Question

A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls the air in and forces it through the resistors where it is heated. If the density of air at the inlet is 1.20 kg/m^3 and 0.95kg/m^3 at the exit, determine the percent increase in the velocity of the air as it flows through the dryer.

Answer 1

% increase = 26.32%

Step-by-step 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%