Question 4.11: An open system consists of a fluid of a single substance kep...

An open system consists of a fluid of a single substance kept between two pistons sliding inside a cylinder with adiabatic walls. Matter enters and exits the cylinder in two specific locations. These two matter flows generate a chemical power P_C. The pressure at the entrance is p^+ and the pressure at the exit is p^−. The pistons exert a mechanical power P_W on the fluid. Since the walls are adiabatic there is a heat transfer through convection but no through conduction, i.e. P_Q = 0 (Fig. 4.2). For this open system, show that the chemical power P_C generated by the matter flow can be written as,

P_C = h^+ \dot{N}^+− h^- \dot{N}^- .

where \dot{N}^+ and \dot{N}^- are the rates of substance entering and exiting the system and h^+ and h^− are the molar enthalpies entering and exiting the system.

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The rates of variation of substance and of volume satisfy the following conservation laws,

\dot{N} = \dot{N}^+ – \dot{N}^-      and        \dot{V} = \dot{V}^+ – \dot{V}^-.

where N is the number of moles of substance in the system, V is the volume of the system, \dot{V}^+ and \dot{V}^- are the volumes entering and exiting the system per unit of time. The molar internal energies u^+ and u^− and the molar volumes ν^+ and ν^− entering and exiting the system per unit of time are defined as,

u^+ = \frac{U^+}{N^+}             u^- = \frac{U^-}{N^-}

ν^+ = \frac{V^+}{N^+}     and  ν^- = \frac{V^-}{N^-}

where U^+ and U^− are the internal energies entering and exiting the system. The time derivative of the internal energy of the system is the difference between the rate of variation of the internal energy due to matter entering the system, i.e. u^+\dot{N}^+, and to matter exiting the system, i.e. −u^− \dot{N}^-. Thus,

\dot{U} = u^+ \dot{N}^+ – u^− \dot{N}^-.

According to relation (2.16), the mechanical power exerted by the environment on the gas is the difference between the mechanical exerted at the entrance, i.e. −p^+\dot{V}^+, and the mechanical power exerted at the exit, i.e. p^−\dot{V}^-. Hence,

P_W = – p\dot{V} .    (reversible process)                        (2.16)

P_W = −p^+\dot{V}^+ p^−\dot{V}^- .

Since there is no conductive heat transfer, the thermal power vanishes, i.e. P_Q = 0. According to the first law (1.28), the chemical power then is written as,

\dot{U} = P_W + P_Q + P_C . (open system)                            (1.28)

P_C = \dot{U} – P_W = u^+ \dot{N}^+ – u^− \dot{N}^- + p^+\dot{V}^+ – p^−\dot{V}^- .

Thus, the chemical power exerted by the matter flow consists of two convective contributions : the convective heat transfers u^+ \dot{N}^+ – u^− \dot{N}^- and the mechanical actions p^+\dot{V}^+ – p^−\dot{V}^-. Taking into account the definition of the entering and exiting volumes per unit of time,

\dot{V}^+ = ν^+ \dot{N}^+    and    \dot{V}^- = ν^- \dot{N}^-.

and the entering and exiting molar enthalpies,

h^+ = u^+ + p^+ν^+     and    h^- = u^- + p^-ν^- .

the chemical power reduces then to,

P_C = h^+\dot{N}^+ – h^-\dot{N}^- .

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