Question

Air at 200 kPa, 528C, and a velocity of 355 m/s enters an insulated duct of varying cross-sectional area. The air exits at 100 kPa, 82°C. At the inlet, the cross-sectional area is 6.57 2 c m 2 .

Answer 1

Answer:

$\bf a)-v_2=220.11\;m/s$

$\bf b)-\sigma_{cv}=0.337\;kw/k$

Explanation:

a) - Taking enthalpy from the ideal air structure that is :

Given,

               $T_1=52\° C$

$Then,\;\;\;\;\;\; T_1=325 k$

$And,\;\;\;\;\;\;\;\;h_1=325.31\; KJ/kg$

               $T_2=82^\circ C$

$Then,\;\;\;\;\;\; T_2=355 k$

$And,\;\;\;\;\;\;\;\;h_2=355.535\; KJ/kg$

Then, we have to apply the equation of the energy rate balance.

After applying that we have:

               $m\left [\left (h_1-h_2 \right )+\frac{v_1^2-v_2^2}{2} \right ]=0$

                $\frac{v_1^2-v_2^2}{2} =h_2-h_1$

                    $v_2^2=\left(V_1^2-2\right)\left(h_2-h_1\right)$

Then, we have to put values in the equation.

                    $v_2^2=\left(330^2\right)-2 \left(355.535-325.31\right)\times10^3$

                    $v_2=\sqrt{108900-60450}=220.11\;m/s$

b) - Here, we have to apply the equation of the continuity.

                $m=\frac{A_1V_1}{V_1}$

$Then,\;\;\;\;\;\;\;\;\;\;\;=\frac{A_1V_1}{\frac{RT_1}{P_1} }$

                    $=\frac{P_1}{RT_1} \times A_1V_1$

Then, we have to put values in the equation.

                   $=\frac{200\times 10^3}{287\left(52+273\right)} \times 16.57 \times 10^{-4}\times330=1.17\:kg/s$

Then, the values are :

              $T_1=52^\circ C=325k$

              $S_1=1.78249\: KJ/s$

and,

             $T_2=82^\circ C=355k$

             $S_2=1.871255\: KJ/s$

             

            $\sigma_{cv}=m[S_2-S_1-R\;ln(\frac{P_2}{P_1}) ]$

Then, we have to put values in the equation.

                  $=1.17\left[1.871255-1.78249-0.287\;ln\left(\frac{100}{200} \right)\right]$

            $\sigma_{cv}=0.337\;KW/k$