Question

An insulated rigid vessel contains 5kg of saturated liquid-vapor mixture of Refrigerant- 134a at 200kPa. At this stage, one-fifth of the mass is in vapor phase and rest is in liquid. Now using an electrical resistance heater, heat is supplied to the contents of the vessel until all of the refrigerant is converted to saturated vapor. Show this process on a P v diagram with respect to saturation lines and determine, 1. volume of the rigid vessel 2. temperature, in °C, at each state of the process 3. pressure of the refrigerant at state 2 of the process 4. amount of electrical energy supplied to the system Insulated State 1 Saturate liquid vapor mixture of R-134a State 2 Phase of R134a is saturated vapor PA 200kPa Wein

Answer 1

Given Pie rookPa m, = m₂ = 5kg =m R-134a Refrigerast mg = 1 thom mpa y hom i ..mg = = 1kg mf = 4 4 24kg. At ; P = rookPa, Tsat from saturated R134a pressure table wf = 0.0007539 m3/kg ng = 6.6993 mbling . Ti= Tsat = -10-10 uf- 36.89 kilkg uga = 221543 kilkg. quality ma mg 20-2 5 174 mgame :. 36.69+ 0.2 (22143 – 36-69) 4s uft aug-up) = 3 4,2 73.578 kJ leg.

Uq= ug=221-43 kJ /kg. Vfemfuf- 4 X 0.00075322 3.0128x10-3 m3 Nga 1x 0-0993 = 00993 m3 (3.0128x10-3) +0.099350-1023128 m3. .. Ve vf tug The system is closed constant volume Vay?V= 0.1023128m3. At state ♡; all the contents are at saturated repour. :. Tez Tiz-10-10.; Px =P = 200kpan Uzug amoont) of By ve usings external eledrical work, the refrigerant which is at liquid state is converted into sabrated rapour. Du thu 2-4) the Q=0 ; insulated vessel) 0=(221:43-73.578) 5 the :. We = 739-26 kg 4,5 of txluguvef) 0,= (0-0007332) 40-70-0993 0.0007552 ,-0.020463 milkg. a Uz of v specific volume)