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 is 1.20 kg/m3 at the inlet and 1.035 kg/m3 at the exit, determine the percent increase in the velocity

Answer 1

Given:

density of air at inlet, $\rho_{a} = 1.20 kg/m_{3}$

density of air at inlet, $\rho_{b} = 1.05 kg/m_{3}$

Solution:

Now,

$\dot{m} = \dot{m_{a}} = \dot{m_{b}}$

$\rho_{a} A v_{a} = \rho _{b} Av_{b}$                        (1)

where

A = Area of cross section

$v_{a}$ = velocity of air at inlet

$v_{b}$ = velocity of air at outlet

Now, using eqn (1), we get:

$\frac{v_{b}}{v_{a}} = \frac{\rho_{a}}{\rho_{b}}$

$\frac{v_{b}}{v_{a}} = \frac{1.20}{1.05}$ = 1.14

% increase in velocity = $1.14\times 100$ =114%

which is 14% more

Therefore % increase in velocity is 14%

Answer 2

Answer:

The percent increase in the velocity is 15.9%.

Explanation:

Given that,

Density of air $\rho_{in}= 1.20\ kg/m^3$

Density of air $\rho_{ex}=1.035\ kg/m^3$

We need to calculate the percent increase in the velocity

Using continuity equation

$\rho_{in}Av=\rho_{ex}Av'$

$\dfrac{\rho_{in}}{\rho_{ex}}=\dfrac{v'}{v}$

Put the value into the formula

$\dfrac{v'}{v}=\dfrac{1.20}{1.035}$

$\dfrac{v'}{v}=1.159$

$v'=1.159v$

For percentage,

$v'=\dfrac{1.159v-v}{v}\times100$

$v'=15.9\%$

Hence, The percent increase in the velocity is 15.9%.