Answer
In this question,
steady state is given so
d$m_{in}$/dt=d$m_{out}$/dt
Now, we need to find,
(specific volume)$v_{1}$=$v_{f}$(approx at 20 degree C) = 1.0018/$10^{3}$ ($m^{3}$/kg)
now ,
we know that mass flow rate is given by ( Area * velocity)/(specific volume)
here D=0.02m and velocity is 40 m/sec .
and specific volume is same as calculated above.
so we get mass flow rate = 12.54 kg/sec.
on comparing mass flow rate at both ends
$A_{1}$$v_{1}$= $A_{2}$$v_{2}$
and diameter at exit is 0.04 m.
so on solvinng we get ,
$v_{2}$=10 m/sec.
Work Step by Step
In this question,
steady state is given so
d$m_{in}$/dt=d$m_{out}$/dt
Now, we need to find,
(specific volume)$v_{1}$=$v_{f}$(approx at 20 degree C) = 1.0018/$10^{3}$ ($m^{3}$/kg)
now ,
we know that mass flow rate is given by ( Area * velocity)/(specific volume)
here D=0.02m and velocity is 40 m/sec .
and specific volume is same as calculated above.
so we get mass flow rate = 12.54 kg/sec.
on comparing mass flow rate at both ends
$A_{1}$$v_{1}$= $A_{2}$$v_{2}$
and diameter at exit is 0.04 m.
so on solvinng we get ,
$v_{2}$=10 m/sec.
