As shown in Fig Q3 an initially evacuated canister of 0002 m

As shown in Fig. Q3, an initially evacuated canister of 0.002 m^3 is connected through a valve to an R-134a supply line. The supplied R-134a is saturated liquid at 0 degree C. The valve is opened, and the canister is filled with the supplied R134a until the pressure in the canister is equal to that in the supply line. As this filling is done very quickly, the process can be assumed to be adiabatic. Find the pressure and internal energy at the final state and determine the quality if it is in the saturation region; Find the final mass of R-134a in the canister and Determine the net change of entropy.

Solution

1)

From R-134a properties at T1 = 0 deg C and quality x1 = sat. liquid we get P0 = 293 kPa, v0 = 0.000772 m^3/kg, u0 = 51.6 kJ/kg, h0 = 51.9 kJ/kg, s0 = 0.204 kJ/kg-K

2)

Flow will happen until P2 = P

Hence, P2 = 293 kPa

Q - W = u2 - u1

Q = 0 since adiabatic

W = -P0*V0

u2 - u1 = P0*v0

u2 = u1 + P0*V0

= h1

= h0

= 51.9 kJ/kg

From R134a properties At P2 = 293 kPa and u2 = 51.9 kJ/kg, we get quality x2 = 0.00137

2)

Since x2 = 0.00137 is very close to zero, we can assume x2 = 0.

From R134a properties At P2 = 293 kPa and x2 = 0, we get v2 = 0.000772 m^3/kg

m = V/v

= 0.002 / 0.000772

= 2.59 kg

3)

From R134a properties At P2 = 293 kPa and u2 = 51.9 kJ/kg, we get quality s2 = 0.205 kJ-kg-K

Entropy change = m*(s2 - s1)

= 2.59*(0.205 - 0.204)

= 0.00259 kJ/K