Respuesta :
Answer:
a).Final temperature, [tex]T_{2}[/tex] = 180°C
b).Initial Volume, [tex]V_{1}[/tex] = 0.713412 [tex]m^{3}[/tex]
Final Volume, [tex]V_{2}[/tex] = 0.33012 [tex]m^{3}[/tex]
c). Initial enthalpy,[tex]H_{1}[/tex] =9129.68 kJ
Final enthalpy, [tex]H_{1}[/tex] =6234.76 kJ
Explanation:
Given :
Total mass, m= 2.8 kg
Initial temperature, [tex]t_{i}[/tex] = 400°C
Initial pressure, [tex]p_{i}[/tex] = 1.2 MPa
Therefore from steam table at 400°C, we can find--
[tex]h_{1}[/tex] = 3260.6 kJ/kg
[tex]v_{1}[/tex] = 0.25479 [tex]m^{3}[/tex] / kg
Now it is mentioned that 28% of the mass is condensed into liquid.
So, mass of liquid, [tex]m_{l}[/tex] = 0.28 of m
= 0.28 m
mass of vapour, [tex]m_{v}[/tex] = 0.72 m
∴ Dryness fraction, x = [tex]\frac{m_{v}}{m_{l}+m_{v}}[/tex]
= [tex]\frac{0.72 m}{0.28 m+0.72 m}[/tex]
= 0.72
a). The final temperature can be evaluated from the steam table at 1.2 MPa,
[tex]h_{2}[/tex] = 2226.7 kJ/kg
[tex]v_{2}[/tex] = 0.1179 [tex]m^{3}[/tex] / kg
Final temperature, [tex]T_{2}[/tex] = 180°C
b). We know [tex]v_{1}[/tex] = 0.25479 [tex]m^{3}[/tex] / kg
∴ Initial Volume, [tex]V_{1}[/tex] = [tex]v_{1}[/tex] x m
[tex]V_{1}[/tex] = 0.25479 x 2.8
[tex]V_{1}[/tex] = 0.713412 [tex]m^{3}[/tex]
We know,[tex]v_{2}[/tex] = 0.1179 [tex]m^{3}[/tex] / kg
∴ Final Volume, [tex]V_{2}[/tex] = [tex]v_{2}[/tex] x m
[tex]V_{2}[/tex] = 0.1179 x 2.8
[tex]V_{1}[/tex] = 0.33012 [tex]m^{3}[/tex]
c). We know,
[tex]h_{1}[/tex] = 3260.6 kJ/kg
∴ Initial enthalpy,[tex]H_{1}[/tex] = [tex]h_{1}[/tex] x m
= 3260.6 x 2.8
= 9129.68 kJ
[tex]h_{2}[/tex] = 2226.7 kJ/kg
∴ Final enthalpy, [tex]H_{1}[/tex] = [tex]h_{2}[/tex] x m
= 2226.7 x 2.8
= 6234.76 kJ
