Explanation Write the energy rate balance equation. 0 = Q ˙ cv − W ˙ cv + m ˙ [ ( h 1 − h 2 ) + ( V 1 2 − V 2 2 ) 2 + g ( z 1 − z 2 ) ] (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. Here, the helium operates in steady state. The inlet, exit of the diffuser is at same elevation and the helium flow adiabatically. Hence, neglect the potential energy i.e. g ( z 1 − z 2 ) = 0 , heat transfer i.e. Q ˙ cv = 0 and the work done is also zero in diffuser i.e. W ˙ cv = 0 . Then the Equation (I) becomes as follows to obtain the exit velocity. ( h 1 − h 2 ) + ( V 1 2 − V 2 2 ) 2 = 0 h 2 − h 1 = 1 2 ( V 1 2 − V 2 2 ) V 1 2 − V 2 2 = 2 ( h 2 − h 1 ) V 2 2 = V 1 2 − 2 ( h 2 − h 1 ) V 2 = V 1 2 + 2 ( h 1 − h 2 ) (II) Write the formula for change in enthalpy ( h 1 − h 2 ) . h 1 − h 2 = c p ( T 1 − T 2 ) Substitute c p ( T 1 − T 2 ) for h 1 − h 2 in Equation (II). V 2 = V 1 2 + 2 [ c p ( T 1 − T 2 ) ] (III) Write the formula for specific heat at constant pressure. c p = k ( R / M ) k − 1 (IV) Here, the specific heat ratio is k , the molar mass of the helium is M and the universal gas constant is R . 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 exit cross-sectional area is A 2 and the exit velocity is V 2 , the pressure is p , universal gas constant is R ¯ , the molar mass of the helium is M , the temperature is T and the suffix 1 indicates the inlet state. The molar mass ( M ) of helium gas is 4.003 lb/lb ⋅ mol and universal gas ( R ) constant is 1.986 Btu ⋅ lb/mol ⋅ ° R . Conclusion: Substitute 1

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Chapter 4.12, Problem 35P

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The mass flow rate of the helium gas through the nozzle, in lb/s.

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