Respuesta :
Solution:
The velocity of the exhaust air flow using continuity equation
[tex]$V=\frac{Q}{\frac{\pi}{4}d^2}$[/tex]
[tex]$V=\frac{0.1}{\frac{\pi}{4}(0.125)^2}$[/tex]
= 8.148 ft/s
Reynolds number,
[tex]$Re=\frac{\rho V d}{\mu}$[/tex]
[tex]$Re=\frac{3\times 10^{-3}\times 8.148\times 0.125}{4.7\times 10^{-7}}$[/tex]
= 6501.06
As Re > 4000, the flow is turbulent.
Now calculating the friction factor of the flow,
[tex]$f=\frac{0.316}{(Re)^{0.25}}$[/tex]
[tex]$f=\frac{0.316}{(6501.06)^{0.25}}$[/tex]
= 0.03519
Calculating the major head loss in the pipe
[tex]$h_{L,major}=\frac{flV^2}{2gd}$[/tex]
[tex]$h_{L,major}=\frac{0.03519\times 9\times (8.148)^2}{2\times 32.174\times 0.125}$[/tex]
= 2.614 ft
Calculating the minor head loss in the pipe
[tex]$h_L=n\left(\frac{k_{L,elbow} V^2}{2g}\right)+\frac{k_{L,muffler}V^2}{2g}$[/tex]
[tex]$h_L=\frac{V^2}{2g}\left(nk_{L,elbow}+k_{L,muffler\right)}$[/tex]
Here, n = number of elbows.
[tex]$h_{L,minor}=\frac{(8.148)^2}{2\times 32.174}\left(7\times 0.3+8.5)}$[/tex]
= 10.936 ft
Now using Bernoulli equation between the entrance of the pipe and the exit of the pipe
[tex]$\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\h_{L,major}+h_{L,minor}$[/tex]
Substitute [tex]$V_2 \text{ with}\ V_1 \text{ and}\ z_2 \text{ with}\ z_1$[/tex], we get
[tex]$\frac{p_1}{\rho g}=\frac{p_2}{\rho g}+h_{L,major}+h_{L, minor}$[/tex]
[tex]$\frac{p_1}{\rho g}-\frac{p_2}{\rho g}=2.614+10.936$[/tex]
[tex]$p_1-p_2=3\times 10^{-3}\times 32.174 \times 13.55$[/tex]
[tex]$p_1-p_2=1.3078 \ lb/ft^2$[/tex]
