Answer:
[tex]\Delta S_{source}>-1.204\frac{kJ}{K}[/tex]
Explanation:
Hello!
In this case, given the initial conditions, we first use the 10-% quality to compute the initial entropy:
[tex]s_1=s_{f,175kPa}+q*s_{fg,175kPa}\\\\s_1=1.4850\frac{kJ}{kg*K} +0.1*5.6865\frac{kJ}{kg*K}=2.0537\frac{kJ}{kg*K}[/tex]
Now the entropy at the final state given the new 40-% quality:
[tex]s_2=s_{f,150kPa}+q*s_{fg,150kPa}\\\\s_2=1.4337\frac{kJ}{kg*K} +0.4*5.7894\frac{kJ}{kg*K}=3.7495\frac{kJ}{kg*K}[/tex]
Next step is to compute the mass of steam given the specific volume of steam at 175 kPa and the 10% quality:
[tex]m_1=\frac{0.028m^3}{(0.001057+0.1*1.002643)\frac{m^3}{kg} } =0.274kg\\\\m_2=\frac{0.028m^3}{(0.001053+0.4*1.158347)\frac{m^3}{kg} } =0.0603kg[/tex]
Then, we can write the entropy balance:
[tex]\Delta S_{source}+\frac{Q}{T_1} -\frac{Q}{T_2} +s_2m_2-s_1m_1-s_{fg}(m_2-m_1)>0[/tex]
Whereas sfg stands for the entropy of the leaving steam to hold the pressure at 150 kPa and must be greater than 0; thus we plug in:
Which is such minimum entropy change of the heat-supplying source.
Best regards!
