For the system shown below the tank initially at time zero c

For the system shown below, the tank initially (at time zero) contains 24 kg of water, there are no reactions taking place Fill in the numerical integration table below for this case. Time interval is = 0.2 min.

Solution

At t=0

M=24 kg=52.9109 lb_m

Hence F_in=(20-0.5*24)kg/min=8kg/min

F_out=4kg/min.

After time interval of 0.2 min

M=24kg+0.2min*8kg/min-0.2min*4kg/min=24.8kg=54.67464 lb_m

Hence F_in=(20-0.5*24.8)kg/min=7.6 kg/min

F_out=4kg/min.

At time 0.4 min

M=24.8kg+0.2min*7.6kg/min-0.2min*4kg/min=25.52kg=56.261969 lb_m

FOR dM and dM/dt

At t=0

dM=M(0.2)-M(0)=0.8 kg=1.7637 lb_m

Hence dM/dt=dM/(0.2-0)=4kg/min=8.81849 lb_m/min

At t=0.2

dM=M(0.4)-M(0.2)=0.72 kg=1.587328 lb_m

Hence dM/dt=dM/(0.4-0.2)=3.6 kg/min=7.93664 lb_m/min