Explanation The figure below shows the flow of refrigerant and air through the heat exchanger, the throttling valve and fan. Figure-(1) The refrigerant flows through the throttling valve (T.V.) from 3 to 4 and then it is allowed to pass through the heat exchanger (H.E.) from 4 to 5 . The air passes through the fan (F) from 1 and then is allowed to flow through the heat exchanger and exits at 2 . The hatched line shows the control volume and the mass flow rate of refrigerant through the throttling valve is constant through it. Write the mass balance equation for the refrigerant through the throttling valve. m ˙ 4 = m ˙ 3 The mass of the refrigerant also remains same for the flow through heat exchanger. Write the mass balance for the refrigerant through the heat exchanger. m ˙ 4 = m ˙ 5 The mass flow rate of air also remains same throughout the fan. Write the mass balance equation for air. m ˙ 1 = m ˙ 2 Write the steady state energy equation for the control volume. W ˙ c v + H 2 = Q ˙ c v + H 1 (I) Here, the rate of work flow through the control volume is W ˙ c v , the rate of enthalpy at the exit is H 2 , the rate of heat flow through the control volume is Q ˙ c v , the rate of enthalpy at the entry is H 1 . There is no heat flow through the control volume that is Q ˙ c v is zero. Write the expression for the enthalpy rate at the entry of the control volume. H 1 = m ˙ a h 1 + m ˙ r h 4 Here, the mass flow rate of the air is m ˙ a and the specific enthalpy of air at 1 is h 1 , the mass flow rate of the refrigerant is m ˙ r and the specific enthalpy at 4 is h 4 . Write the expression for the enthalpy rate at the exit. H 2 = m ˙ a h 2 + m ˙ r h 5 Here, the specific enthalpy of the refrigerant at 5 is h 5 and the specific enthalpy of the air at 2 is h 2 . Substitute m ˙ a h 2 + m ˙ r h 5 for H 2 , m ˙ a h 1 + m ˙ r h 4 for H 1 and 0 for Q ˙ c v in Equation (I) W ˙ c v + m ˙ a h 2 + m ˙ r h 5 = 0 + m ˙ a h 1 + m ˙ r h 4 W ˙ c v = m ˙ a ( h 1 − h 2 ) + m ˙ r ( h 4 − h 5 ) (II) Write the expression for the enthalpy change for the constant pressure process 1 − 2 . h 1 − h 2 = c p ( T 1 − T 2 ) Here, the specific heat at constant pressure for the air is c p , the temperature of the air at 1 is T 1 and the temperature of the air at 2 is T 2 . Substitute c p ( T 1 − T 2 ) for h 1 − h 2 in Equation (II). W ˙ c v = m ˙ a [ c p ( T 1 − T 2 ) ] + m ˙ r ( h 4 − h 5 ) m ˙ a = W ˙ c v + m ˙ r ( h 5 − h 4 ) [ c p ( T 1 − T 2 ) ] (III) Conclusion: Refer to Table A-10E “Properties of Saturated Refrigerant 134a (Liquid-Vapor): Temperature Table” to obtain specific enthalpy at saturated vapor as 49

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 98P

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Chapter 4.12, Problem 97P

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