Question

write a fortran code on helical transcritical compression tube used in pulse tube cryocooler using R407 refrigerant.

it is different than already replied question on chegg. please dont copy paste that answer i want genuine and correct code using fortran 77 only

Answer 1

"Input Data"

R_u = 8.314 [kJ/kmol-K] "universal Gas Constant"

T_Ar = 220 [K]

P_Ar = 5000 [kPa] "Pressure for only Argon in the tank initially."

N_Ar = 1 [kmol]

{N_N2 = 3 [kmol]}

T_mix = 200 [K]

T_cr_Ar=151.0 [K] "Critical Constants are found in Table A.1 of the text"

P_cr_Ar=4860 [kPa]

T_cr_N2=126.2 [K]

P_cr_N2=3390 [kPa]

"Ideal-gas Solution:"

P_Ar*V_Tank_IG = N_Ar*R_u*T_Ar "Apply the ideal gas law the gas in the tank."

P_mix_IG*V_Tank_IG = N_mix*R_u*T_mix "Ideal-gas mixture pressure"

N_mix=N_Ar + N_N2 "Moles of mixture"

"Real Gas Solution:"

P_Ar*V_Tank_RG = Z_Ar_1*N_Ar*R_u*T_Ar "Real gas volume of tank"

T_R=T_Ar/T_cr_Ar "Initial reduced Temp. of Ar"

P_R=P_Ar/P_cr_Ar "Initial reduced Press. of Ar"

Z_Ar_1=COMPRESS(T_R, P_R ) "Initial compressibility factor for Ar"

P_Ar_mix*V_Tank_RG = Z_Ar_mix*N_Ar*R_u*T_mix "Real gas Ar Pressure in mixture"

T_R_Ar_mix=T_mix/T_cr_Ar "Reduced Temp. of Ar in mixture"

P_R_Ar_mix=P_Ar_mix/P_cr_Ar "Reduced Press. of Ar in mixture"

Z_Ar_mix=COMPRESS(T_R_Ar_mix, P_R_Ar_mix ) "Compressibility factor for Ar in mixture"

P_N2_mix*V_Tank_RG = Z_N2_mix*N_N2*R_u*T_mix "Real gas N2 Pressure in mixture"

T_R_N2_mix=T_mix/T_cr_N2 "Reduced Temp. of N2 in mixture"

P_R_N2_mix=P_N2_mix/P_cr_N2 "Reduced Press. of N2 in mixture"

Z_N2_mix=COMPRESS(T_R_N2_mix, P_R_N2_mix ) "Compressibility factor for N2 in mixture"

P_mix=P_R_Ar_mix*P_cr_Ar +P_R_N2_mix*P_cr_N2 "Mixture pressure by Dalton's law. 23800"

"Work is the integral of p dv, which can be done analytically."

w_ideal=R*T[1]*Ln(p[1]/p[2])

"First Law - note that u_ideal[2] is equal to u_ideal[1]"

q_ideal-w_ideal=u_ideal[2]-u_ideal[1]

"Entropy change"

DELTAs_ideal=s_ideal[2]-s_ideal[1]

"***** COMPRESSABILITY CHART SOLUTION ******"

"State 1"

Tr[1]=T[1]/T_critical

pr[1]=p[1]/p_critical

Z[1]=COMPRESS(Tr[1], Pr[1])

DELTAh[1]=ENTHDEP(Tr[1], Pr[1])*R*T_critical "Enthalpy departure"

h[1]=h_ideal[1]-DELTAh[1] "Enthalpy of real gas using charts"

u[1]=h[1]-Z[1]*R*T[1]

"Internal energy of gas using charts"

DELTAs[1]=ENTRDEP(Tr[1], Pr[1])*R "Entropy departure"

s[1]=s_ideal[1]-DELTAs[1] "Entropy of real gas using charts"

"State 2"

T[2]=T[1]

Tr[2]=Tr[1]

pr[2]=p[2]/p_critical

Z[2]=COMPRESS(Tr[2], Pr[2])

DELTAh[2]=ENTHDEP(Tr[2], Pr[2])*R*T_critical "Enthalpy departure"

DELTAs[2]=ENTRDEP(Tr[2], Pr[2])*R "Entropy departure"

h[2]=h_ideal[2]-DELTAh[2] "Enthalpy of real gas using charts"

s[2]=s_ideal[2]-DELTAs[2] "Entropy of real gas using charts"

u[2]=h[2]-Z[2]*R*T[2] "Internal energy of gas using charts"

w_chart=Integral(p,v,v[1],v[2])

"We need an equation to relate p and v in the above INTEGRAL function. "

p*v=COMPRESS(Tr[2],p/p_critical)*R*T[1] "To specify relationship between p and v"

"Find the limits of integration"

p[1]*v[1]=Z[1]*R*T[1] "to get v[1], the lower bound"

p[2]*v[2]=Z[2]*R*T[2] "to get v[2], the upper bound"

"First Law - note that u[2] is not equal to u[1]"

q_chart-w_chart=u[2]-u[1]

"Entropy Change"

DELTAs_chart=s[2]-s[1]

"***** SOLUTION USING EES BUILT-IN PROPERTY DATA *****"

"At state 1"

u_ees[1]=intEnergy(Name$,T=T[1],p=p[1])

s_ees[1]=entropy(Name$,T=T[1],p=p[1])

"At state 2"

u_ees[2]=IntEnergy(Name$,T=T[2],p=p[2])

s_ees[2]=entropy(Name$,T=T[2],p=p[2])

"Work using EES built-in properties- note use of EES Integral funcion to evaluate the integral of

pdv."

w_ees=integral(p_ees, v_ees, v_ees[1],v_ees[2])

"The following equation relates p and v in the above INTEGRAL"PROPRIETARY MATERIAL.

educators for course preparation. If you are a student using this Manual, you are using it without permission.

12-60

p_ees=pressure(Name$,T=T[1], v=v_ees) "To specify relationship between p and v"

"Find the limits of integration"

v_ees[1]=volume(Name$, T=T[1],p=p[1]) "to get lower bound"

v_ees[2]=volume(Name$, T=T[2],p=p[2]) "to get upper bound"

"First law - note that u_ees[2] is not equal to u_ees[1]"

q_ees-w_ees=u_ees[2]-u_ees[1]

"Entropy change"

DELTAs_ees=s_ees[2]-s_ees[1]

"Note: In all three solutions to this problem we could have calculated the heat transfer by

q/T=DELTA_s since T is constant. Then the first law could have been used to find the work.

The use of integral of p dv to find the work is a more fundamental approach and can be

used if T is not constant."

SOLUTION

DELTAh[1]=16.48 [kJ/kg]

DELTAh[2]=91.96 [kJ/kg]

DELTAs[1]=0.03029 [kJ/kg-K]

DELTAs[2]=0.1851 [kJ/kg-K]

DELTAs_chart=-0.4162 [kJ/kg-K]

DELTAs_ees=-0.4711 [kJ/kg-K]

DELTAs_ideal=-0.2614 [kJ/kg-K]

Fluid$='C3H8'

h[1]=-2232 [kJ/kg]

h[2]=-2308 [kJ/kg]

h_ideal[1]=-2216 [kJ/kg]

h_ideal[2]=-2216 [kJ/kg]

M=44.1

Name$='Propane'

p=4000

p[1]=1000 [kPa]

p[2]=4000 [kPa]

pr[1]=0.2165

pr[2]=0.8658

p_critical=4620 [kPa]

p_ees=4000

q_chart=-155.3 [kJ/kg]

q_ees=-175.8 [kJ/kg]

q_ideal=-97.54 [kJ/kg]

R=0.1885 [kJ/kg-K]

R_u=8.314 [kJ/mole-K]

s[1]=6.073 [kJ/kg-K]

s[2]=5.657 [kJ/kg-K]

s_ees[1]=2.797 [kJ/kg-K]

s_ees[2]=2.326 [kJ/kg-K]

s_ideal[1]=6.103 [kJ/kg-K]

s_ideal[2]=5.842 [kJ/kg-K]

T[1]=373.2 [K]

T[2]=373.2 [K]

Tr[1]=1.009

Tr[2]=1.009

T_critical=370 [K]

u[1]=-2298 [kJ/kg]

u[2]=-2351 [kJ/kg]

u_ees[1]=688.4 [kJ/kg]

u_ees[2]=617.1 [kJ/kg]

u_ideal[1]=-2286 [kJ/kg]

u_ideal[2]=-2286 [kJ/kg]

v=0.01074

v[1]=0.06506 [m^3/kg]

v[2]=0.01074 [m^3/kg]

v_ees=0.009426

v_ees[1]=0.0646 [m^3/kg]

v_ees[2]=0.009426 [m^3/kg]

w_chart=-101.9 [kJ/kg]

w_ees=-104.5 [kJ/kg]

w_ideal=-97.54 [kJ/kg]

Z[1]=0.9246

Z[2]=0.6104