + V2/2g)2

Case 2. REAL VALVE: w/ Friction, No Work or added energy: (H + V2/2g)1 = (H + V2/2g)2 + FL

Case 3. Zero Initial Velocity: No Friction: with Work: EG + (H )1 = (H + V2/2g)2

Case 4. Constant Density: Heat Only & No Friction: EG + (H)1 = (H)2

There are too many cases to list. The point is one cannot call (H + V2/2g) a complete description of energy

without misrepresenting Energy Balance Principles. The details will be developed by calculations. The

examples show why either incorporation of thermal head, dU, or the Clausius Equality is a necessity to

balance the different heads, when accounting for friction as thermal energy. Stated in thermo terms :

H=U+ (P/). If either the internal energy term or loss term is neglected the sum of energy terms will

not balance, nor will heat rate be correctly determined. Subsequent tables demonstrate theses points.

Professor Rase, p98-100, details a Newtonian Energy Balance as: Work by exterior force equals work against

friction + change in potential energy + change in Kinetic Energy. He assigns a negative value to work done

by the system to arrive at: Wa = F + (U-Q) + (V2/2g)/J). Expansion work by system between boundaries is

(PV) & Wa -(PV)=F + (U-Q) + (V2/2g)/J) or H + (V2/2g)/J) + F=Q+W. For adiabatic expansion with

zero shaft work, the Newtonian energy balance of Prof. Rase is: H + (V2/2g)/J) – F =0.

29

Energy Balance from Basic State1. {(U+(V2/2g)/J} =(Qh +W) :basic energy balance- steady state without elevation or other energy forms Perry 5-31

W or work is composed of external work & expansion work or W = Ws - (P/) Perry 5-32

2. {U+ (P/) + (V2/2g)/J} = (Qh + Ws), next use Clausius Equality Qh = (TS - FL), Perry p.5-18para.12 & Streeter

3. {U+ (P/) + (V2/2g)/J} = TS - FL +(Ws) & identity:H= U +(P/) with =1/Density=1/, & F is friction

4. {H + (V2/2g)/J}= TS - FL + (Ws =0 for valve)- Friction & heat are zero for an ideal flow nozzle

5. {H + (V2/2g)/J}= (0) Ideal Nozzle, but real nozzle which always has friction, use: TS = -d(P) + H

6. {H + (V2/2g)/J}= -d(P) + H - FL + (Ws =0 for valve) the H cancel, & =ZRT/PM leaving ZRT/M ln(P)

7. {(V2/2g)/J} + (ZRT/M)(dP)/P + FL = Make the integration to arrive at

8. (V2/2g)/J} + {ZRT/M)ln(P) + FL =0, Smith 10-4B p459

If differential velocity head is used, with velocity V = G/, one arrives at {(V2/2g)/J}. So the velocity head

was not placed in differential form to expedite the key point of energy balance with friction.

The reason to present these energy balance equations is pointing out that (H + V2/2g) =C is a very limited

case. Equation, (H + V2/2g) =C does not correctly convey energy balance for applications with friction.

For Case of skin friction the energy balance reduces to

(V2/2g)/J} + {ZRT/M)ln(P) + (4fL/D)(G/LM)2/2g =0 where LM is log mean density based on EOS Z.

This is a sort of adiabatic friction equation where velocity and hence f changes along the flow path. Given

P1, P2, and T1, it is possible to solve or goal for T2 what balances the inlet and outlet energies, which

includes friction effect. All inlet energy terms are known. It is proposed this equation is as accurate as any

enthalpy equation and eliminated determination of heat capacity, and pressure correction of enthalpy.

30

Clausius Equality, Second Thermo Law & Friction Heating

Clausius Equality: non reversible process requires: (Friction) = (TS-(Q)h), Streeter VL 1971 3.8.4; Perry p5.18

{Ws + dP/ +VdV/g + [Pd(1/) + U] - Qh =0} Streeter-3.8.1, General Volume E balance & Perry 5-31/32

{TS = Pd(1/) + (U)} = dH - VdP Streeter (3.7.6): General Entropy, valid between any 2 Closed Equilibrium States,

{ Ws + dP/ +VdV/g + [TS - (Q)h ] =0 } Streeter 3.8.2: Sub of 3.7.6 to 3.8.1

Sub Clausius Equality, 3.8.4 to 3.8.2 : {Ws + dP/ +VdV/g + (F) =0 } Streeter 3.8.5

For case of zero shaft work, obtain: {dP/ +VdV/g + (Losses)}=0, Smith 10-4B

express losses as classic friction term and use V=G/ & dV = -Gd/-2 to obtain:

{dP/ +(1/g)G2d/-3 + (Losses)}=0, multiply by 2 to get: {dP - G2d//g + 2(Losses)}=0. M&Smith 6-59

Use Real Gas density (=PM/ZRT) at average Z & T to get : {(M/ZRT)PdP - G2d//g + 2(Losses)}=0.

Use Appendix identities to get: {M/(2ZRT) P2 - [G2ln(2/1)]/g + 2(Losses)}=0

Use average ZT, ln(2/1) reduces to: {M/(2ZRT)aP2 - G2ln(P2/P1) + a2(Losses)}=0.

Smith takes (Losses) =(4fL/D)(G/LM)2/2g & takes (a/LM)2 =1 & sets Za =1 to get:

{M/(2RTa)P2 - G2ln(P2/P1) + (2fL/D)G2/g} = 0 {Smith 10-4D aka classic “isothermal flow” to solve P2 }

{Ws + (H + V2/2g/778) + {TS =[ Pd(1/)/778 + (U)] } - F = 0, F is friction: F & Ws are in BTU/#

The Clausius Equality validity is proven by the derivation of the “isothermal” flow equation. Also if one

cares to accept simulator answer, heat determined by Clausius Equality more closely matched simulator

result than the “Smith-Van Ness/ Smith” method for SvN X10-4. Perry ChE Handbook taken from Streeter.

31

Clausius Equality, Second Thermo Law & Friction HeatingClausius Equality: non reversible process requires: (Losses or Friction) = (TS-(Q)h), Streeter VL 1971 3.8.4.

{Ws + dP/ +VdV/g +[Pd(1/) +U] - (Q)h =0} Streeter (3.8.1) General E balance for a volume. The term

(U + Pd(1/) + dP/) = U + d(PV) = dH so: dH +VdV/g +{Ws - Qh } = 0. At zero work, dH +VdV/g -(Q)h = 0

Use (Q)h = (TS –F) & dH +VdV/g - (TS –F) = 0. If using pressure head, MEB, with zero shaft work, {TS =

Pd(1/) + (U)} to get, dP/ +VdV/g + TS - TS + F = 0 or dP/ + VdV/g + F = 0. For constant density, the

regular Bernoulli is seen (P/ +V2/2g)1 = (P/ +V2/2g)2 +F.

Example: A Hydraulic Fluid, HF, bypass line on a Delayed Coker Unit (DCU) burns off 7,000 psi prior to

entering a fouled fin fan cooler which is causing overheating in the HF pump. Field survey shows a bypass

away from the hot DCU is possible. The inlet to the bypass valve is 70F and inlet to the cooler is 170F. What

portion of heat is gained by the pipe from the DCU area? Heat Capacity is 0.6 and average SG is 0.85.

Solution: Friction Heat: 7000(#/sq.in)(144Sq.in./SF)*1CF/(0.85*62.4#)(1BTU/778Ft#) = 24.4BTU/#

Friction Temperature Change: (24.4BTU/#)(1R-#/0.6BTU) = 40.7F & Entropy for Liquid is dS=Ci ln(T2/T1) so

TS in Clausius Equality is (460+70/2+170/2)*0.6*ln(630/530) =580 *0.104=60.2BTU/#. Next find (Q)h) by

Clausius Equality as: =(Q)h =TS- F = 60.2 - 24.4 =(Q)h = 35.8 BTU/#.

Since {dP/ + Pd(1/) + U} = H, & Ws =0 rewrite 3.8.1 as {H + V.hd - (Q)h =0} the inlet V is 90fps and

outlet V is 100fps. So {H + (1002-902)/50000 - (Q)h =0} so H=35.8-0.038=35.76BTU/# is the energy

associated with passage thru the DCU hotbox. The total dH=(170-70)*0.6btu/#/F=60BTU/# which is the sum

of friction and sensible gain, 24.4+35.76=60.2, and percent sensible heat gain in hotbox is 35.8/60= 60%. If

a bypass around hotbox is installed the heat duty to cooler is just the friction heat, 24.4btu/#, a 60%

reduction in heat load..

Any thermal solution where there is substantial friction must account for friction heating by the Clausius

Equality. Heat, Friction, and it‟s laws have no respect for the material physical state. However the physical

state does effect the determination of physical properties.

For Smith X10-4 determination of heat by Clausius Equality method leads to dS by T-S diagram: 0.95-0.865 =

0.085BTU/#/R. TdS= (460+70/2+170/2)*0.085=580*.085=49.3, The Friction heat was determined as

19.2BTU/# so (Q)h =49.3-19.2=30.1BTU/#. The V.head difference was determined as 5.1 so heat is:

dH + 5.1 –30.1 => dH= 25BTU/#. Much closer to Simulator result of 24.2BTU/#.

32

The Energy BalancePremise of Energy Balance is Energy Conservation: Classical Physics: total energy is constant

Starting $Energy + $Gains or Work = $Ending Energy + $Losses, most often friction losses.

(P/ + V2/2g + y + 778U)1 + EG = (P/ + V2/2g + y +778U)2 + FL (units: feet of head)

Good, as energy states care not which path is taken. Energy accounting simplifies determination of final

conditions. The idealized thermo paths are visualized as either, constant: Internal energy, Enthalpy, or

Entropy. The actual path is usually some combination of these 3 idealized paths, as indicated on a H-S

diagram. NB: Thermodynamic idealizations are seldom realized in practical application. For applications of

processing equipment the elevation head, y, may be dropped or added to pressure head, leading to:

(P/ + V2/2g)1 + EG = J Cv T + (P/ + V2/2g)2 +FL For liquids Cp=Cv

For gasses, the term P/ is P/(PM/ZRT)= R‟T (but R‟=Cp-Cv) so R‟T = (Cp-Cv)T, & :

(TCp+V2/2g)1+ EG = (CvT–CvT)+(TCp+V2/2g)2+FL

Seen in thermo terms H= TCp or H= (T-T.ref)Cp or use T2=T.ref then H= TCp

For Gases: (H + V2/2g)1 + EG = (H + V2/2g)2 + FL

Case 1. IDEAL NOZZLE/VALVE: No Work, No Friction Losses: (H + V2/2g)1 = (H + V2/2g)2

Case 2. REAL VALVE: w/ Friction, No Work or added energy: (H + V2/2g)1 = (H + V2/2g)2 + FL

Case 3. Zero Initial Velocity: No Friction: with Work: EG + (H )1 = (H + V2/2g)2

Case 4. Constant Density: Heat Only & No Friction: EG + (H)1 = (H)2

There are too many cases to list. The point is one cannot call (H + V2/2g) a complete description of

energy without misrepresenting Energy Balance Principles. The details will be developed by calculations.

The examples show why incorporation of thermal head, dU, is a necessity to balance the different heads,

when accounting for friction as thermal energy. Stated in thermo terms H=U+ (P/). If either the

internal energy term or the loss term is neglected the sum of energy terms will not balance. The following

tables demonstrate the point.

33

L iquid output in out K C v C o

V fps 5.56 5.56 144.1 6 0.38

fric tion Hd ft -1.24 69.21

Work Head ft 271.63 0.00 B HP = > G P M 3.0 33.76

V head ft 0.48 0.48 R e ==> f 259786 0.0045

P head ft 56.90 99.38

T head ft 0.00 158.71 dT pipe fric tion, pipF 0.27

s um 327.77 327.77 dT total, pmpF +pipF 0.87

P ump effic iency 0.416 couls on p.166: TDH 113' vs 113calc

dT P ump 0.60 TD H=WorkHd*effic .

L iquid output in out K C v C o

V fps 3.40 7.66 39.9 170 0.44

fric tion Hd ft -0.41 36.37

Work Head ft 1226.77 0.00 B HP = > G P M 55.8 300.00

V head ft 0.18 0.91 R e ==> f 189300 0.0048

P head ft 163.46 713.46

T head ft 0.00 639.26 dT pipe fric tion, F 0.09

s um 1390.00 1390.00 dT total, pmpF +pipF 0.87

P ump effic iency 0.478 E vans p.150 gives e= 0.479

dT P ump, F 1.65

Liquid Pump Total Energy Balance-a

34

L iquid output in out K C v C o

V fps 1.14 1.14 0.0 48564 0.44

fric tion Hd ft 0.00 0.00

Work Head ft 692.96 0.00 B HP = > G P M 17.5 100.10

V head ft 0.02 0.02 R e ==> f 52583 0.0061

P head ft 46.20 461.98

T head ft 0.00 277.18 dT pipe fric tion, pipF 0.00

s um 739.18 739.18 dT total, pmpF +pipF 0.36

P ump effic iency 0.600 S vN p.457 gives e= 0.6 dTp= 0.35F

dT P ump, F 0.36

The three tables above were calculated with energy boundary condition around pump. These boundary

conditions consider pump work requirements to meet demands of impeller momentum transfer

efficiency, process friction losses, process total pressure head needs and fluid velocity head changes, if

any. In order to balance for irreversible friction effect of impeller momentum transfer, the T head or

internal energy head needs be added. The T head for liquids is heat capacity times fluid temperature

change. This is simply differential temperature necessary to balance the sum of heads terms, CvT. It is

also related to pump impeller efficiency, e, as T=(1/e-1) H/Cp/778. The value 778 ft#/BTU is joule

ratio of heat to work dissipation. Efficiency is defined as ratio of work in to useful energy out. Useful

output is sum of difference between the three heads; Friction, Velocity, and Pressure. Without the T Head

the total energies will not balance.

Liquid Pump Total Energy Balance-b

35

The fallacy of neglecting friction can be easily seen if applied to a metering

orifice. For metering orifice (venturi, flow nozzle, pitot tube or orifice) all

produce some irreversible loss. The pitot or annubar produce the least losses.

An expression for permanent loss is pp =[h(1-Cd)].. For example if meter is

sized at 200 inch w.c. @ line inlet velocity of 200fps, density of 1pcf, T=600R,

air with k=1.4, Cd=0.6 (Branan pp21/2). The line loss on orifice is

200*0.4=80‟‟w.c. or 80‟‟/12‟‟/1ft*(62.4#/cf/144sqin/sf)= 2.9psi. (N.B. !!: were

there zero irreversibility (friction) then all V head produced in orifice would be

recovered back to pressure head.)

Determine T2 with isentropic Expansion:T2=T1(P2/P1)(1-1/k) = The friction term is:

dPf/1= 2.9psi*(144sqin/SF)/(1#/cf)/(778ft#/BTU) =0.5368, then by Bernoulli

heads to get 41.85 = 41.33+0.5368=41.87 or if looked at by enthalpy equation:

0.8238=?0.2855+0.5368=0.8223, clearly the T needs to be less to make perfect

balance.

If try (H+V2/2g/778)1 = (H+V2/2g/778)2 , Line item 15 of In/Out table, clearly

does not balance as 25.472 BTU/# ≠ 24.077BTU/#.

One MUST add friction term to balance. If add fanning friction then is

25.472=24..08+.41=24.49 much smaller error and agree with Energy

Balance . (H+V2/2g/778)1 = (H + V2/2g/778)2 , has limited validity

Summary: A. standard equations which lack nozzle friction term are not

correct for real situations, with friction.

B. Isentropic equations need friction term to balance, dH error less

Correct expression to include friction is:

(H+V2/2g/778)1 = (H + V2/2g/778)2 + HF

HF = CpTF = dP/(J ) or use Fanning Friction term.

LI item in out

1 T,F 140 137.8

2 Psia 221.8 218.9

3 #/hr 4331 4331

4 L'eq 2.635 2.635

5 di" 1.049 1.049

6 MW 29.0 29.0

7 Cp btu/m/R 6.951 6.951

8 u cp air 0.019 0.019

9 Z-RK eos 0.998 0.998

10 f Fanning 0.004 0.004

11 u.jt.rk eos 0.36 0.36

12 V fps 200.000 201.907

13 den pcf 1.002 0.993

14 H btu/# 24.672 24.161

15 H+Vhd 25.472 24.977

16 F. thermo BTU/# 0.555

17 F. fanning BTU/# 0.408

18 F. dE thml BTU/# 0.495

Isentropic dT ME bal

Heads in out I

Phd 40.970 40.819 a

Vhd 0.800 0.815 b

F hd 0.000 0.486 c

Cv dT 0.000 -0.377 d

Total 41.770 41.744 e

Nozzle Efficiency & Friction-I

36

Summary: A. standard equations which lack nozzle friction term are not correct for real situations.

B. Isenthalpic equations balance by including friction head, dH error 0.04% vs 0.06% & using

T2 based on Isenthalpic Equation.

Next: Try Same Conditions but with JT Equations to see if energies balance better.

From prior example Cp = 0.2397 =6.951BTU/mol/R, T1=600R, P1= 221.8psia & P2 = 218.9 or 15.088 atm &

14.891 atm respectively. Determine JT by RK EOS & T2 isenthalpic =T1(P2/P1)(1-1/k)e Table 1 below

P1 dP P2 eff.I dTfrct Ti @1 Ti+f Tjt T2 clcu.jt

221.8 2.90 218.9 0.977 2.239 597.7 599.94 599.9 597.8 0.3169

221.8 5.08 216.72 0.981 3.938 596.0 599.94 599.9 596.1 0.3170

221.8 6.53 215.27 0.980 5.020 594.9 599.9 599.9 595.0 0.3170

221.8 21.80 200.00 0.967 16.83 582.5 599.3 599.5 583.1 0.3176

221.8 40.00 181.80 0.937 30.89 566.9 597.8 599.1 568.9 0.3183

221.8 77.63 144.17 0.876 59.94 530.5 590.44 598.31 538.7 0.3196

Based on (H + V2/2g/778)1 = Hf + (H + V2/2g/778)2H = Cp(T2 - T1 - JT(P2 - P1)) & include friction H as

144dPf/778, Btu/# & T friction= Hf/Cp. For this thermo model it is determined that Tis @e=1 + TF = TJT,

Isentropic Temperature at 100% efficiency plus friction temperature = Temperature calculated by JT

coefficient. The logic is: as JT is based on frictionless expansion and isentropic expansion is based on

expansion with energy conservation from pressure energy to work energy but friction is destruction of work

energy. The mechanical energy balance is (P/ + V2/2g)1 = (P/ + V2/2g)2 + Pf/1 . When ME balance is

used, then rule is 1=2 to balance heads or P/T=C. When using Enthalpy method & neglect JT correction

then is isentropic X.

P1 T1 V1 d1 Ph1 Vh1 Sum1 P2 T2 V2 d2 Ph2 Vh2 fh2 Sum2

221.8 600 200 1.000 41.05 0.7984 41.9 144.2 390.7 200 0.998 26.73 0.801 14.37 41.90

Table 1 to RHS shows

Temperature results of E

balance by enthalpy equation

with JT correction to dH and

inclusion of friction loss.

Table 2, below from Mechanical

Energy Balance, show 1= 2 for

head balance, sum1=sum2

Nozzle Efficiency & Friction-II

37

Nozzles & Useful Gas Information from L. F. Scheel

Nozzles, after Scheel 1.1 Co

Conoidal mouthpiece 0.98

Short Cyl. Round edge 0.92

Short Cyl. Square edge 0.82

Pitot Tube 0.86

Sq. Edge Orifice Plates

d/D from 0.1 to 0.6 0.65

d/D from 0.61 to 0.74 0.68

d/D limit 0.75 0.72

Scheel: ft/sec=CoH

Other useful formula from Scheel are:

•Correction of Nozzle flow for initial velocity:

•Ft/sec = V = 223.8Co {(h)/(1-[A]2/[A]1)} where dh is enthalpy change in BTU/#, 223.8 is root J2g.

Once again the use of Co shows that h1=h2 is valid only for frictionless ideal flow, not for real conditions.

The use of Co, orifice factor , is equalivalent to using friction term in total energy balance. This is easily

verified by dH definition of flow coefficient.

•Choked flow: #/min= Co 6.813 Psia(d”2/Z (k MW/oR) & Choked acfm =73.3d2Co(koR)/MW)

•For generic hole use orifice Coefficient of Co of 0.60, Fliegner Equation factor

•Reynolds Number for Gas: Re=0.105(#/min)/(d” * viscosity-cp) = (#/HR/SF)(D-ft)/(2.42cp)

•Sonic V, ft/sec =224(koR/MW) V, fps=3.06(acfm)/d2 #/sec=MMSCFD(MW/32.8)

•Pc= Po((2/k)+1)(1/(1-1/k)) where k=Cp/Cv for air at 14.7psia the critical pressure ratio is 14.7/27.7 =0.53

•Den #/cf =PM/(10.73ZoR) where T Rankin , P psia, MW is molecular weight

•Gas Valve flow coefficient C.V.=(SCFM){(SG*T)/(520p*Psia.out)} p is psi

•Horsepower = (#/min) (dH BTU/#)/42.5, ideal k =Cp/Cv =Cp/(Cp-1.986), Cp is molar Cp.

-S/P|T = V/T|P (R/P)[1+27Pr/(32Tr3)] & Cp-Cv R[1+27Pr/(32Tr

3)] Berthelot EOS approximation, Walas p58

Gas k Avg T, Scheel Po/Pc

mono 1.67 He Ar 220F 2.04

Diatomic 1.40 Air H2 180F 1.88

Tri 1.30 CO2 Steam 170F 1.85

Poly 1.20 HCCH, nC2, 135F 1.82

Heavy Org 1.10 nC4 Bz 105F 1.67Scheel H is feet of head change, 64.4 =2*32.2; wet steam can have a k low

as 1.12, k=1.3 for super-saturated steam or super heated steam in the 200-

300 psia range, per Faires p 406

38

Generic Method for Gas Expansion with friction SvN EX.10-4

Smith-Van Ness Improved Cp@1 atm.Cp, H&U by expan Coeff

Friction Energy F = (RTa/M)ln(P1/P2) -(V2/2g)/778 F = (G/LM)2(2fL/D)/32.2/778 , fanning f

Total Energy (H+(V2/2g)/778)1 + Q = (H+(V2/2g)/778)2 (U+(V2/2g)/J)1 + Q = (U+(V2/2g)/J)2 or Clausius

dP Friction (G2/g)[ln(1/2)+(2fL/D)] = -M/(2RTa) (P2) (G2/g)[ln(1/2)+(2fL/D)]= -M/(2RZaTa) (P2)

H, , & Cp dH =Cp (T) & =PM/RT dH=Cp (T + (P)JT) & = PM / ZRT

Results of SvN example problem are reviewed by a sum of heads method.

Improvements are possible as the sum of heads do not balance, LI#6. The error is

not in the Z factor. Z at inlet and outlet are about 0.999. It is proposed that SvN

method incorrectly calculates the amount of heat required. They use Q=dH+dVhd.

Determination of heat using SvN for the C3= Joule expansion requires about

7BTU/#, while it is known that Joule expansion is adiabatic & dU=0=dQ. See pg.

Term B TU/# Head in Head out L I

Wrk & F btu/# 29.6 19.0 1

V hd btu/# 1.4 6.5 2

P Hd btu/# 36.3 43.1 3

s um S vN btu/# 67.2 68.6 4

S vN E x.10-4 P 457 5

eror% 0.12% 0.30% 6

V. L. Streeter,5th Ed.‟71, p112 Eqn.3.2.2&.7 & Perry Eqn.5-31, state: dQ - dW = d(U+(V2/2g)/J). This

relationship of heat, work, and internal energy correctly calculates the heat for a Joule Expansion in

absence of work & zero Kinetic Energy change as zero, while dH can be a positive value. The use of dU to

determine heat requirement also finds that Isothermal processes require heat addition. Where-as SvN

incorrectly equate isothermal processes as being constant internal energy processes, which it is not. Smith

carries this incorrect concept over to his Unit Operations book also. The reduction in heat requirement from

using dU more closely matches simulator heat requirement, line item 1 vs. 4 & 5 of attached table. Using

Smith values in the dU equation determines heat, dQ, as (7.1-1.986)*(170-100)/29+(6.5-1.4)=22.8BTU/# vs.

simulator result for identical outlet conditions, L.I. 4,5 of dQ of 24.2. The simulator determines for Smith

heat addition of 29.6 BTU/# the outlet temperature rises to 193F, LI#2, Table B, not 170F as problem states.

The value of heat addition rate may be verified by Clausius Formula: Friction=TdS-Q or TS=T Cv

[(T2/T1)k(P2/P1)

(1-k) ]. Friction by SvN is 19 & TdS calculates as 0.66 at T of 580R, k=1.4, so

Q=19.6+0.66=20.3BTU/#.

If the difference in heat requirement from dH and dU is also looked at as the difference in net friction.

Then one can correctly conclude that friction in gas flow process equate to heat generation.

39

Compare Simulator Results for Air Flow SvN EX10.4 1000#/hr 1.049"Idstl

# Condition Ti To.F P1/P2 P1 dH Qin L' Vin Vo fps dVhe Ta

1 SVN10.4 70 170.0 1.83 34.7 24.50 29.6 36.0 261 570 5.13 120

2 Sim BB 70 192.2 1.70 34.7 29.73 29.6 36.0 261 547 4.60 131

3 Sim BB 70 170.0 1.69 34.7 24.33 24.2 36.0 261 525 4.14 120

4 Sim LM 70 170.0 1.69 34.7 24.33 24.2 36.0 261 525 4.14 120

5 Sim BB 70 170.0 1.83 34.7 24.33 24.2 46.3 261 566 5.04 120

6 Sim LM 70 192.2 1.83 34.7 29.73 29.6 46.3 261 586 5.50 131

7 KKGT 70 170.0 24.20 261 525 4.14 120

8 KKGT 70 192.2 29.53 261 586 5.50 131

'BB=Beggs Brill, LM Lockhart-Martin'li, KK Keys Kenan Gas Table Sim.EOS Soave RK

sim. Z=1.00 dH&dQ btu/# V fps SvN add Vhd to dTCp & over est'mte. Q

Gas in out

T , F 70.0 170.0

P , 'ps ia 34.70 19.00

#/hr 1000.0 1000.0

L pipe E q ft' 36.0 36.0

di inch 1.049 1.049

MW 29.0 29.0

C p btu/#/R 0.245 0.245

u cp 0.019 0.019

Z 1.000 1.000

Gas Expansion w/ friction: SvN Example:10.4 Disagree w/Process Simulator Results

This review of SvN

X10.4. SvN results

do not agree with

Process simulator

results. Conditions

are listed in Green

Table. SvN question

is: what are outlet

pressure & required

heat for listed

values? Results of

X10.4 by SvN are

presented in Line

item 1 of LHS table.

The pipe simulation results (#2) show in case of SvN heat quantity, the resultant

outlet pressure is 20.4 and temperature is 192.2F. It appears SvN method

incorrectly sets dH=dU for heat required. Where-as the simulator method takes dH

as the heat requirement, without adding velocity head. A check of dH based on

temperature was made from Gas Tables for air, LI 7. The check shows agreement

for the dH by all 3 methods , LI 1, 3-5 & 7. SvN acknowledge an “acceptable”

error with their method. A 25% heat error and 14% outlet pressure error indicates a

systematic error by one of the methods. LI-5 shows that heat required to match

SvN outlet pressure is only 24.2BTU/# with 46.3 feet of pipe., for same diameter

of steel pipe. The Simulator results with either Beggs Brill or LM method give near

identical results. It is this recommendation to determine heat based on dU and V

head, which more closely match the Clausius Eqn. F=T*dS-Q, with dH reserved for

work, Which agree better with the H-S method of work.

40

The SvN method is an over specification of a problem with one (1) degree of freedom, Gibbs Phase rule:

2-1+1-2. The SvN problem specify inlet P&T + outlet T + isothermal EOS. Problems of two degree have 2

variable, P&T. Solution is by either one equation & 1 specification or two equations + spec of initial

conditions. One cannot add a constraint (reversible, isothermal, isentropic, plus 2 equation. SvN method

uses not 1, not 2 but three (3) equations to solve a problem with 1 degree of freedom. Thus the SvN

method is questionable.

As dH & dU determine heat requirements. The dH & dU are functions of temperature with minor

variants for pressure on volume. However by also using the isothermal expansion, the variant of process

path is fixed. The proposed equation is: (R.B.Bird 1957 in D.M.Himmelblau, 1974 p291&304 eq4-30):

(U+(V2/2g)/J)1+(Q-W)=(U+(V2/2g)/J)2+(PV)+F.

“A balance on mechanical energy (ME) can be written on a microscopic basis for an elemental volume…

where F represents loss of ME, i.e. irreversible conversion by the flowing fluid of ME to internal energy, a

term which must in each individual process be evaluated by experiment or .. a simular process.. AKA

Bernoulli .. Where friction losses can be evaluated .. from handbooks with aid of friction factors or

orifice coefficients.”. Individual terms may be evaluated as follows: For a constant volume Joule

Expansion with W=0, U0 & (PV) = PdV + VdP = P*0+VdP =+VdP, the equation is

((V2/2g)/J)1+Q=((V2/2g)/J)2+VP/J+F The friction term is: F = (G/LM)2(2fL/D)/32.2/778. The expansion

work is -(RTaln[P2/P1])/MW. The thermal energy Q is enthalpy change, as shown in simulator results or

H=Cp (T + (P)JT) & = PM / ZRT & LM =(2-1)/ln(2/1)

The grand equation is: [(G/)2 + (G/LM)2 (2fL/D)]/50000 – 1.986Ta [LN(P2/P1)]/MW = Cp [T + (P)JT]

This equation for the case of zero heat addition reduces to the familiar isothermal pressure drop

equation:

[M/(2ZRTa)] (P2) - (G2/g)[ln(2/1)+(2fL/D)] = 0

The detailed pressure drop equation without expansion work reduces to Weymouth type (P2) friction

equations. In all events, specification of an outlet temperature leaves 1 degree of freedom, P & one

equation. A spread sheet solution is simple by adjust on P2 to converge solution.

Gas Expansion with friction SvN Example 10.4 Simulator Results not agree w/ SvN Method

4141

The adiabatic flow equation derivation uses ideal gas equation for isenthalpic expansion, and Cp-Cv=R to

arrive at:

(fL/D)M =(4fL/D)F = (1/Mo2 – 1/M2)/k + ½(1+1/k)ln{[(Mo/M)2][(k-1)M2+2]/[(k-1)Mo

2+2]}

Where: M=Moody Friction Factor, F =Fanning Friction factor , M is Mach number, (ratio of thermal to

Kinetic Energy), k is ratio Cp/Cv, Subscript o is initial condition, fL/D is dimensionless skin friction

number, when expressed with Moody Friction factor is same as Crane K, which can also be expressed as

valve coefficient, Cv, as pointed out by this paper. If Y=[(k-1)M2+2], Density, Pressure and Temperature

ratio are

To/T=Y/Yo & o/ =(M/Mo)(Yo/Y) & Po/P =(M/Mo)(Y/Yo).

One problem with the adiabatic equation is it predicts a specific M for a given Cv, where-as it is possible

to select an outlet diameter that forces M=M. Looked at another way a given Cv must have a fixed Mach

ratio, which is a doubtful result. Another approach is to relate T=f(P) by the JTX coefficient or 1-1/k = m

but use Schultz definition of m at zero efficiency, m= (P/T).jtx. This determines a Polytropic k based on

a JTX, which appears more realist compared to dS=0 assumed in the „adiabatic” isentropic approach.

Because any process involving friction is not isentropic. An isentropic temperature drop is greater than a

JTX temperature drop, as indicated by Schultz approach or looking at the H-S diagram, such as for C3=.

Another method is to use real gas energy balance equation, such as: (R.B.Bird 1957 in

D.M.Himmelblau,1974 p291&304 eq4-30): (U+(V2/2g)/J)1+(Q-W)=(U+(V2/2g)/J)2+(PV)+F.

U=Cv(T + TJX) or U=Cv(T + (1/k)TJTX) or H=Cp(T + TJTX) & use real density =PM/(ZRT)

The prior article proves the validity of this total energy approach and the results so obtained do not

introduce ideal gas assumptions. Nor does the total energy equation bias results by introduction of a

predefined thermodynamic path. The TJX & TJTX are just the Joule or JT Expansion temperature

change using the low pressure Cv or Cp and is thermodynamically correct.

Gas Expansion with friction –Adiabatic Flow Equation is an Ideal Gas Isenthalpic Expansion

42

Equation for ideal gas adiabatic compression head: Ha = RT1(k/(k-1)((P2/P1)(1-1/k) – 1) but

Rk/(k-1) = (Cp-Cv)(Cp/Cv)/(k-1) = (k-1)Cp/(k-1) = Cp so rewrite Ha as Ha=T1Cp((P2/P1)(1-1/k) – 1)

& for adiabatic gas T2=T1(P2/P1)(1-1/k) Distribute T1 inside radical of

Ha = Cp(T1(P2/P1)(1-1/k) – T1) = Cp(T2-T1) & dHn= Cp(T2n-T1) efficiency adjust head

The above shows adiabatic compression head is just an expression for ideal gas enthalpy change less the

velocity heads. The ideal compressor or expander does not have velocity heads.. In practice velocity heads

are mitigated by keeping Mach numbers low with selection of inlet/outlet nozzle sizes.

Since adiabatic head is just enthalpy change, it seems more prudent to improve estimates of compression or

expansion temperature change and improve enthalpy evaluation methods. This will more closely match a H-

S chart. Especially so, where gas conditions have large deviations from ideality. The Schultz method is one

attempt to better determine temperature change but failed by double counting efficiency. Enthalpy

calculation is next considered.

Clearly: Power is just product of head and flow, adjusted for driver inefficiency. Some engineering texts double or treble

count the efficiency term for power calculation: examples are: Perry 5th ed Eq.24.20 to 24.24, C.Branan GPC 1976 Eq.3.2&3.4,

Evans, GPC pp42&58, Ludwig GPC 1983 v3 Eqn12-54, Coulson et-al pp94/5 to cite a few readily at-hand, likely there are

others. Once the head has been adjusted for Polytropic efficiency, the only further corrections to shaft power are losses from

driver source power to impeller. Conversely, for expanders, application of double or treble efficiency‟s under estimates the

available gas power. This concept is clearly seen in the H-S diagram, where once head is determined based on efficiency,

power is just product of flow times head. Shaft power adjusted for „bearing & seal friction‟. If this effect is corrected in the

Schultz equations, a much better fit to the H-S power is obtained. Repeat: one does not determine Polytropic head by

Polytropic „n‟, then re-correct Polytropic Head by efficiency or worse still, additionally take another correction of Polytropic

efficiency to the product of head and mass rate: repeat: true head times mass rate = gas power & once head adjusted for

efficiency by use of Polytropic n, ,no further adjustment needed.

Ideal Enthalpy Equation v. Adiabatic head & cautions

43

h = Cp T + (V –T[(V/T)p=c )P (11.17 Eqn. 11-26 Faires & 11-7 Walas 1985), with

JT = {T[(V/T)p=c ]- V)}/Cp so - CpJT = (V –T[(V/T)p=c ) gives h = Cp T + (-CpJT) P

H = Cp (T - JTdP) = Cp (T - TJTX), & U =Cv (T - TJX) C p or C v are C @ 1atm.

The true dH could be also be written in terms of = [(V/ T)p=c ] as JT = (T -1)/(Cp). The above is same

as adding an additional dT.jt. But expansion coefficient were found more accurate than of adjusting Cp for

pressure. Either should define a dh with improved match of H-S chart enthalpy. Calculation of JT coefficient:

JT, by H-S chart is also possible, ie: for C3=, pg94 Scheel, P1=125, P2=20, P=-105, & T1=100F, T2=-35F,

Ta=67.5F T=-135, h=-40BTU/#, Cp=0.35BTU/#/R, & Cp =0.36 at avg. T of 528R =(4.234+.0206R)/42, p270

Ludwig vol.3, or 0.363 SvN p106 so take average of 0.361.

JT= (dTa – dh/Cp)/dPJT =(-135+40/0.361)/-105 = 0.23, about what viral EOS gives if in oR/psi or 3.4R/atm,

see JT comparison page. by Scheel HS chart JT= (T/P)h=c =25/105 = 0.24R/psi =3.5R/atm exactly as given

by BWR EOS.

H & U of real gas as function of JT & J Temperature change:

Callendar Schultz HS chrt*

T2 w/P corr Cp -42.5F -43.3F -38.5

dH w/P corr’n 47.6 btu/# - 40

dH w/Cp @1atm 40.03 btu/# 40.3 btu/# 40

Table on RHS compares T2 & dH calculations.

Using above dH formula w/o Cp pres. correction

gives best dH results. Neither method matches T2

of HS chart. The Table confirms correct Cp is Cp @

1atm as this same equation used to correct dH for

pressure. *P-H CRI 1961 Carrier #144, Not same

as Scheel Chart H-S shown by this document, h

of Scheel chart for P1=125, P2=20, T1=100F,

T2=100F is: -[((20-125)/14.7)* 3.5 ] 0.361 = h =

+9.03 BTU/# H –isothermal. & H isothermal =

137 + 9.03 = 146. For Isentropic at same

pressures: T2= -35F & h=

& h=[(-35-100)-((20-125)/14.7)*3.5] 0.361 =

h=(-135 +25)0.361= -39.71 = h isentropic Vs. chart

of –40BTU/#. For U=0, T2 =93F,

h =0.36[93-100-((20-125)/14.7)*3.5] = +6.5 btu/#.

Sign Convention Check: H=0 & U=0 is always

C(dT-dTx) or dT=dTx & dT is mostly negative

44

Determine Cp by Explicit EOS: (Cp/P)T=c = -(JTCp)/T|p=c

R. L. Callendar EOS: V=RT/P+b+a/T3.33

Cp = 7.2R (P/Pc)/Tr4.33

given without derivation

Or by Berthelot EOS

Cp = 2.5R (P/Pc)/Tr3

given without derivation

Compare Cp Corr Method

y = -16.149Ln(x) + 0.4203

R2 = 0.9957

0.0

5.0

10.0

15.0

20.0

25.0

30.0

0.380 0.480 0.580 0.680 0.780 0.880 0.980Z.rk

dC

p Callendar

Berthelot

Chart2-237

BCdr.avg

Log. (BCdr.avg)

AIR dCp T=-50C Cp P=1atm Cp p=100

ICT Table 0.174 0.238 0.412

dCp Calldr 0.141 0.238 0.379

dCp chart 0.100 0.238 0.338

dCp Bero' 0.096 0.238 0.334

Chart/Table Perry p3.237 & p3.134 BTU/#/R

Tc K / Pc atm 132.8 37.2The comparison table for air

shows the Callendar EOS

provides superior estimation

of heat capacity change with

pressure vs. Reduced

coordinate Chart or the

Berthelot EOS Equation.

It is no surprise that dCp

correlates well with Z. It

appears the Callendar method

of estimating dCp is an

improvement over other

methods. N.B. When using the

corrected Cp then is

necessary to also use

corrected Cp-Cv value to

determine isentropic

exponent to determine T2 of

adiabatic expansion or

compression. Recommend to

use the JT correction for dH.

Note: R. L. Callendar produced the 1st steam tables and Mollier used the

Callendar EOS to make the well known chart for steam,Cira 1925AD. Also,

the Keys & Kay Steam Tables use the Callendar EOS for superheated steam

properties, Wiley Publishers

45

Schultz‟s

method

On page 80 example he

wrongly uses Wa =Wp/e.

Polytropic work Is Actual work

Schultz’s Method For dH & dT Adiabatic, Perry 5th ed. Eq.24.20 to 24.24,

46

Schultz’s X

Chart

47

Schultz’s Y

Chart

48

This result questionable

Not agree w/ Perry

No precise reference

A ?? Mod of Schultz’s dH & dT method

49

Express = Cp-Cv for real gas as z =Cp-Cv

PV=zRT- real gas Law

Define H=U+PV & dH/dT = dU/dT +d(PV)/dT (valid only for reversible processes)

dH/dT = Cp & dU/dT =Cv & d(PV)/dT = d(z T)/dT

Use d(z T)/dT = z + Tdz/dT

Sub the values into: dH/dT = dU/dT +d(PV)/dT to get:

Cp = Cv + z + Tdz/dT or (Cp-Cv) = (z )[1+T/z (dz/dT)]

Cpjt=T/z(dz/dT) or (Cp-Cv) = (z )[1+ Cpjt/2.72] (same as Schultz‟s Eqn.)

a 1st approximation is by neglecting term Cpjt/2.72 is:

Cp - Cv = z

Dimensionless Number Units: Cpjt =(#/CF)(BTU/#/R)(R/atm)(1atm-cf/2.72btu), & H=TCp & U=TCv

Aside: H =U + z(T), =U + PV + VP by steam table @P&T=c, L -> V check:@1atm, 212F:

970.3=897.5 + 72.8 =970.3 & TS=970.3 = H

Compare to Berthelot (1907) approximation for corrected Cp-Cv :

Cp-Cv = R(1+1.7Pr/Tr3). The empirical Berthelot EOS is: Z=1+(1-6/Tr

2)9Pr/(128Tr).

The exact B‟lot value for R correction, d(z T)/dT, is R(1+1.7Pr/2Tr3), hence the word, „approximation‟. B‟lot EOS is a

truncated viral with set to zero. The B‟lot Bo term is 0.07-0.42/Tr2 Compare to Abbott's Bo term of Bo=0.083-0.42/Tr

1.6.

Some simple gases with zero or <0.044, are: the noble gas series, air, O2, N2, H2, CO, and methane. For such gases

the B‟lot EOS should be as accurate as the truncated Viral EOS. The empirical B‟lot EOS is par to vdW or superior at room

T & P<200atm. Walas‟85

50

Review of: Schultz, JM, Polytropic Analysis of Compressors, Trans. ASME J. of Engi. for

Power (84) 1962 Part 1 Jan. p69, part. 2, p222 in Perry’s 5th Ed. Pg.24.34

T2=T1{P2/P1)(m)} m=(Z/Cp)() & =(+X)expansion =(1/ + X)compress

Xs {[(T/V)(V/T)p ] –1} {ln(Z2/Z1)}/ln(Tr2/Tr1) & Ys{(-P/V)(V/P)T } 1-{ln(Z2/Z1)}/ln(Pr2/Pr1)

Now recast X, above in terms of JT function, =(T/V(V/T)p-1)(V/Cp)

Cp /V = {[(T/V)(V/T)p ] –1} = X with 1/V= =( + Cp)expansion

ms=Z(Cp - Cv )/Cp+ZCp/VCp =Z/Cp+Z/V=Z(1-1/) + Z/V

By real gas law P/T=Z/V so m=Z(1-1/) + P/T

At =1 for ideal gas, (Z=1 & =0) reduces to the familiar T2= T1 (P2/P1)(1-1/k) At =0 for ideal

gas, (Z=1 & =0) gives T2=T1 . Few gases are ideal so at small , expect big errors.

For the locked rotor case where =0 gives T2= T1 (P2/P1)(P/T)

Schultz equation results are normal as it defines as: =(P/T)ln(T2/T1)/ln(P2/P1). LF Scheel

shows @ locked rotor case dT= (dP). Schultz says dlnT=(P/T)dP =dT/T=(P/T)dP/P or

dT=(dP) Schultz method gives improved results compared to the ideal gas case, see

following charts.

Since T =-(1/V)(V/P)T & Ys{(-P/V)(V/P)T } then Ys=PT

The term T is the isothermal compressibility aka Cg in petroleum reservoir analysis.

To keep ( + Cp) unit less, when (#/CF)(R/atm)(BTU/#/R), use the conversion of 1CF-atm per 2.72BTU

or 0.73/1.986 = 0.37 or ( + 0.37Cp).

51

ms = (Z/Cp)(+Xs) & m.jt= (1.99Z/Cp)( + 0.73jtCp/1.99) & m.ideal=(1-1/k) & Cp = 4.234+oR/48.54,

btu/mol/R for C3= only, jt per the RK EOS,R/atm.; =PM/(Z10.7oR), #/cf; Z per RK EOS, Pc Tc & Xs as

per data given here-in. k= Cp /Cv .

The ideal case has large error at =0 as it predicts Tin=Tout, vs a JT expansion for locked rotor as

shown by Scheel on the H-S chart for C3=.

The JT correction method without Cp correction is an improvement over other methods, if one lacks H-

S charts for a particular gas.

calc error for Xpander Outlet T

0.000

0.010

0.020

0.030

0.040

0.050

0 0.2 0.4 0.6 0.8 1

Expander Efficiency

ab

sE

rr T

ou

t v

H-S

Err JT.rk w/Cp corr

Err ideal

Err SchulteX w/Cp corr

Err JT.w/o Cp corr

Compare Temperature difference to HS chart of C3= for ideal, Schultz, & JT

52

By Schultz: ns=1/(Y-ms (1+Xs)) & for ideal case n=(1-1/k), k=Cp/Cv ideal.

dHpoly = (-n /(n -1))(ZRT)1((P2/P1)((n-1)/n)-1).

Simpler just calculate dH from JT equation: dh = Cp (dT - JTdP), where - JTdP conveys non-ideality,

per appendix. The above chart shows the lowest accuracy for the ideal case. Were efficiency not double

counted in the Schultz method, it would be the more accurate method. However Schultz method

requires a further 3 parameters over the JT method,

dH @ e=1 is 40BTU/# from HS C3=

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

0 0.2 0.4 0.6 0.8 1efficiency

abs(

dH

clc-

dH

h-s

)

Err JT.rk w/Cp corr

Err ideal

Err SchulteX w/Cp corr

Err JT.w/o Cp corr

Compare Enthalpy difference to HS chart of C3= for ideal, Schultz, & JT

53

Gas in out Gas calc in res ult out.R es ult imed res lt in out

T , F 751.400 445.400 T den pcf 0.190 0.129

P , 'ps ia 88.200 ps ia 22.10 V fps 71.49 0.62

#/hr 4930.000 4930.000 fric tion Hd ft 0.0 81482.6 Ai" s q. 14.499 14.499L pipe E q ft' 10000.000 10000.000 Work Head ft -58413.6 0.0 f 0.0040 0.0040

di inch 4.296 4.296 V head ft 79.4 706.6 R e 381579.63 381579.63

MW- N2 28.000 28.000 P head ft 66786.2 49916.0 Efficiency poly.idl 0.77

C p btu/#/R 0.248 0.248 T hd ft C v 123629.1 0.0 P 2 es timate 44.69

u cp 0.019 0.019 s um 132081.1 132105.2 G #/s ec/s f 13.6

Z 1.000 1.000 Mach 0.04 0.00 2R Tz/M avg 32.4

Anal:C ouls on X3.15:P o=22.1ps ia ,To=445.4F ,poly.E ffic.=0.75&H t=-59950, &this w/e=0.75=>P o=21.6,To=451F &H poly=-57226'

Anal: X3.15 phead tot.=59,950 feet vs ca lc 58414, diff is V H d, us e X3.15 P i/Ti 88.2/751.4 & H w=58414 w/ 77% eff. To/P o445.4/22.1

us e HP =e*(Work-ft)*(#/min)/33000

N2 Expansion Turbine Calc Performance by Pipe Friction

The above example from Coulson X3.15:Po=22.1psia,To=445.4F,poly.Effic.=0.75&Ht=-59950, &

this w/e=0.75=> Po=21.6, To=451F & Hpoly=-57226„

Anal: X3.15 phead tot.=59,950 feet vs calc 58414, diff is V Hd, use X3.15 Pi/Ti 88.2/751.4 &

Hw=58414 w/ 77%eff. To/Po445.4/22.1

When calculating U, or internal energy change, the Joule correction is neglected. This is not

considered a serious omission because jV is small or U=Cv(T + jV) CvT, as seen by how close the

energy balance closes.

54

Cross Correlations of Coefficients I

For ideal gas: JT =0, = 1/T, T =1/p, s =1/(p), & as=1/(s ), Cp- Cv =

For Real gas: JT =(T/P)h , =(V/T)p , T =-(V/P)T , s =-(V/P)s ,

JT = (T/P)s -1/(Cp) = (T -1)/(Cp) = [T(V/T)p -V]/(Cp)= [(P/T)v=c]

Joule Free Expansion: J (T/V)U=c =(-1/Cv )(U/V)T=c =-[T(P/T)V=c – P]/Cv compare to

Joule Thompson Exp.: JT (T/P)H=c =(-1/Cp )(H/P)T=c = [T(V/T)p=c – V]/Cp

Cp - Cv = T2/(T) & Cp/Cv = = T /s & s = (T/P)s & as=(P/)s

s - JT =V/Cp = ZT(1- s/T)/P & = (V/P)T(P/V)s & as(P/)

T =[1/p-(Z/P)/Z] & = [1/T- (Z/T)/Z] & JT =(T2/P)(dZ/dT)/Cp

V/V =[()T + (T)P] & JT = (T/P)s -1/(Cp)

Routines for z & T commonly available using Tr, Pr for, reservoir work, like:

T =(1/(PcPr))(1/(1+(r/Z)(Z/r))) where Z & r from BWR EOS

(Z/r)=[5Ar5+2Br

2 +Cr+2Er2(1+Fr

2 –F2r4)exp(-Fr

2)]/(rTr)

For example at Tr=1.7, Pr=2.8, Pc=361.5psi T =0.001/psi & 1/V = m = (Psia)/(z10.7/oR)

: 1.98BTU/mol/R, 0.7302atmCF/mol/R, 1545ft#/mol/R & J = 778ft#/BTU

55

Cross Correlations of Coefficients II

(Cp/P)T=c = -(JTCp)/T|p=c = -[{T2/P}(Z/T)]/T|p=c for Cp of P1

(dCp)T = -(2T/)dP (Cp/P)T = -TV(2+ ( T)p) & (S/P)T= -(V/T)P

S/P|T=c = -V & S/T|p=c = Cp/T & S/V|T=c = (/T) = (P/T)v

(U/P)T = (PT -T)V = (1+PT - T)V & (U/V)T = {(T/T) – P}

U=(Cp - p/)T + (PT -T/)P = U = Cv(T-JV)

(h/P)T =(1 - T)/ & (h/V)T =(T-1)/T

V/V =()T + (T)P & (/P)T = -(T/T)P & (P/T)v = /T

JT = (T -1)/(Cp) where = [(V/T)p=c ] = [(P/T)v=c ] T & =1/V= P/zT

T =[1/P - (Z/P)|T=c /Z] & = [1/T - (Z/T)|P/Z] & JT={T2/P}(Z/T)/Cp

H = Cp(T) - [T(V/T)p -V]P & U = Cv(T) + [T(P/T)V=c – P]V Perry4-144/151

But (Cp)JT = [T(V/T)p -V] & -J Cv = [T(P/T)V=c – P], so H & U by JT & J is =

H = Cp(T) - [(Cp)JT ]P & U = Cv(T) + [-J Cv](1/V)

The above 2 equations may be used to correct U & H for conditions other than 1atm.

H = Cp(T- JTp) & U = Cv(T-J(1/V)) for gas expansions with H=0 or U=0

TJTX =(JTp) & TJX = (-J(1/V)) AKA isenthalpic w/H=0 & constant U ie U=0

56

Corresponding States Concept for EOS & ZThe corresponding state idea is the basis for Z charts. It is also the basis of any EOS i.e. use Pr & Tr. The idea

is that all gases have identical Z at the critical condition of Pr=Tr=1. This is not always a good approximation,

as seen by the table below. Zc for hydrocarbons and the average for all gasses is 0.27. Thus Zc = 0.27 is the

basis for most Z charts. Polar molecules have smallest Zc, whilst spherical molecules have largest Zc. The

vdW EOS Zc is 0.375, while RK EOS has Zc about 1/3, the BWR form presented here has Zc of 0.27.

Limitations to EOS validity range should be strictly observed. For example, Berthelot and 2 term viral

calculate Zc of 0.65. The recommendation for the 2 term viral is for Vr>2 & below 2 use Pitzer Charts. The

Pitzer Charts also use Zc of 0.27 but adjust for a given gas by . Therefore the potential for severe

departures from true values should Vr<2 or Pr>30 or other range limits set by an EOS. One correction

proposed for H2O, H2, He, Ne, NH4, to improve accuracy for the vdW EOS is the Newton correction. It uses

Tcn=Tc+8K and Pcn=Pc+8atm. Even with these corrections there may be significant error unless Tr>2.5. Faires

pp216/217. Others only recommend Newton correction for H2O & H2 if using Z charts. A vdW gas fitted to

Pitzer viral form has of –0.303. The next idea uses tailored EOS fitted around Zc for a particular gas or gas

type, e.g. BWR given here. Advanced EOS use an accentric factor, , to make adjustment for effect of Zc.

Examples are LK-BWR, S-RW, and Ping-Robinson. All of which use complicated, empirical, and sometimes

proprietary mixing rules. An advanced EOS is needed when needed precision exceeds that of a two

parameter equation. Some examples are: High Concentrations of Polar/Ionic gases, Quantum Gases or

conditions near the critical point. A comparison of results can give an idea of whether to use an advanced

EOS. If the imposed limits are exceeded or a large scatter is seen among results, this may justify additional

investigation. At last count, the number of EOS‟s are hundreds. One of the latest is a 5th order viral, termed

ELIR or extended linear isotherm regularity. ELIR gas is defined by 2nd & 3rd viral coefficients and Boyle

density is family of curves for Z/P|T=c=0 & b=((PV)/P)T=c=0. Boyle Temperature is T at P=0 for which

Z/P|T=c =0

' C ompare Zc to E OS and Z chart C orres ponding S tates : P itzer Viral Zc, 0.24->0.29 & BWR * 0.27

Gas CAN HCN H2O NH3 nC6 Av g Gz BWR* Z chrt C2 C3 CO2 C1 Air H2&He Ne RK v dW

Zc 0.189 0.197 0.233 0.243 0.264 0.270 0.270 0.270 0.270 0.277 0.279 0.290 0.290 0.305 0.310 0.310 0.370

Zc=0.291-A (Yamada Liquid A=1, LK-BWR, A=0.085 & Edminster, A=0.08, Pitzer, A=0.28)

57

Kinetic Theory of Gases

The Kinetic Theory of Gases provides insight to fundamental relations among gas properties like friction,

or momentum, heat, and mass exchange. Kinetic Theory proposes: /=Ð=/Cv=Vrms /3. Stated as,

viscosity over density equals Diffusivity, equals heat conduction coefficient over density times constant

volume heat capacity, which all equal 1/3 of the root mean square velocity of a gas molecule times the

mean free travel path of the molecule. Each of these 3 coefficients also have Dimensionless Numbers as a

basis for engineering calculations. Collectively known as the Colburn analogy: f/2= jH =jM. Friction factor,

f, is the ratio of momentum loss at wall to bulk shear. Likewise jH is ratio of heat transfer at wall to the

bulk available heat. Chilton and Coburn verified the analogy for a wide range of compounds, with errors

typically under 10%. The dimensionless numbers; (DG/), (Cp/), & (/ÐM); are known as Reynolds,

Prandtl, and Schmidt. By Kinetic Theory, = ACv & = A so the Prandtl number is the ratio of heat

capacity, (Cp/Cv). Gambil arranged Prandtl numbers of gasses by molecular structure, 0.67 for mono-

atomic gases, 0.73 for linear Non-polar, 0.79 for non-linear non-polar 0.86 for polar and about 1 for H2O

vapor. Kern made the observation that N. Prandtl is useful for physical properties extrapolation. As

Prandtl values at a mean temperature will be valid for a wide temperature range. This is perhaps simular

to the Cp/Cv value for gases. For gas viscosity, kinetic theory predicts an increase in viscosity

proportional to the root of absolute temperature, such as the Sutherland Equation for temperature

extrapolation of thermal conductivity. Diffusivity and viscosity determinations differ little from

considerations given by gas kinetic theory, corrected for temperature effect. Correlations for these

coefficients are given in various reference texts. See following table for details.

Consider the energy balance in terms of Kinetic Gas Theory for a mono atomic gas:

(P/ + V2/2g)1 + EG = J Cv T + (P/ + V2/2g)2 +FL

Sub JRT for P/=PV & 1.5R for Cv : (JRT + V2/2g)1 + EG = J 1.5R T + (JRT + V2/2g)2 +FL

Sub for T= T2-T1 : (V2/2g)1 + EG = J 1.5R T +J RT + (V2/2g)2 +FL

Add the R T‟s: (V2/2g)1 + EG = J 2.5R T + (V2/2g)2 + FL

Use Mono gas table & Sub for 2.5RT as H: (V2/2g)1 + EG = JH + (V2/2g)2 + FL

Notice result is the same when determined by ideal gas law:

A proper energy balance MUST account for friction to specify the result.

58

Kinetic Theory of Gases

Item Viscosity Heat Flow Diffusion

Coefficient /=Vrms/3=74.3(0R/M)= /Cv= Vrms /3 = Ð=Vrms/3=74.3(0R/M)= Vrms /3

term [d(mv)/dt]/A= dv/dr Q/A= dT/dr J=dm/dt=DA dm/dr

property * =27(MT)/[Vb (1+Tc/Tb)] =(Cv+9/mw) Ð*AA1.57Vc

-0.67(0R/MW) Perry 1973

Analogy f/2=0.023/(DG/)0.2 = jH=(h/CpG)(Cp/)0.67= . jM =(M/G)(/ÐM)0.67

Analogy (P).wall/(bulk momentum) = Q.wall/Bulk Q= =m.wall/bulk mass= = Constant

-d # name Reynolds Number Prandtl Number Schmidt Number Lewis #

-d # ratio (DG/), inertia/viscous (Cp/) (/ÐM) = (/kd) (/ÐMCp)

application PD/(4LV2/g)=f/2 jH=(T/t)(a/A)(Cp/)0.67 jd=(P/p)(a/A)(/kd)0.67 Kern 1950

This analogy is limited to Reynolds numbers above 5k and is exclusive to wall drag. It is also limited to moderate

temperatures to eliminate radiant heat transfer.

Ð*AA Sq.ft/hr Critical Volume in CF/Mol. T is degrees of Rankin temperature. These expressions are used to demonstrate

that viscosity and thermal conductivity are proportional (MT) to while diffusivity and sonic velocity are proportional to

(T/M). Improved methods make better accounting for the mean free path or collision integral as in, Perry Ch.3, Kern

pg344. The sonic velocity is predicted by gas kinetic theory as C=Vrms(Cp /Cv/3).

Another dimensionless number, Lewis Number, is basis for humidification or circumstances of co-current heat & mass

transfer. It is the ratio of Prandtl to Schmidt Numbers, known as the Lewis Number or (/ÐM)/(Cp/). Kinetic Theory

predicts a value of 1 for the Lewis number. Experimental evaluation determined Lewis numbers to range from 0.9 & 1.1

for air -water between 150F and 950F.

A consideration is that boiling temperature is 63% of the Critical temperature and gas molecular radius in Angstroms is

estimated at 3.32 times cube root of critical volume in cf/mol, as applied in estimation of diffusion constant above.

59

Dimensionless Numbers of Gas Expansion

The dimensionless Number, CpJT , is a ratio of gas expansion temperature change to friction

temperature change, for dP & JT 0, as TF = PF/(J.) = TJT = (JTCp)P. Which is true only when

0h=CpT+(-CpJT)P. Recognizable as the isenthalpic case using Schultz Temperature relation for the

general isentropic case.

To keep (Cp) unit less, when (#/CF)(R/atm)(BTU/#/R), use the conversion of 1CF-atm per 2.72BTU or R/R

=0.73/1.986 = 0.37CF-atm/BTU or (0.37CpJT)=NJT.

Another dimensionless Number, the Joule Number or NJ = (J/P)CvJ , is associated with the Joule Free

Expansion. It is the ratio of dTw =P(dV/J) and dT.jx = (dV)CvJ , the dV‟s cancel leaving: NJ = (J/P)CvJ .

A 3rd dimensionless number may be formed by taking the ratio of the 1st two, namely:

NJA =(NJT/NJ) = (0.37CpJT)/((J/P)CvJ) = (0.37/J) (P/V)(CpJT)/(CvJ) from pressure & volume change is:

NJA =EJT /EJ = (CpTH=c)/(CvTU=c) = (PCpJT)/(VCvJ) which can be related to the ratio of (H/U) as

the ratio number relates to how to adjust Enthalpy and Internal Energy changes from standard Conditions

of 1 atmosphere to pressure either higher or lower than 1 atmosphere.

H = Cp (T- PJT) & U = Cv (T- VJ) for at H =0 & U =0, then TJX = VJ & TJTX = PJT

Furthermore it seems NJA =(NJT/NJ) must be related to acoustic velocity at real conditions. Sonic velocity is

determined as a =(P/) =-(V/)(P/V)|s ? -(V/)N{(CpJT)/(CvJ)/|s]

This is just a presentation of talking points. The main point is to better understand what Schultz did in

development of his method.

60

Kinetic Theory of Gases provides insight to the relations among gas properties such as, friction, heat,

and mass transfer. Kinetic Theory proposed: R=Nk and also adds the quantumization to account for

energy stored by vibration and rotational modes. Quantization implies molecules have discrete energy

levels. Quantum responses are discrete step functions activated at specific temperature intervals. The

resulting expressions of various monatomic gas (such as He, Ne) properties are expressed in Table 1,

below. Some of the Quantum functions, (Qi) , are as follows:

(Qt1)= (2MkT)1.5V/h3N & (Qr2)= (eT) / & (Qv3)=x/(ex-1) & (Qv4)=ex(x/(ex-1))2 & (Qv5)=Qv3-ln(1-1/ex)

where X = ch/kT. The various terms are: temperature, T , absolute, & , is vibration frequency.

Example from Daniels & Alberty Pg. 577 (1975) is for H2O Cp at 1000K which is listed as having 3 normal

vibration modes at 1000K of 1595, 3652, & 3756 inverse cm. The calculated Cv is:

R(5/2+3/2+(0.66+0.146+0.133)) = 9.81 vs. experimental of 9.77. The terms were modified for 3 degrees

of freedom based on the diatomic system. T.1 Mono

T.2 Diatomic

Kinetic Theory of Gases

Degree disassociation

Pnot

Uconsidered

Cv CP H S

Translation RT/V 1.5RT 1.5R 2.5R 2.5RT 2.5Rln(Qt1)

Rotation RT R R RT Rln(Qr2)

Vibration RT (Qv3) R(Qv4) R(Qv4) RT (Qv3) R(Qv5)

Summary: Gas energy by Kinetic Theory is an exclusive function of temperature. Gas Pressure and

Entropy are related to temperature, specific volume. /=Ð=/Cv=VRMS (/3).

P RT/V

U 1.5RT

Cv 1.5R

Cp/Cv 1.67R

H 2.5RT

S 2.5Rln(Qt1)

A formula for CP is R(1+f/2) & CP/Cv =1+2/f. With f = number of degree of freedom. For the example

H2O vapor at 1000K the degrees of freedom activated are: 2(9.81/1.986-1)= 7 plus 88% of the 8th

degree, thus k=1+2/7.88=1.25 At room temperatures k mono=1.67, diatomic=1.4 and poly at.=1.33.

Kinetic theory Temperature is proportional to the square of Vrms . Thermal equilibrium between 2

objects of different temperature is achieved when they have identical RMS velocities.

61

Properties of Common Gasesgas MW k60F Tc Pc Zc a.vdw b.vdw w.actr A B Zac u.jt.v u.j.v Cp

Air 29 1.41 239.0 37.2 0.284 345 0.585 0.036 6.737 3.97E-04 0.998 0.43 0.08 6.96

N2 28 1.41 227.0 33.5 0.291 346 0.618 0.040 6.839 2.13E-04 0.999 0.42 0.08 6.96

O2 32 1.40 278.0 50.1 0.290 348 0.506 0.021 6.459 1.02E-03 0.996 0.46 0.08 7.03

CO2 44 1.30 548.0 72.9 0.276 926 0.686 0.225 6.075 5.23E-03 0.962 1.16 0.17 9

C1 16 1.32 344.0 45.8 0.290 581 0.685 0.007 4.877 6.77E-03 0.987 0.68 0.11 8.67

C2 30 1.22 550.0 48.2 0.284 1410 1.041 0.091 3.629 1.68E-02 0.943 1.22 0.17 13

C2= 28 1.25 510.0 50.5 0.270 1158 0.922 0.086 3.175 1.35E-02 0.957 1.20 0.17 10.7

C3= 42 1.16 657.0 45.6 0.275 2129 1.315 0.148 4.234 2.06E-02 0.893 1.57 0.23 15.8

H2 2 1.41 59.8 12.8 0.304 63 0.426 0.000 6.662 4.17E-04 1.005 -0.05 0.01 6.9

R12 121 1.13 692.0 39.6 0.272 2718 1.595 0.158 8.399 1.59E-02 0.853 1.84 0.29 17.3

NH3 17 1.32 730.0 111 0.242 1076 0.598 0.250 6.219 4.34E-03 0.934 1.47 0.22 8.65

Toul 92 1.10 1065.1 40.6 0.264 6285 2.394 0.257 1.869 4.30E-02 0.315 2.97 3.11 25.9

Ne 20 1.67 79.9 27.2 0.311 53 2.68 0.000 5.000 0.0E+00 1.003 -1.32 0.02 5

Tc,R Pc atm, Cp BTU/mol/R=A+B(F+460), a atm(CF/m)^2 b cf/m u.jt R/atm u.j R-mol/cf 125psia 560R

Basis of a & b from vdW is Tc & Pc. Joule Thompson & Joule Coefficient Calc by vdW formula. Z calc by viral

equation with Abbott eqn4B at P=125psia & T= 100F. Volume for C3= at these conditions is

0.893*10.72*560/125 = 42.9cf/mol & at 20psia V=0.96*10.72*556/20=286cf/mol, P=(125-20)/14.7 = 7.14

atm & Cv=15.8-1.986=13.81BTU/mol/F. EX: For C3= expanded from 125psia to 20 psia, from initial

temperature of 100F, the resulting temperature is

TJTX = (JTP) =1.57*7.14/ = -11.2R & TJX = J V2 ((1/V)) =0.23*432*(1/43-1/245)= -8.4R

62

Review of C3= Gas eXpansion by thermodynamic states

This table shows results for the thermodynamic state changes of C3= expansion by various processpaths. The values used to determine these changes were tabulated from Carrier Chart #144, as best itcould be read. The initial condition was taken at 125psia and 100F.

The comparison is made for: Isothermal X (JX), Isenthalpic (JTX), and Isentropic (IEX) Expansion paths.The focus is to compare JX expansion to other expansions. Some argue that JX type expansions requireheat addition. No more so than the original Joule expansion required heat to be isothermal. For apedagogical ideal gas only, Isothermal X is just a Joule X aka constant internal energy X. Heat need notbe added to get a JX any more than work need be done to get a IEX. The classic IEX is thru a deLaValnozzle, where about 96% of available energy is converted to kinetic energy, typically for turbineutilization. The G function indicates spontaneity or degree of organization needed. While –dG indicatesmaximum work potential. The S function indicates the probability of the process. Both these quantities,G & S, show that IEX requires more organization than JTX, than JX. Square edged turbulentexpansion favors a JX compared to highly machined curved surfaces with streamlined flows to favor anIEX. The JTX has minimal friction losses and minimal energy conversion to kinetic energy, as comparedto IEX- deLaVal expansion). For JX case: calculate Friction=TS-Q & Q=0 so F=560*0.089=48 BTU/#.For JTX F=TS-Q = 548*0.075=41, for IEX, F=TS+0=560*0=0. So Friction Loss JX>JTX>IEX.Isothermal, ITX, represents a case where heat addition is required.

Gas X C3= initial: 125psia,100F basis: P-H Carrier #144chart S1=1.42 H1=137 Z1=0.932

Condition -dG dH dU dS dT dPsi Cp avg dTjx T2 P2 V2/V1 S2

++ThermX 443.1 51.0 78.3 0.235 200.0 -110 0.425 -7.6 300 15.0 12.1 1.66

IsoThermX 39.6 9.5 2.5 0.094 0.00 -105 0.375 -7.7 100 20.0 6.6 1.51

FreeX dU=0 31.5 6.5 0.00 0.089 -7.8 -105 0.374 -7.8 92 20.0 6.5 1.51

Isenth.X dH=0 4.6 0.00 -5.5 0.075 -25.0 -105 0.369 -7.8 75 20.0 6.3 1.50

Isentropic X -67.7 -40.0 -37.5 0.00 -135.0 -105 0.342 -7.9 -35 20.0 4.9 1.42

Chilled X -117.9 -51.0 -49.0 0.019 -180.0 -118 0.331 -7.6 -80 7.0 12.1 1.44

63

Review of C3= Gas eXpansion by thermodynamic statesThis chart plots results of enthalpy

and internal energy changes for

C3= expansion from 125psia &

100F to 20 psia. The values used

to determine these changes were

tabulated from Carrier Chart

#144. The Z values were

determined and averaged by BWR

and Viral EOS‟s. These two EOS

showed the best JTX

temperatures. The dU was

calculated from dH-d(PV), with

d(PV)= d[(Psia*144)10.73ZT/(Psia

42)/778=1.986d(ZT)/42. The

temperatures for dH or dU =0,

determined by the plot, are the

temperatures for JX and JTX

expansion. The T for JTX is 75F

and 75.5F for JX, by the plot.

-2.0

-1.0

0.0

1.0

2.0

3.0

4.0

5.0

6.0

60.0 65.0 70.0 75.0 80.0

BT

U/#

12

5p

si to

20

psi

T, F @20psia C3= expansion

C3= Exp 125psia/100F to 20psia Carrier #144 Z1=0.932 dU=dH-d(PV)

dU btu/#

dH btu/#

This chart plots results show dT for a JTX of -25F, same as reported in the presented Scheel Chart of this

paper. The dT for JX from this plot is -24.5F. The dT.jx does not agree with the rule: dT.jx=k(dT.jtx). By

this rule, dT.jx =(15.77/(15.77-1.986))(-25)= (-28.6)F.

The prior table was based on using VdW for initial estimate of temperature. The VdW grossly under

estimates temperature change in many cases.

64

Review of Math forms used here:

1. d(uV) = Vdu + udV for real gas Z=f(T,P) with Tc, Pc constant

- ie real gas law PV=Z T, (PV)/T = Z + T(z/T)

- ie real gas law PV=Z T, (V)/T = Z /P+ T(z/T)/P

2 : d(1/Z) = -(½)/Z2 ie dP/dV =Z T[-(½ )/V2] + Z T[dZ/dV]/V

3. Natural logs: -du/u =d(ln[u]) & du=u[dln(u)]

4. Integrated forms: du/u = ln(u2/u1); du = u2-u1; vdv = ½ v2

5. Implicit: (g)/v = - (f/v) / (f/g) where f =F(g,v)

6. Chain Rule: dy/dx = (dy/dv) (dv/du)(du/dx) & dZ2=2Z[d(Z)]

7. Total differential df= (f/z)dZ+ (f/x) dX+(f/y)dY

8. PdV = -VdP+ (PV) & -VdP = PdV - (PV) & d(PV)=PdV +VdP

9. Log differentials: PV=ZRT or dP/P = dZ/Z + dT/T – dV/V

10. ln(X) 1-X for X close to 1:

11. Partial fractions a/((X+a)(X+b)) = m/(X+a) + n/(X+b)

65

The empirical Berthelot EOS is: Z = 1 + 9Pr/(128Tr) - 54Pr/(128Tr3) = PV/RT so V & P are:

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

V= RT/P +(RT/P)9Pr/(128Tr) –(RT/P)54Pr/(128Tr3) & P= RT/V +(RT/V)9Pr/(128Tr) -(RT/V)54Pr/(128Tr

3)

JT (T/P)h= -(h/p)T/Cp = {TV/T|p - V}/Cp & J (T/V)U= (-1/Cv )(U/V)T = (-1/Cv){TP/T|V – P} ________________________________________________________________________________________________________________________________________________________

{TV/T|p - V} =T(RT/P +(RT/P)9Pr/(128Tr) –(RT/P)54Pr/(128Tr3) )/T|p – V

{TV/T|p - V} =(RT/P + 0 +(RT/P)108Pr/(128Tr3) – (RT/P +(RT/P)9Pr/(128Tr) –(RT/P)54Pr/(128Tr

3)

{TV/T|p - V}=(0+ 0 +(RT/P)108Pr/(128Tr3) +(-0-(RT/P)9Pr/(128Tr) +(RT/P)54Pr/(128Tr

3)

{TV/T|p - V}=RT/Pc{1.27/Tr3 - 1/(14.2Tr)} = & T = P RT/Pc{1.27/Tr

3 - 1/(14.2Tr)}/Cp

JT (T/P)h= RT/Pc{1.27/Tr3 - 1/(14.2Tr)}/Cp & TJTX = {[PR]}(RT/CP){1.27/Tr

3 - 1/(14.2Tr)}----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

{TP/T|V - P} =T(RT/V+(RT/V)9Pr/(128Tr) –(RT/V)54Pr/(128Tr3) )/T|p – P

{TP/T|V- P} =T(R/V+(0) +(R/V)108Pr/(128Tr3) – P

{TP/T|V - P} =(RT/V + (RT/V)108Pr/(128Tr3) +(-RT/V -(RT/V)9Pr/(128Tr) + (RT/V)54Pr/(128Tr

3))

{TP/T|V- P} =(0+ (RT/V)108Pr/(128Tr3) +(-0 - (RT/V)9Pr/(128Tr) + (RT/V)54Pr/(128Tr

3))

{TP/T|V - P}(-1/Cv)=(RT/V){1.266/(Tr3) - 1/(14.2Tr) }(Pr)(-1/Cv) but RT/V =P for Z defined by this EOS

J (T/V)U= (P){1.27/(Tr3) - 1/(14.2Tr) } (P/Pc) (-1/Cv ), but dV =-(RT/P/P)dP

TJX = (P){1.27(Tr3) - 1/(14.2Tr)} (P/Pc) (-1/Cv ) -(RT/P/P)dP, the P*P/P/P cancel out leaving (dP/Pc)

TJX = (PR)(RT/Cv ){1.266/(Tr3) - 1/(14.2Tr)} & also Cv TJX = Cp TJTX

Cv TJX = (Cp TJTX ) , so TJX = k TJTX

Berthelot EOS for Joule & JT Coefficients

66

Based on results for both Berthelot and Viral EOS‟s it appears the ratio for (Tjx /TJTX ) is just the ratio of

(Cp/Cv). The slide examines this concept from another view. That is to use basis of the definition of

enthalpy and internal energy, as pressure corrected by Joule temperatures:

H = Cp(T- JTp) = Cp(T- TJTX), & : U = Cv(T-J(1/V)) = Cv(T-TJX) . For JTX H =0;

H =0 = Cp(TJTX - TJTX) = {U + (PV)}={U + (PV)2 – (PV)1} = {U + R(ZT)}, if use avg Z

H =0 = Cp(TJT - TJT) = U + ZR(TJT) = Cv(TJTX - TJX) + ZR(TJTX) = 0, but ZR= Cp – Cv

Cv(TJTX - TJX) + (Cp - Cv)(TJTX) = 0 & distribute: 0= (Cv TJTX - Cv TJX) + (Cp TJTX - Cv TJTX)

0 = (0 - Cv TJX) + (Cp TJTX - 0) = 0 or: Cp TJTX - Cv TJX or: Cv TJX = Cp TJTX

This ratio of (Cp/Cv) = (TJX)/TJTX) is the same result arrived at by the Viral & Berthelot EOS‟s

(Tjx /TJTX )=(Cp/Cv) {f(B.JX)/f(B.JTX)} =(Cp/Cv) {1, for B'lot or Viral EOS} =(Cp/Cv)

Temperature change of a free expansion from same initial conditions to same final conditions is always

greater than the temperature change for a JT expansion? Why because Cp/Cv is always >1. These results

appear confirmed by steam table analysis as per below, one of few listed dU values.

A further result is that if one accurately determines temperature change of one expansion, the other

expansion path temperature change can be determined by the k value of the gas.

This is very surprising as the literature calls a free expansion an isothermal expansion, where-as what this

work shows is that a Free or Joule Expansion produces lower temperatures than a Joule Thompson

Expansion.

Repeat: ratio of Joule Free Expansion temperature change to Joule Thompson temperature change is ratio of

Constant Pressure heat capacity to constant volume heat Capacity: Tjx = (Cp/Cv)TJTX = (k)TJTX

Ratio of dT Joule & to dT Joule-Thompson by Energy Equations

Sat.Steam: 325 to 162.5psia k=1.18 & k rule 1.15U BTU/# H BTU/# T1, F T2 H=C T2 U=C dT H=C dT U=C

1117.5 1203.5 424.5 375.5 368.0 49.0 56.5

67

Use results that the ratio for (Tjx /TJTX ) is just the ratio of (Cp/Cv) or k. The slide looks at a constant

enthalpy expansion from a PV standpoint: with proof from H-S chart of (Tjx /TJTX) = (Cp/Cv) = k

H = Cp(T- JTp) = Cp(T- TJTX), & U = Cv(T-J(1/V)) = Cv(T-TJX) & for JTX

H =0 = Cp(TJTX - TJTX) = {U + (PV)}={Cv(TJTX - TJX) + (PV)} = 0

H =0= {U + (PV)}={Cv(TJTX - TJX) + (PV)} = {Cv(TJTX - kTJTX) + (PV)} =0 & ZR= Cp – Cv

H =0= {Cv(TJTX - kTJTX) + (PV)} = {Cv TJTX (1- k) + (PV)}= {TJTX (Cv - Cp) + (PV)} =0

H =0 = {TJTX (Cv - Cp) + (PV)} = {-ZRTJTX + (PV)} =0 or {(PV)JTX}/J= ZRTJTX

The term (PV) implies work, but an isenthalpic expansion is a locked rotor or zero work expansion. The

d(PV) work is internal expansion energy. Below Table is worked example based on Scheel‟s C3= H-S. The

equations offers thermo dynamic proof of presented equations. A TJTX from EOS or chart is needed. The

term, (PV)/J= ZRT, is valid any gas, but arriving at this by use of, Tjx =kTJTX validates that Tjx =kTJTX

.

Simplified Look at Isenthalpic & Joule Expansions

C3= 1 2

Psia 125 20

T, F 100 75

Z 0.94 0.94

M 42 42

V cf/# 1.08 6.42

PV btu/# 24.9 23.8

d(zRT) -1.11 EX.1

d(PV)/J -1.11 JTX

diff/err -9E-06 C3=

dH For A) Joule Expansion, {U=0} & use Scheel‟s chart from 100F & 125 to 20psia,

followed by; B). constant pressure expansion to JT condition, 75F. With k= 1.149;

Cp=15.31BTU/mol/F; TJTX = -25F. & TJT = -25F x 1.149= -28.7F.

A) HJX = {U + (PV)}={0+(PV)}={Cp(TJX - TJTX)}=Cp(k TJTX -TJTX)

H =15.31/42(-25)(1.149-1)= -1.36BTU/#

B).From the constant pressure Joule Expansion conditions, no JT correction is needed

as the expansion is at constant pressure so HJX/JTX ={Cp(TJTX -TJX}=0.365(75-71.3F)

=1.35 BTU/#.

C) The sum of these 2 expansions is HJTX = -1.36+1.35=0.01≈0. As no heat is added at

any step perhaps one could say the difference between a Joule expansion and a JTX is

with limited friction energy, just equal to the dH between JX & JTX? For brevity the

d(PV) calc is left off but closes with use of Z, as shown in Table to right, which

validates (Tjx /TJTX) = (Cp/Cv) = k by using JX- JTX rules: arrive at (PV)/J= ZRT.

68

Why is the ratio for (Tjx /TJTX ) just the ratio of (Cp/Cv) or k? The slide looks at this concept from the EOS

thermo concept using primary definition of the terms:

JTX (T/P)h = -(H/P)T/Cp = {T(V/T)p - V} /Cp or dTJTX Cp = {T(V/T)p - V} dP

JX (T/V)U = -(U/V)T /Cv = {T(P/T)V – P} /Cv or dTJX Cv = {T(P/T)V – P} dV

What is said in prior 2 slides is: dTJTX Cp = dTJX Cv so {T(V/T)p - V} dP = {T(P/T)V – P} dV

Using PV=RT, V=RT/P & P=RT/V, & T(V/T)p = RT/P & T(P/T)V =RT/V & dP = -RT/P2 so dV= dP RT/P2

Make the sub‟s {T(V/T)p - V} dP = {T(P/T)V – P} dV & {RT/P - V} dP = {RT/V – P} -dP RT/P2

The dP‟s cancel & multiply both by P to get {RT - PV} = {RT/V – P} RT/P

But PV=RT & divide by RT, plus RT/V =P so {RT - RT} /RT = {P– P} /P or 1-1 =1-1 or 0 = 0 , perfect

All terms are equal terms, just expressed differently, as proved by 3 prior slides, Berthelot & Viral EOS,

Enthalpy definitions, Steam Table Evaluation of temperature for dU =0 and temperature for dH=0 and

subsequent temperature changes for constant U & H, plus the evaluation of C3= from the H-S diagram.

dTJTX Cp = dTJX Cv k dTJTX = dTJX

Other considerations are when variables separate, then correct is dT/X(T) = dP/Y(P) is correct integral form.

Also it is strange but these findings contradict the concept that a free expansion is very close to an

Isothermal expansion, when in fact a free JX produces T‟s lower than JTX. However a free X is an X into a

vacuum. Which for practical applications such is not possible but vacuum conditions just mean that P1>>Po

and minimal work, so how significant is that?

Specification of 2 variables make a complete specification of a single component system. The combination

metrics are PV, PT, & VT. The PV spec is used to define phases, PT spec for common process specifications &

to denote ideality departure, while VT spec is common to define work, as in TS or HS diagrams. The use of

and work in P & T equations are preferred as they are directly measurable. Volume or density is less

common and more difficult measurement. Thus was the concept at looking at JTX coefficient not in terms of

volume but in basic terms of pressure and temperature, which then found: (Tjx /TJTX )= (Cp/Cv)

Ratio of dT Joule & to dT Joule-Thompson: Part 3

69

Multi Stage Gas Valve Field Trial Part 1Here are some details of the field trial previously discussed. Notice the mixed

water/ice on valve discharge, confirming 32F outlet T. A downstream

thermowell showed 40F with small ice/water on line, 32F IR thermometer is

the accepted outlet temperature.

Vendor B

gas Fd vol frac

N2 0.0048

CO2 0.0094

H2S 0.0000

C1 0.6775

nC2 0.1661

nC3 0.1028

iC4 0.0101

nC4 0.0238

iC5 0.0028

nC5 0.0026

nC6 0.0001

H2O 0.0001

sum 1.0000

Gas Lab Anal

Psia.1 990.00

Psia.2 195.00

T1, F 95.00

The picture, in upper RH corner, is a

depiction of valve trim for Vendor A,

who predicted a more correct

temperature drop, closer to the field

result. Vendor B was selected for the

trial. After the field trial, B

recalibrated his model to better match

field results. The middle graph of a

CFD output for vendor B‟s warranty

outlet temperature, 56F is provided.

Because the valve failed to achieve the

guaranteed outlet temperature and

Vendor B‟s valve experienced complete

plugging from dirty gas, the valve was

returned to vendor. The field trial

showed a 10F to 12F increase in outlet

temperature with application of high

friction multistage trim. Economics

may support use of pre-filter on gas in

some instances where dirty gas is

processed. Vendor A claims his trim is

less prone to blockage. Repair of

plugged trim is near impossible.

Sundry vendors‟ multistage trims have

experienced erosion and/or plugging

when processing pipeline gas. A few

ppb of “Black Powder” in a gas pipeline

can cause significant problems for

multistage valves.

70

Multi Stage Gas Valve Field Trial Part 2

Here are some details of the field trial previously discussed. Vendor A

determined a JTX. Temperature closest to what was experienced under

field conditions. Vendor B was selected for the trial. As result of trial, B

recalibrated his model to better match field results. A check of the PR-LK

constant enthalpy simulator model was made. Object of re-run was to

check predicted pressure drop to achieve 32F outlet temperature. The

predicted outlet pressure was 83psi higher than field measured. This is a

significant variance and exceeds any calibration error of field

instruments. As discussed on prior picture, the measured outlet

temperature of 32F indicates frictional heating that exceeds constant

enthalpy predictions.

Vendor A

CFD Real Gas Model Results

Vendor A CFD Real Gas Model Results

Source.F/100psid MW dPSI dF/100psid

Field Measured 23.3 770 7.79

HySys- w/Lab Gz 23.3 795 9.43

BWR.JT w/LabGZ 23.3 770 10.90

Vendor A ++stage 23.0 770 7.79

Vendor B 22 stage 23.0 795 4.78

Vendor C H-S/JTX 20.6 730 7.54

RK.JTX Dsgn Gas 20.3 730 7.75

RK.JTX Lab Gas 23.0 795 11.40

Source MW dPSI dF/100psid

PR.-w/LabGz 23.3 795 9.53

PR-LK.-w/LabGz 23.3 795 9.30

SRK.-w/LabGz 23.3 795 9.50

PR-LK./LGz-32F.o 23.3 712 8.85

71

Multi Stage Gas Valve Field Trial Part 3A What is Black Powder (BP)?Heather to fore, BP was considered exclusively an issue for “product pipelines”. This test, with a raw gas

pipeline, shows BP can exist in other parts of a gas system. The following quote is from www.Wiki “Black

Powder Gas Pipeline”, Dec.09/2012

“Black powder is a solid contamination in finished product pipelines. The material may be wet and have a

tar-like appearance, or dry and be a very fine powder, sometimes like smoke. Black powder can cause a

range of problems, including product contamination, erosion wear in compressors, instrument and filter

clogging and equipment contamination for product consumer, erosion and sealing problems for valves, and

flow reduction. Black powder may be mechanically mixed or chemically combined with any number of

contaminants such as water, liquid hydrocarbons, salts, chlorides, sand, or dirt. Chemical analyses of the

material have revealed that it consists mainly of a mixture of iron oxides and iron sulfides.” The following

slide details upstream testing showing BP in gas pipelines exists from wellhead to final gas product lines.

Forms of Black Powder in Gas pipeline

72

Multi Stage Gas Valve Field Trial Part 3B What is Black Powder?

Black powder is a solid contamination in natural gas pipelines. The median average particle diameter is

about 45 m. Separation of a sludge sample into water and oil components, indicates the darker brown BP

results from water phase, whilst the tar portion of sludge was far darker. The water constituent of this

sludge sample had iron as the predominate (50%) non oxygen element. The hydrocarbon portion was more

black with 87% carbon. The oil fraction contained significant silica and sulfur.

The abrasiveness of this fine powder can be seen in the eroded valve trim and cage, plus the erro0ded

compressor seal. The tested high friction valve trim has a more convoluted flow pattern, as opposed to

straight flow path and larger diameter holes in a conventional valve. The high friction valve trim was

rapidly plugged off due to convoluted flow pattern and small opening diameter of ports.

Vendor A CFD Real Gas Model Results

73

Multi Stage Gas Valve Field Trial Part 4a Process Simulator Comparison A

Thermo=> PR-LK Sim Valve model u.jtx

in out delta F/100psi

T,F 95.0 21.1 -73.9 9.30

P psia 990.0 195.0 -795

H btu/# -35.20 -35.19 0.001

S btu/#F -0.2714 -0.1981 0.0733

Z 0.7672 0.9284 d(TS)=> 55.32

Cp/Cv 1.67 1.322 dPV=> 1.78

Cp btu/#/F 0.701 0.4855 dPfrict=> 59.99

mw 23.31 23.31 Qc => -4.67

den 5.05 0.95 #/CF

TS -150.63 -95.306 BTU/#

V 27.15 147.55 fps

P head 35.74 35.19 BTU/#

V head -0.12 -2.25 BTU/#

Frict. Head 0.00 2.68 BTU/#

Total Head 35.62 35.62 BTU/#

H(32F,1atm) -35.20 -35.19 BTU/#

V head -0.12 -2.25 BTU/#

F.hd 0 2.12 BTU/#

Total Head -35.32 -35.32 BTU/#

Thermo=> PRLK32FoutSim Valve model u.jtx

in out delta F/100psi

T,F 95.0 32.0 -63 8.85

psia 990.0 278.0 -712

H btu/# -35.20 -36.12 -0.925

S btu/#F -0.2714 -0.2187 0.0527

Z 0.7672 0.905 d(TS)=> 43.03

Cp/Cv 1.67 1.356 dPV=> 1.92

Cp mass 0.701 0.5086 dPfrict=> 47.02

mw 23.31 23.15 Qc => -4.00

den 5.05 1.35

TS -150.63 -107.600 43.0266

V 27.15 105.84 fps

P head 35.74 36.18 BTU/#

V head -0.12 -1.31 BTU/#

Frict. Head 0.00 0.75 BTU/#

Total Head 35.62 35.62 BTU/#

H(32F,1atm) -35.20 -36.12 BTU/#

V head -0.12 -1.31 BTU/#

F.hd 0 2.12 BTU/#

Total Head -35.32 -35.32 BTU/#

74

Multi Stage Gas Valve Field Trial Part 4b Process Simulator Comparisons B

Thermo=> PR Sim Valve model u.jtx

in out delta F/100psi

T,F 95.0 19.2 -75.8 9.53

psia 990.0 195.0 -795

H btu/# -38.93 -36.81 2.12

S btu/#F -0.2714 -0.2023 0.0691

Z 0.7208 0.9117 d(TS)=> 53.68

Cp/Cv 1.79 1.3623 dPV=> 3.43

Cp mass 0.6972 0.4787 dPfrict=> 57.36

mw 23.31 23.13 Qc => -3.67

den 5.38 0.96 #/CF

TS -150.63 -96.942 BTU/#

V 27.15 146.96 fps

P head 33.58 34.68 BTU/#

V head -0.12 -2.23 BTU/#

Frict. Head 0.00 1.00 BTU/#

Total Head 33.45 33.45 BTU/#

H(32F,1atm) -38.93 -36.81 BTU/#

V head -0.12 -2.23 BTU/#

F.hd 0 -0.02 BTU/#

Total Head -39.05 -39.05 BTU/#

Thermo=> SRK Sim Valve model u.jtx

in out delta F/100psi

T,F 95.0 19.5 -75.55 9.50

psia 990.0 195.0 -795

H btu/# -36.22 -36.21 0.006

S btu/#F -0.2714 -0.1996 0.0718

Z 0.7547 0.9206 d(TS)=> 54.93

Cp/Cv 1.656 1.35 dPV=> 1.92

Cp mass 0.6594 0.4775 dPfrict=> 59.11

mw 23.31 23.31 Qc => -4.18

den 5.13 0.96 #/CF

TS -150.63 -95.698 BTU/#

V 27.15 147.04 fps

P head 35.16 34.77 BTU/#

V head -0.12 -2.23 BTU/#

Frict. Head 0.00 2.49 BTU/#

Total Head 35.03 35.03 BTU/#

H(32F,1atm) -36.22 -36.21 BTU/#

V head -0.12 -2.23 BTU/#

F.hd 0 2.10 BTU/#

Total Head -36.34 -36.34 BTU/#

75

Multi Stage Gas Valve Field Trial P4c Outlet T prediction by Clausius EBBelow is outlet temperature calculation using Berthelot EOS. The Left table calculates outlet temperature

based on Friction as just TdS. TdS is calculated as dH-(dP)/Density in BTU/#. Density is based on log mean

formula. Pressure corrected dH is determined by Berthelot JT temperature drop, low pressure Cp and actual

dT as presented in this paper. The vdP is reciprocal log mean density times pressure drop. The pressure head

is determined by ZRTln(P/14.7) which is correct for a real gas. Gas composition is used to determine critical

temperature and pressure. Temperature of outlet is adjusted until inlet and outlet energy are in balance. For

a real process TdS should be positive, increased entropy, as calculated in table below. The right hand table

uses mechanical friction as determined by gas valve CV and prior formula from Crane TP-410 relating CV and

K. Friction is K times velocity head. Friction Velocity Head is based on simple average velocity. A formula for

Thermodynamic Friction, F=-(H+ (V2/2Jg ) –TdS), is arrived at by the Streeter Clausius energy balance

method, Perry & Streeter. Setting F=TdS is specification of Total Thermo energy of zero. With 2 unknowns,

either simplification or Mechanical F equation is required for solution. The B‟lot EOS with F=TdS arrives at

same T drop as one vendors‟ CFD calc & ME B‟lot seems a tad high. Measured outlet temperature was 32F.

inlet outlet units Bertholet Z EOS Friction by

MW 23.35 23.35 #/mol item/ u inlet outlet Clausius TdS

Z 0.795 0.934 v/v Pc psia 664.4 664.4 10.65 Cp

Press 990.0 195.0 Psia Tc R 426.9 426.9 8.66 Cv

Temp. 95.00 25.58 F Pr 1.49 0.29 1.23 k

Den 4.885 0.935 #/cf Tr 1.30 1.14 -69.42 dT

Flow 9.000 9.000 mmscfd Z B'lot 0.795 0.934 -77.03 dTjx

id sch 80 2.9 3.826 inch SG 0.805 0.805 7.60 dTu

Velocity 28.6 85.9 ft/sec Cp mass 0.471 0.441 2.82 dU

B'lot EOS Therm Energy Balance Z LM 0.86 B'lot -69.42 dT

in out units dP 795.00 psi -62.66 dT.jtx

P hd 157.92 99.74 btu/# P LM 489.32 psi -6.77 dTh

Vhd 0.02 0.15 btu/# T LM 52.91 F -3.09 dH

Frict 0.00 58.05 btu/# den LM 2.41 #/cf -61.14 vdP

Sum 157.94 157.94 btu/# Frict 58.05 btu/# 58.05 F=TdS

Bertholet dT JTX, degree F -62.66 Outlet T: B'lot JTX 32.34 P hd calc as:

Friction =TdS = dH-vdP, use log mean density for v (ZRT)ln(P/14.7)

inlet outlet units Bertholet EOS Z,dT.jtx Friction by

MW 23.35 23.35 #/mol item/ u inlet outlet Clausius TdS

Z 0.795 0.942 v/v Pc psia 664.4 664.4 10.65 Cp

Press 990.0 195.0 Psia Tc R 426.9 426.9 8.66 Cv

Temp. 95.00 43.61 F Pr 1.49 0.29 1.23 k

Den 4.885 0.894 #/cf Tr 1.30 1.18 -51.39 dT

Flow 9.000 9.000 mmscfd Z B'lot 0.795 0.942 -77.03 dTjx

id sch 80 3.826 3.826 inch SG 0.805 0.805 25.63 dTu

Velocity 16.4 89.9 ft/sec Cp mass 0.471 0.441 9.51 dU

B'lot EOS Therm Energy Balance Z LM 0.87 B'lot -51.39 dT

in out units dP 795.00 psi -62.66 dT.jtx

P hd 157.92 104.34 btu/# P LM 489.32 psi 11.26 dTh

Vhd 0.01 0.16 btu/# T LM 66.00 F 5.14 dH

Frict 0.00 53.43 btu/# den LM 2.34 #/cf -62.97 vdP

Sum 157.93 157.93 btu/# Frict.T 62.81 btu/# 68.11 TdS

Bertholet dT JTX, degree F -62.66 Outlet T: B'lot JTX 32.34 P hd calc as:

Friction =-(dH+dVhd-TdS), &= KV2/(J2g) => 53.43 (ZRT)ln(P/14.7)

76

Multi Stage Gas Valve Field Trial P4d Outlet T prediction by Clausius EB

Below is outlet temperature calculation using Ping Robinson Lee Kessler EOS (on Left side) and BWR EOS on

RH side. Both calculations use the Mechanical Energy method to determine valve friction. Friction is

determined by mechanical formula of K times velocity head. The K is determined by Crane TP410 formula

mentioned prior in this paper, using valve Coefficient, CV. The gas valve CV is determined by formula: CV =

scfm{SG (R/520)/psi/psi.inlet}0.5 . The temperature, Rankin, is simple average of inlet and valve outlet, SG

being gas gravity relative to air. The PR-LK EOS result arrears to give a temperature slightly in excess of the

measured outlet temperature of 32F.

The BWR EOS fitted to Standing Katz hydrocarbon Z data calculates an outlet temperature of 29.23F. This EOS

with the Mechanical Friction Formula of Crane TP-410 gives a result that most closely matches the field

measured temperature. This BWR EOS validity is listed down to reduced temperature >1.05 & Pr<30.

inlet outlet units PR-LK Z EOS Friction by

MW 23.35 23.35 #/mol item/ u inlet outlet Clausius TdS

Z 0.767 0.923 v/v Pc psia 664.4 664.4 10.18 Cp

Press 990.0 195.0 Psia Tc R 426.9 426.9 8.19 Cv

Temp. 95.00 39.89 F Pr 1.49 0.29 1.24 k

Den 5.060 0.919 #/cf Tr 1.30 1.17 -55.11 dT

Flow 9.000 9.000 mmscfd Z prlk 0.767 0.923 -91.94 dTjx

id sch 80 3.826 3.826 inch SG 0.805 0.805 36.82 dTu

Velocity 15.9 87.4 ft/sec Cp mass 0.436 0.436 12.92 dU

B'lot EOS Therm Energy Balance Z LM 0.84 -55.11 dT

in out units dP 795.00 psi -74.00 dT.jtx

P hd 152.45 101.52 btu/# P LM 489.32 psi 18.89 dTh

Vhd 0.01 0.15 btu/# T LM 63.51 F 8.23 dH

Frict 0.00 50.79 btu/# den LM 2.41 #/cf -60.98 vdP

Sum 152.46 152.46 btu/# Frict 60.83 btu/# 69.22 TdS

PR-LK dT JTX, degree F -74.00 Outlet T: PRLK JTX21.00 P hd calc as:

Friction =-(dH+dVhd-TdS), &= KV2/(J2g) => 50.79 (ZRT)ln(P/14.7)

inlet outlet units BWR EOS Z,dT.jtx Friction by

MW 23.35 23.35 #/mol item/ u inlet outlet Clausius TdS

Z 0.754 0.929 v/v Pc psia 664.4 664.4 10.48 Cp

Press 990.0 195.0 Psia Tc R 426.9 426.9 8.50 Cv

Temp. 95.00 29.23 F Pr 1.49 0.29 1.23 k

Den 5.145 0.933 #/cf Tr 1.30 1.15 -65.77 dT

Flow 9.000 9.000 mmscfd Z BWR 0.795 0.929 -106.1 dTjx

id sch 80 3.826 3.826 inch SG 0.805 0.805 40.34 dTu

Velocity 15.6 86.1 ft/sec Cp mass 0.471 0.427 14.68 dU

B'lot EOS Therm Energy Balance Z LM 0.84 -65.77 dT

in out units dP 795.00 psi -86.00 dT.jtx

P hd 149.93 100.00 btu/# P LM 489.32 psi 20.23 dTh

Vhd 0.00 0.15 btu/# T LM 55.80 F 9.08 dH

Frict 0.00 49.78 btu/# den LM 2.46 #/cf -59.80 vdP

Sum 149.93 149.93 btu/# Frict.t 59.65 btu/# 68.89 TdS

BWR dT JTX, degree F -86.00 Outlet T: BWR.jtx 9.00 P hd calc as:

Friction =-(dH+dVhd-TdS), &= KV2/(J2g) => 49.78 (ZRT)ln(P/14.7)

77

Multi Stage Gas Valve Field Trial P4e Outlet T prediction by Clausius EB

Below is outlet temperature calculation using RK EOS on left side. Table to right is comparison of methods

and the field trial result measured temperature. It appears the BWR EOS fitted to the Standing Katz

Compressibility data gives a match closest to the field observed result for this sweet hydrocarbon gas, with

relatively low CO2 component. The Joule Thompson temperature should be lower than the temperature

determined when accounting for mechanical friction. This is in agreement with all principles repeated

routinely in the paper. Friction produces heat and increased entropy. The Clausius heat balance is the correct

energy balance to account for friction heating. Few if any process simulators treat friction heating properly,

instead they rely upon a JT outlet temperature. Measured outlet temperature was 32F, with ice and water on

valve outlet.

inlet outlet units RK EOS Z & T.jtx Friction by

MW 23.35 23.35 #/mol item/ u inlet outlet Clausius TdS

Z 0.768 0.933 v/v Pc psia 664.4 664.4 10.48 Cp

Press 990.0 195.0 Psia Tc R 426.9 426.9 8.50 Cv

Temp. 95.00 34.72 F Pr 1.49 0.29 1.23 k

Den 5.051 0.919 #/cf Tr 1.30 1.16 -60.28 dT

Flow 9.000 9.000 mmscfd Z RK 0.795 0.933 -112.2 dTjx

id sch 80 3.826 3.826 inch SG 0.805 0.805 51.9 dTu

Velocity 15.9 87.4 ft/sec Cp mass 0.471 0.427 18.89 dU

RK EOS Therm Energy Balance Z LM 0.85 -60.28 dT

in out units dP 795.00 psi -90.93 dT.jtx

P hd 152.72 101.53 btu/# P LM 489.32 psi 30.6 dTh

Vhd 0.01 0.15 btu/# T LM 59.89 F 13.76 dH

Frict 0.00 51.04 btu/# den LM 2.41 #/cf -60.94 vdP

Sum 152.73 152.73 btu/# Frict.T 60.79 btu/# 74.70 TdS

RK dT JTX, degree F => -90.93 Outlet T: RK JTX 4.1 P hd calc as:

Friction.T = -(dH+dVhd-TdS), &= KV2/(J2g)=> 51.04 (ZRT)ln(P/14.7)

Calculation Summary & Field Measure valve

JTX Mechanical Friction BTU/#

EOS T2.deg.F T2. deg. F Mech Thermo

B'lot 32.3 43.6 53.4 62.8

PR-LK 21.0 39.9 50.8 60.8

BWR 9.0 29.2 49.8 59.7

RK 4.1 34.7 51.0 60.8

Measured 32F T2 outlet T

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The gas concept: adiabatic, (without heat) has been hijacked to be equated with isentropic (IS, -

reversible or frictionless, aka fiction, aka imaginary, zero entropy change). This begs the issue: What is

name of adiabatic non reversible process? Is it isothermal, (IT)? No, because IT requires heat addition in

real world applications. Is it isentropic? No as isentropic is zero dS and dS=0 only holds for reversible non

friction process. Adiabatic change is any process in any insulated format. While the term IT requires heat

transfer. Isobaric (IB) is a constant pressure processes. The term IT is counter intuitive. For when IT is

applied to, Constant Volume or (IM), isometric (ideal liquids), it conveys meaning of adiabatic. At other

end of scale is a constant internal energy process, dU=0, another hijacked term, (see SvN textbook)

coined to mean isothermal. A dU=0 process is only IT for a few, ideal gasses. Bad thinking produces bad

math, as the math is just a means to quantity a concept. The constant internal energy concept is so rare,

that one is hard pressed to find a Latin equalivalent word. The closest concept is the oft forgotten, Joule

(Free) Expansion, JX, not to be confused with a Joule Thompson Expansion, JTX. Both expansions are

neither IM nor isobaric (IB Constant Pressure) nor IS, as dS>0. A JX (dU=0) is related to constant volume

heat capacity, whilst JTX is related to constant pressure heat capacity, Cp. Both JTX & JX are zero work

expansion (W=0), and zero heat, (Q=0) expansion. The difference between a JX & a JTX expansion can

only be see if viewed from a real gas perspective. For Ideal gasses both JX & JTX have zero temperature

change. Looking at adiabatic Joule expansion, JX, on H-S diagram for C3=, from 125psia and 100F to

20psia, gives a negative dH, BTU/#. And for any JX, dH is negative, while dH of JTX is zero. If heat is

calculated as dH then a JX would say heat is needed for this expansion. However the Joule experiment

and the definition of heat, dQ=dU=0 for a JX. Making correction by Clausius Equality Method is the correct

accounting of frictional heat, for either gas or liquid flow. The correction of dH & dU for pressure is easily

implemented by the Joule or JT coefficients.

This paper determined that temperature change for a JX expansion should be greater than the

temperature change of a JTX. The JTX temperature change multiplied by the constant pressure heat

capacity and divided by the constant volume heat capacity is the approximate temperature change for a

Joule Expansion. It is an abuse of nomenclature to call an isothermal expansion a Joule or “Free

Expansion” as a JX has a greater temperature change than the JTX. The increase of temperature for a

JTX is likely the result of frictional heating, as detailed in the article so titled.

SUMMARY