Respuesta :
Explanation:
The given data is as follows.
Mass of refrigerant, m = 10 kg
Volume of the refrigerant, V = 1.595 [tex]m^{3}[/tex]
Formula for specific volume of the refrigerant is as follows.
v = [tex]\frac{V}{m}[/tex]
= 0.1595 [tex]m^{3}/kg[/tex]
So, at [tex]-26.4^{o}C[/tex] specific volume will be within [tex]v_{f}[/tex] and [tex]v_{g}[/tex] and pressure is constant.
The fluid will be in super-heated state at temperature [tex]100^{o}C[/tex] and at T = [tex]-26.4^{o}C[/tex] pressure 1 bar = 0.1 MPa.
According to super-heated tables, the specific volume is v = 0.30138 [tex]m^{3}/kg[/tex].
Hence, the final volume will be calculated as follows.
[tex]V_{f} = v \times m[/tex]
= [tex]0.30138 m^{3}/kg \times 10 kg[/tex]
= 3.0138 [tex]m^{3}[/tex]
Thus, we can conclude that final volume of the R-134a is 3.0138 [tex]m^{3}[/tex].
