1. KV
WORKED EXAMPLES
{for energy conversion}
Keith Vaugh BEng (AERO) MEng

2. The flow system used to test a centrifugal
pump at a nominal speed of 1750 rpm is
shown in the figure. The liquid water enters
the pump through a 90° bend at 20 °C after
being drawn from the sump through a hinged
disk foot valve along 7 m long pipe. This fluid
is then transferred along a 250 m long pipe
and passes through a second standard 90°
elbows as shown. The suction and discharge
pipes diameters are 120 mm. Develop an
expression to represent the Pressure at the
Pump and an expression to represent the
required Head at the Pump
PUMP SYSTEM EXAMPLE 2
Pd
Ps
Zd
Zs
Elev. Surface of Water in
Sump 6 m
Elev. Eye of
Impeller 8.5 m
Height
Foot Valve with
Hinge Disk
Sudden
Enlargeme
nt
Standard
90° Elbows
Elev. Surface of
Water in Tank 60 m

3. GIVEN: Pump and piping system
FIND
Conduct an engineering assessment must
develop expressions for the Pressure at the
Pump and an expression to represent the Head
at the Pump, calculate pressure at Pump (eye of
the impeller) the required head at the pump for
the system, the NPSHA, select pumps, examine
the impact ageing of pipe have on the system
flow and plot the System Curves vs. Pump
Curve
ASSUMPTIONS
• Steady and incompressible flow
•Uniform flow at each section
• U1 = U2 = 0, Upipe found from Volumetric flow
• P1 = P2 = Patm
Pd
Ps
Zd
Zs
Elev. Surface of
Water in Tank
Elev. Surface of Water in
Sump
Elev. Eye of Impeller
Height
Foot Valve with
Hinge Disk
Sudden
Enlargement
Standard
90° Elbows

4. GOVERNING EQUATIONS
The energy equation for steady incompressible pipe flow can be written as;
The governing equations given that in represents the inlet and out the outlet of the system
𝑃
𝜌𝑔
+
𝑈2
2𝑔
+ 𝑧
𝑠𝑢𝑐𝑡𝑖𝑜𝑛
=
𝑃
𝜌𝑔
+
𝑈2
2𝑔
+ 𝑧
𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒
+ ℎ𝐿 − 𝐻
Total head loss is the summation of the major and minor losses in the system
ℎ𝐿 = 𝑓
𝐿
𝐷
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
+ ∑𝑓
𝐿𝑒
𝐷
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
+ ∑𝐾
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
𝐻 =
ℎ𝑝
𝑔
and
𝑃
𝜌𝑔
+
𝑈2
2𝑔
+ 𝑧
𝑠𝑢𝑐𝑡𝑖𝑜𝑛
=
𝑃
𝜌𝑔
+
𝑈2
2𝑔
+ 𝑧
𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒
+ 𝑓
𝐿
𝐷
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
+ ∑𝑓
𝐿𝑒
𝐷
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
+ ∑𝐾
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
− 𝐻
𝑓 = −1.8𝑙𝑜𝑔10
𝜖
𝐷
3.7
1.11
+
6.9
𝑅𝑒
−2
Friction factor

5. Develop an expression to represented the total Pressure at the Pump
𝑃𝑖𝑛
𝜌𝑔
+
𝑈𝑖𝑛
2
2𝑔
+ 𝑧𝑖𝑛 =
𝑃𝑝𝑢𝑚𝑝
𝜌𝑔
+
𝑈𝑝𝑢𝑚𝑝
2
2𝑔
+ 𝑧𝑝𝑢𝑚𝑝 + 𝑓
𝐿
𝐷
+
𝐿𝑒
𝐷
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
+ 𝐾
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
Pin = Patm, Uin = 0, Upump = 0 i.e. the fluid is entering the eye of the impeller
𝑧𝑖𝑛 −
𝑃𝑝𝑢𝑚𝑝
𝜌𝑔
+ 𝑧𝑝𝑢𝑚𝑝 = 𝑓
𝐿
𝐷
+
𝐿𝑒
𝐷
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
+ 𝐾
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
𝑧𝑖𝑛 −
𝑃𝑝𝑢𝑚𝑝
𝜌𝑔
− 𝑧𝑝𝑢𝑚𝑝 = 𝑓
𝐿
𝐷
+
𝐿𝑒
𝐷
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
+ 𝐾
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
−
𝑃𝑝𝑢𝑚𝑝
𝜌𝑔
= 𝑧𝑝𝑢𝑚𝑝 − 𝑧𝑖𝑛 + 𝑓
𝐿
𝐷
+
𝐿𝑒
𝐷
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
+ 𝐾
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
𝑃𝑝𝑢𝑚𝑝 = −𝜌𝑔 𝑧𝑝𝑢𝑚𝑝 − 𝑧𝑖𝑛 + 𝑓
𝐿
𝐷
+
𝐿𝑒
𝐷
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
+ 𝐾
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
.
.
.
.
.
.
.
steps omitted
.
.
steps omitted

6. Develop an expression to represented the required head at the pump
𝑃
𝜌𝑔
+
𝑈2
2𝑔
+ 𝑧
𝑠𝑢𝑐𝑡𝑖𝑜𝑛
=
𝑃
𝜌𝑔
+
𝑈2
2𝑔
+ 𝑧
𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒
+ 𝑓
𝐿
𝐷
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
+ ∑𝑓
𝐿𝑒
𝐷
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
+ ∑𝐾
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
− 𝐻
Pin = Pout = Patm, Uin = Uout = 0
𝑧𝑖𝑛 − 𝑧𝑜𝑢𝑡 = 𝑓
𝐿
𝐷
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
+ ∑𝑓
𝐿𝑒
𝐷
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
+ ∑𝐾
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
− 𝐻
𝑧𝑖𝑛 − 𝑧𝑜𝑢𝑡 = 𝑓
𝐿
𝐷
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
+ 𝑓
𝐿𝑒
𝐷
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
+ 𝐾
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
− 𝐻
𝐻 = 𝑧𝑜𝑢𝑡 − 𝑧𝑖𝑛 + 𝑓
𝐿
𝐷
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
+ 𝑓
𝐿𝑒
𝐷
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
+ 𝐾
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
𝐻 = 𝑧𝑜𝑢𝑡 − 𝑧𝑖𝑛 + 𝑓
𝐿
𝐷
+
𝐿𝑒
𝐷
+ 𝐾
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
.
.
.
.
.
.
.
steps omitted
.
.
steps omitted

7. Determine the total pressure at the eye of the impeller and the required head at the pump
Given Data:
𝐷 = 12𝑐𝑚
𝑃𝑎𝑡𝑚 = 101.3𝑘𝑃𝑎
𝑉
·
= 600𝐿/𝑚𝑖𝑛 and 𝜖 = 0.046𝑚𝑚 taken from table
From relevant tables water at 20 °C
𝜌 = 998𝑘𝑔/𝑚3
𝑃𝑣𝑎𝑝𝑜𝑢𝑟 = 2.34𝑘𝑃𝑎
𝜈 = 1.01 × 10−6
𝑚2
/𝑠
At the specified flow rate, the speed of the fluid is:
𝑉
·
= 𝐴𝑈𝑝𝑖𝑝𝑒 ⟶ 𝑈𝑝𝑖𝑝𝑒 =
𝑉
·
𝐴
=
4𝑉
·
𝜋𝐷2
= 0.884𝑚/𝑠
𝑅𝑒 =
𝑈𝑝𝑖𝑝𝑒𝐷
𝜈
=
0.844𝑚/𝑠 × 0.12𝑚
1.01 × 10−6𝑚2/𝑠
= 1.05 × 105
𝜖
𝐷
=
0.046𝑚𝑚
0.12𝑚
= 3.8 × 10−4
𝑓 = −1.8𝑙𝑜𝑔10
3.8 × 10−4
3.7
1.11
+
6.9
1.05 × 105
−2
= 0.0194
1
𝑓
= −1.8𝑙𝑜𝑔10
𝜖
𝐷
3.7
1.11
+
6.9
𝑅𝑒
Therefore using
Note for Clarification - kinematic Viscosity, 𝜈 =
𝜇
𝜌

8. KV
• Saturation temperature Tsat: The temperature at which a pure
substance changes phase at a given pressure.
• Saturation pressure Psat: The pressure at which a pure substance
changes phase at a given temperature.
• Vapor pressure (Pv): The pressure exerted by its vapour in phase
equilibrium with its liquid at a given temperature. It is identical to the
saturation pressure Psat of the liquid (Pv = Psat).
• Partial pressure: The pressure of a gas or vapor in a mixture with
other gases. For example, atmospheric air is a mixture of dry air
and water vapour, and atmospheric pressure is the sum of the
partial pressure of dry air and the partial pressure of water vapour.
VAPOUR
PRESSURE AND
CAVITATION

9. KV
• There is a possibility of the liquid pressure in liquid-flow
systems dropping below the vapour pressure at some
locations, and the resulting unplanned vaporisation.
• The vapour bubbles (called cavitation bubbles since they
form “cavities” in the liquid) collapse as they are swept
away from the low-pressure regions, generating highly
destructive, extremely high-pressure waves.
• This phenomenon, which is a common cause for drop in
performance and even the erosion of impeller blades, is
called cavitation, and it is an important consideration in
the design of hydraulic turbines and pumps.
Cavitation damage on a 16-mm by 23-mm
aluminium sample tested at 60 m/s for 2.5 h.
The sample was located at the cavity collapse
region downstream of a cavity generator
specifically designed to produce high damage
potential.

10. CONSIDER THE LOSSES (Major & Minor)
Pd
Ps
Zd
Zs
Elev. Surface of
Water in Tank
Elev. Surface of Water in
Sump
Elev. Eye of Impeller
Height
Foot Valve with
Hinge Disk
Sudden
Enlargement
Standard
90° Elbows
Table 1: Representative Dimensionless Equivalent
Lengths for Values and Fittings
Fitting Type Equivalent Length,
Valves (Fully Open)
- Gate Valve 8
- Globe Valve 340
- Angle Valve 150
- Ball Valve 3
- Lift Check Valve: Globe Lift 600
- Lift Check Valve: Angle Lift 55
- Foot Valve with Strainer: Poppet Disk 420
- Foot Valve with Strainer: Hinged Disk 75
Standard Elbow: 90º 30
Standard Elbow: 45º 16
Return bend, close pattern 50
Standard Tee: Flow Through Run 20
Standard Tee: Flow Through Branch 60

11. CONSIDER THE LOSSES (Major & Minor)
Pd
Ps
Zd
Zs
Elev. Surface of
Water in Tank
Elev. Surface of Water in
Sump
Elev. Eye of Impeller
Height
Foot Valve with
Hinge Disk
Sudden
Enlargement
Standard
90° Elbows

12. The total pressure at the eye of the impeller
𝑃𝑝𝑢𝑚𝑝 = −𝜌𝑔 𝑧𝑝𝑢𝑚𝑝 − 𝑧𝑖𝑛 + 𝑓
𝐿
𝐷
+
𝐿𝑒
𝐷
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
+ 𝐾
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
From part (c) above
𝑃𝑝𝑢𝑚𝑝 = −𝜌𝑔 8.5𝑚 − 6𝑚 + 0.0194
7𝑚
0.12𝑚
+
105 × 0.12𝑚
0.12𝑚
0.884 2
𝑚/𝑠
2 × 9.81𝑚/𝑠2
+ 0.78 ×
0.884 2
𝑚/𝑠
2 × 9.81𝑚/𝑠2
𝑃𝑝𝑢𝑚𝑝 = −26015𝑃𝑎(𝑔𝑎𝑢𝑔𝑒) = −26𝑘𝑃𝑎(𝑔𝑎𝑢𝑔𝑒)
The Required head at the Pump from equation developed in (d)
𝐻 = 𝑧𝑜𝑢𝑡 − 𝑧𝑖𝑛 + 𝑓
𝐿
𝐷
+
𝐿𝑒
𝐷
+ 𝐾
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
𝐻 = 60𝑚 − 6𝑚 + 0.0194
257𝑚
0.12𝑚
+
135𝑚 × 0.12𝑚
0.12𝑚
+ 1.78
0.884𝑚 2
2 × 9.81𝑚/𝑠2
𝐻 = 55.83𝑚
Hinged Foot Valve plus
One 90º Elbow
Just Re-entrant
Re-entrant plus
Sudden Enlargement

13. The Net Positive Suction Head Available (NPSHA)
𝑁𝑃𝑆𝐻𝐴 =
𝑃𝑝𝑢𝑚𝑝 + 𝑃𝑎𝑡𝑚 − 𝑃𝑣𝑎𝑝𝑜𝑢𝑟
𝜌𝑔
=
−26𝑘𝑃𝑎 + 101.3𝑘𝑃𝑎 − 2.34𝑘𝑃𝑎
998𝑘𝑔/𝑚3 × 9.81𝑚/𝑠2
A pump would be selected by finding one for which the NPSHR is less than the NPSHA. Based on the
data and the information in the pump selection chart, a 5AE8N or a 3AE14 pump would be capable of
supplying the required head at the given flow rate. The pump should be operated at a speed of between
1750 and 3500 rpm, but the efficiency may not be acceptable. One should consult a complete catalog to
make an informed decision.
𝑁𝑃𝑆𝐻𝐴 = 7.4𝑚
Select a pump suitable for this application and provide a justification

14. 1 foot - 0.3048 meters
1 gpm - 0.0038 m3/min
1 gpm - 3.785 l/min
Select these pumps
for this case

15. Develop and format appropriately an excel worksheet which
calculates the;
•Pump head for a range of volumetric flow rates
•calculates the impact that the ageing of the pipes have on the
Pump Head at twenty years and forty years of service
Assume a head at 600 L/min for 40 year old pipes is 80% of the
maximum head of the pump, and that the pump curve has the
form 𝐻 = 𝐻𝑜 − 𝐴𝑉
·
2
. Using the calculations from the Excel
worksheet, plot the Pump Curve and the system curves for new
pipes, pipes at 20 years service and pipes at 40 years service.
PUMP SYSTEM EXAMPLE Pt 2

16. Table 2 - Data given in question or sourced from fluids tables
Given Data Value Units Source
Water at 20 Degrees
Pipe Diameter 12 cm
ε 4.6E-05 mm
Patm 101.3 kPa
Kinematic Viscosity 1.01E-06 m2/s Tables
Pvapour 2.34 kPa Tables
Density 998 kg/m3
z1 6 m
z2 60 m
Lsuction 7 m Side of pump
Ldelivery 250 m Side of pump
LT 257 m
Equivalent Lengths
Hinged Disk Foot Valve 75 Tables
Angle Lift Valve 0 Tables
Gate Valve 0 Tables
Standard 90 deg Elbow 30 2 of these Tables
Le 135 Note 2 Elbows
K
Reentrant 0.78 Tables
Sudden Expansion 1 Tables
KT 1.78
20 Years 5 Tables
40 Years 8.75 Tables
Summarise the Data provide in a table.
Begin by summarising all the relevant data from the
question into a table. Data that is not provided in the
question should be sourced from the relevant tables
and resources available in the essential reading text
book (Fundamentals of Thermal Fluid Sciences from
semester I) or from online sources. Pay particular
caution to units and ensure these are corrected to the
SI unit system.
The data in the summary table should be linked to the
calculations in the data sheets developed from the
governing equations created previously in the question

17. Table 3: Calculate results
Volumetric Flow
Rate (L/Min)
Vel (m/s) Reynolds
Number, Re
Friction
Factor,
f
New Pipes
(m)
20 Year
Pipe (m)
40 Year
Pipe (m)
Pump Curve
(m)
0 0.000 0.000 0.0000 54.00 54.00 54.00 86.88
200 0.295 35017.589 0.0234 54.24 55.19 56.07 84.95
400 0.589 70035.178 0.0207 54.86 58.20 61.32 79.16
600 0.884 105052.768 0.0194 55.83 62.89 69.50 69.52
800 1.179 140070.357 0.0187 57.15 69.22 80.54 56.03
1000 1.474 175087.946 0.0182 58.80 77.19 94.43 38.68
1200 1.768 210105.535 0.0179 60.78 86.78 111.15 17.47
1400 2.063 245123.125 0.0177 63.11 97.99 130.69 -7.60
1600 2.358 280140.714 0.0175 65.77 110.81 153.04 -36.52
1800 2.653 315158.303 0.0173 68.76 125.24 178.20 -69.29
2000 2.947 350175.892 0.0172 72.09 141.29 206.17 -105.93
𝐻600 = 69.5𝑚 𝐻𝑜 =
69.5𝑚
0.8
= 86.875𝑚
𝐻 = 𝐻𝑜 − 𝐴𝑉2
·
⟶ 69.5𝑚 = 86.875𝑚 − 𝐴 6002 𝐴 =
86.875𝑚 − 69.5𝑚
6002
= 4.82 × 10−5
𝑚/(𝐿/𝑚𝑖𝑛)
Assume that the heat at 600 L/min for 40 year old pipe is 80% of the maximum head for the
pump, and that the pump curve has the form 𝐻 = 𝐻𝑜 − 𝐴𝑉2
·
𝐻 = 𝐻𝑜 − 𝐴𝑉2
·

18. 50.00
54.00
58.00
62.00
66.00
70.00
74.00
78.00
82.00
86.00
90.00
200 300 400 500 600 700 800 900 1000
Pump
Head
(m)
Volumetric Flow Rate in L/min
New Pipes (m) 20 Year Pipe (m)
40 Year Pipe (m)