QuestionDetails: One kilogram of R-134a fills a 0.14-m 3 weightedpiston-cylinder
ID: 1815609 • Letter: Q
Question
QuestionDetails: One kilogram of R-134a fills a 0.14-m3 weightedpiston-cylinder device at a temperature of -26.4 degrees C. Thecontainer is now heated until the temperature is 100 degrees C.Determine the final volume of the R-134s
(answer: 0.3014 m3 )
mass= 1 kg temp initial= -26.4 degrees C temp final = 100 degrees C V initial = 0.14 m3 P = 100 kPa
--> refrigerant is in saturatedregion
QuestionDetails: One kilogram of R-134a fills a 0.14-m3 weightedpiston-cylinder device at a temperature of -26.4 degrees C. Thecontainer is now heated until the temperature is 100 degrees C.Determine the final volume of the R-134s
(answer: 0.3014 m3 ) QuestionDetails:
mass= 1 kg temp initial= -26.4 degrees C temp final = 100 degrees C V initial = 0.14 m3 P = 100 kPa
--> refrigerant is in saturatedregion
Explanation / Answer
if you look up the table for Saturated refrigerant 134a(pressure)... not sure where for your specific book...but under100kPa the Saturation Temp is -26.4. Now under the specific volume column you see that your 0.14m^3/kgis between v(f) and v(g) which means that the 100kPa isrelevant. Since Pressure stays the same ie. P1=P2.... @ 100 degrees C it issuperheated.. go to that table at 100kPa @ 100 degreesC----------> v = .30138m^3/kg since V = v/m then v = V*m =(0.30138m^3/kg)(1 kg) = 0.30138 m^3 Which answers the question.......This is just a problem to get youused to looking up tables.. hope this helps
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