Conditie de suprafata

1. m vapori =6 t/h

mAT= = = = =136,3 t/h

136,3 t/h *1000=37,86 kg/s

Np = =22,72

=22,72

Conditia de amestec

mAT= = = = =

= =71,97 t/h

71,97 t/h *1000= 19,99 kg/s

Np= =11,9948

=11,9940

2. mv=100 Kg/h=0.02778Condensatorul de suprafata

mAR= = =2272,08 Kg/h0.63 Kg/s

Kp= =22,72

Condensatorul de amestec

mAR= = = = =

=1203,95 Kg/h /3600 =0,33 Kg/s

Np= =12,04

Condensatorul de amestec este mai eficient, Up mai mic,ΔT mai mare,debit de AR mai micMv=2400 Kg/h=0,6667 Kg/s

mAT= = = =54594,98 Kg/h

Qp=0Qced= Qprim=mv(i’’-i’)=mAT*CpAT(Tf-Ti)

Q=0,6667*2380 *4185,25

ΔTM=echivalent

ΔTM= = = = =13,966 K

K= = = = =1200,48[W/m2 *k]

λotel carbon(<1,5 %)(50 °C)= 20,85[W/m2*k)

A= = =94,64 m2

1586666,67 1586666,67

20 °C

45 °C

Tcond=50 °C