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A hot-water stream at 80°C enters a mixing chamber with a mass flow rate of 0.46 kg/s where it is mixed with a stream of cold water at 20°C. If it is desired that the mixture leave the chamber at 42°C, determine the mass flow rate of the cold-water stream. Assume all the streams are at a pressure of 250 kPa. The enthalpies are 335.02 kJ/kg, 83.915 kJ/kg, and 175.90 kJ/kg. The saturation temperature at a pressure of 250 kPa is 127.41°C.

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fahadisahadam

Answer:

[tex]m_{2} =[/tex] [tex]0.795 kg/s[/tex]

Explanation:

[tex]m_{1} h_{1} +m_{2} h_{2}=m_{3} h_{3}[/tex]

  • m means mass
  • T means temperature

[tex]m_{1} T_{1} +m_{2} T_{2} =T_{3} (m_{1} +m_{2} )[/tex]

make [tex]m_{2}[/tex] the subject of the formula

[tex]m_{2} T_{2} -m_{2} T_{3} =m_{1} T_{3} -m_{1} T_{1}[/tex]

[tex]m_{2} (T_{2}- T_{3} )=m_{1} (T_{3}- T_{1} )[/tex]

divide both side by [tex]T_{2} -T_{3}[/tex] to find

[tex]m_{2} =m_{1} *\frac{T_{3}-T_{1} }{T_{2}-T_{3} }[/tex]              [tex]m_{1} =0.46,T_{1} =80,T_{2}=20,and,T_{3}=42[/tex]

[tex]m_{2} =0.46 *\frac{42-80 }{20-42 }[/tex]

[tex]m_{2} =0.46*\frac{-38}{-22}[/tex]

[tex]m_{2} =0.46*1.73[/tex]

[tex]m_{2} =0.795 kg/s[/tex]  

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