Question

For this problem you should apply the methodology for solving thermodynamics problems that has been emphasized in the course, including clearly defining all systems and subsystems used for analysis. You should also begin analyses with general equations and simplify them to a form appropriate for the specific problem that you ae solving, stating your assumption.

Refrigerant 22 enters a valve at 1 MPa and +20 degrees celsius and is throttled to a temperature of -26 degrees celsius at the exit.

Determine:

a.) the exit pressure [bar]

b.) the exit internal inergy [kJ/kg] of the refrigerant.

c.) Show the inlete state [1] and exit state [2] as well as the process between 1 and 2 on a P-V diagram

Answer 1

Given data: R-22 P = 1 MPa = 1000 kPa T = +20°c { Compressed liquid Pe = ? T2 = -26°C To be found: a). Pe (bar) b). Uz (kJ/kg) c). Process carve on P-v diagram. solution: For Tsat = 20c, we have Psat = 0.91 MPa Since P, > Psat, we have compressed liquid. In the absence of compressed liquid data it is okay to treat it as saturated liquid at the given temperature.

For T = +20°C, we have P = 0.91 MPa 7, = 0,8263x10 m = 2. = 68-33 ki = h,= 69.09 b = h Since R-22 is throttled, we have be= he For state-2, we have T2 = -26°c h = hy = 69.09 kr - 0.7327 milliy = 0.7327x1ől % = 0.1159 m3/kg mp = 15.31 kJ/kg uq = 217.11 kJ/kg hf = 15.45 kJ/kg hq = 239.53 kilks The <hechy

ha= bett (ng-hp) 69.09 = 15-45 + 22 (239.53-15.45) X = 0.2394 [x2 = 23.94% Uz = f t 22 (Ag-Up) 4₂ = 15.31 + 0.2394 (217- 11 - 15.31) amb = 63.62 het ons P = Peat - 1.9845 kas Lo Exit pressure Az = 63.32 kolko Exit internal energy Vi = 0.7827X163 mm + 0.2394 [0,159 - 071827X163 mm = 0.028804 m/BA

Pa ---- - Pe = 1.9345 bar IP, = 10 bar Vi = 0-8263x10 m V = 0.02830 4