Answer:
% increase = 26.32%
Explanation:
From conservation of mass, we can say that;
Mass flow rate at inlet = mass flow rate at exit.
Thus;
m'1 = m'2
Formula for mass flow rate is;
m' = ρV'
Where V' is volumetric flow rate = Av
Thus;
m' = ρAv
Where;
ρ is density
A is area
v is velocity
Therefore from m'1 = m'2, we can say that;
ρ1•A1•v1 = ρ2•A2•v2
Since the duct has a constant diameter, then A1 = A2
Thus, we now have;
ρ1•v1 = ρ2•v2
Making v2 the subject, we have;
v2 = ρ1•v1/ρ2
Now, since we want to find the percent increase in the velocity of the air as it flows through the dryer,we would use;
% increase = ((v2 - v1)/v1) × 100%
We have v2 = ρ1•v1/ρ2
Thus;
% increase = ((ρ1•v1/ρ2) - v1)/v1) × 100%
Factorizing v1 out, we have;
% increase = ((ρ1/ρ2) - 1)/1) × 100%
We are given;
ρ1 = 1.2 kg/m³
ρ2 = 0.95 kg/m³
Thus;
% increase = ((1.2/0.95) - 1)/1) × 100%
% increase = 26.32%
