A cooler in an air conditioner brings 0.5 kg/s air at 35°C to 5°C, both at 101 kPa and it then mix the output with a flow of 0.25 kg/s air at 20°C, 101 kPa sending the combined flow into a duct. Find the total heat transfer in the cooler and the temperature in the duct flow.

Question

Accepted Answer

C.V. Cooler section (no [latex]\dot{ W }[/latex] )

Energy Eq.4.12: [latex]\dot{ m } h_{1}=\dot{ m } h_{2}+\dot{ Q }_{ cool }[/latex]

[latex]\dot{ Q }_{ cool }=\dot{ m }\left( h _{1}- h _{2} ight)=\dot{ m } C _{ p }\left( T _{1}- T _{2} ight)=0.5 imes 1.004 imes(35-5)=15.06 kW[/latex]

C.V. mixing section (no [latex]\dot{ W }, \dot{ Q }[/latex] )

Continuity Eq.: [latex]\dot{ m }_{2}+\dot{ m }_{3}=\dot{ m }_{4}[/latex]

Energy Eq.4.10: [latex]\dot{ m }_{2} h _{2}+\dot{ m }_{3} h _{3}=\dot{ m }_{4} h _{4}[/latex]

[latex]\dot{ m }_{4}=\dot{ m }_{2}+\dot{ m }_{3}=0.5+0.25=0.75 kg / s[/latex]

[latex]\dot{ m }_{4} h _{4}=\left(\dot{ m }_{2}+\dot{ m }_{3} ight) h _{4}=\dot{ m }_{2} h _{2}+\dot{ m }_{3} h _{3}[/latex]

[latex]\dot{ m }_{2}\left( h _{4}- h _{2} ight)+\dot{ m }_{3}\left( h _{4}- h _{3} ight)=\varnothing[/latex]

[latex]\dot{ m }_{2} C _{ p }\left( T _{4}- T _{2} ight)+\dot{ m }_{3} C _{ p }\left( T _{4}- T _{3} ight)=\varnothing[/latex]

[latex]T _{4}=\left(\dot{ m }_{2} / \dot{ m }_{4} ight) T _{2}+\left(\dot{ m }_{3} / \dot{ m }_{4} ight) T _{3}=5(0.5 / 0.75)+20(0.25 / 0.75)= 1 0 ^{\circ} C[/latex]

............................

Eq.4.12 : [latex]\dot{Q}_{ C.V. }+\dot{m}\left(h_{i}+\frac{ V _{i}^{2}}{2}+g Z_{i} ight)=\dot{m}\left(h_{e}+\frac{ V _{e}^{2}}{2}+g Z_{e} ight)+\dot{W}_{ C.V. }[/latex]

Eq.4.10 : [latex]\dot{Q}_{ C.V. }+\sum \dot{m}_{i}\left(h_{i}+\frac{ V _{i}^{2}}{2}+g Z_{i} ight)=\sum \dot{m}_{e}\left(h_{e}+\frac{ V _{e}^{2}}{2}+g Z_{e} ight)+\dot{W}_{ C . V. }[/latex]

Question 4.101: A cooler in an air conditioner brings 0.5 kg/s air at 35°C t...

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