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

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

`\bf b)-\sigma_(cv)=0.337\;kw/k`

Step-by-step 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 )+(v_1^2-v_2^2)/(2) \right ]=0`

`(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)*10^3`

`v_2=√(108900-60450)=220.11\;m/s`

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

`m=(A_1V_1)/(V_1)`

`Then,\;\;\;\;\;\;\;\;\;\;\;=(A_1V_1)/((RT_1)/(P_1) )`

`=(P_1)/(RT_1) * A_1V_1`

Then, we have to put values in the equation.

`=(200* 10^3)/(287\left(52+273\right)) * 16.57 * 10^(-4)*330=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((P_2)/(P_1)) ]`

Then, we have to put values in the equation.

`=1.17\left[1.871255-1.78249-0.287\;ln\left((100)/(200) \right)\right]`

`\sigma_(cv)=0.337\;KW/k`