Question

1. (100 points) A counterflow heat exchanger operates at steady state to transfer heat between air and refrigerant 134a. The air enters at 22°C and 0.1 MPa, then exits at 7°C. The R-134a enters at 0°C at a rate of 30 kg/h by mass, and is a saturated vapor upon exit. Both streams experience no notable change in pressure. (a) Determine the heat transfer rate and associated rate of exergy transfer for the R-134a stream in kJ/h. (b) Evaluate the flow exergy change for both of the two streams in kJ/h. You may assume conditions of the surroundings to be P = 0.1 MPa and T. = 22°C.

Answer 1

Counter flow heat exchanges. faze Air 22°C Air ote on MPa 0.1 Saturated vapor R-134a - öc 30 kg/h Heat transfer rate cal 3 - mairy Cp x 4T = inrefy the Lewyth) orsaturarted liquid R-134 a from two-phase reaches single ph liquid phase (saturated veupor). 4 = 242.54 kJlKg Sa= o. gigo ko lng-k 247.23 Assumptions. Refrigerant enters at one as saturated hquid. ha= 50.02 kJ/kg, sz= 0.1970 kilug-k in â = 30 kg (247.23 - 50.02) =15816.3 kJ/h ! Po= 0.1 MPa, To=220 - 295k - For R-134a: no=271.768 kilug - so = 1.08876 kJ/kg-k i specific energy at starte 3, 4 z= the-ho - they tho) - To (43-50) Neglecting kinetic and potential energy effect) - 150.02 - 271.768) - 295 x (0.1990 - 1.08876) - 41.3212 rulkg :. 44 = (247.23-271768) - 29541o.glgo-1.08876) - 25.5412 kglug Rate of evergy transfer mirefa (42-42) OM C 47 3.4 kith

273+22) b) for air 4, = (hi-ho) - To (si-so) (As surrounding condition and starte 1 identical 42 = (ha-ho) - To (32-so) Cp of air - cp (Ta - To) - Tox Up in (T 1 ) = 1.005 kJ lieg-K Canume) = 1.005x (7-22) - 295x1005xlu ( 273 + 7 ) & - 15.075 + 15.47177 = 0.39677 kileg À - in air Co OT = 5916.3 5916. 3 - mair 100sx (22-7 = 1.005 (22-7 = 392.45 77 kg/h - Flow energy change of cur - mair & ( 42-41) - 1 155.715 kJ IN R-134 a = 47 3.4 kJ (4, flow energy change of
Flow exergy change for R-134a will be state 4 - state 3 exergy. So, -473.4 kJ/h.

Rate of exergy transfer for R-134a is same as this value.