Question:A closed vial with a volume of 0.01 m3 stands in an
atmospheric environment at 90...
Question
A closed vial with a volume of 0.01 m3 stands in an
atmospheric environment at 90...
A closed vial with a volume of 0.01 m3 stands in an
atmospheric environment at 90 kPa pressure and 27 oC temperature. A
valve on the neck of the bottle is opened and atmospheric air is
allowed to enter the bottle. As a result of the air trapped inside
the bottle and the heat transfer of the bottle wall with the
atmosphere, the thermal balance is achieved with the atmosphere.
Throughout this process, the valve is left open until the air
trapped in the bottle reaches a mechanical balance with the
atmosphere.
a) Calculate the amount of entropy production that occurs
during the filling process. (7 points)
Queria Giveni - Air Patm= go kPa Tatm=To=27°C Staten V= = 0.0lm 3 Statez P₂ = Patm=sokia V2=vi T2=To=3ook atm air Patm= 90 kPa To = 27C = 300k sola"!- Mass Consuvation for Control volum - min-mout = (M₂-m.) min (M2-m.) Energy balance for Control volumes Ein- Eout m₂ - mu, men (hif-mout (ho) - Q = myYz- min, Scanned With CS CamScanner
of vial is initially empty then mi=0 so from mass Consovation - min = m₂-mi min = m₂ from emingy consuvation > minnit Q= malz-smin Q = Mr Ug - minha fi) since when thermal and mechanical balance greachs, P₂= Patma go kPa Tz=To = 300k V2=V, = 0.01 m3 (closed) veal let us at T = 3ook say for air R=0.287 kJ /ks K (p=1.005 KJKSK (r=0.718 Kolkok from ain table min P2 V2 gox 0.01 = m₂= 0.01045 kg RT2 0.2878 300 from equ" () > Q = min Cv Tz - min Cp Te Q = min (Cv -(p) Tz Q = 0.01045 (0.718-1005)X 300 Q =-0.8997 = -0.9 kJ cs Scanned, withign indicates that is transfered to the surrounding, CamScanner
Entropy production in Sgen (Asl system (AS) Burged unding Sgen [ms - mis, - misa Q mis [S₂ - Sai] - Sgin = 0.9 Sgin o Since Sq=Si miz + 300 same state Sgen = 0.9 30o 3x103 kJ/K = 0.003 Kolk = Sgen = 3 Ilk CS Scanned with CamScanner