Question

An exhaust system is modeled as 9 ft of 0.125 ft diameter smooth pipe with the equivalent of (7) 90○ elbows and a muffler. The muffler has a measured loss coefficient of 8.5. The average flowrate is 0.1 ft3 /s with a fluid density of 0.003 slug/ft3 and a dynamic viscosity of 4.7 ⋅ 10−7 lbf-s/ft2 . (a) Determine the system resistance. (b) Compute the pressure drop through the exhaust system.

Answer 1

Solution:

The velocity of the exhaust air flow using continuity equation

$V=\frac{Q}{\frac{\pi}{4}d^2}$

$V=\frac{0.1}{\frac{\pi}{4}(0.125)^2}$

   = 8.148 ft/s

Reynolds number,

$Re=\frac{\rho V d}{\mu}$

$Re=\frac{3\times 10^{-3}\times 8.148\times 0.125}{4.7\times 10^{-7}}$

    = 6501.06

As Re > 4000, the flow is turbulent.

Now calculating the friction factor of the flow,

$f=\frac{0.316}{(Re)^{0.25}}$

$f=\frac{0.316}{(6501.06)^{0.25}}$

  = 0.03519

Calculating the major head loss in the pipe

$h_{L,major}=\frac{flV^2}{2gd}$

$h_{L,major}=\frac{0.03519\times 9\times (8.148)^2}{2\times 32.174\times 0.125}$

             = 2.614 ft

Calculating the minor head loss in the pipe

$h_L=n\left(\frac{k_{L,elbow} V^2}{2g}\right)+\frac{k_{L,muffler}V^2}{2g}$

$h_L=\frac{V^2}{2g}\left(nk_{L,elbow}+k_{L,muffler\right)}$

Here, n = number of elbows.

$h_{L,minor}=\frac{(8.148)^2}{2\times 32.174}\left(7\times 0.3+8.5)}$

             = 10.936 ft

Now using Bernoulli equation between the entrance of the pipe and the exit of the pipe

$\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\h_{L,major}+h_{L,minor}$

Substitute $V_2 \text{ with}\ V_1 \text{ and}\ z_2 \text{ with}\ z_1$, we get

$\frac{p_1}{\rho g}=\frac{p_2}{\rho g}+h_{L,major}+h_{L, minor}$

$\frac{p_1}{\rho g}-\frac{p_2}{\rho g}=2.614+10.936$

$p_1-p_2=3\times 10^{-3}\times 32.174 \times 13.55$

$p_1-p_2=1.3078 \ lb/ft^2$