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 refrigerant operates in steady state. The inlet, exit of the diffuser is at same elevation and the diffuser is insulated well. 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 enthalpy ( h 2 ) of diffuser. h 2 = h 1 + V 1 2 2 − V 2 2 2 (II) When, the exit velocity is negligible i.e. V 2 = 0 . Substitute 0 for V 2 in Equation (II). h 2 = h 1 + V 1 2 2 − ( 0 ) 2 2 = h 1 + V 1 2 2 (III) At inlet: The refrigerant is in the state of saturated vapor

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118412930

Author: Michael J. Moran, Howard N. Shapiro, Daisie D. Boettner, Margaret B. Bailey

Publisher: WILEY

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Chapter 4.12, Problem 39P

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The exit pressure of refrigerant 134a from the insulated diffuser, in lbf/in.2.

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Chapter 4.12, Problem 38P

Chapter 4.12, Problem 40P

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Fundamentals of Engineering Thermodynamics

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The exit pressure of refrigerant 134a from the insulated diffuser, in lbf/in . 2 .

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The exit pressure of refrigerant 134a from the insulated diffuser, in lbf/in.2.

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