Question

Qoutl condenser Condenser A standard 4-component vapor-compression cycle using R-134a is shown in the figure to the right. The cycle is used as a refrigeration cycle to cool a refrigerator at 5 °C with a cooling capacity of 200 W, with a heat transfer to a kitchen at 20 °C. Assume that the pressure drops in the evaporator and condenser are negligible, and that the compressor and expansion valve are adiabatic. Take the boundary temperature for heat transfer into the evaporator to be 5 °C, and the boundary temperature for heat transfer out of the condenser to be 20 °C. Assume the compressor suction state is P1 = 2.4 bar and saturated vapor, the compressor discharge pressure is P2 = 7.0 bar, the compressor isentropic efficiency is 70%, and the condenser outlet state is saturated liquid. Determine: valve compressor Kil Evaporator 1 TQin (a) the mass flow rate of the cycle, in g/s (b) the compressor input power, in W (c) the compressor discharge temperature (°C) (d) the COP of the cycle Now consider replacing the expansion valve with an adiabatic turbine. Use the same cooling capacity of 200 W and assume a turbine isentropic efficiency of 70%. For this new architecture, determine: (e) the mass flow rate of the cycle, in g/s (f) the COP of the new cycle HW-37: 2nd Law Analysis of Vapor Compression Cycles Consider the initial refrigeration cycle from HW 36 (i.e., using an expansion valve, not a turbine) and assess the following: (a) determine the rate of entropy generation for each of the components, in W/K (b) determine the total rate of entropy generation using a single-closed system entropy balance for the entire cycle and compare the entropy generation rate from part (a), in W/K (c) plot this cycle on a T-s diagram using these operating conditions

Please help with HW 37

Answer 1

HW-37: (c) Draw the T-s diagram of the given VCRS cycle as follows: TA 20°C 3/ 7 bar 2.4 bar 5°C 14 Process 1-2s: Compression Process 2s-3: Condenser Process 3-4: Expansion Valve (Isenthalpic process) Process 4-1: Evaporator Refer to the saturated refrigerant R-134a pressure table to obtain the following properties: At 2.4 bar (240 kPa): h = h = 247.28 kJ/kg Si = s, = 0.93458 kJ/kg-K his = 202.62 kJ/kg Sjg = 0.75664 kJ/kg . K h 4 = 44.66 kJ/kg Sy = 0.17794 kJ/kg-K

At 7 bar (700 kPa): hz = h, = 88.82 kJ/kg h = 265.03 kJ/kg Sz = S, = 0.33230 kJ/kg. K Se=0.91994 kJ/kg . K he = 176.21 kJ/kg Sig = 0.58763 kJ/kg-K At 20°C: Specific heat of saturated refrigerant vapor, ca = 0.9976 kJ/kg.K. Note: Expansion (process 3-4) is an isenthalpic process which means ha =h . HW-36: Calculate the mass flow rate of cycle as follows: Cooling capacity (RC) is given by refrigeration effect (RE) times the mass flow rate. RC = mx RE = mx (hh) 200 = mx(247.28–88.82)x10² J/kg m=1.257x10-kg/s =1.257 g/s

(b) Calculate the compressor input power as follows: Si = S2 = Sg + Ca, In 0.93458=0.91994 +0.9976 In- T (273+20) K T, = 297.33 K = 24.33°C From superheated refrigerant R-134a, at 700 kPa and 24.33º: h = 265.03 kJ/kg his-h Compressor isentropic efficiency, nien = h-m lisen 07_265.03 – 247.28 hs - 247.28 his = 272.637 kJ/kg Compressor input power, Win=mx(hs - h) = 1.257x10- kg/sx(272.637- 247.28)x10 J/kg W. = 31.88 W Calculate the Compressor Discharge Temperature as follows:

hs = hq + pv (T25 -12) 272.637 = 265.03 +0.9976(T25 – 297.33) Tzs = 304.95 K = 31.95°C|| Calculate the COP of the cycle as follows: RC COP = RC 200 31.88 COP = 6.273 HW-37: Rate of Entropy Generation for each component. For Compressor: S = $ +calan =0.91994 +0.9976x In 31.88+273 30+273 =0.9261 kJ/kg . K

d(S)system m( Son – Sout) + Seen= di Son = 0 – m(530-) Son=-1.257x10- kgx(0.92611-0.93448) Sce=0.0105 W/K For Expansion Valve: hz = h = h 4 + xhi94 88.82 = 44.66+x(202.62) x=0.218 S4 = S, +XSE = 0.17794 +(0.218x 0.75664) = 0.34288 kJ/kg.K d(S)gistem m(son – Sout) + Sgen=- Sgen = 0–m(s; – sa) Sem = 1.257x10- kg x(0.33230 -0.34288) kJ/kg. K Sco=0.0133 W/K For Condenser:

= w[69 -->): "A =1.257x10°((0.3323 -0.9261) + 272,637–88.827 5. = 0.0161 W/K For Evaporator: ..-m[4-5)- hand = 1.257x10- (0.93458–0.34288) – 247,28 – 88.82] Spon = 0.0272 W/K 5+273 Total rate of entropy generation in a full cycle, (Sgen) = 0.0272+0.0161 +0.0133+0.0105 = 0.0671 W/K (b) Rate of Entropy Generation for the cycle: 88.82-272.637) x1.257x10- 21 [ 247.28-88.82 278 (Sgen Daten = 0.0721 W/K 293