A 0.5-m3 rigid tank contains refrigerant-134a initially at 200 kPa and 40 percent quality. Heat is transferred now to the refrigerant from a source at 358C until the pressure rises to 400 kPa. Determine (a) the entropy change of the refrigerant, (b) the entropy change of the heat source, and (c) the total entropy change for this process.

(a)

(b)

(c)

Explanation:

Hello,

(a) In this case, such refrigerant, we can notice that at the given conditions, the initial entropy from property tables (Cengel 7th ed) is:

Now, for the final condition, we first need to compute the initial specific volume as it remains the same (rigid tank) after the thermodynamic process:

Then, at 400 kPa we evaluate the given volume that is also between the liquid and vapor specific volume, thus, we calculate the quality at the end of the process:

With it, we compute the final entropy:

Finally, entropy change for the refrigerant turns out:

(b) In this case, by using the first law of thermodynamics we compute the acquired heat by the refrigerant from the heat source by computing the initial and final internal energy respectively (no work is done):

Hence:

Finally, the entropy change of the heat source (which release the heat, therefore it is negative):

(c) Then, the total entropy change or the entropy generation for the process is:

Which has thermodynamic agreement as it is positive

Regards.

c8h10 is your answerr