Question

Water flows through a very long 52 mm ID (Internal Diameter) drawn copper pipe. The pipe is horizontal and has an internal roughness height of ε=0.0015 mm. The pressure drop along a 200m long section of the pipe is 100 kPa. Find the friction factor f, Reynolds number and average speed of the flow in the pipe. Is the pipe flow laminar or turbulent?

Answer 1

friction factor is 0.01819

average speed v = 1.69 m/s

Reynolds number = 98786.7

and flow is turbulent

Step-by-step explanation:

given data

diameter d = 0.052 m

roughness ε=0.0015 mm

length l = 200 m

ΔP = 100 kPa

to find out

friction factor f, Reynolds number Re and average speed v

solution

we solve this by hit and try assume flow is turbulent

so we can say pipe flow

`( pressure drop )/(\rho g)  = (flv^2)/(2gd)` ...............1

here f is friction factor and l is length and d is diameter and ρ is density

so

`(100*10^3)/(10^3 g)  = (f*200*v^2)/(2*g*0.052)`

f v² = 0.052 ................2

we know Reynolds number =
`(\rho v d)/(u)`

here u is viscosity of water that is 8.9 ×
`10^(-4)`

Reynolds number =
`(10^3 v 0.052)/(8.9*10^(-4))`

put here v from equation 2

Reynolds number =
`(10^3 v 0.052)/(8.9*10^(-4))` ×
`\sqrt{(0.052)/(f) }`

Reynolds number =
`(13323.398)/(√(f) )` .................3

we know that 1/√f is

`(1)/(√(f)) = -2 log( (2.51)/(Re√(f) ) + (E)/(3.7d))`

`(1)/(√(f)) = -2 log( (2.51)/(13323.398 ) + (0.0015*10^(-3))/(3.7*0.052))`

`(1)/(√(f)) = 7.41466`

f = 0.01819

friction factor is 0.01819

and from equation 2

f v² = 0.052

0.01819 v² = 0.052

average speed v = 1.69 m/s

and from equation 3

Reynolds number =
`(13323.398)/(√(f) )`

Reynolds number =
`(13323.398)/(√(0.01819) )`

Reynolds number = 98786.7

so flow is turbulent