Respuesta :
Explanation:
The given data is as follows.
[tex]T_{1} = 105^{o}C[/tex], x = 0.85
[tex]V_{1}[/tex] = 1 L = 0.001 [tex]m^{3}[/tex] (as 1 L = 0.1 [tex]m^{3}[/tex])
[tex]V_{2}[/tex] = 1.5 L = 0.0015 [tex]m^{3}[/tex],
[tex]d_{p}[/tex] = 150 mm = 0.15 m, K = 100 N/mm = 100 kN/m
[tex]P_{3}[/tex] = 200 kPa
According to the table,
[tex]P_{1}[/tex] = 120.8 kPa, [tex]v_{f}[/tex] = 0.001047 m/kg
[tex]v_{fg} = 1.41831 [tex]m^{3}/kg[/tex]
So, we have
[tex]v_{1} = v_{f} + xv_{fg}[/tex]
= 0.001047 + 0.85 \times (1.41831 m^{3}/kg)[/tex]
= 1.2066 [tex]m^{3}/kg[/tex]
Now, we will calculate [tex]m_{1}[/tex] as follows.
[tex]m_{1} = \frac{V_{1}}{v_{1}}[/tex]
= [tex]\frac{0.001}{1.2066}[/tex]
= 0.000828 kg
Now, Area = [tex]\frac{\pi}{4} dp^{2}[/tex]
= [tex]\frac{\pi}{4} (0.15)^{2}[/tex]
= 0.0176 [tex]m^{2}[/tex]
As we have,
[tex]v_{2} = v_{1} \times (\frac{V_{2}}{v_{1}})[/tex]
= [tex]1.2066 \times \frac{0.0015}{0.001}[/tex]
= 1.8099 [tex]m^{3}/kg[/tex]
From table, [tex]T_{2} = 203.5^{o}C[/tex] for [tex]P_{2}[/tex] = 120.8 kPa and [tex]v_{2} = 1.8099 m^{3}/kg[/tex].
[tex]P_{3} = P_{2} + (\frac{K}{A^{2}})m (v_{3} - v_{2})[/tex]
200 = [tex]120.8 + (\frac{100}{(0.0176)^{2}})(0.000828)(v_{3} - 1.8099)[/tex]
[tex]v_{3} = 2.106 m^{3}/kg[/tex]
Therefore, from the table
T = [tex]600^{o}C[/tex], v = 2.01297 [tex]m^{3}/kg[/tex]
T = [tex]700^{o}C[/tex], v = 2.2443[tex]m^{3}/kg[/tex]
[tex]T_{3} = 600 + \frac{2.106 - 2.01297}{2.2443 - 2.01297} (700 - 600)[/tex]
= [tex]640.2^{o}C[/tex]
Thus, we can conclude that the cylinder temperature when the pressure reaches 200 kPa is [tex]640.2^{o}C[/tex].
