Draw Pnu diagram and locate each of the following conditions

Draw P-nu diagram and locate each of the following conditions and determine the values of the specified properties (a) For refrigerant 134a at T = 60 degree C and nu = 0.072 m^3/kg, determine P in kPa and h in kJ/kg (b) For ammonia at P = 8 bar and nu = 0.005 m^3/kg, determine T in degree C and u in kJ/kg (c) For refrigerant 22 at T = -10 degree C and u = 200 kJ/kg, determine P in bar and nu in m^3/kg. (d) For water at 15 MPa and 100 degree C, determine the nu in m^3/kg and h in kJ/kg (e) For refrigerant 22 at T = 30 degree C and P = 2000 kPa, determine the nu in m^3/kg and h in kJ/kg.

Solution

a) Refrigerant 134 a

At superheated steam tables at T=60^oC and v=0.072

At T=60C

v = v₁ + x ( v₂ - v₁)

0.072 = 0.08106+ x ( 0.06405- 0.08106)

x= 0.532

P = P₁ + x ( P₂ - P₁)

P = 320 + 0.532 * (400-320)

P=362.56 kPa

h = h₁ + x ( h₂ - h₁)

h = 302.72 + 0.532 * (301.51-302.72)

h = 302.07 KJ/kg

b.) For amonia

At 8 bar saturated tables

T= 17.84 ^oC

v_f =1.6302 * 10^-3 m³/kg

v_g = 0.1596  m³/kg

u_f = 262.64 KJ/Kg

u_g = 1330.64 KJ/Kg

v = v_f + x (v_g - v_f)

0.005 = 1.6302 * 10^-3 + x ( 0.1596 - 1.6302 * 10^-3)

x = 0.0213

u = u_f + x (u_g - u_f)

u =  262.64 + 0.0213* ( 1330.64- 262.64)

u = 285.42 KJ/Kg

c) For R22

At saturated R22 tables T=-10 C

P = 3.5485 bar

u_f = 33.27 KJ/Kg

u_g = 223.02 KJ/Kg

v_f = 0.7606 * 10^-3 m³

v_g = 0.0652 m³/Kg

u = u_f + x (u_g - u_f)

200 = 33.27 + x ( 223.02 - 33.27)

x = 0.878

v = v_f + x (v_g - v_f)

v=  0.7606 * 10^-3 + 0.878 * ( 0.0652 - 0.7606 * 10^-3 )

v = 0.0573 m³/Kg

d) For Water

From Compressed liquid water tables at 15 MPa and 100 C

v = 1.0361 * 10^-3 m³/Kg

h = 430.28 KJ/Kg

e) For R22

using compressed liquid R22 tables

v = 0.00085 m³/kg

h = 81.7 KJ/kg

pressure	320 KPa	400KPa
volume	0.08106	0.06405
enthalapy	302.72	301.51