A REPORT ON
Numerica's with Formula and Schematic Representation Based on
Laminar Flow/Turbulent Flow etc./Boundary Layer Theory etc.
Bachelor of Technology
Chemical Engineering
School of Technology, GSFC University
Vadodara
AY 2020-21
Submitted by: Submitted to:
Sheth Jay Rajnikant Ms. Priyanka Pandya
DtoD 3 rd Sem Assistant Professor
TABLE OF CONTENTS
Sr. No. Numerical Page No.
1 Example 10.3 3
2 Example 10.4
3 Example 10.5
4 Example 10.6
5 Example 10.7
6 Example 10.8
7 Example 10.9
8 Example 10.10
9 Example 10.11
10 Example 10.12
Numerical
Example 10.3.
A crude oil of viscosity 0.9 poise and relative density 0.9 is flowing through a
horizontal circular pipe of diameter 120 mm and length 12 m. Calculate the
difference of pressure at the two ends of the pipe, if 785 N of the oil is collected in
a tank in 25 seconds.
Given Data.
∴ Viscosity of the crude oil, μ = 0.9 poise = 0.09 Ns/m2
∴ Relative density = 0.9
∴ Weight density = 0.9 × 9810 = 8829 N/m3
∴ Diameter of the pipe, D = 120 mm = 0.12 m
∴ Length of the pipe, L = 12 m
∴ Weight of the oil collected in 25 s = 785 N
Equations.
𝑄 𝑄
Average velocity, u = = 𝜋
𝐴𝑟𝑒𝑎 ( 4 )×𝐷2
𝜌𝑉𝐷
Reynolds number, Re =
𝜇
32 𝜇 ∪ 𝐿
Difference in pressure for viscous or laminar flow is given by, (𝑝1 - 𝑝2 ) =
𝐷2
Example 10.4.
A liquid with a specific gravity 2.8 and a viscosity 0.8 poise flows through a
smooth pipe of unknown diameter, resulting in a pressure drop of 800 N/m2 in 2
km length of the pipe. What is the pipe diameter if the mass flow rate is 2500 kg/h.
Given Data.
∴ Sp. Gravity = 2.8
∴ μ = 0.8 × 0.1 = 0.08 Ns/m2
∴ ∆p = 800 N/m2
∴ L = 2 km = 2000 m
∴ m = 2500 kg/h = 2500/3600 = 0.6944 kg/s
∴ D = Diameter of pipe ( ? ) m
Equations.
m=𝜌𝐴∪
or
𝑚
∪=
𝜌𝐴
32 𝜇 ∪𝐿
Assuming flow to be laminar we have, ∆𝑝 =
𝐷2
𝜌∪𝐷
Re =
𝜇
Example 10.5.
A fluid of viscosity 8 poise and specific gravity 1.2 is flowing through a circular
pipe of diameter 100 mm. The maximum shear stress at the pipe wall is 210 N/m2
Find: (i) The pressure gradient,
(ii) The average velocity, and
(iii) Reynolds number of flow.
Given Data.
∴ Solution. Viscosity of fluid, μ = 8 poise = 0.8 Ns/m2
∴ Specific gravity = 1.2
∴ Mass density, ρ = 1.2 × 1000 = 1200 kg/m3
∴ Diameter of the pipe, D = 100 mm = 0.1 m
∴ Maximum shear stress, τ0 = 210 N/m2
Equations.
𝜕𝑝 𝑁
(i) The pressure gradient, , ( )
𝜕𝑥 𝑚2
𝜕𝑝 𝑅
𝜏0 = − ×
𝜕𝑥 2
1 𝑚
(ii) The average velocity, ∪ = × 𝑢𝑚𝑎𝑥 , ( )
2 𝑠
𝜌∪𝐷
(iii) Reynolds number, Re =
𝜇
Example 10.6.
A fluid of density 1200 kg/𝑚3 and viscosity 0.5 poise is flowing at a rate of
5𝑚3 /min in a circular pipe of cross-section of 1𝑚2 . Is the flow laminar or
turbulent? Can you predict the maximum velocity of the fluid in the pipe ?
Given Data.
∴ ρ = 1200 kg/m3
∴ μ = 0.5 poise = 0.5 × 1/10 = 0.05 Ns/𝑚2
𝑄 5 𝑚⁄
∴ 𝑉𝑎𝑣 = ∪ = = 𝑠
𝐴 60
𝜋
∴A= × 𝐷2
4
∴ D = 1.1128 m
Equations.
𝜌∪𝐷
Reynolds number, Re =
𝜇
5
∴ 𝑢𝑚𝑎𝑥 = 2 ∪ = 2 × = 0.1667 𝑚⁄𝑠
60
