A REPORT ON
Numerical 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:
Panchal Hetkumar Ms. Priyanka Pandya
DtoD 3rd Sem Assistant Professor
TABLE OF CONTENTS
Sr. No. Numerical Page No.
1 Example 6.18 1
2 Example 6.24 2
3 Example 6.25 3
4 Example 6.26 5
5 Example 11.5 6
6 Example 11.11 7
Numerical
Example 6.18: the closed tank of a fire engine is partly
filled with water, the air space above being under
pressure. A 6 cm bore connected to the tank discharge
on the roof of building 2.5 m above the level of water in
the tank. The friction losses are 45 cm of water.
Determine the air pressure which must be maintained in
the tank to deliver 20 lit/sec. on the roof.
Solution: Data: Diameter of hose pipe d = 6 cm = 0.06 m
Friction, hf = 45 cm = 0.45 m
2
Roof
Closed tank 2.5 m
Hose pipe
AIR 1 (d= 6 cm)
Water
𝑄
Velocity V,= where, Q= Volumetric Flow Rate,m3/s
𝐴
A= Area of hose pipe, m2
Bernoulli’s Equation
𝑃1 𝑣₁² 𝑝2 𝑣₂²
+ 2𝑔 + 𝑧₁ = 𝑤 + 2𝑔 + 𝑧₂ + ℎ𝑓
𝑤
𝑝
Where, = Pressure head
𝑤
𝑣²
= Velocity head
2𝑔
𝑧 = Potential head
ℎ𝑓 = head loss due to friction
Example 6.24: Fig. shows a pump P pumping 72
liters/sec of water from tank
(i) What will be the pressure at points L & M when the
pump delivers 12 kw of power to the flow? Assume the
losses in the system to be negligible.
(ii) What will be the pressure at M when the loss in the
inlet pump is negligible and between the pump and the
point M, loss equal to 1.8 times the velocity head at B
takes place.
Solution: Q = 72 lit/sec = 0.0072m3/s
Power P = 12 KW M
1.3 m
16 cm dia.
3.5 m 10 cm
L P pump dia.
L
With the help of Continuity Equation
𝑄 = 𝐴˪𝑉˪ = 𝐴ₘ𝑉ₘ
Where, Q = Discharge
𝐴˪= area at point L
𝑉˪= Velocity at point L & same for 𝐴ₘ & 𝑉ₘ
Power P = wQHp
Where, w = Weight density
Hp= Head delivered by pump
Bernoulli’s equation at two points
𝑝₁ 𝑣₁² 𝑝₂ 𝑣₂²
+ + 𝑧₁ + 𝐻𝑝 = + + 𝑧₂ + 𝑙𝑜𝑠𝑠𝑒𝑠
𝑤 2𝑔 𝑤 2𝑔
𝑝
Where, = Pressure head
𝑤
𝑣²
= Velocity head
2𝑔
𝑧 = Potential head
Example 6.25: Fig. shows a pump drawing a solution
(specific gravity =1.8) from a storage tank through an 8
cm steel pipe in which the flow velocity is 0.9 m/s. The
pump discharges through a 6 cm steel pipe to an
overhead tank, the end of discharge is 12 m above the
level of the solution in the feed tank. If the friction
losses in the entire piping system are 5.5 m & pump
efficiency is 65 percent, determine:
(i) Power rating of the pump.
(ii) pressure developed by the pump
Solution: Data: d₂= 8 cm = 0.08 m, d₃= 6 cm= 0.06 m
V₂= 0.9m/s & ⴄ𝑝𝑢𝑚𝑝= 65%
From Continuity Equation
𝐴₂𝑉₂ = 𝐴₃𝑉₃
Where, 𝐴₂ = Area at point 2 & 𝐴₃= area at point 3
V₂= velocity at point 2 & V₃= velocity at 3
. 4
6 cm Overhead Tank
Dia.
1 12 m
8 cm pump
2 3
Storage delivery pipe
tank
𝑝₁ 𝑣₁² 𝑝₂ 𝑣₂²
+ + 𝑧₁ + 𝐻𝑝 = + + 𝑧₂ + 𝑙𝑜𝑠𝑠𝑒𝑠
𝑤 2𝑔 𝑤 2𝑔
𝑝
Where, = Pressure head
𝑤
𝑣²
= Velocity head
2𝑔
𝑧 = Potential head
Hp= Head delivered by pump
Example 6.26: A pump is 2.2 m above the water level
in the sump and has a pressure of 20 cm of mercury
at the suction side. The suction pipe is 20 cm
diameter and the delivery pipe is short 25 cm
diameter pipe ending in a nozzle of 8 cm diameter. If
the nozzle is directed vertically upward at an
elevation of 4.2 m above the water sump level,
determine
(i) The discharge.
(ii) The power input into the flow by pump.
(iii) The elevation above the water sump level, to
which the jet would reach. Neglect all losses.
Solution:
4.
h
Jet
3.
Nozzle
20 cm dia. 25 cm
2. dia.
pump
2.2 m 4.2 m
1.
Water
Sump
Applying Bernoulli’s equation at two points
Bernoulli’s equation at two points
𝑝₁ 𝑣₁² 𝑝₂ 𝑣₂²
+ + 𝑧₁ + 𝐻𝑝 = + + 𝑧₂ + 𝑙𝑜𝑠𝑠𝑒𝑠
𝑤 2𝑔 𝑤 2𝑔
𝑝
Where, = Pressure head
𝑤
𝑣²
= Velocity head
2𝑔
𝑧 = Potential head
Hp= Head delivered by pump
𝑄 = 𝐴₂𝑉₂
Where, 𝑄 = discharge in m3/s
From Continuity Equation
𝐴₂𝑉₂ = 𝐴₃𝑉₃
Where, 𝐴₁ = Area at point 1 & 𝐴₂= area at point 2, m
V₁= velocity at point 1 & V₂= velocity at 2 m/s
Example 11.5: In a pipe of diameter 100 mm, the
velocities at the pipe center and 30 mm from the pipe
center are found to be 2.5 m/s and 2.2 m/s respectively.
Find the wall shearing stress.
Solution: Data - radius R= 50 mm = 0.05 m
𝑚
Velocity 𝑢ₘₐₓ = 2.5
𝑠
Velocity at 30 mm from the center = 2.2 m/s
Wall shearing stress, τ₀:
𝑢ₘₐₓ−𝑢 𝑅
= 5.75 log10 (𝑌 )
𝑢𝑓
Where, 𝑢ₘₐₓ= maximum velocity in pipe
𝑢 = velocity at some distance from pipe
𝑢𝑓 = shear frictional velocity
R = radius of pipe
Y= distance from the pipe
Now, from relation:
τ₀
𝑢𝑓 =√ 𝜌
Where, τ₀ = shearing stress
ρ = density of fluid
Example 11.11: In a pipe of diameter 300 mm the
center-line velocity and the velocity at a point 100
mm from the center, as measured by pitot tube, are
2.4 m/s and 2.0 m/s respectively. Assuming the flow
in the pipe to be turbulent, find:
(i) Discharge through the pipe,
(ii) Co-efficient of friction, and
(iii) Height of roughness projections.
Solution: Diameter of pipe, D = 300 mm = 0.3 m
Radius of pipe R= 300/2 = 150 mm = 0.15 m
Center-line velocity, 𝑢ₘₐₓ= 2.4 m/s
(i) Discharge through the pipe, Q:
𝑢ₘₐₓ−𝑢 𝑅
= 5.75 log10 (𝑌 )
𝑢𝑓
Where, 𝑢ₘₐₓ= maximum velocity in pipe
𝑢 = velocity at some distance from pipe
𝑢𝑓 = shear frictional velocity
R = radius of pipe
Y= distance from the pipe
Using equation
𝑢−Ū 𝑅
= 5.75 log10 ( ) + 3.75
𝑢𝑓 𝑌
Where, Ū = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
(ii) Co-efficient of friction, 𝑓:
𝑓
𝑢𝑓 = Ū√
2
Where, 𝑓 = 𝐶𝑜 − 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛
(iii) Height of roughness friction, k:
1 𝑅
As we know, = 2.0 log10 𝑘 + 1.74
√4𝑓
Where 𝑘 = 𝐻𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑟𝑜𝑢𝑔ℎ𝑛𝑒𝑠𝑠 𝑝𝑟𝑜𝑗𝑒𝑒𝑐𝑡𝑖𝑜𝑛
