Question

Estimate pressure drop for an estimate of pipe diameter Pressure drop is a function of flow rate, length, diameter, and roughness. Either iterative methods OR equation solvers are necessary to solve implicit problems. For a first guess of a 1 ft diameter pipe, what is the fluid velocity? V = 5.67 ft/s What is the Reynolds number? Re = 96014 What is the pipe relative roughness?

Answer 1

Answer:

Explanation:

By using Bernoulli's Equation:

$\frac{P_1}{P_g}+\frac{v_1^2}{2g}+z_1=\frac{P_2}{P_g}+\frac{v_2^2}{2g}+z_2+f\frac{L}{D}\frac{v^2}{2g}$

where;

$z_1 = z_2 \ and \ v_1 = v_2$

$P_1 - P_2 = f \frac{L}{D}\frac{1}{2}\rho v^2$

$P_1-P_2 = \frac{5 \ lb}{in^2}( 144 \frac{in^2}{ft^2})$

$P_1-P_2 = 720 \frac{lb}{ft^2}$

$V = \frac{Q}{A} \\ \\ V = \frac{6.684 \ ft^2/s}{\frac{\pi}{4}D^2} \\ \\V = \frac{8.51}{D^2}$

Density of gasoline $\rho = 1.32 \ slug/ft^3$

Dynamic Viscosity $\mu$ = $6.5*10^{-6} \frac{lb.s}{ft}$

$P_1-P_2 = f \frac{L \rho V^2}{2D}$

$720 = f \frac{L(100)(1.32)}{2D}(\frac{8.51}{D^2})^2$

D = 1.46 f

$Re, = \frac{\rho VD}{\mu} = \frac{1.32 *\frac{8.51}{D^2} D}{6.5*10^{-6}}$

$= \frac{1.72*10^6}{D}$

$\frac{E}{D}= \frac{0.00015}{D}$

However; the trail and error is as follows;

Assume ; f= 0.02 → D = 0.667ft$\left \{ {{Re=2.576*10^6} \atop {\frac{E}{D}=0.000225}} \right.$   $\right \{ {{f=0.014} \atop {\neq 2}}$

f = 0.0145  → D = 0.0428 ft $\left \{ {{Re=4.018*10^6} \atop {\frac{E}{D}=0.00035}} \right.$ $\right \{ {{f=0.015} \atop {\neq 0.0145}}$

f = 0.0156  → D = 0.43 ft $\left \{ {{Re=4.0*10^6} \atop {\frac{E}{D}=0.000348}} \right.$ $f = 0.0156$

∴ pipe diameter d = 0.43 ft

Given that:

D = 1 ft

$V = \frac{Q}{A} \\ \\ V = \frac{6.684}{\frac{\pi}{4}(1)^2} \\ \\ V = 8.51 \ ft/s$

$Re = \frac{\rho \ V \ D}{\mu } \\ \\ Re = \frac{1.32 *8.51 *1 }{6.5*10^{-6}}$

$Re = 1.72 *10^6$

$\frac{E}{D} = \frac{0.00015}{1} \\ \\ = 0.00015$

$f = 0.0136$

$P_2-P_1 = \frac{fL \rho V^2 }{2D}$

$P_2-P_1 = \frac{0.036(100)(1.32)(8.51)^2 }{2*1}$

$P_2-P_1 = 65 \frac {lb}{ft^2}$  to psi ; we have:

$P_2-P_1 = 0.45 \ psi$