MMAD BIScerns ollege of aanteat t of Mec Problem 2 (5 Mark/30) A closed, rigid t

Question

MMAD BIScerns ollege of aanteat t of Mec Problem 2 (5 Mark/30) A closed, rigid tank contains a two-phase liquid-vapor mixture of with a quality of 50.36%. Energy transfer by final pressure of 6 har -2oc Refrigerant-22 nitially at heat into the link occurs unta the refrigerant is at a a) Determine the final temperature, in C does the the final state is in the superheated vapor region, at what temperature, in tank contain only saturated vapor State 1 Stats 2 Pconstan m coustant R-22 R-22 0.5036 Therefore,

Explanation / Answer

Initial temperature,T1 = -20 C Quality, x1 = 0.5036

From the property tables of R22.The saturation properties at T = -20 C are

vf = 0.0007409 m^3/kg and vg = 0.09284 m^3/kg, Therefore

v1 = (1-x1)*vf + x*vg = v1 = 0.04712 m^3/kg

P1 = Psat= 2.5 bar

Final properties of the referigerant are: P = 6 bar

Volume is constant since it is closed tank, therfore v2 = v1 = 0.04712 m^3/kg

Now saturation properties at P = 6 bar are

vf = 0.0007917 m^3/kg and vg = 0.03879

Since v2 > vg The state is SUPERHEATED VAPOR

Now at P2 = 6 bar and v2 = 0.04712 m^3/kg

T2 = 47.9 C

Final Temperature is 47.9 C

The temperature where vg = 0.04712 m^3/kg corresponds to saturated vapor

at T = 0 C, vg = 0.04712

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