A 0.5 m3 rigid tank contains R-134a initially at 200 kPa and 40% quality. Heat is transferred to the refrigerant from a source at 35oC until the pressure rises to 400 kPa. Determine…
a.) The entropy change of the R-134a.
b.) The entropy change of the heat source.
c.) The entropy change of the universe for this process.
2 old comments
5 comments:
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HINTS:
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* The specific volume is the same in states 1 and 2.
* The heat source can be treated as a true thermal reservoir, so deltaS can be determined from the definition of entropy.
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ANSWERS:
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deltaS_univ = 0.4496 kJ/K
ANOTHER HINT: Since the tank is a rigid, closed system, there is no mass transfer or volume change. (Therefore, V1 = V2).
once we find the delta S for the R-134a, how do we go about finding deltaS for the heat source??
figured it out
Dave 2:31
Just so everyone else knows, the key to determining the entropy change of the reservoir is the assumption that it is a true reservoir and is isothermal. As a result, DSreservoir = Q/Treservoir. Just be careful with the SIGN on Q !
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