1 Machine frigorifique :

23
R143a
−5◦C5,30 bar
35
16,23 bar
−5 2◦C
16,23 bar 2−(−5) = 7
h1
35
h1= 254,43 KJ/Kg
5,30 bar
2◦C h3= 391,27 KJ/Kg
˙
Qf= 8,2KW ˙
Qf= ˙m(h3−h2)
˙m=˙
Qf
h3−h2
=8,2
391,27 −254,43 = 0,0599 Kg/s
60 g/s
h2=h1= 254,43 KJ/Kg

−5
h0=h0(−5 ) = 192,71KJ/Kg h” = h” (35 ) = 383,68KJ/Kg
x=h2−h0
h”−h0= 0,323
˙
Wu= ˙m(h4−h3)=0,0599.(424,35 −391,27) = 1,981 KW
2KW
s4= 1,7330 KJ/Kg.K
s3= 1,7135 KJ/Kg.K
s4> s3
T1= 10 ϕ1= 70% ˙ma=
0,106 Kg/s 1bar
T2= 20
100 3
η3= 0,014 Kg/Kg a.s.
10
Ps1=Ps(10 ) = 0,01227 bar
Pv1=ϕ1.Ps1= 0,7∗0,01227 = 0,008589 bar
η1=me
ma
=Me
Ma
×Pv
P−Pv
= 0,622 ×0,008589
1−0,008589 = 0.00539 Kg/Kg a.s
5,4g Kg
η2=η1et Pv2=Pv1

h=h0+cpa (T−T0) + η[cpv (T−T0) + Lv0]
h1=cpa t1+η[cpv t1+Lv0]=5.64 Kcal/Kg = 23.61 KJ/Kg a.s.
t1=T1−T0= 10
cpa = 0,24 Kcal.Kg−1K−1cpv =
0,46 Kcal.Kg−1K−1
0Lv0= 597 Kcal/Kg
h2=cpa t2+η[cpv t2+Lv0] = 8,06 Kcal/Kg = 33,76 KJ/Kg a.s.
˙
Q= ˙ma(h2−h1)=0,106 (33,76 −23.61) = 1,076 KW
3
h3=h2+hv(η3−η2) = 56,79 KJ/Kg a.s.
hv
hv= 639,1Kcal/Kg = 2674.6KJ/Kg
h0= 0
0
T3=h3−η3Lv0
cpa +η3cpv
=56,79 −0,014 ×597 ×4,185
(0,24 + 0,014 ×0,46) 4,185 = 21.15
KJ
Pv3=η3P
0,622 + η3
= 0.02201 bar

3
2 ∆ ˙me= ˙me3−˙me2
∆ ˙me= ˙me3−˙me2=η3˙ma−η2˙ma= ˙ma(η3−η2) = 0,0009128 Kg/s
0,91 g/s 3,286 Kg/heure
η4
η4=∆ ˙me
˙ma
+η2= 0.0183
h4=h2+hv(η4−η2) = 68.31 KJ/Kg a.s.
2
3 2 −3

isentropique