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

Step-by-step explanation:

By using Bernoulli's Equation:

`(P_1)/(P_g)+(v_1^2)/(2g)+z_1=(P_2)/(P_g)+(v_2^2)/(2g)+z_2+f(L)/(D)(v^2)/(2g)`

where;

`z_1 = z_2 \ and \ v_1 = v_2`

`P_1 - P_2 = f (L)/(D)(1)/(2)\rho v^2`

`P_1-P_2 = (5 \ lb)/(in^2)( 144 (in^2)/(ft^2))`

`P_1-P_2 = 720 (lb)/(ft^2)`

`V = (Q)/(A) \\ \\ V = (6.684 \ ft^2/s)/((\pi)/(4)D^2) \\ \\V = (8.51)/(D^2)`

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

Dynamic Viscosity
`\mu` =
`6.5*10^(-6) (lb.s)/(ft)`

`P_1-P_2 = f (L \rho V^2)/(2D)`

`720 = f (L(100)(1.32))/(2D)((8.51)/(D^2))^2`

D = 1.46 f

`Re, = (\rho VD)/(\mu) = (1.32 *(8.51)/(D^2) D)/(6.5*10^(-6))`

`= (1.72*10^6)/(D)`

`(E)/(D)= (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 {(E)/(D)=0.000225}} \right.`
`\right \{ {{f=0.014} \atop {\\eq 2}}`

f = 0.0145 → D = 0.0428 ft
`\left \{ {{Re=4.018*10^6} \atop {(E)/(D)=0.00035}} \right.`
`\right \{ {{f=0.015} \atop {\\eq 0.0145}}`

f = 0.0156 → D = 0.43 ft
`\left \{ {{Re=4.0*10^6} \atop {(E)/(D)=0.000348}} \right.`
`f = 0.0156`

∴ pipe diameter d = 0.43 ft

Given that:

D = 1 ft

`V = (Q)/(A) \\ \\ V = (6.684)/((\pi)/(4)(1)^2) \\ \\ V = 8.51 \ ft/s`

`Re = (\rho \ V \ D)/(\mu ) \\ \\ Re = (1.32 *8.51 *1 )/(6.5*10^(-6))`

`Re = 1.72 *10^6`

`(E)/(D) = (0.00015)/(1) \\ \\ = 0.00015`

`f = 0.0136`

`P_2-P_1 = (fL \rho V^2 )/(2D)`

`P_2-P_1 = (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`