Question

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.

Answer 1

Answer:

$m_{2} =$ $0.795 kg/s$

Explanation:

$m_{1} h_{1} +m_{2} h_{2}=m_{3} h_{3}$

• m means mass
• T means temperature

$m_{1} T_{1} +m_{2} T_{2} =T_{3} (m_{1} +m_{2} )$

make $m_{2}$ the subject of the formula

$m_{2} T_{2} -m_{2} T_{3} =m_{1} T_{3} -m_{1} T_{1}$

$m_{2} (T_{2}- T_{3} )=m_{1} (T_{3}- T_{1} )$

divide both side by $T_{2} -T_{3}$ to find

$m_{2} =m_{1} *\frac{T_{3}-T_{1} }{T_{2}-T_{3} }$              $m_{1} =0.46,T_{1} =80,T_{2}=20,and,T_{3}=42$

$m_{2} =0.46 *\frac{42-80 }{20-42 }$

$m_{2} =0.46*\frac{-38}{-22}$

$m_{2} =0.46*1.73$

$m_{2} =0.795 kg/s$