Tema de Casa 1 La TT I
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Date initiale Masa molara i= 26 Gaze O2= 32 kg/kmol A= oxigen O2 H20= 18 kg/kmol B= vapori de apa H20 CO2= 44 kg/kmol C= dioxid de carbon CO2 m 10^3 [kg] n 10^3 [kmol VN 10^3 [m3N] V1 10^3 [m3] a b 13.8 0.36 7.2 17.4 22.6 2.54 a) Compozitia amestecului, masa molara, aparenta a amestecului si constanta specifica a amestecului Compozitia amestecului mH2O= 0.00648 kg nO2= 0.00043125 Kmol VnO2= 0.009666038 mCO2= 0.0141429 kg nH2O= 0.00036 Kmol VnH2O= 0.00806904 mO2= 0.0138 kg nCO2= 0.00032142857 Kmol VnCO2= 0.0072 mtotal= 0.0344229 ntotata= 0.00111267857 Kmol Vntotal= 0.024935078 Participatile masice gO2= 0.4008964 = 40.0896414 % gH20= 0.188247 = 18.8247012 % gCO2= 0.4108566 = 41.0856574 % total 100 Participatile volumice rO2= 0.3876482 = 38.7648184 % rH2O= 0.323602 = 32.3601962 % rCO2= 0.2887499 = 28.8749854 % total 100 Masa molara a amestecului Mam= 30.934571 kg/kmol
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Transcript of Tema de Casa 1 La TT I
Date initialeMasa molara Temperatura gazelor
i= 26Gaze O2= 32 kg/kmol tO2= 5A= oxigen O2 H20= 18 kg/kmol tH2O= 100B= vapori de apa H20 CO2= 44 kg/kmol tCO2= 200C= dioxid de carbon CO2
m 10^3 [kg] n 10^3 [kmol] VN 10^3 [m3N] V1 10^3 [m3] a b d13.8 0.36 7.2 17.4 22.6 2.54 2.24
a) Compozitia amestecului, masa molara, aparenta a amestecului si constanta specifica a amesteculuiCompozitia amestecului
mH2O= 0.00648 kg nO2= 0.00043125 Kmol VnO2= 0.009666038 m3NmCO2= 0.01414286 kg nH2O= 0.00036 Kmol VnH2O= 0.00806904 m3NmO2= 0.0138 kg nCO2= 0.00032142857 Kmol VnCO2= 0.0072 m3Nmtotal= 0.03442286 ntotata= 0.00111267857 Kmol Vntotal= 0.024935078 m3NParticipatile masice
gO2= 0.40089641 = 40.08964143 %gH20= 0.18824701 = 18.8247012 %gCO2= 0.41085657 = 41.08565737 %
total 100Participatile volumice
rO2= 0.38764818 = 38.76481836 %rH2O= 0.32360196 = 32.36019619 %rCO2= 0.28874985 = 28.87498545 %
total 100Masa molara a amestecului
Mam= 30.9345708 kg/kmol
Constanta specifica a amesteculuiConstante specifice
RM= 8314Ram= 268.74032 J/kgK RO2= 259.8125
RH2O= 461.8889RCO2= 188.9545
b) Temperatura si presiunea amestecului in starea initiala
259.8125 J/kgK RM= 8314461.888889 J/kgK188.954545 J/kgK
k= 1.33 pentru gaz triatomic CO2, H2Ok= 1.4 pentru gaz biatomic O2, N2k= 1.6 pentru gaz monoatomic
CMVO2= 20785 J/kmolk cvO2= 649.53125 J/kgKCMVH2O= 25193.9394 J/kmolk cvH2O= 1399.6633 J/kgKCMVCO2= 25193.9394 J/kmolk cvCO2= 572.5895317 J/kgK
CMPO2= 29099 J/kmolk cpO2= 909.34375 J/kgKCMPH2O= 33507.9394 J/kmolk cpH20= 1861.552189 J/kgKCMPCO2= 33507.9394 J/kmolk cpCO2= 761.5440771 J/kgK
CMVam= 23484.822 J/kmolk cpam= 1027.859708 J/kgKCMpam= 31798.822 J/kmolk cvam= 759.1193879 J/kgK
Kam= 1.35401588
RO2=
RH20=RCO2=
Temperatura in starea initiala
t1= 98.4030602 grade C
T1= 371.55306 K
Presiunea in starea initiala
p1= 197538.312 Pa= 1.975383122 bari
c) Parametri de stare p, V, T in punctele caractreristice ale ciclului (prezentare recapitulativa tabelara)
I Transfoarmarea 1-2: T =constant (izoterma)
p2= 44.6436586 bari
T2=T1= 371.55306 K
V2= 0.7699115 10^3 m3
II Transformarea 2-3: V=constant (izocora)
T3= 943.744773 K
V3=V2= 0.7699115 10^3 m3
p3= 113.394893 bari
III Transformarea 3-4:P constant (izobara)
p3=p4= 113.394893 bari
T4= 2113.98829 K
V4= 1.72460177 10^3m3
IV Transformarea 4-5: destindere politropa
V5=V1= 17.4 10^3 m3
p5= 4.46436586 bari
Exponentul politropic
n= 1.39943132
T5= 839.709916 K
Prezentare recapitulativa tabelara
Starea p[Pa]*10^5 V[m3] 10^3 T [K]1 197538.312 0.0174 371.55306022 4464365.86 0.0007699 371.55306023 11339489 0.0007699 943.7447734 11339489 0.0017246 2113.9882925 446436.586 0.0174 839.7099161
d) Schimbul energetic sub forma lucrului mecanic si caldura, pentru fiecare transformare in parte si pentru intregul ciclu
I Transformarea 1-2: T1=T2
-10716.913 J/CC-10716.913 J/CC
II Transformarea 2-3: V2=V3
0
14951.975 J/CC
III Transformarea 3-4: p3=p4
L1-2=
Q1-2=L1-2=
L2-3=
Q2-3=
10825.7 J/CC
41405.4016 J/CC
IV Transformarea 4-5
29512.224 J/CC cn= 86.31206256 J/kgK
-3786.0184 J/CC
V Transformarea 5-1: V5=V1
0
-12233.434 J/CC
Lc= 29621.0107 J/CCQc= 29621.0107 J/CC
e) Variatiile energiei intarne, entalpiei si entropiei pentru fiecare transformare in parte si pentru intregul ciclu.
I Transformarea 1-2: T1=T2
L3-4=
Q3-4=
L4-5=
Q4-5=
L5-1=
Q5-1=
T=const. DU1-2= 0 JDH1-2= 0 J
DS1-2= 28.8435611 J/K
II Transformarea 2-3: V2=V3
DU2-3= Q2-3= 14951.975 J
DH2-3= 20245.2116 J
DS2-3= 24.3584339 J/K
III Transformarea 3-4: p3=p4
DU3-4= 30579.7015 J
DH3-4= Q3-4= 41405.402 J
DS3-4= 13.6135131 J
IV Transformarea 4-5 destindere politropa
DU4-5= -33298.242 J
DH4-5= -45086.349 J
DS4-5= -45.509163 J
V Transformarea 5-1 racire izocora
DU5-1= -12233.434 J
DH5-1= -16564.264 J
DS5-1= -21.306345 J
DUtotal= 0
DHtotal= 0
DStotal= 0
grade Cgrade Cgrade C
pentru gaz triatomic CO2, H2Opentru gaz biatomic O2, N2pentru gaz monoatomic
