Answer:
34°C
36.9 KJ/kg
Explanation:
solution:
the initial enthalpy is determined by the following equation:
[tex]h_{1}[/tex]=338.404 kJ/kg
next the final enthalpy is determined from the energy balance:
[tex]h_{1} +\frac{v_{1} ^{2}}{y} =h_{2} +\frac{v_{2} ^{2}}{y}+q\\h_{2}=h_{1} +v_{1} ^{2}-{v_{2} ^{2}/2-q\\=338.404 kJ/kg+{35^{2}-{\frac{240 ^{2}}{2} } -3kJ/kg\\=307.6125 kJ/kg\\[/tex]
from the final enthalpy the final temperature is determined by the following equation:
[tex]T_{2} =T_{1} +\frac{T_{2}-T_{1}}{h_{2}-h_{1}} (h_{2}-h_{1})\\=305 K+\frac{310-305}{310.24-305.22}(307.215-305.22)K\\ =307 K\\=34 C[/tex]
The energy destroyed calculated from the enthalpy generation
[tex]x_{dest} =T_{0} s_{gen} \\=T_{0}(c_{p} ln\frac{T_{2}}{T_{1}} -Rln\frac{P_{2}}{P_{1}}+\frac{q}{T_{0}}\\ =36.9 KJ/kg[/tex]
