Answer:
[tex]0.01196lb/ft^3[/tex]
Explanation:
To deduce the density of the fluid we start from the concept related to the conservation of Flow. In which the mass of air and the mass of the fuel is equal to the total mass expelled.
In this way,
[tex]\dot{m_{EX}}= \dot{m_A}+\dot{m_F}[/tex]
For definition we know that the mass flow is given by,
[tex]\dot{m} = \rho A \bar{\upsilon}}[/tex]
Where,
[tex]\rho =[/tex]Density
A = Cross-sectional area
[tex]\bar{\upsilon}=[/tex] Average velocity
Replacing in the mass flow in the Exhaust:
[tex]\rho A \bar{\upsilon}}= \dot{m_A}+\dot{m_F}[/tex]
[tex]\rho = \frac{\dot{m_A}+\dot{m_F}}{A \bar{\upsilon}}}[/tex]
We have that in our problem that
[tex]\dot{m_A} = 60lbm/s[/tex]
[tex]\dot{m_F} = 0.58lbm/s[/tex]
[tex]A = 3.3ft^2[/tex]
[tex]\bar{\upsilon}= 1535ft/s[/tex]
Replacing,
[tex]\rho = \frac{\dot{m_A}+\dot{m_F}}{A \bar{\upsilon}}}[/tex]
[tex]\rho = \frac{60+0.58}{3.3 \bar{1535}}}[/tex]
[tex]\rho = 0.01195lb/ft^3[/tex]
Therefore the denisty of the exhaust gases is [tex]0.01196lb/ft^3[/tex]
