Answer:
Explanation:
Step one:
given data
T_{wi} = 20^{\circ}C
T_{Ai}=1000K
T_{Ae}= 400kPa
P_{Wi}=200kPa
P_{Ai}=125kPa
P_{We}=200kPa
P_{Ae}=100kPa
m_A=2kg/s
m_W=0.5kg/s
We know that the energy equation is
[tex]m_Ah_{Ai}+m_Wh_W=m_Ah_{Ae}+m_Wh_{We}[/tex]
making [tex]h_{We}[/tex] the subject of formula we have
[tex]h_{We}=h_{Wi}+\frac{m_A}{mW}(h_A-h_{Ae})[/tex]
from the saturated water table B.1.1 , corresponding to [tex]T_{wi}= 20c[/tex]
[tex]h_{Wi}=83.94kJ/kg[/tex]
from the ideal gas properties of air table B.7.1 , corresponding to T=1000K
the enthalpy is:
[tex]h_{Ai}=1046.22kJ/kg[/tex]
from the ideal gas properties of air table B.7.1 corresponding to T=400K
[tex]h_{Ae}=401.30kJ/kg[/tex]
Step two:
substituting into the equation we have
[tex]h_{We}=h_{Wi}+\frac{m_A}{mW}(h_A-h_{Ae})[/tex]
[tex]h_{We}=83.94+\frac{2}{0.5}(2046.22-401.30)\\\\h_{We}=2663.62kJ/kg[/tex]
from saturated water table B.1.2 at [tex]P_{We}=200kPa[/tex] we can obtain the specific enthalpy:
[tex]h_g=2706.63kJ/kg[/tex]
we can see that [tex]h_g>h_{Wi}[/tex], hence there are two phases
from saturated water table B.1.2 at [tex]P_{We}=200kPa[/tex]
[tex]T_{We}=120 ^{\circ} C[/tex]
