91Ó°ÊÓ

In thermodynamics, an ideal gas mixture is a theoretical combination of different gases, where each gas behaves as an ideal gas. This means that each gas in the mixture obeys the ideal gas law, which is: \[ PV = nRT \] Here, P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature. For an ideal gas mixture, each component gas's partial pressure is proportional to its mole fraction in the mixture. Additionally, in isentropic processes (where entropy remains constant), the relationships between pressures and temperatures for ideal gases simplifies calculations. It is assumed that the gases do no work and there are no heat exchanges with the surroundings. This makes the calculations straightforward, as shown in the exercise where we compute isentropic flow properties through a converging nozzle.

To find the exit temperature of a gas in isentropic flow, we use the relation that connects the total temperature (\[T_o\]) and the static temperature (\[T_{exit}\]). The equation is: \[ T_{exit} = T_o \times \bigg( \frac{p_{exit}}{p_o} \bigg)^{\frac{k-1}{k}} \] Here, \[k\] is the adiabatic index, \[p_{exit}\] is the exit pressure, and \[p_o\] is the total pressure. For our given problem, \[T_o=680 \text{ K}\], \[p_o=4 \text{ bar}\], and \[p_{exit}=1 \text{ bar}\]. After substituting these values into the formula, we get: \[ T_{exit} = 680 \times \big( \frac{1}{4} \big)^{0.242} = 680 \times 0.728 = 495.04 \text{ K} \] Thus, the exit temperature \[T_{exit} \] is 495.04 K.

To calculate the exit velocity in an isentropic flow, the following expression is used: \[ V_{exit} = \big( 2 \times c_p \times (T_o - T_{exit}) \big)^{0.5} \] Here, \[c_p\] is the specific heat at constant pressure. To find \[c_p\], we use the relation: \[ c_p = \frac{k \times R}{k-1} \] where \[R\] is the specific gas constant calculated as \[ R = \frac{8314}{M} \] Given \[M=23 \], we have \[ R = 361.48 \]. Therefore, \[ c_p = \frac{1.32 \times 361.48}{1.32-1} = 1588.42 \]. Now substituting \[ c_p \], \[ T_o = 680 \], and \[ T_{exit} = 495.04 \], we calculate: \[ V_{exit} = \big( 2 \times 1588.42 \times (680 - 495.04) \big)^{0.5} = \big( 2 \times 1588.42 \times 184.96 \big)^{0.5} = 766 \text{ m/s} \] Hence, the exit velocity of the gas \[V_{exit} \] is 766 m/s.

The mass flow rate in a steady isentropic flow through a nozzle can be determined using the following formula: \[ \text{mass flow rate} = \frac{p_o \times A_{exit}}{R \times T_o} \times \bigg( \frac{2k}{k+1} \bigg)^{\frac{k+1}{2(k-1)}} \] Given: \[ p_o = 4 \text{ bar} = 400000 \text{ Pa}\] \[ A_{exit} = 35 \text{ cm}^2 = 0.0035 \text{ m}^2 \] \[ R = 361.48 \] \[ T_o = 680 \text{ K} \] and \[ k = 1.32 \] Substituting these values: \[ \text{mass flow rate} = \frac{400000 \times 0.0035}{361.48 \times 680} \] \[ \times \bigg( \frac{2 \times 1.32}{1.32 + 1} \bigg)^{\frac{1.32 + 1}{2 \times (1.32 - 1)}} = \frac{1400}{245806.2} \times \bigg( \frac{2.64}{2.32} \bigg)^{1.5517} = 0.0057 \times 1.265 = 0.0072 \text{ kg/s} \] The mass flow rate \[ \text{mass flow rate} \] is 0.0072 kg/s.