PLAIN AND CIVIL: EXAMPLE 6.2.5. APPLICATION OF MOODY DIAGRAM

Friday, 19 April 2019

Determine the size of a steel pipe to be installed if it is to carry 500,000 gpd of heavy fuel oil at 75 degrees Fahrenheit at a distance of 1 mile with an allowable pipe friction loss of 50psi.

Convert the given Q:

$Q=500,000\frac{gals}{day}x\frac{1day}{24hours}x\frac{1hour}{3600secs}x\frac{0.133681ft^{3}}{gal}=0.7736cfs$

$L=1mile\simeq 5280ft$

To obtain the other needed parameters, determine the properties of the fluid, heavy fuel oil.

SG = 0.906

$\upsilon =0.00135\frac{ft^{2}}{s}$

Pipe friction loss should also be converted into head, hf.

$h_{f}=2.3059\frac{P}{SG}=2.3059\frac{50}{0.906}=127.2572ft$

Use a trial velocity to solve for the Reynold's number and identify the flow.

Try    $v=5\frac{ft}{s}$

The area (with the diameter) can be solved.

$A=\frac{Q}{v}$

   $\frac{\pi }{4}D^{2}=\frac{0.7736}{5}$

$D=0.4438ft$

Take D=0.50ft or 6 inches.

To check the trial section, determine Reynold's number:

   $A=\frac{\pi }{4}0.5^{2}=0.1963ft^{2}$

Determine velocity from the given discharge,

$v=\frac{Q}{A}=\frac{0.7736cfs}{0.1963ft^{2}}=3.9409\frac{ft}{s}$

Then,

   $R_{e}=\frac{vD}{\upsilon }=\frac{3.9409*0.5}{0.00135}=1459.5926$

So,

   $f=\frac{64}{R_{e}}=\frac{64}{1459.5926}=0.0438$

Check for allowable head loss:

   $h_{f}=f\frac{L}{D}\frac{V^{2}}{2g}$

$h_{f}=0.0438\frac{5280}{0.5}\frac{3.9409^{2}}{2*32.20} =111.5432ft<127.2572(allow)OK!$

So take:

${\color{Red} \mathbf{D=0.50ft\simeq 6inches}}$

RELATED TOPIC:

6.2. GENERAL METHOD OF DETERMINING DARCY WEISBACH'S FRICTION FACTOR- THE MOODY DIAGRAM

PLAIN AND CIVIL

Pages

Friday, 19 April 2019

EXAMPLE 6.2.5. APPLICATION OF MOODY DIAGRAM

1 comment: