Question

Just want to check to make sure im doing this right

Refrigerant = R-134a (sorry, though it had this info in it)

11.1) (25 points) A refrigeration system, as shown in Figure 1, needs to remove heat at a rate of 25 kW to maintain a temperature of -12°C inside a freezer located in an environment with an ambient temperature of 40°C. The maximum pressure in the system is 14 bar and the minimum pressure is 1.2 bar. The isentropic efficiency of the compressor is 85%. Determine (a) The temperature, pressure, enthalpy, and entopy at each state (b) The mass flow rate of the working fluid (c) The coefficient of performance for the system (d) The amount of work required to operate the compressor (e) The second law efficiency of the evaporator (f) Sketch a T-s diagram of the working fluid TH Warm environment 2 (3 1 0 Condenser Expansion valve Evaporator Compressor 1 T: Cold environment Figure 1: Basic refrigeration cycle

Answer 1

As there is no mention of state of refrigerant at 1 and 3 so assuming ideal Vapour compression cycle conditions ५०० To Mc=0.85 sor Condensos Qin = 25 kW compressos Ti = -12 expansio valve Win To = - 40°c teva pere to Qin -12°c Drawing T's diagram 3 Process 1-2' is ideal process for compressor Process 1-2 is actual procer Proces 3-4 is iso enthalpis expansin ie hz=ht shown dotted lines 14 bas [shower hy dated? 4 1.2 bas) [A-12 233.86 kulio from saturated table of R-134a At 1.2 bar (0.12 MPa) Tsat = = -22.36°C 1 hg shi-d12.54 kilig + 21:32 Sg - S, -0.9354 KI/kg-k S, as,' = 0.9354 kJ lig-a from superheated table [A-13] for S2 = 0.9354 & P = 14 bar By interpolation Now, 1.4 mla; Tsat = "= 52.43° c 0.9354 -0.9297 ha - 283.1 295.31 - 283.1 so, ha = 285.0d8 kalig 0.9658 -0.9294 Sg = = 0.9003 kJ/19.

S-S2 = Cpln * J/mk Cp of R-134 9 = 84.008 1 - - Borg (p = 1.2468 kJ/18-k 0.9354 - 0.9003 - 1.2468 * in So. T2 = 334.722K = 61.72°C 325.43 Now, Ne hi-hi ha-hi Mc = 0.85 >> 0.8.5 = 285.028 - 233,86 h2 233.86 T2-TI 61.72 – (-22:36) 12 - 2.236) h2 - 294,057 Kulis So, T₂ = 76.557 °C -> At 14 bar (1.4 MPa) T3 - T2 = 52.43°c ho = hg = 125.26 kJ/ly h3=h4 = 125.26 KJ/kg) S3 = 0.4453 kJ /eg-k Ti = TA = -22.36°c At 1.2 baro ha = hqta a Chg-ho) 750.489 195.26 = 21.32 + 2x 212.54 so, S4-S4 + ($9-$8)=0.0879+ 0.489 X(0.9354-0.0879) so. S4 = 0.5023 kJ/kg-k So, we have bind te mos peratuse , Pressure , enthalpy and entropy at each state.

b) Qin = m Chi-he] * 25 - m [233-86-125-26) so, in=0.23 kgls. - Mass flow rate of refrigerant, in = 0.23 kals (0) COP = Refrigerant effect compressor work So, compressor work, Win = m ( h2-hi] = 0.23% [294.05 7 - 233.86 Win = 13.857 KJ/ - 13.857 KW 1.804 COP=1.804 сора 13.857 d) compressor work, Win = 13.857 km Maximum COP = (COP) cannot (873-12) 273+40) - @73-12) Torti (COP) 261 raimon 52 -5.019 soi COP 1.804 5.019 -0.35 943 Nsecond (COP) man In second = 35.543 %