Answer :
Given:
density of air at inlet, [tex]\rho_{a} = 1.20 kg/m_{3}[/tex]
density of air at inlet, [tex]\rho_{b} = 1.05 kg/m_{3}[/tex]
Solution:
Now,
[tex]\dot{m} = \dot{m_{a}} = \dot{m_{b}}[/tex]
[tex]\rho_{a} A v_{a} = \rho _{b} Av_{b}[/tex] (1)
where
A = Area of cross section
[tex]v_{a}[/tex] = velocity of air at inlet
[tex]v_{b}[/tex] = velocity of air at outlet
Now, using eqn (1), we get:
[tex]\frac{v_{b}}{v_{a}} = \frac{\rho_{a}}{\rho_{b}}[/tex]
[tex]\frac{v_{b}}{v_{a}} = \frac{1.20}{1.05}[/tex] = 1.14
% increase in velocity = [tex]1.14\times 100[/tex] =114%
which is 14% more
Therefore % increase in velocity is 14%
Answer:
The percent increase in the velocity is 15.9%.
Explanation:
Given that,
Density of air [tex]\rho_{in}= 1.20\ kg/m^3[/tex]
Density of air [tex]\rho_{ex}=1.035\ kg/m^3[/tex]
We need to calculate the percent increase in the velocity
Using continuity equation
[tex]\rho_{in}Av=\rho_{ex}Av'[/tex]
[tex]\dfrac{\rho_{in}}{\rho_{ex}}=\dfrac{v'}{v}[/tex]
Put the value into the formula
[tex]\dfrac{v'}{v}=\dfrac{1.20}{1.035}[/tex]
[tex]\dfrac{v'}{v}=1.159[/tex]
[tex]v'=1.159v[/tex]
For percentage,
[tex]v'=\dfrac{1.159v-v}{v}\times100[/tex]
[tex]v'=15.9\%[/tex]
Hence, The percent increase in the velocity is 15.9%.