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Two rigid tanks are connected by a valve. Tank A is insulated and contains 0.3 m3 of steam at 400 kPa and 60% quality. Tank B is uninsulated and contains 2 kg of steam at 200 kPa and 250°C. 3 The valve is now opened, and steam flows from tank A to tank B until the pressure in tank A drops to 200 kPa. During this process, 300 kJ of heat is transferred from tank B to the surroundings at 17°C. Assuming the steam remaining inside tank A to have undergone a reversible adiabatic process, determine (a) final temperature in each tank and (b) entropy generated during this process.

Respuesta :

xero099

Answer:

a) [tex]T = 120.21\,^{\textdegree}C[/tex], b) [tex]S_{gen} = 1.186\,\frac{kJ}{K}[/tex]

Explanation:

a) Let find the initial properties of the fluids on each rigid tank:

Tank A - Initial state (Liquid-Vapor Mixture)

[tex]P = 400\,kPa[/tex]

[tex]T = 143.61\,^{\textdegree}C[/tex]

[tex]\nu = 0.2779\,\frac{m^{3}}{kg}[/tex]

[tex]u = 1773.56\,\frac{kJ}{kg}[/tex]

[tex]s = 4.8480\,\frac{kJ}{kg\cdot K}[/tex]

[tex]x = 0.6[/tex]

Tank B - Initial state (Superheated Vapor)

[tex]P = 200\,kPa[/tex]

[tex]T = 250\,^{\textdegree}C[/tex]

[tex]\nu = 0.47443\,\frac{m^{3}}{kg}[/tex]

[tex]u = 2723.8\,\frac{kJ}{kg}[/tex]

[tex]s = 7.2725\,\frac{kJ}{kg\cdot K}[/tex]

The system is modelled after the First Law of Thermodynamics:

[tex]-Q_{out} = (m_{B}+m_{A})\cdot u_{A+B}-m_{B}\cdot u_{B} -m_{A}\cdot u_{A}[/tex]

The final internal energy of the system is:

[tex]u_{A+B} = \frac{m_{B}\cdot u_{B} + m_{A}\cdot u_{A} - Q_{out}}{m_{B}+m_{A}}[/tex]

[tex]u_{A+B} = \frac{(2\,kg)\cdot (2723.8\,\frac{kJ}{kg} )+\left(\frac{0.3\,m^{3}}{0.2779\,\frac{m^{3}}{kg} }\right)\cdot (1773.56\,\frac{kJ}{kg} )-300\,kJ }{2\,kg + \frac{0.3\,m^{3}}{0.2779\,\frac{m^{3}}{kg} }}[/tex]

[tex]u_{A+B} = 2293.28\,\frac{kJ}{kg}[/tex]

Tanks A + B - Final state

[tex]P = 200\,kPa[/tex]

[tex]T = 120.21\,^{\textdegree}C[/tex]

[tex]u = 2293.28\,\frac{kJ}{kg}[/tex]

[tex]s = 6.4721\,\frac{kJ}{kg\cdot K}[/tex]

[tex]x = 0.883[/tex]

The final temperature in each tank is:

[tex]T = 120.21\,^{\textdegree}C[/tex]

b) The entropy generated during the process is derived of the Second Law of Thermodynamics:

[tex]-\frac{Q_{out}}{T_{surr}} + m_{A} \cdot s_{A} + m_{B} \cdot s_{B} - (m_{A} + m_{B})\cdot s_{A+B} + S_{gen} = 0[/tex]

The generated entropy is:

[tex]S_{gen} = \frac{Q_{out}}{T_{out}} + (m_{A}+m_{B})\cdot s_{A+B} - m_{A}\cdot s_{A} - m_{B}\cdot s_{B}[/tex]

[tex]S_{gen} = \frac{300\,kJ}{290.15\,K} + \left(\frac{0.3\,m^{3}}{0.2779\,\frac{m^{3}}{kg}}+2\,kg\right)\cdot (6.4721\,\frac{kJ}{kg\cdot K} )-\left( \frac{0.3\,m^{3}}{0.2779\,\frac{m^{3}}{kg}}\right)\cdot (4.8480\,\frac{kJ}{kg\cdot K} )-(2\,kg)\cdot (7.2725\,\frac{kJ}{kg\cdot K} )[/tex]

[tex]S_{gen} = 1.186\,\frac{kJ}{K}[/tex]