Given & Find:     a) R-134a at T = 60^oC and v = 0.072 m³/kg.  Find P and h.

                                b) Ammonia at P = 8 bar and v = 0.005 m³/kg. Find T and u.

                                c) R-22 at T = -10^oC and u = 200 kJ/kg.  Find P and v.

Analysis:

Answer:

(a) R-134a, T = 60oC, v = 0.072 m3/kg From Table v>vg,

so the R-134a is a superheated vapor. Investigation of Table indicates that we will need to interpolate between p = 3.2 bar and 4.0 bar (at 60oC) This yields p = 3.63 bar, h = 302.08 kJ/kg

(b) Ammonia, p = 8 bar, v = 0.005 m3/kg At 8 bar (Table A-14),

vf = 0.0016302 m3/kg, and vg = 0.1596 m3/kg So, vf < v < vg , and this is a saturated mixture. Therefore, T = Tsat = 17.84oC.

x = (v-vf)/(vg -vf) = 0.0213

u = xug + (1-x)uf ug = 1330.64 kJ/kg

uf = 262.64 kJ/kg u = 285.39 kJ/kg

(c) R-22, T = -10oC, u = 200 kJ/kg

From Table A-7, uf = 33.27 kJ/kg, ug = 223.02 kJ/kg

So, this is a saturated mixture (uf < u < ug) x = (u-uf)/(ug-uf) = 0.8787

p = psat = 3.5485 bar v = xvg + (1-x) vf = 0.0574 m3/kg