Answer:
[tex]\bf a)-v_2=220.11\;m/s[/tex]
[tex]\bf b)-\sigma_{cv}=0.337\;kw/k[/tex]
Explanation:
a) - Taking enthalpy from the ideal air structure that is :
Given,
[tex]T_1=52\° C[/tex]
[tex]Then,\;\;\;\;\;\; T_1=325 k[/tex]
[tex]And,\;\;\;\;\;\;\;\;h_1=325.31\; KJ/kg[/tex]
[tex]T_2=82^\circ C[/tex]
[tex]Then,\;\;\;\;\;\; T_2=355 k[/tex]
[tex]And,\;\;\;\;\;\;\;\;h_2=355.535\; KJ/kg[/tex]
Then, we have to apply the equation of the energy rate balance.
After applying that we have:
[tex]m\left [\left (h_1-h_2 \right )+\frac{v_1^2-v_2^2}{2} \right ]=0[/tex]
[tex]\frac{v_1^2-v_2^2}{2} =h_2-h_1[/tex]
[tex]v_2^2=\left(V_1^2-2\right)\left(h_2-h_1\right)[/tex]
Then, we have to put values in the equation.
[tex]v_2^2=\left(330^2\right)-2 \left(355.535-325.31\right)\times10^3[/tex]
[tex]v_2=\sqrt{108900-60450}=220.11\;m/s[/tex]
b) - Here, we have to apply the equation of the continuity.
[tex]m=\frac{A_1V_1}{V_1}[/tex]
[tex]Then,\;\;\;\;\;\;\;\;\;\;\;=\frac{A_1V_1}{\frac{RT_1}{P_1} }[/tex]
[tex]=\frac{P_1}{RT_1} \times A_1V_1[/tex]
Then, we have to put values in the equation.
[tex]=\frac{200\times 10^3}{287\left(52+273\right)} \times 16.57 \times 10^{-4}\times330=1.17\:kg/s[/tex]
Then, the values are :
[tex]T_1=52^\circ C=325k[/tex]
[tex]S_1=1.78249\: KJ/s[/tex]
and,
[tex]T_2=82^\circ C=355k[/tex]
[tex]S_2=1.871255\: KJ/s[/tex]
[tex]\sigma_{cv}=m[S_2-S_1-R\;ln(\frac{P_2}{P_1}) ][/tex]
Then, we have to put values in the equation.
[tex]=1.17\left[1.871255-1.78249-0.287\;ln\left(\frac{100}{200} \right)\right][/tex]
[tex]\sigma_{cv}=0.337\;KW/k[/tex]
