Answer
We know that ,
mass flow rate can be given as = (Av)/(specific volume)
where , A and v are area and velocity respectively.
here , in question we are given area and velocity at the inlet and the outlet. and we are given the data of temperature and pressure, so get the specific volume from the table.
$mass_{in}$/sec =Av/specific volume= (0.01*20)/(0.19962)
= 1 kg/sec
and $mass_{out}$/sec= (0.006*1)/(0.001127)
= 5.3 kg/sec.
so rate of mass contained in the tank is (1-5.3)=(-4.3) kg/sec.
Work Step by Step
We know that ,
mass flow rate can be given as = (Av)/(specific volume)
where , A and v are area and velocity respectively.
here , in question we are given area and velocity at the inlet and the outlet. and we are given the data of temperature and pressure, so get the specific volume from the table.
$mass_{in}$/sec =Av/specific volume= (0.01*20)/(0.19962)
= 1 kg/sec
and $mass_{out}$/sec= (0.006*1)/(0.001127)
= 5.3 kg/sec.
so rate of mass contained in the tank is (1-5.3)=(-4.3) kg/sec.
