Explanation Write the equation of energy rate balance. Q ˙ cv − W ˙ cv + m ˙ [ ( h 1 − h 2 ) + ( V 1 2 − V 2 2 ) 2 + g ( z 1 − z 2 ) ] = 0 (I) Here, the rate of heat transfer is Q ˙ cv , rate of work transfer is W ˙ cv , rate of mass flow is m ˙ , the enthalpy is h , velocity is V , gravitational acceleration is g , the elevation is z and the suffixes cv, 1 and 2 indicates control volume, initial and final state of the system respectively. It is given that the inlet velocity is small compared to the exit velocity and the heat transfer between the turbine and its surroundings, potential energy effects are negligible i.e. g ( z 1 − z 2 ) = 0 , Q ˙ cv = 0 . The Equation (I) becomes as follows. − W ˙ cv + m ˙ [ ( h 1 − h 2 ) + ( V 1 2 − V 2 2 ) 2 ] = 0 (II) When the inlet velocity is small compared to the exit velocity, neglect the initial velocity i.e. V 1 = 0 . Substitute 0 for V 1 in Equation (II). − W ˙ cv + m ˙ [ ( h 1 − h 2 ) + ( 0 2 − V 2 2 ) 2 ] = 0 − W ˙ cv + m ˙ [ ( h 1 − h 2 ) − V 2 2 2 ] = 0 W ˙ cv = m ˙ [ ( h 1 − h 2 ) − V 2 2 2 ] (III) Rearrange the equation (III) to obtain the mass flow rate. W ˙ cv = m ˙ [ ( h 1 − h 2 ) − V 2 2 2 ] m ˙ = W ˙ cv ( h 1 − h 2 ) − V 2 2 2 (IV) Here, the air expands through the turbine at steady state. Hence, the inlet and exit mass flow rates are equal. m ˙ 1 = m ˙ 2 Write the formula for exit mass flow rate in terms of ideal gas equation. m ˙ 2 = A 2 V 2 p 2 ( R ¯ / M ) T 2 (V) Here, the cross sectional area is A , the velocity is V , the pressure is p , universal gas constant is R ¯ , the molar mass of the nitrogen is M , the temperature is T and the suffix 2 indicates the exit state. Rewrite the Equation (V) to obtain the exit area. A 2 = m ˙ 2 ( R ¯ / M ) T 2 V 2 p 2 (VI) At inlet: The air is at state of superheated water vapor

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Chapter 4.12, Problem 43P

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Chapter 4.12, Problem 42P

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The mass flow rate of air, in kg/s and the exit area, in m 2 .

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