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Q.4) Refrigerant R134a enters the throttling valve of a refrigerator as saturated liquid at 1 MPa and is throttled to a pressure of 0.32 MPa. In this process enthalpy remains constant. Determine the a) Temperature and enthalpy of the initial state (state 1). b) Quality of the last state (state 2). c) Temperature of the last state (state 2). d) Show the process in P-h diagram

          Q.4) Refrigerant R134a enters the throttling valve of a refrigerator as saturated liquid at 1 MPa and is throttled to a pressure of 0.32 MPa. In this process enthalpy remains constant.
Determine the
a) Temperature and enthalpy of the initial state (state 1).
b) Quality of the last state (state 2).
c) Temperature of the last state (state 2).
d) Show the process in P-h diagram
        
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Q.4) Refrigerant R134a enters the throttling valve of a refrigerator as saturated liquid at 1 MPa and is throttled to a pressure of 0.32 MPa. In this process enthalpy remains constant.
Determine the
a) Temperature and enthalpy of the initial state (state 1).
b) Quality of the last state (state 2).
c) Temperature of the last state (state 2).
d) Show the process in P-h diagram

Added by Michaela G.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Q.4) Refrigerant R134a enters the throttling valve of a refrigerator as saturated liquid at 1 MPa and is throttled to a pressure of 0.32 MPa. In this process, enthalpy remains constant. Determine the Throttling a) Temperature and enthalpy of the initial state (state 1). b) Quality of the last state (state 2). c) Temperature of the last state (state 2). h = h1 d) Show the process in P-h diagram.
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Thermodynamics: An Engineering Approach


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Transcript

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00:02 So carnot refrigeration cycle so let us draw two diagrams over here one is tv diagram and one is ts diagram so here v this side t here this is 4 3 2 1 theta 2 theta 4 20 degree celsius minus 8 degree celsius now from this side it's 1 2 3 and 4 20 degree celsius minus 8 degree celsius this side s this side t so at 20 degree celsius as f equal to 0 .30063 kg per kg k and sg equal to 0 .92234 kg per kg k now s3 equal to s2 as process 2 3 is isentropic process now s2 equal to sg at 20 degree celsius therefore s2 equal to 0 .9 2234 now s2 equal to sf plus x2 sfg at…
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