Text 1.6 – EES code and solution
pc=p_crit(STEAM_NBS) {critical pressure}
Tc=t_crit(STEAM_NBS) {critical temperature}
{calculate reduced pressure and temperature}
pr=p1/pc
Tr=T1/Tc
{estimate Z from reduced temperature and pressure using compressibility chart}
Z=0.7
p1=5000/14.7*101
T1=(1000+460)/1.8
v1=Z*R_bar*T1/(Mw*p1) {estimate from ideal gas law with compressibility correction}
Mw=18 {molecular weight of water}
R_bar=R#
v11=volume(STEAM_NBS,p=p1,T=T1) {from (EES) steam tables}
K1=(100/(2.54*12))^3/2.2
v1eng=v1*K1
v11eng=v11*K1
V=100
m1=V/v1eng {estimated mass}
m11=V/v11eng {actual mass}
Solutions
K1=16.05 [(ft^3/lbm)/(m^3/kg)]
m1=816 [lbm]
m11=757.5 [lbm] - estimated mass is about 7% low, might have been better with more careful
reading of chart.
Mw=18 [kg/kmole]
p1=34354 [kPa]
pc=22064 [kPa]
pr=1.557 [-]
R_bar=8.314 [kJ/kmol-K]
T1=811.1 [K]
Tc=647.1 [K]
Tr=1.253 [-]
V=100 [ft3]
v1=0.007634 [m3/kg]
v11=0.008224 [m^3/kg]
v11eng=0.132 [ft3/lbm]
v1eng=0.1225 [ft3/lbm]
Z=0.7
Text 2.1 – EES code and solution
{Ideal Rankine Cycle Text 2-1 use SI units with T in K}
T2=300+273 {boiler temperature}
T4=40+273 {condenser temperature}
{determine states}
{state 2 is the boiler outlet, turbine inlet}
h2=enthalpy(STEAM_NBS,T=T2,x=1)
p2=pressure(STEAM_NBS,T=T2, x=1)
s2=entropy(STEAM_NBS,T=T2,x=1)
s3=s2
p3=p4
p4=pressure(STEAM_NBS,T=T4,x=0)
h3=enthalpy(STEAM_NBS, p=p3, s=s3)
h4=enthalpy(STEAM_NBS, T=T4, x=0)
s4=entropy(STEAM_NBS, T=T4, x=0)
s1=s4
p1=p2
h1=enthalpy(STEAM_NBS,p=p1, s=s1)
{cycle performance}
wp=h4-h1
wt=h2-h3
qin=h2-h1
wnet=wp+wt
eta=wnet/Qin
{plant performance}
m_dot=125
P=m_dot*wnet/K1
K1=1000
Solution
Variables in Main
eta=0.3755 [-]
h1=175.5 [kJ/kg]
h2=2749 [kJ/kg]
h3=1774 [kJ/kg]
h4=166.9 [kJ/kg]
K1=1000 [kW/MW]
m_dot=125 [kg/s]
P=120.8 [MW]
p1=8566 [kPa]
p2=8566 [kPa]
p3=7.323 [kPa]
p4=7.323 [kPa]
qin=2574 [kJ/kg]
s1=0.5703 [kJ/kg-K]
s2=5.705 [kJ/kg-K]
s3=5.705 [kJ/kg-K]
s4=0.5703 [kJ/kg-K]
T2=573 [K]
T4=313 [K]
wnet=966.3 [kJ/kg]
wp=-8.61 [kJ/kg]
wt=974.9 [kJ/kg]
Text 2.5 – EES code and solution
This is a copy of the EES code. The codes are identical for water, Freon and propane except that the
working fluid is changed.
{Ideal Rankine Cycle Text 2-5}
T2=(200+460)/1.8 {boiler temperature in K}
T4=(70+460)/1.8 {condenser temperature in K}
{determine states}
{state 2 is the boiler outlet, turbine inlet}
h2=enthalpy(PROPANE,T=T2,x=1)
p2=pressure(PROPANE,T=T2, x=1)
s2=entropy(PROPANE,T=T2,x=1)
v2=volume(PROPANE,T=T2,x=1)
s3=s2
p3=p4
p4=pressure(PROPANE,T=T4,x=0)
h3=enthalpy(PROPANE, p=p3, s=s3)
v3=volume(PROPANE, p=p3, s=s3)
h4=enthalpy(PROPANE, T=T4, x=0)
s4=entropy(PROPANE, T=T4, x=0)
s1=s4
p1=p2
h1=enthalpy(PROPANE,p=p1, s=s1)
{Working fluid thermodynamic performance}
C_p_liq=cp(PROPANE,p=p1, s=s1)
h_fg=enthalpy(PROPANE,p=p2, x=1)-enthalpy(PROPANE,p=p2, x=0)
R1=h_fg/C_p_liq
T_H_av=(h2-h1)/(s2-s1)
eta_carnot=1-T4/T_H_av
{cycle performance}
wp=h4-h1
wt=h2-h3
qin=h2-h1
wnet=wp+wt
eta=wnet/Qin
{plant performance}
P=100
P=m_dot*wnet
K1=2.2
K2=1/(.3048)^3
{turbine flows}
m_dot=V_dot_in/v2
m_dot=V_dot_out/v3
{flows in English units}
m_dot_eng=m_dot*K1
V_dot_inENG=V_dot_in*K2
V_dot_outENG=V_dot_out*K2
The solutions were copied from EES and then pasted into Excel and rearranged for
discussion in the table below.
Freon 12 Propane Water
C_p_liq
0.9554
2.641
4.183 [kJ/kg-K]
h_fg
74.32
100.4
2274 [kJ/kg]
R1
77.8
38.02
543.5 [K]
T_H_av
346.4
343.9
361.8 [K]
eta_carnot
0.15
0.1439
0.1863 [-]
eta
0.15
0.1439
0.1863 [-]
h1
58.76
261.8
89.33 [kJ/kg]
h2
212.3
606.9
2665 [kJ/kg]
h3
186.6
550.9
2185 [kJ/kg]
h4
56.1
255.5
89.25 [kJ/kg]
K1
2.2
2.2
2.2 [kW/MW]
K2
35.31
35.31
35.31 ft^3/m^3
m_dot
4.341
2.013
0.2084 [kg/s]
m_dot_eng
9.551
4.429
0.4585 [lb_m/s]
P
100
100
100 [kW]
p1
2973
4012
80.02 [kPa]
p2
2973
4012
80.02 [kPa]
p3
587.8
865.5
2.533 [kPa]
p4
587.8
865.5
2.533 [kPa]
qin
153.5
345.1
2576 [kJ/kg]
s1
0.2119
1.193
0.3146 [kJ/kg-K]
s2
0.6551
2.196
7.434 [kJ/kg-K]
s3
0.6551
2.196
7.434 [kJ/kg-K]
s4
0.2119
1.193
0.3146 [kJ/kg-K]
T2
366.7
366.7
366.7 [K]
T4
294.4
294.4
294.4 [K]
v2
0.004767 0.007088
2.087 [m^3/kg]
v3
0.02771 0.04648
45.83 [m^3/kg]
V_dot_in
0.0207 0.01427
0.435 [m^3/s]
V_dot_inENG 0.7308
0.5039
15.36 [m^3/s]
V_dot_out
0.1203 0.09358
9.553 [m^3/s]
V_dot_outENG 4.249
3.305
337.3 [m^3/s]
wnet
23.03
49.67
479.8 [kJ/kg]
wp
-2.655
-6.286 -0.07763 [kJ/kg]
wt
25.69
55.95
479.9 [kJ/kg]
In these cells we are looking at the thermodynamic performance
h_fg/C_p_liq gives a measure of the fraction of heat added at the highest T
For a reversible heat addition T_av = (h2-h1)/s2-s1)
Based on average temperature of heat addition
Reversible Rankine efficiency same as Carnot based on T_H_av and T_L_av
Water has the highest efficiency because it has the highest average
temperature of heat addition and the same heat rejection temperature
Water has lowest mass flow
Water has lowest pressures
Because of the low p's the specific volume is much higher for water
Even though the mass flow is lowest for water the high specific volume
leads to very high volume flows. For the same velocity in the duct the
area for water would have to be
102 times greater for water
than for propane