A crude oil of viscosity 0.97 poise and relative density 0.9 is flowing through a horizontal circular pipe of diameter 100 mm and length 10m. Calculate the difference of pressure at the two ends of the pipe, if 100 kg of the oil is collected in a tank in 30 seconds.

 A crude oil of viscosity 0.97 poise and relative density 0.9 is flowing through a horizontal circular pipe of diameter 100 mm and length 10m. Calculate the difference of pressure at the two ends of the pipe, if 100 kg of the oil is collected in a tank in 30 seconds.

Solution:- 

Given:-

μ=0.97$\mu =0.97$ poise

μ=0.9710=0.097NS/m2$\mu ={\displaystyle \frac{0.97}{10}}=0.097NS/m^2$

ρo=0.9×1000=900kg/m3${\rho }_o=0.9\times 1000=900kg/m^3$ (As density of water =1000kg/m3$1000kg/m^3$)

D=100mm=0.1m

L=10m

Mass, m=100kg

T=30seconds

(i) Pressure difference (P1−P2$P_1-P_2$)

=32μu¯¯¯LD2$={\displaystyle \frac{32\mu \overline{u}L}{D^2}}$

∴u¯¯¯=QA$\therefore \overline{u}={\displaystyle \frac{Q}{A}}$

Mass of oil/sec = 10030kg/s…………………(1)${\displaystyle \frac{100}{30}}kg/s\dots \dots \dots \dots \dots \dots \dots (1)$

=ρo×Q$={\rho }_o\times Q$

=900×Q……………..(2)$=900\times Q\dots \dots \dots \dots \dots ..(2)$

Equate (1) and (2), we get

10030=900×Q${\displaystyle \frac{100}{30}}=900\times Q$

∴Q=10030×1900$\therefore Q={\displaystyle \frac{100}{30}}\times {\displaystyle \frac{1}{900}}$

Q=0.0037m3/s$Q=0.0037m^3/s$

∴u¯¯¯=QA$\therefore \overline{u}={\displaystyle \frac{Q}{A}}$

u¯¯¯=0.0037π4D2$\overline{u}={\displaystyle \frac{0.0037}{{\displaystyle \frac{\pi }{4}}{D}^2}}$

u¯¯¯=0.0037π4(0.1)2$\overline{u}={\displaystyle \frac{0.0037}{{\displaystyle \frac{\pi }{4}}(0.1{)}^2}}$

u¯¯¯=0.471m/s$\overline{u}=0.471m/s$

Reynolds Number,

Re=ρVDμ$R_e={\displaystyle \frac{\rho VD}{\mu }}$ (ρ=900kg/m3,V=u¯¯¯=0.471$\rho =900kg/m^3,V=\overline{u}=0.471$)

Re=900×0.471×0.10.097=436.91$R_e={\displaystyle \frac{900\times 0.471\times 0.1}{0.097}}=436.91$

As Re<2000$R_e<2000$, the flow is laminar,

∴P1−P2=32×0.097×0.471×10(0.1)2$\therefore P_1-P_2={\displaystyle \frac{32\times 0.097\times 0.471\times 10}{(0.1{)}^2}}$

=1462.28NS/m2=1462.28×10−4N/cm2$=1462.28NS/m^2=1462.28\times {10}^{-4}N/c{m}^2$

∴P1−P−2=0.1462N/cm2