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Example 3 :
A liquid with relative density 0.9 and dynamic viscosity of 0.06 Pas
flows in a cast iron pipe of length 100 m and diameter 120 mm.
Determine the friction factor using formula method, and then
calculate the head loss using the answer obtained then using
friction factor formula when the velocity is:
a) 1 ms-1
b) 3 ms-1

Sagot :

The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

How to determine the friction factor

Using the formula

μ = viscosity = 0. 06 Pas

d =  diameter = 120mm = 0. 12m

V =  velocity = 1m/s and 3m/s

ρ = density = 0.9

a. Velocity = 1m/s

friction factor = 0. 52 × [tex]\frac{0. 06}{0. 12* 1* 0. 9}[/tex]

friction factor = 0. 52 × [tex]\frac{0. 06}{0. 108}[/tex]

friction factor = 0. 52 × 0. 55

friction factor [tex]= 0. 289[/tex]

b. When V = 3mls

Friction factor = 0. 52 × [tex]\frac{0. 06}{0. 12 * 3* 0. 9}[/tex]

Friction factor = 0. 52 × [tex]\frac{0. 06}{0. 324}[/tex]

Friction factor = 0. 52 × 0. 185

Friction factor [tex]= 0.096[/tex]

Loss When V = 1m/s

Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter

Head loss = 0. 289 × [tex]\frac{1}{2*6. 6743 * 10^-11}[/tex] × [tex]\frac{1^2}{0. 120}[/tex] × [tex]\frac{1}{100}[/tex]

Head loss =  1. 80 × 10^8

Head loss When V = 3m/s

Head loss = [tex]0. 096[/tex] × [tex]\frac{1}{1. 334 *10^-10}[/tex] × [tex]\frac{3^2}{0. 120}[/tex] × [tex]\frac{1}{100}[/tex]

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss  when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

Learn more about friction here:

https://brainly.com/question/24338873

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