FLUIDS MECHANICS (a letter to a friend) UPDATED+

For flow through noncircular pipes, the Reynolds number is based on the hydraulic
diameter Dh defined as

Dh

UNIFORM AND NON UNIFORM FLOW

1.Uniform flow
Occurs when flow characteristics do not change from point to point Flow
characteristics include velocity, discharges, cross section etc

WHERE C is flow characteristics & S is path

2.Non uniform flow
Occurs when flow characteristics change from point to point Most flow
are non-uniform because

C is flow characteristics & S is path

STEADY AND UN-STEADY FLOWS

3.Steady flow
Steady flow-flow characteristics at a given point do not change with
time…

4.Unsteady flow
Unsteady flow –flow characteristics at given point change with time

Musadoto felician Deus

 Let’s say we’re not dealing with a system open to the atmosphere
(e.g., a pipe vs. a pond).There’s no storage potential, so Q1 = Q2, a
mass balance equation. For essentially incompressible fluids such as
water, the equation becomes
V1A1 = V2A2,; where V = velocity (m/s) and A = area (m2)
This Can be used to estimate flow velocity along a pipe, especially
where constrictions are concerned.

Self-check bro Deus!

Solve the following self-check question using the above concept

(answer V2 = 5m/s)

QN If one end of a pipe has a diameter of 0.1 m and a flow rate of 0.05
m/s, what will be the flow velocity at a constriction in the other end
having a diameter of 0.01 m? (hints: V1A1 = V2A2)

PIPE NETWORKS

‘Pipe flow’ generally refers to fluid in pipes and appurtenances
flowing full and under pressure .

Examples are Water distribution in homes, industry, cities; irrigation

SYSTEM COMPONENTS

1.Pipes
2.Valves
3.Bends
4.Pumps and turbines
5.Storage (often unpressurized, in reservoirs, tanks, etc.)

Musadoto felician Deus

ENERGY RELATIONSHIPS IN PIPE SYSTEMS

Energy equation between any two points:

Analysis involves writing expressions for hL in each pipe and for each
link between pipes (valves, expansions, contractions), relating
velocities based on continuity equation, and solving subject to system
constraints (Q, p, or V at specific points).

MAJOR LOSES IN PIPE SYSTEMS

Source:

1.Frictions due to pipe material
2.Length of the system
3.Diameter/Cross-sectional Area

ENERGY LOSSES IN PIPING SYSTEMS

A quantity of interest in the analysis of pipe flow is the pressure drop
∆P since it is directly related to the power requirements of the fan or
pump to maintain flow. We note that dP/dx = constant, and integrating
from x = x1 where the pressure is P1 to x = x1 + L where the pressure is P2
gives

Musadoto felician Deus

………………………….eqn1
Substituting Eqn1 into the Vavg expression (see below) the pressure drop
can be expressed as

Laminar flow

The symbol ∆ (is typically used to indicate the difference between the
final and initial values, like (∆y = y2 - y1). But in fluid flow, ∆P is
used to designate pressure drop, and thus it is P1 - P2. A pressure drop
due to viscous effects represents an irreversible pressure loss, and it
is called PRESSURE LOSS
∆PL to emphasize that it is a loss (just like the head loss hL, which is
proportional to it).

NOTE that, from the ∆P equation above the pressure drop is proportional
to the viscosity of the fluid, and ∆P would be zero if there were no
friction. Therefore, the drop of pressure from P1 to P2 in this case is
due entirely to viscous effects, and ∆P Eqn above represents the pressure
loss ∆PL when a fluid of viscosity flows through a pipe of constant
diameter D and length L at average velocity Vavg.
In practice, it is found convenient to express the pressure loss for all
types of fully developed internal flows (laminar or turbulent flows,
circular or noncircular pipes, smooth or rough surfaces, horizontal or
inclined pipes)

PRESSURE LOSS:

Where
is the dynamic pressure and f is the Darcy friction factor

It is also called the Darcy–Weisbach friction factor, named after the
Frenchman Henry Darcy (1803–1858) and the German Julius Weisbach (1806–
1871),the two engineers who provided the greatest contribution in its
development. It should not be confused with the friction coefficient Cf
[also called the Fanning friction factor, named after the American
engineer John Fanning (1837–1911)], which is defined as

Cf = = f /4.

Setting P2 and ∆P eqns equal to each other and solving for f gives the
friction factor for fully developed laminar flow in a circular pipe,

Musadoto felician Deus

Circular pipe, laminar

This equation shows that in laminar flow, the friction factor is a
function of the Reynolds number only and is independent of the roughness
of the pipe surface.

In the analysis of piping systems, pressure losses are commonly expressed
in terms of the equivalent fluid column height, called the HEAD LOSS hL.
Noting from fluid statics that ∆P = gh and thus a pressure difference of
∆P corresponds to a fluid height of h =∆P/ g, the pipe head loss is
obtained by dividing ∆PL by g to give

OR

Where

hL = Head loss due to friction, m [ft]
f = Moody friction factor
L = Pipe length, m [ft]
V = Velocity, m/s [ft/sec]
g = Gravitational acceleration, 9.81 m/sec2 [32.2 ft/sec2]
D = Inside diameter, m [ft]

IMPORTANT bro!

THE MOODY CHART

The friction factor in fully developed turbulent pipe flow depends on the Reynolds
number and the relative roughness  /D. which is the ratio of the mean height of
roughness of the pipe to the pipe diameter. The functional form of this dependence
cannot be obtained from a theoretical analysis, and all available results are
obtained from painstaking experiments using artificially roughened surfaces (usually
by gluing sand grains of a known size on the inner surfaces of the pipes). Most such
experiments were conducted by Prandtl’s student J. Nikuradse in 1933, followed by
the works of others. The friction factor was calculated from the measurements of the
flow rate and the pressure drop.

The experimental results obtained are presented in tabular, graphical, and
functional forms obtained by curve-fitting experimental data. In 1939, Cyril F.
Colebrook (1910–1997) combined the available data for transition and turbulent flow
in smooth as well as rough pipes into the following implicit relation known as the
Colebrook equation:

Musadoto felician Deus

We note that the logarithm in Equation above is a base 10 rather than a
natural logarithm. In 1942, the American engineer Hunter Rouse (1906–
1996) verified Colebrook’s equation and produced a graphical plot of f as

a function of Re and the product √ . He also presented the laminar flow
relation and a table of commercial pipe roughness.

Two years later, Lewis F. Moody (1880–1953) redrew Rouse’s diagram into
the form commonly used today. The now famous Moody chart is shown below.
It presents the Darcy friction factor for pipe flow as a function of the
Reynolds number and e/D over a wide range. It is probably one of the most
widely accepted and used charts in engineering. Although it is developed
for circular pipes, it can also be used for noncircular pipes by
replacing the diameter by the hydraulic diameter.

Musadoto felician Deus

OBSERVATIONS FROM THE MOODY CHART

• For laminar flow, the friction factor decreases with increasing
Reynolds number, and it is independent of surface roughness.

• The friction factor is a minimum for a smooth pipe and increases with
roughness. The Colebrook equation in this case ( = 0) reduces to the
Prandtl equation.

• The transition region from the laminar to turbulent regime is
indicated by the shaded area in the Moody chart. At small relative
roughnesses, the friction factor increases in the transition region
and approaches the value for smooth pipes.

• At very large Reynolds numbers (to the right of the dashed line on
the Moody chart) the friction factor curves corresponding to
specified relative roughness curves are nearly horizontal, and thus
the friction factors are independent of the Reynolds number. The flow
in that region is called fully rough turbulent flow or just fully
rough flow because the thickness of the viscous sublayer decreases
with increasing Reynolds number, and it becomes so thin that it is
negligibly small compared to the surface roughness height. The
Colebrook equation in the fully rough zone reduces to the von Kármán
equation.

Note Bro!

At very large Reynolds numbers, the friction factor curves on the
Moody chart are nearly horizontal, and thus the friction factors are
independent of the Reynolds number.

IMPORTANT NOTE BRO!

In calculations, we should make sure that we use the actual internal
diameter of the pipe, which may be different than the nominal diameter.

Musadoto felician Deus

TYPES OF FLUID FLOW PROBLEMS

1.Determining the pressure drop (or head loss) when the pipe length
and diameter are given for a specified flow rate (or velocity)

2.Determining the flow rate when the pipe length and diameter are
given for a specified pressure drop (or head loss)

3.Determining the pipe diameter when the pipe length and flow rate
are given for a specified pressure drop (or head loss)

To avoid tedious iterations in head loss, flow rate, and diameter
calculations, these explicit relations that are accurate to
within 2 percent of the Moody chart may be used.

Musadoto felician Deus

QUIZ 1

Compare the velocity and pressure heads for typical
conditions in a street main: V = 1.5 m/s; D = 0.5 m;
p = 500 kPa

(hints)

 p  500 kPa  1000 N/m2 kPa  51.0 m  V 2  1.5 m/s2  0.115 m

 9800 N/m3 2g 2 9.8 m/s2

Answer

If f = 0.02, hL for each 0.5 m of pipe is 2% of the
velocity head, or 0.0023 m, corresponding to 0.0045%
of the pressure head.

QUIZ 2

A 20-in-diameter galvanized pipe (e = 0.0005 ft) 2 miles long carries 4
cfs at 60oF. Find hL using (a) the Moody diagram and (b) the Colebrook
eqn.(use the mood chart in above notes)

a)

B) Colebrook equation

1  2 log  D  2.71 
f  Re f 
3.7

Musadoto felician Deus

TYPICAL PIPE FLOW PROBLEM >>(repeated, unique anyway)

• Type II: Pipe properties (e, D, l) and hL known, find V.
• Guess V, determine f and hL as in Type I, iterate until hL equals

known value, or
• Solve Colebrook and DW eqns simultaneously to eliminate V, yielding:

SOLVING TYPE II PIPE PROBLEMS
Iterative Approach

Musadoto felician Deus

Rearranged D-W eqn:

V  2 2gDhL log  D  2.51 l
l  3.7 D 2gDhL 

Example

For the pipe analyzed in the preceding QUIZ 2, what is the
largest flow rate allowable if the total frictional head loss
must remain <8 ft?

SOLUTION

V  2 2gDhL log  D  2.51 l
l  2gDhL 
3.7 D

Substituting known values,

Type III.

ANOTHER SIMPLE QUIZ bro! DEUS

QN What diameter galvanized pipe would be required in the preceding
QUIZ 2 if a flow rate of 10 cfs was needed, while keeping the total
frictional headloss at <8 ft?

(HINTS SOLUTION)

Musadoto felician Deus

DEPENDENCE OF HL ON D AND V
1. In laminar region:

2. In turbulent region, when f becomes constant

Under typical water distribution conditions, hL in a given
pipe can be expressed as kQn with n slightly <2.

QUIZ 2 CONTINUES……..
For the systems analyzed in the QUIZ 2, what value of n
causes the data to fit the equation hL = kQn?

Musadoto felician Deus

ALTERNATIVE EQUATIONS FOR FLOW
Headloss Relationships in Turbulent Pipe Flow
 Hazen-Williams equation – widely used for hL as function
of flow parameters for turbulent flow at typical
velocities in water pipes:

NOTE: Coefficients shown are for SI units; for BG units, replace
0.849 by 1.318 and 10.7 by 4.73.

COMPARISON OF EQUATIONS FOR TRANSITIONAL AND TURBULENT CURVES ON
THE MOODY DIAGRAM

*Coefficients shown are for SI units (V in m/s, and D and Rh in m); for
BG units (ft/s and ft), replace 0.849 by 1.318; 0.354 by 0.550; 0.278 by
0.432; 10.7 by 4.73; 1/n by 1.49/n; 0.397 by 0.592; 0.312 by 0.465; and

10.3 by 4.66.

Musadoto felician Deus

ENERGY LOSSES IN BENDS, VALVES, AND OTHER TRANSITIONS
(‘MINOR LOSSES’)

Minor headlosses generally significant when pipe sections are
short (e.g., household, not pipeline) Caused by turbulence
associated with flow transition; therefore, mitigated by
modifications that ‘smooth’ flow patterns Generally much greater
for expansions than for contractions.

Often expressed as multiple of velocity head:

Where, K is the ratio of energy lost via friction in the device
of interest to the kinetic energy of the water (upstream or
downstream, depending on geometric details)

ENERGY LOSSES IN CONTRACTIONS

Note bro!:
all pictures by Fluid
Mechanics With Engineering
Applications10th Edition
By E. Finnemore and Joseph
Franzini Copyright: 2002

Musadoto felician Deus

ENERGY LOSSES IN EXPANSIONS

ENERGY LOSSES IN PIPE
FITTINGS AND BENDS

Musadoto felician Deus

Bro! Deus….Use the idea above to solve following the
following EXAMPLE

A 5-in-diameter pipe with an estimated f of 0.033 is 110
feet long and connects two reservoirs whose surface
elevations differ by 12 feet. The pipe entrance is
flushed, and the discharge is submerged.

a. Compute the flow rate.
b. How much would the flow rate change if the last 10

ft of the pipe were replaced with a smooth conical
diffuser with a cone angle of 10o?

Any idea bro? about the question

SOLUTION

Musadoto felician Deus

From graph, for a smooth, 10o cone, kcone = 0.175

BRO! if your following, Mr Materu’s slides ‘I’ end here.

Musadoto felician Deus

See questions next pages……………..

REVISION QUESTIONS

1. A mountain lake has a maximum depth of 40m, the barometric pressure
at the surface is 598mm Hg. Determine the absolute pressure (in
Pascal) at the deepest part of the lake. Given that the density of
water and mercury are 1000kg/m3 and 13558kg/m3 respectively
(Answer=471.9KPa)

2. A manometer is attached to a tank containing three different fluids
as shown in figure below, what will be the difference in elevation
of mercury column in the manometer (y)? (Answer =0.626m)

3. Determine the new differential reading along the inclined leg of
the mercury manometer of Figure below, if the pressure in pipe A
is decreased 10 kPa and the pressure in pipe B remains unchanged.
The fluid in A has a specific gravity of 0.9 and the fluid in B is
water. (Answer=0.212m)

4. Viscosity of liquid decrease with increase in temperature while for
a gas the viscosity increase with the increase in temperature.
Explain why???

HINTS bro! to use

1.Molecular structure
2.Cohesive force
3.Momentum exchange (gas)
5. What is the viscous force of the fluid on a 30.48m length pipe of
0.305m diameter if shear stress is 0.0262N/m2?
See hints below

Musadoto felician Deus

6. Determine the Torque and power required to run a 300 mm diameter
shaft at 400 rpm in journals with uniform oil thickness of 1 mm.
Two bearings of 300 mm width are used to support the shaft. The
dynamic viscosity of oil is 0.03N-s/m2
(Answer Torque = 15.995Nm and Power =670W)

BRO! (KIM DE FEL) ATTACK THE FOLLOWING ADDITIONAL QUESTIONS

Question 1
a)Find the capillary rise in the tube shown in figure F1 below, the
air- water-glass interface ( o  0 ) and tube radius is 1mm at 20ºC
temperature. Given that the surface tension of water at 20ºC is
0.0728N/m (Answer 14.8mm)

b)Assume the liquid is mercury with air-mercury-glass interface is 130º
and density of mercury is 13.570kg/m3 while surface tension of
mercury is 0.514N/m, calculate capillary rise, explain why the
results is negative and draw the figure to represent the result
(Answer -5mm)

Figure F1

Question 2
a)A liquid compressed in a cylinder has a volume of 1000cm3 at 1MN/m2
and volume of 995cm3 at 2MN/m2. What is bulk modulus of elasticity
(answer 200MPa)
b)The glass tube in figure F2 is used to measure Pressure (P1) in the
water tank, the tube diameter is 1mm and water surface tension is
0.0712N/m. the tube is reading 17cm of height, what is the true
height of water after correcting the effect of surface tension
(density of water is 1000kg/m3) [answer 2.9cm]

Figure F2

Musadoto felician Deus

Question 3

Water flows into larger tank as shown in figure F3 at the rate of
0.11m3/s, the water leaves the tank through 20 holes in the bottom of the

tank, each hole has a diameter of 10mm. Determine equilibrium height (h)
for the steady state operation (answer 2.50m)

Figure F3

Question 4
The volume of fluid is found to be 0.00015m3, if the specific gravity of
this fluid is 2.6. Calculate the weight of fluid (Answer 3.82N)

Question 5
If the specific weight of a substance is 8.2KN/m3, calculate its density
[Answer: 836kg/m3]

Question 6
A vertical cylindrical tank with a diameter of 12m and depth of 4m is
filled to the top with water at 20°C. If water is heated to 50°C, how
much water will spill over? Given that density of water at 20°C and 50°C
is 999kg/m3 and 989kg/m3 respectively. [Answer 4.6m3]

Question 7
If bulk modulus of elasticity for water is 2.2GPa, what pressure is
required to reduce a volume of water by 0.6% [Answer: 13.2MPa]

Question 8
The mercury manometer of Figure indicates a differential reading of 0.30
m when the pressure in pipe A is 30-mm Hg vacuum. Determine the pressure
in pipe B. (ANS; 4171.28Pa)

Musadoto felician Deus

Figure

Question 9
For the stationary fluid shown in Figure , the pressure at point B is 20
kPa greater than at point A. Determine the specific weight of the
manometer fluid (Answer = 7100N/m3)

Figure

Question 10
Water flows through the pipe contraction as shown in Figure . For the
given 0.2m difference in manometer level, determine the flow rate as a
function of the diameter of the small pipe, D. [Answer Q=1.56D2 where Q
(m3/s) and D (m)]

Figure

Musadoto felician Deus

Question 11
The fluid in figure M8 is water, determine the manometer reading (h)
[Answer=0.37m)]

Figure
Question 11
Determine the elevation difference h between the water levels in the two
open tanks shown in figure M9 [Answer = 0.040m]

Figure

Question 12
Determine the flowrate through the pipe shown in figure below
(Answer = 0.0111m3/s)

Figure

Musadoto felician Deus

Question 13
A swimming pool is 18 m long and 7m wide, determine the magnitude and
location of the resultant force of the water on the vertical end of the
pool where depth is 2.5m (Answer, FR= 214KN, YR= 1.67)

Question 14
The vertical cross section of closed storage tank as shown in figure
contains ethyl alcohol, the air pressure is 40KPa. Determine the
magnitude of the resultant fluid force acting on one end of the tank.
Given that specific weight of ethyl alcohol is 7.74 KN/m3
(Answer=FR=847KN)

Figure1

Question 15
The rectangular gate CD of figure Q2 is 1.8m wide and 2m long, assuming
the material of the gate to be homogenous and neglecting friction at the
hinge C, determine the weight of the gate necessary to keep the gate
shunt until the water level rises to 2m above the hinge (Answer = 180KN)

Figure

Musadoto felician Deus

Question 16
A 4m long curved gate is located in the side of the reservoir containing
water as shown in figure Q3 below. Determine the magnitude of the
horizontal, vertical and resultant forces of water on the gate. Will
resultant force pass through point O? Explain! (Answer, FH=882KN,
FV=983.67KN, FR=1321.8KN)

Figure
Question 17
The rigid gate OAB of figure G1 below is hinged at O and rests against a
rigid support at B. What minimum horizontal force P is required to hold
the gate closed if its widht is 3m? Neglet the weight of the gate and
friction in the hinge, the back of the gate exposed to the atmosphere.
(Answer 436KN)

Figure
Question 18
A dam of 20m long retain 7m of water as shown in figure below, find the
total resultant force acting on the dam and location of the centre of
pressure. Given that the angle between water and dam at the surface is
60° (Answer 5550.6KN, Centre of pressure is 4.667m below the surface)

Musadoto felician Deus

Question 19
An inclined circular gate with water on one side shown in figure G2
below, Determine the total resultant force acting on the gate and
location of the Centre of pressure (Zcp). Assume specific weight of water
at 20°C is 9.79KN/m3 (Answer Resultant force =14.86KN and Zcp =2.26m)

Figure
Question 20
The 4-m-diameter circular gate in figure G3 below is located in the
inclined wall of a large reservoir containing water. The gate is mounted
on a shaft along its horizontal diameter. For a water depth of 10 m above
the shaft determine

(a) The magnitude of the resultant force exerted on the gate by water
(Answer= 1.23MN)

(b) Location in y-axis (YR) of the resultant force (Answer 11.6m)

Musadoto felician Deus

Figure
Question 21
Consider figure G4 below, if atmospheric pressure is 101.03KPa and
absolute Pressure at the bottom of the tank is 231.3KPa. What is the
density of olive Provided that the density of mercury, water and SAE-30
oil are 13570kg/m3, 1000kg/m3 and 800kg/m3 respectively?

Figure
Question 22
Determine the pressure heads at A and B in meter of water in figure G5
below and explain your answer (Answer HA= -2.38m H2O and HB= -0.51m H2O)

Figure

Musadoto felician Deus

Question 23
A large open tank contains a layer of oil floating on water as shown in
figure M2, the flow is steady and fluid is incompressible, Determine
a)Height, h, to which the water will rise
b)Water velocity in the pipe
c)Pressure in the horizontal pipe.

Figure
Question 24
Determine the flow rate through the pipe in M3 below (Answer = 0.0111m3/s)

Figure
Question 25
The specific gravity of the manometer fluid shown in figure , determine
the volume flowrate, Q, if the flow is incompressible and the flowing
fluid is water (density of water is 1000kg/m3)

Musadoto Figure

felician Deus

Question 26
Determine the elevation difference hbetween the water levels in the two
open tanks shown in figure F6 [Answer = 0.040m]

Figure
BRO! ARE YOU TIRED IS YES STOP HERE FOR TODAY? IF YES SEE MY GIFT TO YOU

SAMPLE QUESTIONS ON FLUID FLOW IN CLOSED CONDUITS

Question 1

Lubricating Oil at a velocity of 1 m/s (average) flows through a pipe of

100 mm; determine whether the flow is laminar or turbulent. Also

determine the friction factor (f) and the pressure drop over 10 m length
in Pa or N/m2. Given that Density = 930 kg/m3and Dynamic viscosity μ= 0.1
Ns/m2

Solution

Given

Velocity (v) =1m/s

Pipe diameter (D) =100mm = 0.1m

Pipe length = 10m
Density = 930kg/m3
Dynamic viscosity (µ) = 0.1Ns/m2

Type of flow: Re  vD  930 *1* 0.1  930
 0.1

Since Re<2000, then the flow is laminar

Friction factor: since the flow is laminar, then f  64  64  0.06882

Re 930

Pressure drop:

hf  fLV 2  0.06882*10*12  0.351m
2Dg 2 * 0.1* 9.81

P   gh  930*9.81*0.351

P  3200Pa

Musadoto felician Deus

Question 2

Use mood diagram to find friction factor for the following data: Diameter
of the pipe =0.305m, Kinematic viscosity of the fluid = 1.3 x 10-6,

Velocity =0.043m/s and Internal pipe roughness =0.00061m

Solution

Relative roughness (e/D)  0.00061  0.002

0.305

Reynolds number  DV  0.305 * 0.043  10000
v 4.3 106

From moody diagram, f = 0.034 (check!! IT BRO! DEUS F)

QUESTION 3

Calculate the energy head loss due to friction in a pipe of length 1000m,

diameter of 0.25m and roughness of 0.0005m given that the fluid of
kinematic viscosity of 1.306x10-6m2/s flow in the pipe at the rate of
0.051m3/s.

Solution

Given

Pipe length =1000m

Diameter = 0.25m

Roughness = 0.0005m
Kinematic viscosity= 1.306x10-6m2/s
Flow rate = 0.051m3/s

Friction head loss

hf  fLV 2            (a)
2Dg

 Velocity

(V )  Q  4Q
A  D2

(V )  4 * 0.051  1.039m / s
 *(0.25)2

Friction factor

 Re  DV  0.25 *1.039  200000
v 1.036 106

 e D  0.0005  0.002
0.25

From moody diagram, f=0.0245 (check!! BRO!)

Head loss (from equation a)

hf  fLV 2  0.0245*1000*(1.039)2
2Dg 2 * 0.25 * 9.81

hf  5.39m

Hence head loss due to friction is 5.39m

Musadoto felician Deus

Question 4
Oil (specific weight = 8900N/m3 & viscosity = 0.10Ns/m2 flows through
horizontal 23mm diameter tube as shown in figure Q1 below, a differential
U-tube manometer is used to measure the pressure drop along the tube,
determine the range of value for h for laminar flow (Answer h<0.509m)

Figure
Question 5
Oil of SG = 0.87 and kinematic viscosity (v) = 2.2 x10-4m2/s flows through
a vertical pipe as shown in figure at flow rate of 4x10-4m3/s. Determine
the manometer reading h, also determine the magnitude and direction of
flow rate which will cause h to be zero. (Answer h=18.5m and Q=0)

Figure

Question 6
Oil with density of 900 kg/m3 and kinematic viscosity of 0.0002 m2/s
flows upward through an inclined pipe as shown in figure Q3. The pressure
and elevation are known at sections 1 and 2, 10 m apart. Assuming the
flow is steady laminar, calculate

a)Friction head loss between (hf)point 1 and 2
b)Discharge (Q)
c)Velocity (V)
d)Reynolds number (Re)
[Answers: hf=4.9m, Q=0.0076m3/s, V=2.7m/s and Re=810]

Musadoto felician Deus

Figure

Question 7
From figure Q4 below, find the diameter of the pipe which connects two
reservoirs given that the length of the pipe is 304.8m, the flow is
0.013m3/s, roughness (e) is 0.001m and kinematic viscosity is 1.31x10-
6m2/s (Answer=0.15m)

Figure
Solution
This is the kind of engineering problem that you are suppose to find the
diameter of pipe to accommodate a given fluid flow with other available
information. Find everything in term of diameter

Musadoto felician Deus

 Relative roughness in term of D

e  0.001
DD
 Velocity in term of D

V  Q  4Q
A
 D2

V  4 * 0.013  0.017 D2

 D2

 Reynolds number in term of D

 Re  VD 
v
0.017 *D
D2

1.31106

Re  12977
D

 Head loss in term of D

hL  fLV 2
2Dg

15.2  fLV 2
2Dg

 15.2  
f 304.8* 0.017 
D2

2 *9.81* D

 f

15.2 
0.088
D4

19.62D

1

D  0.295 f 5

f can be solved by trial and error, try different values of f until the

trial value of f converges with f values of moody chart as shown in table

below

f-trial D e/D Re f-chart Remarks

0.025 0.0141 0.007 92035 0.034 Try again

0.034 0.15 0.0067 86513 0.034 Converges

Since the f-trial converges with f from moody chart, then the diameter of
the pipe is 0.15m

Question 8
Oil with density of 900 kg/m3 and kinematic viscosity of 0.00001m2/s,
flows at 0.2 m3/s through 500m of 200-mm diameter cast-iron pipe.

Determine (a) the head loss and (b) the pressure drop provided that the

pipe slopes down at 10° in the flow direction and caste iron pipe has

roughness of 0.26mm (Answers: hf=117 and change in pressure =265KPa)

Question 9
Oil with density of 950kg/m3 and kinematic viscosity of 0.00002m2/s, flow

through 30cm diameter pipe 100m long with a head loss of 8m. The relative

roughness (e/D) of the pipe is 0.00002. Find the average velocity and
flow rate (Answer V=4.84m/s and Q=0.342m3/s)

Musadoto felician Deus

Solution
This type of problems requires iteration process or computer software
(solvers) because velocity (or flow rate) appears in both the ordinate
and the abscissa on the Moody chart, iteration for turbulent flow is
nevertheless quite fast, because f varies so slowly with Re.
First solve V in term of f

hf  fLV 2
2Dg

f  hf 2Dg
LV 2

f  8* 2*0.3*9.81
100V 2

fV 2  0.471

V  0.471
f

To get started, you only need to guess f, compute V from equation above,
then get Re (Re=VD/v), compute a better f from the Moody chart and
repeat. The process converges fairly rapidly. A good first guess is to
assume the flow is ‚fully rough‛ e.g. the value of f for (e/D) 0.0002 is
f =0.014. Calculation can be done as shown in table below

f-trial Velocity (V) Re (VD/v) f-chart Remarks

0.014 5.80 87000 0.0195 Try again

0.0195 4.91 73700 0.0201 Try again

0.0201 4.84 72600 0.0201 Converges

Since the f-trial converges with f from moody chart, then the velocity of
flow is 4.84m/s

Flow rate

Flow rate (Q)
Q=AV
Q=  D2V

4
Q   *(0.3)2 * 4.84  0.342m3 / s

4

Hence the flow is 0.342m3/s

Question 10
Oil with density of 950kg/m3 and kinematic viscosity of 0.00002m2/s, flow
through a pipe of unknown diameter, the length of pipe is 100m long with
a head loss of 8m. The relative roughness (e/D) of the pipe is 0.00002
and the flow is 0.342m3/s. calculate the pipe diameter (Answer: 30cm)

Musadoto felician Deus

Question 11
Consider figure Q5, the total pressure drop PA-PB = 150,000 Pa, and the
elevation drop ZA- ZB=5 m. The pipe data are shown in table 1 below. The
fluid is water, with density of 1000kg/m3 and kinematic viscosity of
1.02x10-6m2/s. Calculate the flow rate Q in m3/h through the system.

Figure
Table : Pipes data

REST FOR A WHILE BRO KIM DE FEL BEFORE TAKING
YOU TO PUMPS

Musadoto felician Deus

PUMP

PUMP is machine used to lift fluid from one point to another.It converts
mechanical energy into hydraulic energy. Pump may need to lift liquids
from some heights below the pump and push them to some height above the
pump.

PUMP CLASSIFICATION
 Pumps are classified according to its working principles.
 Pumps are classified into three main groups:

Positive displacement pumps (Static pumps)
Kinetic pumps
Electric pumps (electromagnetic pumps)
Basic groups are sub-divided more into several types.

Musadoto felician Deus

POSITIVE DISPLACEMENT PUMPS (STATIC TYPE)
Fluid is pumped into & out of a chamber by changing the volume of the
chamber. Pressures & works done are a result of static forces rather
than dynamic effects. Common examples are tire pump, human heart,
gear pump etc

KINETIC PUMPS
It involve a collection of blades, buckets, flow channels around
an axis of rotation to form a rotor. It is dominated by dynamic
force. Rotor’s rotation produces a dynamic effect that adds energy
to the fluid. A good example of kinetic pump is the one that used
in deep wells (i.e. centrifugal pumps). Depending on direction of
the fluid motion relative to the rotor’s, kinetic pump classified
into axial-flow, mixed-flow & radial-flow

1. Radial flow pump
It involves a substantial radial-flow component at the rotor
inlet or exit. Dominated by the action of centrifugal forces

Musadoto felician Deus

2. AXIAL FLOW PUMPS
The pumped fluid maintains a significant axial-flow direction
from the inlet to outlet. The flow entering the pump inlet
axially and discharge nearly axially. These pumps has low head
(H) but with larger discharges (Q)

CENTRIFUGAL PUMPS
Is the radial flow pump. Has two main components which are an
impeller & stationary casing (housing).As the impeller rotates,
fluid is sucked in through the hub (eye)

Musadoto felician Deus

The casing shape (increase in area in direction of flow) is
designed to reduce the velocity so as increase in pressure.

PUMP CHARACTERISTICS

There is theoretical head (hi) –provided by manufactures. There is
actual head (ha) – field performance

This is because of Losses (hL)

hL=(friction losses, minor losses, other losses)

ACTUAL HEAD (SHOULD BE USED FOR DESIGN)

 Actual head should be obtained for design. Engineer should find
it from available information e.g. flow rate. Sometimes pumps
characteristics are obtained by experiments

Musadoto felician Deus

Energy at the inlet is less than the energy at exit, the difference
is energy added by pump/energy gained by fluid (ha)

This (ha) is the net head rise, Can be converted to Power gained by
fluid by (in Watts) or (in horse power)
In watts

In horsepower:

CHARACTERISTIC PERFORMANCE CURVE

Musadoto felician Deus

Is the graph which gives the information about the
characteristics of pump

The characteristics includes:-

a.Discharge
b.Efficiency
c.Pressure requirements (head)
d.Power consumption etc.

All are important information to engineer. All characteristics are
drawn Vs flow rate.

DISCHARGE (Q)

Is the volume of liquid pumped per unit time. Head curve raises
as the flow rate (Q) decrease. To lift water up to higher
building the flow rate should be small

ACTUAL HEAD

Is the net work done on a unit weight of water –done by pump Head
at zero discharge is called the shutoff head No flow.Efficiency is
zero.

OPERATING POINT OF PUMP

Is the point which gives the head and flow rate that satisfies both
system equation and pump equation. Can be obtained by plotting both
curves on the same graph. The intersection point is operating point

STOP HERE AND PRACTISE THE FOLLOWING EXAMPLE BRO!

QN Water is pumped from a deep well to students hostel’s tanks, the
centrifugal pump used for pumping has characteristics which is given by
equation H = 22.9+10.7Q-111Q3 ,but one Agricultural engineering students
tried to find the system demand and she obtained the equation
H = 15 +85Q2 .Where Q is the flow in m3/s and H is head in m. Determine
the operating point of the pump (Flow rate and head)

[Answer: Q=0.23 and H=19.49]

POWER

Is the rate at which work is done on a liquid by pump. Is when a
unit volume of liquid is raised through a given height.

Musadoto felician Deus

Is power required by water to be lifted. As the discharge increase
then the power requirement also increases
Efficiency
1.Pumps receives power from motors through shaft (BP)
2.Motor receives power from electricity (LP)
3.Pump develops power into fluid (fluid required power)

Neither motor (M) nor Pump (P) operates in 100% efficiency.
Efficiency of the motor (ηm)

Efficiency of the Pump (ηp)

SIMPLE QUIZ BRO! DEUS

QN The pump is used to increase the pressure of water flow rate of
0.2m3/s from 200KPa to 600KPa. If the overall efficiency of the
pump is 85%, how much electrical power is required to pump the
water? The suction tank is 10cm below the centre line of the pump
and delivery tank is 10cm above the centerline of the pump. Assume
the inlet and exit diameters are equal and velocities at suction
and delivery can be neglected. Also minor and friction losses can
be neglected. [Answer P=187KW]

The overall pump efficiency is affected by the

1.Hydraulic losses in the pump due to friction loss and
minor losses..

2.Mechanical losses in the bearings and seals
3.Volumetric losses due to leakage of the fluid.

Efficiency is important characteristics of pump performance.

Musadoto felician Deus

NET POSITIVE SUCTION HEAD (NPSH)
Is expression of the suction capability of the pump. It is used to
calculate the inlet pressure needed at the pump to avoid
cavitation. Inlet pressure must be equal or higher than designed
(Requirements)

There are two types of NPSH
1.NPSHR-(Required NPSH)

 Is required suction head of the chosen pump
 It is given by the manufacturer
2.NPSHA-(Available NPSH)
This is true (available) suction head at the pump location Can be
estimated mathematically or experimentally. NPSHA must be greater
than NPSHR –otherwise it may cause cavitation.
Consider the figure below.

Total head at the suction side

Liquid vapour pressure

To avoid cavitation

Musadoto felician Deus

NOTE THE FOLLOWING,BRO! from the above ,last expression.

1.If z1 is increased, the NPSHA is decreased
2.There is some critical value of Z1 which the pump cannot

operate without cavitation
3.What is the effect to NPSHA if source is above the pump?

Bro! Use the above idea to solve the following example

QN The pump is installed to pump water from Mazimbu well as shown
in figure F1 below, determine the critical elevation (Z1) where the
pump can be situated above water surface of suction without
experiencing cavitation. Given that the diameter of the pump is
240m, pumping rate is 250m3/hr and NPSH value for discharge is
7.4m. Use atmospheric pressure of 101Kpa and Vapour pressure of
1666Pa. [Answer = 2.72m]

These are the Laws which gives the relations between the following…
i. Volumetric flow rate (Q)

ii. Head (H)
iii. Power requirements

iv. Diameter (D)
v. Shaft rotation speed (N)

Musadoto felician Deus

There are two affinity laws.
Affinity law 1
In this case, the point of interest is to investigate how change in
operating speed (N) affects pumps characteristics like

1.Discharge (Q)
2.Head (H)
3.Required Power (P)
Discharge versus rotational speed

Head versus rotational speed

Power versus rotational speed

Affinity law 2
In this case, the point of interest is to investigate how change
in diameter (D) affects pumps characteristics like…
1.Discharge (Q)
2.Head (H)
3.Required Power (P)
Test your IQ bro!
Is it possible to increase impeller diameter?
Discharge versus Impeller diameter

Head versus Impeller diameter

Power versus Impeller diameter

Musadoto felician Deus

PUMP INSTALLATION

Sometime one pump is not enough to meet discharge demand /Head
required. Hence two or more pumps may be connected in series or
parallel. The choice of connection depend on weather you want to
increase discharge or head.

PUMPS IN SERIES

 The discharge form first pump is piped into the inlet side of
the second pump.

 Each pump adding more energy to the fluid.
 Only head is increased.
 Discharge remains the same.

PUMPS IN SERIES

 Applicable in deep wells pumping or higher building water
pumping….

 The combined head (total head) is equal to the sum of
individual heads…

Musadoto felician Deus

PUMPS IN PARALLEL
Two or more pumps draw water from the sources and individual flows
are discharged into a single pipeline. Pumps in parallel operates
approx. in the same head. The total discharge is equal to the sum
of individual discharge

QUIZLET

QN The pump of Figure below is to increase the pressure of 0.2 m3
/s of water from 200 kPa to 600 kPa. If the pump is 85%
efficient, how much electrical power will the pump require? The
exit area is 20 cm above the inlet area. Assume inlet and exit
areas are equal.

Do you like 3D view answers? Hahahaha! See below

Try the following important question

1.A water pump has one inlet and two outlets as shown in Figure below,
all at the same elevation. What pump power is required if the pump is
85% efficient? Neglect pipe losses.

Musadoto felician Deus

2.A vehicle with a mass of 5000 kg is traveling at 900 km/h. It is
decelerated by lowering a 20-cm wide scoop into water a depth of 6 cm
(Figure below). If the water is deflected through 180°, calculate the
distance the vehicle must travel for the speed to be reduced to 100
km/h.

3.A 1-kW motor drives the rotor shown in Figure below at 500 rad/s.
Determine the flow rate neglecting all losses. Use = 1.23 kg/m3.

4.Air enters the centrifugal-type air pump of a leaf blower through the
blue area shown in Figure below. The 10-cm-diameter 1.2-m-long tube
has an attached nozzle with a 30-cm2 exit area.The exit velocity is
240 km/h.
a.Calculate the discharge.
b.If the overall loss coefficient is 1.2, estimate the pump head.
c.What power must the pump supply to the air?
d.If the pump is 65% efficient, what is the required horsepower of
the gasoline engine?
e.Estimate the pressure at the tube entrance (just downstream of
the pump).
f.If the 10-kg blower hangs from a strap, what force must be
applied at the handle located 30 cm above the nozzle? The center
of gravity is 70 cm above and 120 cm to the left of the exit.

Musadoto felician Deus

IMPORTANT QUESTIONS

Question 1

Water is to be pumped from one large, open tank to a second tank
open tank as shown in figure Q1 below. The pipe diameter throughout
is 152.4mm and the total length of the pipe between the pipe
entrance and exit is 61m. Minor loss coefficients for the entrance
(KL1), exit (KL2), and the elbow (KL3) are shown on the figure, and
the friction factor for the pipe can be assumed constant and is
equal to 0.02. Calculate the flow rate in term of head gained by
the fluid from the pump (hp) then draw H-Q curve. If someone
installs a pump with characteristics as shown in figure Q2 (in ft
and gal/min), what can you say about this pump? Do you thing this
pump is good choice?

Figure Q1

Figure Q2

Solution

From energy equation

P1V12  z1  hp  P2  V22  z2  hL
2g  2g


 P1  P2  Patm ,
 V1  V2  0

z1  hp  z2  hL

hp  z2  z1  hL                    (1)

Musadoto felician Deus

But

 hL   Friction losses   Minor losses

 hL  flV 2  KV 2
2Dg 2g

 hL  flV 2  V2 (KL1  KL2  KL3 )
2Dg 2g

Hence

hp  z2  z1   flV 2  V2 (KL1  KL2  
 2Dg 2g KL3 )



hp  z2  z1  V2  fl  (KL1  KL2  KL3 ) 
2g  D 

hp  3.05  2 V2  0.02 * 61  (0.5 1.5 1)
* 9.81  0.152

hp  3.05  0.562V 2

But

V  Q  4Q
A  D2

V 2  3037Q2

Therefore

hP  3.05 1706.8Q2

This equation reveals how much actual head the fluid will need to
gain from the pump to maintain a certain flow rate (flow rate in
term of head gained by fluid)

The H-Q curve

The hp-Q curve can be drawn from the equation above as shown in
figure below, this curve represent the actual system performance,
hence the choice of the pump can be based on this information.

To compare this information with the pump installed or any pump in
the market, then this graph is plotted on the same graph of pump
characteristics, the intersection of the two curves represents the
operating point for the pump and the system. Figure below shows the
system curve and pump characteristics curve intersection, the
corresponding actual head gained equal to 20m

Musadoto felician Deus

Question 2

Water is pumped between 2 reservoirs in a pipeline with the
following characteristics, Length of the pipe is 70m, diameter of
the pipe is 300mm, friction factor of the pipe is 0.025 and total
minor loss coefficient . The radial flow pump characteristics curve
is approximated by the formula where Q is discharge in m3/s and H
is head in m. Determine the discharge and head when Z1-Z1=15
[Answer Q=0.29m3/s and H=22.2m]

Question 3

Determine the specific speed of pump required to deliver water at
flow of 1800l/min with pressure head of 448KPa. Assume the
rotational speed of the pump is 3600rev/min. [0.69]

Question 4

A centrifugal pump is to be placed over a larger open tank and is
to pump water at a rate of 3.5 x 10-3m3/s, at this flow rate the
value of NPSH given by the manufacturer is 4.5m, if the atmospheric
pressure is 101KPa and vapor pressure is 1666Pa, determine the
maximum height that the pump can be located above the water surface
without cavitation. Assume the major head loss between tank and the
pump is due to filter at the pipe inlet having a minor loss
coefficient of 20. Neglect all other losses; the pipe diameter at
the suction side of the pump is 10cm [5.43m]

Question 5

The (NPSH)min for a pump given by manufacturer is 7m. This pump is
being used to pump water from a reservoir at a rate of 0.2832m3/s.
The water level in the reservoir is 1.280m below the pump.
Atmospheric pressure is 98.62KN/m2 and vapor pressure is 2340N/m2.
Assume the head loss in the suction pipe is 1.158m. Determine
whether or not the pump is safe from cavitation effects [NPSHA is
7.40m, hence no problem of cavitation]

Question 6

Determine the elevation at which the 240-mm-diameter pump can be
situated above the water surface of the suction reservoir without
experiencing cavitation. Water at 15°C is being pumped at 250 m3/h.
Neglect losses in the system. Use patm =101 kPa.

Musadoto felician Deus