Effect of Friction (Joule Effect) Heating in Gas Flow

8/3/2019 Effect of Friction (Joule Effect) Heating in Gas Flow

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Considerations on Real Gas Expansion: Otis P. Armstrong P.E., -Dec/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 ofgas 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 decreasein gas expander mechanical efficiency is accompanied by a subsequent increase in outlettemperature. A result of friction between gas and machinery surfaces. This is noticed on aH-S diagram by a counter-clockwise rotation away from the minimum isentropictemperature towards isothermal operation. Also for compression, friction will increase

outlet temperature and result in reduced efficiency, the same is true for nozzle flow.Likewise for valves: gas friction increases the outlet temperature. This is noticed on theH-S diagram by a counter-clockwise rotation away from the minimum isentropictemperature towards the Free Expansion temperature. The Joule or Free Expansiontemperature is lower than isothermal expansion temperature and proportional to theJoule Coefficient, J.Many thermodynamic concepts of gas expansions are just idealization concepts. Actual

equipment will always have friction effect, which will produce heat. For real gasexpansion, 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 .pptdocument 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 ofthermodynamics 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 areoutlined on pages 14 and 15, plus other discussions. The term fL/D is also the Crane K factor for flowapertures. Also Crane K, (TP410) is related to valve coefficient as detailed on page 14. Details for skinfriction are: K f L/D = (29.84d2/CV)2 = (1/Kd)

2 = 1/ = 1/( n)The Clausius Equality {F=(T S- Qh)} is correct accounting of frictional heat, for either gas or liquid flow withfriction. Dissipation of energy by Friction to heat always increases temperature relative to a reversibleprocess. A correct heat balance must always calculate zero heat addition for Both a Joule Expansion {dU=0}anda 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 temperaturechange are presented in detail. U =Cv( T - TJX) & H = Cp ( T - TJTX) : The Cs are low pressure heatcapacity. The temperature sign convention is easily remembered. For at either U or H = 0, thetemperature change will be negative for all but 3 quantum gasses, H2 for example. This is detailed in thesection Inversion Point. These rules offer simple EOS thermodynamic consistency checks for simulationwork.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 theJT Coefficient, in Rankin/atmosphere, Cp in BTU/#/F, and density in #/CF as: X=(0.37 Cp) . The details areon page 40. Compressor head and Power are determined by use of real gas enthalpy change, as explainedabove 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 theratio of specific heats: ( TJTX/ TJX) = (Cv/Cp). The temperature drop associated with a Joule Free expansionis greater than a JT, the difference may be attributed to friction effect.Black Powder is found in raw gas pipelines andsales 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


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IntroductionDiscussion with some engineers indicated a lack of graspon the heating effect of friction in gas flow. Additionalreview of literature showed a lack of presentation, thatdissipation 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 ineither compression or expansion. It shows that frictionincreases outlet temperature in both cases.

Even-so, some engineering routines perpetuate the cooling myth by misleading statements like: inliquids friction produces heat but for gasses friction is converted to internal energy, hello?? Justwhat 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 myththat friction pressure drop in gases does not heat a gas is routinely perpetuated. True: a normal gasexpansion cools, but friction in flow apertures reduces the amount of cooling. Why?, because frictionacts to reduce the output of useful work by energy dissipation to heat production. Heat generationalways 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 theexpansion resulting from pressure reduction. In BOTH cases, flow friction acts to produce heat. It isjust that the heating effect of flow friction in gas is masked by the normal cooling effect of gasexpansion.

A field test with a high friction valve demonstrated that friction in gas flow acts to counter theexpansion 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, ctd:Dissipation of work or momentum via friction will produceheat, irrespective of the medium in which friction isdissipated. The TS diagram on right shows friction effect ineither compression or expansion. Friction increases outlet

temperature, irrespective of expansion or compression.The result of friction or irreversible losses causes theoutlet temperature to INCREASE to point 2. This is abovethe reversible expansion temperature, point 2.

There are an accumulation of routines to perpetuate the cooling myth of gas friction. Looking at frictionon a molecular level, a gas molecule which losses momentum to a stationary object, gives up translationenergy. By conservation of energy, this organized momentum energy is translated to un-organized thermalenergy, so dS must increase. As correctly depicted in the above diagrams. The Clausius inequality for anyprocess involving friction requires: d(losses)>(Tds-d(Q)h). Likewise for flowing gas, Momentum lost tointermolecular 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)]. Thereading 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 istermed 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 aluminummaterials for the achieved velocity. For a gas flowing inside a tube, the boundary layer volume to totalexpanded volume is a small fraction, (momentum transfer efficiency). The small ratio of boundary layervolume to a much greater bulk volume gives a net decreases in bulk of temperature due to gas volumeexpansion. The effect of friction in gas is not a special case but just the nature of friction. Friction alwaysproduces heat, irrespective of the involved media. Some practical aspects are: increase flare linemomentum loss to maximum limits to minimize cooling, improve SRVs by use of high turbulence valves togain maximum possible temperature in flare flow lines.


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H-S C3=

Unzip compressed

folder to storage.

JT calculator is the

HTML file & executesin 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 withzero 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 outperformed the conventional rule. By the 7/100 rule the outlet T calculates to about 8F.

Small ports are required to obtainhigh skin friction losses. These portscan result in valve trim plugging or

erosion for dirty gases. Port pluggingduring this trial indicated gas filterswere required due to dirty gas.Looking at long term service, filtercosts may prove economical, ifoutlet gas heat is of value or ifhydrate inhibitors are used.

Test Results show a 70% increase inoutlet temperature compared to JTisenthalpic Performance. This is aresult of friction heating. Onemethod to create friction heating isto increase turbulence by squareedge port trim.

prop unit viral R K actual

W/mw ac entric 0.0720 20.3000 Hi

C p btu/mol/R 11.6 11.6 F riction

Tc Kelvin 213.9 213.9 Valve

P c Pc,atm 45.2 45.2 70%eff

T in T1R ankine 520.0 520.0 30%dH

P in P1atm 62.3 62.3 to heatPout P2atm 11.2 11.2

ujt.clc Rankin/at 0.823 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 frictionheating in gasses. Instead placing emphasis on idealizedthermodynamic conditions. Such ideal conditions are notmeet in practice.*

As shown to right, effect of friction is to heat gas. Step 1 to2 is the idealized thermodynamic temperature drop of anozzle. Step 1 to 2 is actual temperature drop. Nozzleefficiency 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 ofdischarge. Since kd is tabulated for many physical configurations it is possible to calculate actual outlettemperatures 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 meteringnozzle situated inside a constant diameter pipe as an idealized throttling process. the discharge from the nozzle, with itshigh kinetic energy, swirls about and dissipates its KE, thus KE is reconverted to internal energy and Ws=- H=0. This is onlytrue in pedagogical circles. For practicing engineers, permanent pressure losses are the reality of actual operatingequipment. Because not all velocity head is converted back to pressure head, some velocity head is dissipated to what aretermed permanent pressure losses to be lost as low grade thermal energy. The engineering equation that correctlydescribesa real process is: [H + V2/(2gJ)]1= [H + V2/(2gJ)]2 + F. Where F are friction losses. The pedagogical concept focus on JT coolingneglects 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 =(Cp JT) P, H =0 & if P2V1 then T20 T2=T1 & S>>0 & dQ by Clausius Equality

Polythermal Temperature: T 0 , PV=ZRT & dQ either >0 or


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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 inform of Crane K, CV, or Co, or efficiency are all identical forms of irreversible friction process whichproduce heat. Refer to Perry 5th Ed. Page 5-18 No friction term occurs in this expression since frictionrepresents a conversion of mechanical energy into HEAT.Caution: Knowledgeable texts on friction place emphasis on manufacturer rating of equipment. One reasonfor such caution is different types of energy dissipation. The above deal only with skin friction, otherforms are vortex &/or centrifugal. For example the Twister process which gives more cooling than JT bymitigating friction heating.Valve Flow Coefficient, CVCV, gpm (SG/ psi), & Crane pressure loss is 144( psi)/ =KV2/2g, Equate psi for CV, & Crane K, usingwater density of 62.4#/CF & SG= /62.4 & gpm=2.45Vd2. When making these substitutions & eliminate thedensity, SG, Velocity, & flow terms one arrives at CV,=29.84d2/ K or K= (29.84d2/CV)2. The constant for theCrane 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 isaccurately represented by fL/D= Crane K, so also is relation between CV, and Crane K.Gas Valve Flow CoefficientThe 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 haveextended formula for choked flow. For friction, the above is proposed to best reps total energy concept.

Summary: K= fL/D = (29.84d

2

/C

V

)

2

= (1/Co)

2

= 1/ = 1/( n)Revised slide

Friction losses for valves, throttling processes and many flow apertures are well documented, CraneTP410, and easily calculated. The Crane method defines friction loss as multiple of velocity head, the Kmethod. Where Cranes 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 AperturesSummary: 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 Coefficient

5. Joule Coefficient by VdW EOSlimited utility6. Joule Coefficient by RK EOS7. Joule Coefficient by Viral EOS8. Joule Coefficient by BWR EOS9 Summary of Equations for JT Coefficient, JT10 JT EOS COMPARE RESULTS11 Van der Waals Simplified JT Coefficient

12. Van der Waals Exact JT Coefficient13. JT by the Two Term Viral EOS14. Real gas JT coefficient as function of Z15. Forms & Solutions of Redlich Kwong EOS16 Determine ( Z/ oR) by Redlich Kwong h

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

EOS

18. 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

Correctedfor 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. useof Joule Coefficient.28. Review of Polytropic Head Calculation Methods7pp The Schultz Method29. Express = Cp-Cv for real gas as z = Cp-Cv30. Compare Gas X Calculation Methods 131. Compare Gas X Calculation Methods 2

32. 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 Balance

40 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/Cv45. Field Trial Details: MultiStage Hi Friction Valve46. Black Powder in Raw Gas Pipelines

47 Summary


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Friction is not fiction, nor is it imaginary. Friction is the law of the universe: you cant win, you cantbreak even and you cant get out of the game. A frictionless device is parlance for a non-existent RubeGoldberg perpetual motion device. Ditto for the words: ideal, reversible, and isentropic. Such equipmentis 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 transportmeans from sliding out of control. It also reduces impacts when control is lost. But for all this, FatherFriction, extracts a price of additional exerted energy. Motion has resistance which is mostly friction. Itwill 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 werethe observations of Joule and Rumford. About 1850, Mr. Joule quantified the equality of work and heat as778ft#=1BTU. A few BTU = lots of work. The work of getting into the bath tub will not substantially changewater temperature but, a few degree of extra heat will quickly end a bath. Friction is associated withincrease in entropy, dS. A P-V process or cycle with zero Entropy, S, change is isentropic aka reversibleadiabatic, whilst the isenthalpic ( H=0) process is between has positive S. The most probable state hasthe highest positive S. The word adiabatic is another ideal concept, meaning zero heat transfer. Allisentropic changes (dS=0) must be adiabatic, but not all adiabatic changes need follow the dS=0 path. TheH-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:

T S> Q & TdS = (dU + PdV) = (dH VdP) & Friction = (T S - Q)

e S=(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 Stirringexperiments) 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 andrepulsion terms. The cooling effect of gas pressure reduction arises when the gas molecules are dominatedby 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, Cp JT , is the ratio of gas expansion temperature change to frictiontemperature change, for dP & JT 0, as TF = PF/( J)= TJT = ( JTCp) P. Which is true only for

Joule Thompson Expansion: 0 H=Cp T+(Cp JT) P. While for Joule Expansion: 0 U=Cv T+(Cv J) 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, thetemperature 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 freeexpansion friction dissipation caused heating on the B sphere just equal to the cooling of the A sphere. Thenet result for the free expansion was temperature decrease too small for Joule to detect. Joule and LordKelvin devised the JT experiment with a porous plug. The ratio of Joule Thompson expansion dT to the ratioJoule Free expansion dT is proportional to Cv/Cp. On the C3= H-S diagram, the adiabatic JT Expansion froman initial pressure of 125Psia to 62.5psia at JT Temperature has a zero dH with no heat added. A Jouleexpansion follows a constant internal energy line. A Joule or Free Expansion is not an isothermal

expansion. Lord Kelvin & Joule both agreed Joules device (& gas generally) was subject to frictionalheating. C W Smith 1977 Kent, Review of Lord Kelvins 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 sphereswas air, surrounded by aninsulated 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 quantumchange happens so the dominate force between gas molecules is repulsive. Inversion temperature forvarious gases at 1atm are as follows: He, 40K; Ne; 231K, air 659K, N2, 621K; and O2, 764K. When a gas isexpanded 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 ofthe 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; H 2O, CO2, H2S, to prevent gashydrate formation in the equipment.

The phenomena of expansion is unrelated to friction heating. Just the opposite, careful measurement isneeded to eliminate friction heating to accurately measure ( JT). Mr. Joules original experiment was inreply 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 betweenspheres. Mr. Joule re-made the experiment by placing each sphere in individual water baths. After againequalizing pressures found the water bath on pressure side had a decrease in temperature just equal totemperature increase of the evacuated spheres side. Joule & Thompson revised the experiment usingflowing gases and a highly compressed cotton plug between 2 thermometers. The porous plug reduced flowenough to mitigate friction effect. Also the relative mass effect of Joules method was eliminated. Atemperature decrease was measured for air at room temperature and an increase in temperature forhydrogen 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 combinedmass of copper spheres and water bath and friction effects. C W Smith 1977 Kent, Review of Lord Kelvins 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 gasesinversion point developed by Miller, presented on page 105 Walas ,1985 is: Pr = 24.21-18.54/Tr-0.825Tr2.


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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)/Cv3. Viral 2 term form:

Tjx (PR){1.09/TR1.6 - 0.139 +0.89 /TR

4.2 - 0.083}( TC)/Cv for Pr>0.8 & TR>1This Viral form is limited to Vr


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Isothermal Joule Coefficient of a Reversible Free Expansion(not to be confused with JT-Xwhich 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 = -0.5 oR

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 (0.5 oR/(8.54) (#/CF)= -0.06 oR/ (CF/#)

For O2 298K & 10Bar to vac.o

R = - 4.9o

R, V=10.72(536)/147=39CF/mol, & 390CF/moR =-(1/39-1/390) (mol/CF) (348atm(CF/mol)2)(2.72Btu/atm/CF)(mol-oR/5Btu) = -4.9 oR

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

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

U = Cv T + (T( P/ T)V=c -P ) V = Cv T - 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 Ts, so, TJX & TJTX is normally negative and is identical to dTfor 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 tolook on P-H chart for CO2 and conditions are identified by internal energy line on which corresponding tothe 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


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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/(bCv T)] [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/(2 T(V+b)V) (form kind of dV/V2)

dT=(-1.5a/(Cv T))[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)[lnVln(V+b)] |21 =(1/b)[ln{V/(V+b)}|21]

Tjx= [-1.5a/(bCv T)] {ln([V/(V+b)]2/[V/(V+b)]1)} by RK EOS integrated formfor CO2 from 8.54CF/# to 17.1CF/# & Cv=6.71btu/mol/R, a.vdw=926atm/(CF/mol)

2, T 60F,b.vdw=0.686cf/mol, 2.72Btu/atm-CF, a.rk =21677atm(CF/mol)2, b.rk =0.48CF/mol or 0.0109CF/#, MW=44, Tc=548R, left apprentices is -1195.3R and right apprentices, which is 0.0064unit-less

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

dT=(-1157.5)(0.00064)= -0.75Rslightly 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 addw/V

J, Isothermal Joule Coefficient For Free Expansion by RK eos


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

1.6 +0.72 /Tr4.2)-B }

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

4.2)-(0.083-0.42/Tr1.6+ (0.139-0.172/Tr4.2)}

T( P/ T)-P=(Pr Tc/V){(0.67/Tr1.6 +0.72 /Tr4.2)+(-0.083+0.42/Tr1.6 - 0.139+ 0.172/Tr4.2)}T( P/ T)-P=(Pr Tc/V){1.09/Tr

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

cannot do integral dV/V as P & V related, use PV=Z T or P=(Z T/V), so (Pr Tc /V)=(Z T/V) Tc/(V Pc)The terms in brackets { } are dimensionless as & Tr , are dimensionless, as is Z. The integral ofdV/V2 is(1/V) & term RTc/Cv has units ofoR ((btu/mol)(mol-R/btu)=R), the term (1/V) Z T/(Pc) when using atm, comes

to V=ZRT/P & 1/V=P/(ZRT)avg so (1/V)=(P2-P1)/(ZRT)avg so (1/V) Z T/(Pc)=(Pr2Pr1)(ZRT)avg/(ZRT)avg.

Tjx { (Pr)}{1.09/Tr1.6

- 0.139 +0.89 /Tr4.2

-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


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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/(Tr2 ra)](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


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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))averageZT2 is numerically evaluated at average Pr and the average Tr estimated by JT for thegiven 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/Tr4.2)}, Pr>0.8 & Tr>1

This Viral form is limited to Vr


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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, Blot &RK method are better forestimate of JouleThompson coefficient.

This result is not

surprising. Because BWREOS is a tailored EOS andthe BWR form used hereis based on light HCgasses. Possibly, themore involved Lee-Kessler BWR EOS is

superior 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 Resul t Compare Viral /BWR/RK/vdW/B'lot EOS v. Tabul ated 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 perry5th 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


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The VDW method for JT is used to make a 1st estimate of dT. From this expansion temperaturethe 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 Waals 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 temperaturechange for either JX or JTX.

Count Rumfords observation thatwork produced heat via friction.Joules Jar which established ratiobetween heat or friction and workat about 778 BTU/Ft-#.


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

JT T/ P = {-b/Cp}{[4 3 - 3 (3 -1)2]/[4 3 - (3 -1)2]} & =Tr, =Pr

JT: Van der Waals 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}{[4 3 - 3 (3 -1)2]/[4 3 - (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 fair

job 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-b T)V2-4abV+2ab2=0Likewise, Boyle curve of a vdW gas is any point defined by (a-b T)V2-2abV+ab2=0,

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


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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 EOSs 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


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JT: JT Coefficient as function of Z any EOS

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

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

Td(z T/P)/dTV=T[z /P +(RT/P)(dz/dT)]-z T/P ={ T 2/P}(dZ/dT)

JT =( T2/P)(dZ/dT)/Cp = 1.99(oR2/Patm)( Z/ oR)/Cp {Schaums 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=Z T so T/P = V/Z & d(lnX) = 1/X dX

JT = (VT/Z)(dZ/dT)/Cp =(V)(dlnZ/dlnT)/Cp =(Z T/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. Callendars 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)?


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Forms & Solutions of Redlich Kwong EOS

A. M-N Exact Cubic Solution (after LF Scheel GPC 72 p144/103 & WC Edminster GPC62)

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

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

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

H/RT = 1.5 ln(1+h) + (1-Z) & S/T = 0.5 ln(1+h)-ln(Z-hZ) & ln =( S- H/T)/RMixing 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=0.0371 b(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;


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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 Tr3.5

z2 Tr3.5(1-(K/(zTr))2)Zrk- z2 Tr3.5(1+ K/(zTr)) + z2 Tr3.5 J(K/(zTr)(K/(zTr))2)/(Tr1.5)

(Zrk3Tr3.5 ZrkTr1.5K2) - (Zrk2 Tr3.5 + Zrk1 Tr2.5 K) + Z2 Tr3.5 J(K/(zTr)(K/(zTr))2)/(Tr1.5)(Zrk3 Tr3.5 ZrkTr1.5K2) - (Zrk2 Tr3.5 + Zrk1 Tr2.5 K) + Z2 Tr2 J(K/(zTr)(K/(zTr))2)(Zrk3 Tr3.5 ZrkTr1.5K2) - (Zrk2 Tr3.5 + Zrk1 Tr2.5K) + J(Z Tr K (K2))

(Zrk3 Tr3.5 ZrkTr1.5K2) - (Zrk2 Tr3.5 + 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 Tr2.5 1.5K2 ZrkTr0.5) - (3.5z2 Tr2.5 + 2.5K ZrkTr1.5) + ZrkJK

( f/ Tr)= (3.5Zrk3 Tr2.5 + KZrk(J 1.5KTr0.5 - 2.5Tr1.5) - (3.5z2 Tr2.5)

( f/ Z)= (3z2

Tr3.5

Tr1.5

K2

) - (2z1

Tr3.5

+ Tr2.5

K) + JK(Tr)( f/ Z)= Tr3.5(3z22z1 ) + K(J Tr - Tr2.5 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

D i (

Z/ R) b BWR R d d D i EOS


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Z =0.27Pr/(Tr r) & Z/ T= ( Z/ Tr)( Tr / T) =-0.27Pr /(Tr 2 ra)(1/ Tc )Solve reduced density at average Z by Benedict-Webb-Rubin EOS

f( r) = A r6 + B r3 +C r2 +D r + (E r3)(1+F r2)exp(-F r2) -G

f( r) = 6A r5 + 3B r2 +2C r1 +D + (E r2)[3+F r2 (3-2F r2)] exp(-F r2)

( 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)IThis 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 commonmethods to estimate Z. Valid for 1.05


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Energy Balance SummaryThis section is written because energy balance for gasses was found to be misrepresented by a prolific andwidely used author of Chemical Engineering texts. The same misrepresentation is oft found by internetsearch 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 andcorrect 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 temperaturechange 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] - Q h =0} & Qh = (T S - 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)2Case 2. REAL VALVE: w/ Friction, No Work or added energy: (H + V2/2g)1 = (H + V2/2g)2 + FLCase 3. Zero Initial Velocity: No Friction: with Work: EG + (H )1 = (H + V2/2g)2Case 4. Constant Density: Heat Only & No Friction: EG + (H)1 = (H)2There are too many cases to list. The point is one cannot call (H + V2/2g) a complete description of energywithout misrepresenting Energy Balance Principles. The details will be developed by calculations. Theexamples show why either incorporation of thermal head, dU, or the Clausius Equality is a necessity tobalance 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 willnot 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 againstfriction + change in potential energy + change in Kinetic Energy. He assigns a negative value to work doneby 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.

E B l f B i St t


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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 = W s - (P/ ) Perry 5-32

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

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

4. {H + (V2/2g)/J}= T S - 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: T S = - 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 headwas 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 limitedcase. 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. GivenP1, P2, and T1, it is possible to solve or goal for T2 what balances the inlet and outlet energies, whichincludes friction effect. All inlet energy terms are known. It is proposed this equation is as accurate as anyenthalpy equation and eliminated determination of heat capacity, and pressure correction of enthalpy.


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Clausius Equality, Second Thermo Law & Friction Heating

Clausius Equality: non reversible process requires: (Friction) = (T S- (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

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

{ Ws + dP/ +VdV/g + [T S - (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)a P2 - 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

- G2

ln(P2/P1) + (2fL/D)G2

/g} = 0 {Smith 10-4D aka classic isothermal flow to solve P2 }

{Ws + (H + V2/2g/778) + {T S =[ 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 onecares to accept simulator answer, heat determined by Clausius Equality more closely matched simulatorresult than the Smith-Van Ness/ Smith method for SvN X10-4. Perry ChE Handbook taken from Streeter.

Clausius Equality Second Thermo Law & Friction Heating


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Clausius Equality, Second Thermo Law & Friction HeatingClausius Equality: non reversible process requires: (Losses or Friction) = (T S- (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 = 0Use (Q)h = (T S F) & dH +VdV/g - (T S F) = 0. If using pressure head, MEB, with zero shaft work, {T S =Pd(1/ ) + (U)} to get, dP/ +VdV/g + T S - T S + F = 0 or dP/ + VdV/g + F = 0. For constant density, theregular 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 toentering a fouled fin fan cooler which is causing overheating in the HF pump. Field survey shows a bypassaway from the hot DCU is possible. The inlet to the bypass valve is 70F and inlet to the cooler is 170F. Whatportion 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=C i ln(T2/T1) soT S in Clausius Equality is (460+70/2+170/2)*0.6*ln(630/530) =580 *0.104=60.2BTU/#. Next find (Q)h) byClausius Equality as: = (Q)h =T S- 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 andoutlet V is 100fps. So { H + (1002-902)/50000 - (Q)h =0} so H=35.8-0.038=35.76BTU/# is the energyassociated with passage thru the DCU hotbox. The total dH=(170-70)*0.6btu/#/F=60BTU/# which is the sumof friction and sensible gain, 24.4+35.76=60.2, and percent sensible heat gain in hotbox is 35.8/60= 60%. Ifa 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 ClausiusEquality. Heat, Friction, and its laws have no respect for the material physical state. However the physicalstate 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 as19.2BTU/# 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/#.

The Energy Balance


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The Energy BalancePremise of Energy Balance is Energy Conservation: Classical Physics: total energy is constantStarting $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, orEntropy. The actual path is usually some combination of these 3 idealized paths, as indicated on a H-Sdiagram. 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)= RT (but R=Cp-Cv) so RT = (Cp-Cv)T, & :

(TCp+V2/2g)1+ EG = (Cv TCv T)+(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)2Case 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 ofenergy 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 theinternal energy term or the loss term is neglected the sum of energy terms will not balance. The followingtables demonstrate the point.

Liquid Pump Total Energy Balance a


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L iquid output in out K C v C o

V fps 5.56 5.56 144.1 6 0.38

friction Hd ft -1.24 69.21

Work Head ft 271.63 0.00 B H P = > 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 ienc y 0.416 c oulson p.166: TDH 113' vs 113calc

dT P ump 0.60 TDH=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 H P = > 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


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

Liquid Pump Total Energy Balance b


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Liquid output in out K C v C o

V fps 1.14 1.14 0.0 48564 0.44

friction Hd ft 0.00 0.00Work Head ft 692.96 0.00 B H P = > 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


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 boundaryconditions consider pump work requirements to meet demands of impeller momentum transferefficiency, process friction losses, process total pressure head needs and fluid velocity head changes, ifany. In order to balance for irreversible friction effect of impeller momentum transfer, the T head orinternal energy head needs be added. The T head for liquids is heat capacity times fluid temperaturechange. This is simply differential temperature necessary to balance the sum of heads terms, Cv T. It isalso related to pump impeller efficiency, e, as T=(1/e-1) H/Cp/778. The value 778 ft#/BTU is jouleratio of heat to work dissipation. Efficiency is defined as ratio of work in to useful energy out. Usefuloutput is sum of difference between the three heads; Friction, Velocity, and Pressure. Without the T Headthe total energies will not balance.

Liquid Pump Total Energy Balance-b

LI item in out


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The fallacy of neglecting friction can be easily seen if applied to a meteringorifice. For metering orifice (venturi, flow nozzle, pitot tube or orifice) allproduce 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 is200*0.4=80w.c. or 80/12/1ft*(62.4#/cf/144sqin/sf)= 2.9psi. (N.B. !!: werethere zero irreversibility (friction) then all V head produced in orifice would berecovered 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 Bernoulliheads 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 perfectbalance.If try (H+V2/2g/778)1 = (H+V2/2g/778)2 , Line item 15 of In/Out table, clearlydoes not balance as 25.472 BTU/# 24.077BTU/#.One MUST add friction term to balance. If add fanning friction then is25.472=24..08+.41=24.49 much smaller error and agree with EnergyBalance . (H+V2/2g/778)1 = (H + V2/2g/778)2 , has limited validity

Summary: A. standard equations which lack nozzle friction term are notcorrect for real situations, with friction.

B. Isentropic equations need friction term to balance, dH error lessCorrect expression to include friction is:

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

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

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 fFanning 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 24.16115 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

Headsin 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

N l Effi i & F i ti II


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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 cl u.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

221.821.80

200.00 0.967 16.83 582.5 599.3 599.5 583.1 0.3176221.8 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)2 H = 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 JTcoefficient. The logic is: as JT is based on frictionless expansion and isentropic expansion is based onexpansion 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 isused, then rule is 1= 2 to balance heads or P/T=C. When using Enthalpy method & neglect JT correctionthen 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 showsTemperature results of Ebalance by enthalpy equationwith JT correction to dH andinclusion of friction loss.

Table 2, below from MechanicalEnergy Balance, show 1= 2 forhead balance, sum1=sum2

Nozzle Efficiency & Friction-II

Nozzles & Useful Gas Information from L F Scheel


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Nozzles & Useful Gas Information from L. F. Scheel

Nozzles, after Scheel 1.1 Co

Conoidal mouthpiece 0.98

Short Cyl. Round edge 0.92Short 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=Co H

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 easilyverified 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 factorReynolds 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 =0.53Den #/cf =PM/(10.73ZoR) where T Rankin , P psia, MW is molecular weight

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

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

Term BTU/# Head in Head out L I


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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 isnot in the Z factor. Z at inlet and outlet are about 0.999. It is proposed that SvNmethod 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.

W rk & 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). Thisrelationship of heat, work, and internal energy correctly calculates the heat for a Joule Expansion inabsence of work & zero Kinetic Energy change as zero, while dH can be a positive value. The use of dU todetermine heat requirement also finds that Isothermal processes require heat addition. Where-as SvNincorrectly equate isothermal processes as being constant internal energy processes, which it is not. Smithcarries this incorrect concept over to his Unit Operations book also. The reduction in heat requirement fromusing 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 Smithheat 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 T S=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.

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


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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 1204 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

G as in out

T , F 70.0 170.0

P , 'ps ia 34.70 19.00

#/hr 1000.0 1000.0

Lpipe E q ft' 36.0 36.0

di inc h 1.049 1.049

MW 2 9.0 29 .0

C p btu/#/R 0.245 0.245

u c p 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 SvNX10.4. SvN resultsdo not agree withProcess simulator

results. Conditionsare listed in GreenTable. SvN questionis: what are outletpressure & requiredheat for listedvalues? Results ofX10.4 by SvN are

presented in Lineitem 1 of LHS table.

The pipe simulation results (#2) show in case of SvN heat quantity, the resultantoutlet pressure is 20.4 and temperature is 192.2F. It appears SvN methodincorrectly sets dH=dU for heat required. Where-as the simulator method takes dHas the heat requirement, without adding velocity head. A check of dH based ontemperature 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 acceptableerror with their method. A 25% heat error and 14% outlet pressure error indicates asystematic error by one of the methods. LI-5 shows that heat required to matchSvN outlet pressure is only 24.2BTU/# with 46.3 feet of pipe., for same diameterof steel pipe. The Simulator results with either Beggs Brill or LM method give nearidentical results. It is this recommendation to determine heat based on dU and Vhead, 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.

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


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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 2variable, P&T. Solution is by either one equation & 1 specification or two equations + spec of initialconditions. One cannot add a constraint (reversible, isothermal, isentropic, plus 2 equation. SvN methoduses 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 minorvariants for pressure on volume. However by also using the isothermal expansion, the variant of processpath 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 volumewhere 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.. AKABernoulli .. Where friction losses can be evaluated .. from handbooks with aid of friction factors ororifice coefficients.. Individual terms may be evaluated as follows: For a constant volume JouleExpansion with W=0, U 0 & (PV) = PdV + VdP = P*0+VdP =+VdP, the equation is((V2/2g)/J)1+Q=((V2/2g)/J)2+V P/J+F The friction term is: F = (G/ LM)2(2fL/D)/32.2/778. The expansionwork 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 dropequation:

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

The detailed pressure drop equation without expansion work reduces to Weymouth type (P2) frictionequations. In all events, specification of an outlet temperature leaves 1 degree of freedom, P & oneequation. A spread sheet solution is simple by adjust on P2 to converge solution.

p p g

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


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

(fL/D)M =(4fL/D)F = (1/Mo2 1/M2)/k + (1+1/k)ln{[(Mo/M)2][(k-1)M2+2]/[(k-1)Mo2+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 frictionnumber, when expressed with Moody Friction factor is same as Crane K, which can also be expressed asvalve coefficient, Cv, as pointed out by this paper. If Y=[(k-1)M2+2], Density, Pressure and Temperatureratio 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 possibleto select an outlet diameter that forces M=M. Looked at another way a given Cv must have a fixed Machratio, which is a doubtful result. Another approach is to relate T=f(P) by the JTX coefficient or 1-1/k = mbut use Schultz definition of m at zero efficiency, m= (P/T) .jtx. This determines a Polytropic k based ona 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 aJTX 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 notintroduce ideal gas assumptions. Nor does the total energy equation bias results by introduction of apredefined thermodynamic path. The TJX & TJTX are just the Joule or JT Expansion temperaturechange using the low pressure Cv or Cp and is thermodynamically correct.

p q p p

Ideal Enthalpy Equation v. Adiabatic head & cautions


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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 thevelocity heads. The ideal compressor or expander does not have velocity heads.. In practice velocity headsare 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 orexpansion 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 oneattempt to better determine temperature change but failed by double counting efficiency. Enthalpycalculation is next considered.

Clearly: Power is just product of head and flow, adjusted for driver inefficiency. Some engineering texts double or treblecount 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 areothers. Once the head has been adjusted for Polytropic efficiency, the only further corrections to shaft power are losses fromdriver source power to impeller. Conversely, for expanders, application of double or treble efficiencys under estimates theavailable 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 theSchultz equations, a much better fit to the H-S power is obtained. Repeat: one does not determine Polytropic head byPolytropic n, then re-correct Polytropic Head by efficiency or worse still, additionally take another correction of Polytropicefficiency to the product of head and mass rate: repeat: true head times mass rate = gas power& once head adjusted forefficiency by use of Polytropic n, ,no further adjustment needed.

Ideal Enthalpy Equation v. Adiabatic head & cautions

H & U of real gas as function of

JT & J Temperature change:


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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 - Cp JT = (V T[( V/ T)p=c ) gives h = Cp T + (-Cp JT) 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 sameas adding an additional dT.jt. But expansion coefficient were found more accurate than of adjusting Cp forpressure. 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, p270Ludwig 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 givenby 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 corrn 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. correctiongives best dH results. Neither method matches T2of 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 sameas Scheel Chart H-S shown by this document, hof 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 samepressures: 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 alwaysC(dT-dTx) or dT=dTx & dT is mostly negative

Determine Cp by Explicit EOS: ( C / P) ( C )/ T|


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Determine Cp by Explicit EOS: ( Cp/ P)T=c = - ( JTCp)/ T|p=cR. L. Callendar EOS: V=RT/P+b+a/T3.33

Cp = 7.2R ( P/Pc)/Tr4.33 given without derivationOr 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

dCp 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 airshows the Callendar EOSprovides superior estimationof heat capacity change withpressure vs. Reduced

coordinate Chart or theBerthelot EOS Equation.It is no surprise that dCpcorrelates well with Z. Itappears the Callendar methodof estimating dCp is animprovement over other

methods. N.B. When using thecorrected Cp then isnecessary to also usecorrected Cp-Cv value todetermine isentropicexponent to determine T2 ofadiabatic expansion orcompression. Recommend to

use the JT correction for dH.

Note: R. L. Callendar produced the 1st steam tables and Mollier used theCallendar EOS to make the well known chart for steam,Cira 1925AD. Also,the Keys & Kay Steam Tables use the Callendar EOS for superheated steamproperties, Wiley Publishers

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


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Schultzs

method

On page 80 example he

wrongly uses Wa =Wp/e.Polytropic work Is Actual work


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Schultzs X

Chart


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Schultzs YYs {(-P/V)( V/ P)T } then Ys=P T

A ?? Mod of Schultzs dH & dT method


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This result questionable

Not agree w/ PerryNo precise reference

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


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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)]

Cp jt=T/z(dz/dT) or (Cp-Cv) = (z )[1+ Cp jt/2.72] (same as Schultzs Eqn.)

a 1st approximation is by neglecting term Cp jt/2.72 is: Cp - Cv = z

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

Aside: H = U + z ( T), = U + P V + V P by steam table @P&T=c, L -> V check:@1atm, 212F:970.3=897.5 + 72.8 =970.3 & T S=970.3 = H

Compare to Berthelot (1907) approximation for corrected Cp-Cv :Cp-Cv = R(1+1.7Pr/Tr

3). The empirical Berthelot EOS is: Z=1+(1-6/Tr2)9Pr/(128Tr).

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

truncated viral with set to zero. The Blot 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


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, , y p y p , gPower (84) 1962 Part 1 Jan. p69, part. 2, p222 in Perrys 5 th 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+Z Cp/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 idealgas, (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 Scheelshows @ 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, seefollowing charts.

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

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 ( + 0.37 Cp).

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


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

R/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 asper 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 asshown 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

absErrToutv

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

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


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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 doublecounted 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

Effect of Friction (Joule Effect) Heating in Gas Flow

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