Explanation Write the energy rate balance equation for one inlet and one outlet system. [ Q ˙ 1 + W ˙ 1 + m ˙ ( h 1 + V 1 2 2 + g z 1 ) ] − [ Q ˙ 2 + W ˙ 2 + m ˙ ( h 2 + V 2 2 2 + g z 2 ) ] = Δ E ˙ system (I) Here, the rate of heat transfer is Q ˙ , the rate of work transfer is W ˙ , the enthalpy is h and the velocity is V , the gravitational acceleration is g , the elevation from the datum is z and the rate of change in net energy of the system is Δ E ˙ system ; the suffixes 1 and 2 indicates the inlet and outlet of the system. Consider the nozzle operates at steady state. Hence, the rate of change in net energy of the system becomes zero. Δ E ˙ system = 0 Since, the gas passes through the adiabatic nozzle, the heat transfer rate is negligible i.e. Q ˙ = 0 . Also neglect the work transfer rates i.e. W ˙ = 0 . The inlet, outlet are at same elevation, the potential energy becomes negligible i.e. g z = 0 . The Equations (I) reduced as follows to obtain the exit velocity. [ 0 + 0 + m ˙ ( h 1 + V 1 2 2 + 0 ) ] − [ 0 + 0 + m ˙ ( h 2 + V 2 2 2 + 0 ) ] = 0 m ˙ ( h 1 + V 1 2 2 ) − m ˙ ( h 2 + V 2 2 2 ) = 0 m ˙ ( h 1 + V 1 2 2 ) = m ˙ ( h 2 + V 2 2 2 ) h 1 + V 1 2 2 = h 2 + V 2 2 2 V 2 2 2 − V 1 2 2 = h 1 − h 2 V 2 2 − V 1 2 = 2 ( h 1 − h 2 ) V 2 = V 1 2 + 2 ( h 1 − h 2 ) (II) It is given that the initial velocity is low. Hence, neglect the initial velocity i.e. V 1 = 0 . Substitute 0 for V 1 in Equation (II). V 2 = ( 0 ) 2 + 2 ( h 1 − h 2 ) = 2 ( h 1 − h 2 ) (III) Write formula for enthalpy departure factor ( Z h ) . Z h = ( h ¯ ideal − h ¯ ) T , P R u T cr (IV) Here, the molar enthalpy at ideal gas state is h ¯ ideal , the molar enthalpy and normal state is h ¯ , the universal gas constant is R u , and the critical temperature is T cr ; The subscripts T , P indicates the correspondence of given temperature and pressure. Rearrange the Equation (I) to obtain h ¯ . h ¯ = h ¯ ideal − Z h R u T cr (V) Refer Equation (V) express as two states of enthalpy difference (initial–final). h ¯ 1 − h ¯ 2 = ( h ¯ 1 − h ¯ 2 ) ideal − ( Z h 1 − Z h 2 ) R u T cr (VI) Here, the molar enthalpy difference is h ¯ 2 − h ¯ 1 . The enthalpy difference is expressed as follows. h 1 − h 2 = h ¯ 1 − h ¯ 2 M CO 2 (VII) Refer Table A-1, “Molar mass, gas constant, and critical-point properties”

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Chapter 12.6, Problem 75P

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The exit velocity of the carbon dioxide using generalized enthalpy departure chart.

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