Consider a mixing tank initially containing 2000 lbm of liquid water. The tank is fitted with two inlet pipes, one delivering hot water at a mass flow rate of 0.8 lbm/s and the other delivering cold water at a mass flow rate of 1.2 lbm/s. Water exits through a single exit pipe at a mass flow rate of 2.5 lbm/s. Determine the amount of water [lbm] in the tank after one hour.

[tex]\frac{m_{cv}}{dt} = \sum\dot{m_i}-\sum\dot{m_e}\\\frac{m_{cv}}{dt} = \dot{m_1}+\dot{m_2}-\dot{m_3}\\dm_{cv}=(\dot{m_1}+\dot{m_2}-\dot{m_3})dt[/tex]

We integrathe this expresión to have the initial and final mass.

[tex]\int\limit^{final}_{initial} = \int\limit^{t=1hr}_{t=0}(\dot{m_1}+\dot{m_2}-\dot{m_3})dt[/tex]

[tex]m_{final}=m_{initial}+(\dot{m_1}+\dot{m_2}-\dot_{m_3})t[/tex]

Here we can note remark that

[tex]\dot{m_1},\dot{m_2}=[/tex] Mass flow rate of hot water and cold water

[tex]\dot{m_3} =[/tex] Mass flow rate of liquit water

Substituting the values

[tex]m_{initial}=2000lb\\\dot{m_1}=0.8lb/s\\\dot{m_2}=1.2lb/s\\\dot{m_3}=2.5lb/s\\t=1hr[/tex]

[tex]m_{final}=m_{initial}+(\dot{m_1}+\dot{m_2}-\dot{m_3})\\m_{final}=2000+(0.8+1.2-2.5)(1hr*3600s/1hr)\\m_{final}=200lb\\[/tex]