Explanation Here, the air flows at steady state through the compressor. Hence, the inlet and exit mass flow rates are equal. m ˙ 1 = m ˙ 2 (I) Write the formula for inlet mass flow rate in terms of ideal gas equation. m ˙ 1 = A 1 V 1 p 1 ( R ¯ / M ) T 1 (II) 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 air is M , the temperature is T and the suffix 1 indicates the inlet state. The volumetric flow rate is expressed as AV . 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 ) ] (III) 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. Neglect the kinetic and potential energy effects i.e. ( V 1 2 − V 2 2 ) 2 = 0 and g ( z 1 − z 2 ) = 0 . Then the Equation (III) becomes as follows to obtain the exit enthalpy ( h 2 ) . 0 = Q ˙ cv − W ˙ cv + m ˙ ( h 1 − h 2 ) m ˙ ( h 2 − h 1 ) = Q ˙ cv − W ˙ cv h 2 − h 1 = Q ˙ cv m ˙ − W ˙ cv m ˙ h 2 = Q ˙ cv m ˙ − W ˙ cv m ˙ + h 1 (IV) Here, the rate of heat transfer from the compressor to the surrounding occurs at the rate of 9.7 Btu per lb of mass flow of air. Q ˙ cv m ˙ = − 9.7 Btu/lb At inlet: Refer Table A-22E, “Ideal Gas Properties of Air”. The enthalpy ( h 1 ) corresponding to the temperature of 80 ° F ( 540 ° R ) is 129.06 Btu/lb (using linear interpolation). The molar mass of air is 28.97 lb/lbmol . Conclusion: Substitute 18 ft 3 /s for A 1 V 1 , 14

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118832301

Author: SHAPIRO

Publisher: JOHN WILEY+SONS,INC.-CONSIGNMENT

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Chapter 4.12, Problem 58P

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Fundamentals of Engineering Thermodynamics

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The exit temperature of the air, in °F.

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