Question

In an aircraft cooling system, air enters the compressor at 1 bar and 4°C and is compressed to 3 bar with an isentropic efficiency of 72%. The compressed air is cooled in a heat exchanger at constant pressure to 55°C and expands in an expander to 1 bar with an isentropic efficiency of 78%. The low temperature air absorbs heat load of 3 tonnes of refrigeration at constant pressure before re- entering the compressor. The compressor is driven by the air turbine. Determine the COP of refrigerator, the power required and air mass flow rate.

Answer 1

Guion-Refrigeration cycle 22 T 3ban. 1-2--I sentropie Compression ladb> Azbucal compresion. 2-3-> isobaric heat rejection 3-4-→ Ixentropic axpansion : 3-41-> Actual expansion 4-1-1 isobaric Heat odelitron Gwen Thar 4 4. S P = Pelbar To=4°C -277 le. Pa = P2 = 3bcam nomp- 78y. T3=55" lisenemp = 78 y. Qin = 3 tonnes. - Boeckg3 tonne of Retigerations 3* 3:5 Kw = lastew solution T2 Т) Assumption! - Aly i's ideal gas Ys 1-4, (p= 1:0055/kak R=0.287K5lkyle.

Gave detailed LILY Ta 277 ( 3 ) lit. T2=277 (3) 09 14 31784-10) Te 0:4 Tз T4 (3) (2369/an 273755 - Ти T4 = 229.66K 0-723 372 24-277 comp = To - Ti Tz'- 277 T2 T TA': 418-44K 333 -7 ) nexp= T3-TA' T3-T4 0.78 - 338 - 239.64 TA'=261-27% @ Cup = Qin Whet. Qin - ñ 4 (T, - TA') => 10-5= mx1005( 277-261-27) M20.664 Kg's 0.664 x 1005 1005( 418-44- 277) (Wnet) outwal = cp (Ta'-TI) (Whet out: = 94.38 kw. COP= 10.5 - O-111 94:38 COP=0.11) Power required 94: 3810 0.6bakgls mass flow rate=0 solution, please upvote It doubt please comment