Question

a crude of oil 0.97 poise and relative density 0.9 is flowing through a horizantal pipe of diameter 100 mm and of the length 10 m. Calculate the difference of pressure at the two ends of the pipe. if 100kg of oil is collected in a tank in 30 secondd​

Answer 1

Answer:

Data : μ=0.97poise=0.9710=0.097Ns/m2

Relative density = 0.9 (of 0:1)

∴Densityρo=0.9×1000=900kg/m3 (of oil)

Dia. of pipe D = 100 mm = 0.1 m

Length L = 10 m

Mass of Oil = m = 100 kg.

and time t = 30 seconds.

P1−P2=?

The difference of pressure for viscous flow is

p1−p2=32mŪLD2 Avg. velocity Ū=QArea

Now, mass of oil per second = 10030 kg/s

= ρ×Q=900×Q

∴10030×Q

∴Q=0.0037m3/s

Now, Ū=QA=0.0037π×D2=0.0037π×0.12=0.471m/s

For Laminar/viscous flow, Re < 2000

Now, Re=ρVDμ

ρ=ρo=900,v=Ū=0.471m/s

D = 0.1m, μ = 0.097

Re=900×0.471×0.10.097=436.91

∵ Value of Re < 2000, hence flow is laminar.

∴p1−p2=32μŪLD2=32×0.097×0.471×10(0.1)2

p1−p2=1462.28N/m2

= 1462.28×10−4N/cm2

= p1−p2=0.1462N/cm2