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 is 1.20 kg/m3 at the inlet and 0.955 kg/m3 at the exit, determine the percent increase in the velocity of air as it flows through the dryer.

The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.

m1=m2

Now remember that mass flow is given by the product of density, cross-sectional area and velocity

(α1)(V1)(A1)=(α2)(V2)(A2)

where

α=density

V=velocity

A=area

Now we can assume that the input and output areas are equal

(α1)(V1)=(α2)(V2)

[tex]\frac{V2}{V1} =\frac{\alpha1 }{\alpha 2}[/tex]

Now we can use the equation that defines the percentage of increase, in this case for speed

[tex]i=(\frac{V2}{V1} -1) 100[/tex]

Now we use the equation obtained in the previous step, and replace values

[tex]i=(\frac{\alpha1 }{\alpha 2} -1) 100\\i=(\frac{1.2}{0.955} -1) 100=25.65[/tex]

the percent increase in the velocity of air is 25.65%