T3a - Finding the operating point of a pumping system 2023.pptx

Keith Vaugh
Keith VaughSTEM Education & Design um MAGVA Design + Letterpress
KV
WORKED EXAMPLES
{Pumping Systems Example 2}
Keith Vaugh BEng (AERO) MEng
PUMPING SYSTEM OPERATING POINT
A centrifugal pump pumps water at 25 °C through a cast iron pipes in the system as illustrated. The pump
has an impeller of 200 mm diameter and a shutoff head H0 = 7.6 m off water when operated at 1170 rpm.
The best efficiency occurs at a volumetric flow rate of 68m3/h where the head H is 6.7m for this speed.
Given these conditions it can be shown that the parabolic equation representing this pump system is given
by; 𝐻 = 7.6 − 1.95 × 10−4
𝑉
·
2
If the pump is scaled to 1750 rpm, the parabolic equation can be shown to be 𝐻 = 17 − 1.95 ×
10−4 𝑉
·
2
For this case;
• Develop an algebraic expression for the general shape of the system resistance curve.
• Calculate and plot the system resistance curve.
• Solve graphically for the system operating point.
PUMPING SYSTEM OPERATING POINT
0.6 m 900 m
250 mm diameter 200 mm diameter
Given
Pump operating at 1750 rpm with a 𝐻 = 𝐻0 − 𝐴𝑉
·
2
where H0 = 17 m and 𝐴 =
1.95 × 10−4
𝑚/ (𝑚3
/ℎ)2
. The pipe from the first reservoir to the pump has a
length of 0.6 m and a diameter of 250 mm, and the pipe from the pump to the
second reservoir has a length of 900 m and a diameter of 200 mm. Water at 25
degrees celsius is transferred horizontally between the reservoirs and the level of
water in each is at the same height.
Find:
(a) A general algebraic expression for the system head curve
(b) The system head curve by direct calculation
(c) The system operating point using a graphical solution
Solution
Apply the energy equation to the flow system
Total head loss is the summation of the major and minor losses in the system
𝑓 = −1.8𝑙𝑜𝑔10
𝜖
𝐷
3.7
1.11
−
6.9
𝑅𝑒
−2
𝑃𝑖𝑛
𝜌𝑔
+
𝑈𝑖𝑛
2
2𝑔
+ 𝑧𝑖𝑛 =
𝑃𝑜𝑢𝑡
𝜌𝑔
+
𝑈𝑜𝑢𝑡
2
2𝑔
+ 𝑧𝑜𝑢𝑡 + 𝑓
𝐿
𝐷
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
+ ∑𝑓
𝐿𝑒
𝐷
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
+ ∑𝐾
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
− 𝐻
𝐻 =
ℎ𝑝
𝑔
ℎ𝐿 = 𝑓
𝐿
𝐷
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
+ ∑𝑓
𝐿𝑒
𝐷
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
+ ∑𝐾
𝑈𝑝𝑖𝑝𝑒
2
2𝑔
𝑃𝑖𝑛
𝜌𝑔
+
𝑈𝑖𝑛
2
2𝑔
+ 𝑧𝑖𝑛 =
𝑃𝑜𝑢𝑡
𝜌𝑔
+
𝑈𝑜𝑢𝑡
2
2𝑔
+ 𝑧𝑜𝑢𝑡 + ℎ𝐿 − 𝐻
Friction factor
Total head loss is the summation of the major and minor losses in the system
The energy equation for steady incompressible pipe flow can be written as;
𝑁𝑃𝑆𝐻𝐴 =
𝑃𝑝𝑢𝑚𝑝 + 𝑃𝑎𝑡𝑚 − 𝑃𝑣𝑎𝑝𝑜𝑢𝑟
𝜌𝑔
Net Positive Suction Head Available
GOVERNING EQUATIONS
GOVERNING EQUATIONS
Piping system -
𝑃0
𝜌𝑔
+
𝑈0
2
2𝑔
+ 𝑧0 + 𝐻𝑎 =
𝑃3
𝜌𝑔
+
𝑈3
2
2𝑔
+ 𝑧3 +
ℎ𝑙𝑡
𝑔
Friction Factor - 𝑓 = −1.8𝑙𝑜𝑔10
𝜖
𝐷
3.7
1.11
+
6.9
𝑅𝑒
−2
where z0 and z3 are the surface levels for the supply and discharge reservoirs respectively
ASSUMPTIONS
P0 = P3 =Patm
U0 = U3 = 0
z0 =z3
𝐻𝑎 =
ℎ𝑙𝑡
𝑔
=
ℎ𝑙𝑇01
𝑔
+
ℎ𝑙𝑇23
𝑔
= 𝐻𝑙𝑇
Simplifying the Governing Equation
where section ① and ② are located just upstream and downstream from the pump, respectively.
ℎ𝑙𝑇23
= 𝑓2
𝐿2
𝐷2
𝑈2
2
2
+ 𝐾𝑒𝑥𝑖𝑡
𝑈2
2
2
= 𝑓2
𝐿2
𝐷2
+ 𝐾𝑒𝑥𝑖𝑡
𝑈2
2
2
ℎ𝑙𝑇01
= 𝐾𝑒𝑛𝑡
𝑈1
2
2
+ 𝑓1
𝐿1
𝐷1
𝑈1
2
2
= 𝐾𝑒𝑛𝑡 + 𝑓1
𝐿1
𝐷1
𝑈1
2
2
① ② ③
⓪
The total heads losses are the sum of the
major and minor losses, so
From continuity, 𝑈1𝐴1 = 𝑈2𝐴2 therefore 𝑈1 = 𝑈2
𝐴2
𝐴1
= 𝑈2
𝐷2
𝐷1
2
Hence
𝐻𝑙𝑇
= 𝐾𝑒𝑛𝑡 + 𝑓1
𝐿1
𝐷1
𝐷2
𝐷1
4
+ 𝑓2
𝐿2
𝐷2
+ 𝐾𝑒𝑥𝑖𝑡
𝑈2
2
2𝑔
𝐻𝑙𝑇
=
ℎ𝑙𝑡
𝑔
= 𝐾𝑒𝑛𝑡 + 𝑓1
𝐿1
𝐷1
𝑈2
2
2𝑔
𝐷2
𝐷1
4
+ 𝑓2
𝐿2
𝐷2
+ 𝐾𝑒𝑥𝑖𝑡
𝑈2
2
2𝑔
Simplifying this equation results in an equation representative of the Total Head Loss in the pipes as
consequence of the Major and Minor Head Losses
At the operating point as per the simplified governing equation 𝐻𝑎 =
ℎ𝑙𝑡
𝑔
=
ℎ𝑙𝑇01
𝑔
+
ℎ𝑙𝑇23
𝑔
= 𝐻𝑙𝑇
the head loss is equal to
the head produced nay the pump given by 𝐻 = 𝐻0 − 𝐴𝑉
·
2
where 𝐻0 = 17𝑚 and 𝐴 = 1.95 × 10−4
𝑚/ (𝑚3
/ℎ)2
𝐻𝑙𝑇
=
ℎ𝑙𝑡
𝑔
= 𝐾𝑒𝑛𝑡 + 𝑓1
𝐿1
𝐷1
𝑈1
2
2𝑔
+ 𝑓2
𝐿2
𝐷2
+ 𝐾𝑒𝑥𝑖𝑡
𝑈2
2
2𝑔
𝑈1 = 𝑈2
𝐴2
𝐴1
= 𝑈2
𝐷2
𝐷1
2
ℎ𝑙𝑇23
= 𝑓2
𝐿2
𝐷2
𝑈2
2
2
+ 𝐾𝑒𝑥𝑖𝑡
𝑈2
2
2
= 𝑓2
𝐿2
𝐷2
+ 𝐾𝑒𝑥𝑖𝑡
𝑈2
2
2
ℎ𝑙𝑇01
= 𝐾𝑒𝑛𝑡
𝑈1
2
2
+ 𝑓1
𝐿1
𝐷1
𝑈1
2
2
= 𝐾𝑒𝑛𝑡 + 𝑓1
𝐿1
𝐷1
𝑈1
2
2
Pipe on the Left side of the pump
Pipe on the Right side of the pump
𝐻𝑙𝑇
= ℎ𝑙𝑇01
+ ℎ𝑙𝑇23
Add these two equations to get the total head loss in the system
But from continuity
Gives
𝑈1
2
= 𝑈2
𝐷2
𝐷1
2 2
= 𝑈2
2
𝐷2
𝐷1
4
𝐻𝑙𝑇
=
ℎ𝑙𝑡
𝑔
= 𝐾𝑒𝑛𝑡 + 𝑓1
𝐿1
𝐷1
1
2𝑔
𝑈2
2
×
𝐷2
𝐷1
4
+ 𝑓2
𝐿2
𝐷2
+ 𝐾𝑒𝑥𝑖𝑡
𝑈2
2
2𝑔
Table 1 - Data given in question or sourced from fluids tables
Given Data Value Units Source
Water at 25 Degrees
Pipe Diameter D1 25 cm
Pipe Diameter D2 20 cm
ε 2.6E-04 m Tables
Patm 101.3 kPa
Kinematic Viscosity 8.96E-07 m2/s Tables
Density 997 kg/m3
z1 0 m
z2 0 m
Lsuction 0.6 m Side of pump
Ldelivery 900 m Side of pump
LT 900.6 m
For minor losses, K
Reentrant 0.5 Tables
Sudden Expansion 1 Tables
KT 1.5
Compile all available data into a table
Some data needs to be sourced from
reference tables, databases etc…
EXAMPLE Table: Roughness for pipes of common Engineering Materials
Pipe Roughness, ε (mm)
Riveted steel 0.9 - 9
Concrete 0.3 - 3
Wood Stave 0.2 - 0.9
Cast Iron 0.26
Glavanised Iron 0.15
Asphalted Cast Iron 0.12
Commercial Steel or Wrought
Iron
0.046
Dran Tubing 0.0015
Generate a data table populated with the calculations distilled from the formulas developed
Table 2 - Data Tabulation and Analysis to determine the operating point of the system
Volumetric
Flow Rate
(m3/hr)
U1 (m/s)
Reynolds
Number, Re
Friction
Factor, f1
U2 (m/s)
Reynolds
Number, Re
Friction
Factor, f2
New Pipes
(m)
Pump Curve
(m)
0 0.00 0.00 0.0000 0.00 0.00 0.00 0.00 17.00
25 0.14 39457.07 0.0246 0.22 49321.34 0.0246 0.28 16.88
50 0.28 78914.14 0.0226 0.44 98642.68 0.0230 1.04 16.51
75 0.42 118371.21 0.0218 0.66 147964.02 0.0224 2.28 15.90
100 0.57 157828.28 0.0214 0.88 197285.35 0.0221 4.00 15.05
125 0.71 197285.35 0.0211 1.10 246606.69 0.0219 6.19 13.95
150 0.85 236742.42 0.0209 1.33 295928.03 0.0217 8.86 12.61
175 0.99 276199.49 0.0208 1.55 345249.37 0.0216 12.01 11.03
200 1.13 315656.57 0.0207 1.77 394570.71 0.0215 15.63 9.20
225 1.27 355113.64 0.0206 1.99 443892.05 0.0215 19.73 7.13
250 1.41 394570.71 0.0205 2.21 493213.38 0.0214 24.31 4.81
Table 2 - Data Tabulation and Analysis to determine the operating point of the system
Volumetric
Flow Rate
(m3/hr)
U1 (m/s)
Reynolds
Number, Re
Friction
Factor, f1
U2 (m/s)
Reynolds
Number, Re
Friction
Factor, f2
New Pipes
(m)
Pump Curve
(m)
0 0.00 0.00 0.0000 0.00 0.00 0.00 0.00 17.00
25 0.14 39457.07 0.0246 0.22 49321.34 0.0246 0.28 16.88
50 0.28 78914.14 0.0226 0.44 98642.68 0.0230 1.04 16.51
75 0.42 118371.21 0.0218 0.66 147964.02 0.0224 2.28 15.90
100 0.57 157828.28 0.0214 0.88 197285.35 0.0221 4.00 15.05
125 0.71 197285.35 0.0211 1.10 246606.69 0.0219 6.19 13.95
150 0.85 236742.42 0.0209 1.33 295928.03 0.0217 8.86 12.61
175 0.99 276199.49 0.0208 1.55 345249.37 0.0216 12.01 11.03
200 1.13 315656.57 0.0207 1.77 394570.71 0.0215 15.63 9.20
225 1.27 355113.64 0.0206 1.99 443892.05 0.0215 19.73 7.13
250 1.41 394570.71 0.0205 2.21 493213.38 0.0214 24.31 4.81
𝐻𝑙𝑇
= 𝐾𝑒𝑛𝑡 + 𝑓1
𝐿1
𝐷1
𝐷2
𝐷1
4
+ 𝑓2
𝐿2
𝐷2
+ 𝐾𝑒𝑥𝑖𝑡
𝑈2
2
2𝑔
𝐻 = 17 − 1.95 × 10−4 𝑉
·
2
Plot the relevant curves and identify the operating point for the pumping system.
0.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
16.00
18.00
20.00
0 50 100 150 200 250
Pump
Head
(m)
Volumetric Flow Rate in m^3/hr
New Pipes (m)
Where curves cross is the
optimum operating point for
the system. The graphical solution is
shown on the plot. At the
operating point H ≈ 11.4 m
and the volumetric flow
rate, 170 m3/h.
As pipes age with time, build up forms on the inner walls. This must be taken into account when
designing the system. Typical multipliers are available that can be applied to the calculations
Table 2 - Data Tabulation and Analysis to determine the operating point of the system
Volumetric
Flow Rate
(m3/hr)
U1 (m/s)
Reynolds
Number,
Re
Friction
Factor, f1
U2 (m/s)
Reynolds
Number,
Re
Friction
Factor, f2
New Pipes
(m)
Age pipes
10 years
Age pipes
20 years
Pump Curve
(m)
0 0.00 0.00 0.0000 0.00 0.00 0.00 0.00 17.00
25 0.14 39457.07 0.0246 0.22 49321.34 0.0246 0.28 16.88
50 0.28 78914.14 0.0226 0.44 98642.68 0.0230 1.04 16.51
75 0.42 118371.21 0.0218 0.66 147964.02 0.0224 2.28 15.90
100 0.57 157828.28 0.0214 0.88 197285.35 0.0221 4.00 15.05
125 0.71 197285.35 0.0211 1.10 246606.69 0.0219 6.19 13.95
150 0.85 236742.42 0.0209 1.33 295928.03 0.0217 8.86 12.61
175 0.99 276199.49 0.0208 1.55 345249.37 0.0216 12.01 11.03
200 1.13 315656.57 0.0207 1.77 394570.71 0.0215 15.63 9.20
225 1.27 355113.64 0.0206 1.99 443892.05 0.0215 19.73 7.13
250 1.41 394570.71 0.0205 2.21 493213.38 0.0214 24.31 4.81
Add two new
columns to
the table
Table 2: Typical Multipliers applied to friction
factors, with ageing pipes
Pipe
Age (Years)
Small Pipes,
100 - 250
mm
Large Pipes,
300 - 1500
mm
New 1.00 1.00
10 2.2 1.60
20 5.00 2.00
30 7.25 2.20
40 8.75 2.40
50 9.6 2.86
60 10.0 3.70
70 10.1 4.70
From the table locate the multipliers
for the ageing pipe
Multiply these values by the fraction
factor for that condition
Given that each pipe is ≤ 250 mm diameter the multiplier for 10 years is 2.2 and for 20 years is 5
Table 3 - Data Tabulation and Analysis to determine the operating point of the system
Volumetric
Flow Rate
(m3/hr)
U1 (m/s) Reynolds
Number, Re
Friction
Factor, f1
U2 (m/s) Reynolds
Number,
Re
Friction
Factor, f2
New
Pipes (m)
Ageing
pipes 10
years
Ageing
pipes 20
years
0 0.00 0.00 0.0000 0.00 0.00 0.00 0.00 0.00 0.00 17.00
25 0.14 39457.07 0.0246 0.22 49321.34 0.0246 0.28 0.61 2.41 16.88
50 0.28 78914.14 0.0226 0.44 98642.68 0.0230 1.04 2.28 9.02 16.51
75 0.42 118371.21 0.0218 0.66 147964.0
2
0.0224 2.28 4.99 19.76 15.90
100 0.57 157828.28 0.0214 0.88 197285.3
5
0.0221 4.00 8.74 34.62 15.05
125 0.71 197285.35 0.0211 1.10 246606.6
9
0.0219 6.19 13.54 53.61 13.95
150 0.85 236742.42 0.0209 1.33 295928.0
3
0.0217 8.86 19.37 76.72 12.61
175 0.99 276199.49 0.0208 1.55 345249.3
7
0.0216 12.01 26.25 103.95 11.03
200 1.13 315656.57 0.0207 1.77 394570.7
1
0.0215 15.63 34.16 135.30 9.20
225 1.27 355113.64 0.0206 1.99 443892.0
5
0.0215 19.73 43.12 170.77 7.13
0.00
2.50
5.00
7.50
10.00
12.50
15.00
17.50
20.00
0 75 150 225 300
Pump
Head
(m)
Volumetric Flow Rate (m^3/hr)
New Pipes (m) Ageing pipes 10 years Ageing pipes 20 years
Plot the relevant curves and identify the operating point for the ageing pipes in the pumping system.
The graphical solution is
shown on the plot. At the
operating point for new
pipes H ≈ 11.4 m and the
volumetric flow rate, 170
m3/h, for 10 year old pipes
H ≈ 13.75 m and the
volumetric flow rate, 127
m3/h, and for 10 year old
pipes H ≈ 16 m and the
volumetric flow rate, 67
m3/h.
1 von 16

Recomendados

PROBLEMA 3 von
PROBLEMA 3 PROBLEMA 3
PROBLEMA 3 yeisyynojos
150 views3 Folien
T3b - MASTER - Pump flow system - operating point 2023.pptx von
T3b - MASTER - Pump flow system - operating point 2023.pptxT3b - MASTER - Pump flow system - operating point 2023.pptx
T3b - MASTER - Pump flow system - operating point 2023.pptxKeith Vaugh
1.5K views18 Folien
Sizing of water piping system von
Sizing of water piping systemSizing of water piping system
Sizing of water piping systemTajudeen Ibrahim
1.2K views14 Folien
MATHEMATICAL MODELLING FOR ANALYSIS OF CHANGE IN SHAPE OF SUCTION MANIFOLD TO... von
MATHEMATICAL MODELLING FOR ANALYSIS OF CHANGE IN SHAPE OF SUCTION MANIFOLD TO...MATHEMATICAL MODELLING FOR ANALYSIS OF CHANGE IN SHAPE OF SUCTION MANIFOLD TO...
MATHEMATICAL MODELLING FOR ANALYSIS OF CHANGE IN SHAPE OF SUCTION MANIFOLD TO...ijiert bestjournal
35 views11 Folien
Automatic Generation Control von
Automatic Generation ControlAutomatic Generation Control
Automatic Generation ControlMINAL RADE
1.1K views44 Folien
Pwm Control Strategy for Controlling Of Parallel Rectifiers In Single Phase T... von
Pwm Control Strategy for Controlling Of Parallel Rectifiers In Single Phase T...Pwm Control Strategy for Controlling Of Parallel Rectifiers In Single Phase T...
Pwm Control Strategy for Controlling Of Parallel Rectifiers In Single Phase T...IJERA Editor
391 views8 Folien

Más contenido relacionado

Similar a T3a - Finding the operating point of a pumping system 2023.pptx

Episode 40 : DESIGN EXAMPLE – DILUTE PHASE PNEUMATIC CONVEYING (Part 2) von
Episode 40 : DESIGN EXAMPLE – DILUTE PHASE PNEUMATIC CONVEYING (Part 2)Episode 40 : DESIGN EXAMPLE – DILUTE PHASE PNEUMATIC CONVEYING (Part 2)
Episode 40 : DESIGN EXAMPLE – DILUTE PHASE PNEUMATIC CONVEYING (Part 2)SAJJAD KHUDHUR ABBAS
826 views19 Folien
Episode 40 : DESIGN EXAMPLE – DILUTE PHASE PNEUMATIC CONVEYING von
Episode 40 :  DESIGN EXAMPLE – DILUTE PHASE PNEUMATIC CONVEYINGEpisode 40 :  DESIGN EXAMPLE – DILUTE PHASE PNEUMATIC CONVEYING
Episode 40 : DESIGN EXAMPLE – DILUTE PHASE PNEUMATIC CONVEYINGSAJJAD KHUDHUR ABBAS
6.1K views19 Folien
Chapter_9_Instrument.pdf von
Chapter_9_Instrument.pdfChapter_9_Instrument.pdf
Chapter_9_Instrument.pdfAnshuChandola1
4 views25 Folien
Momentum equation.pdf von
 Momentum equation.pdf Momentum equation.pdf
Momentum equation.pdfDr. Ezzat Elsayed Gomaa
179 views43 Folien
LiquidHeatControl von
LiquidHeatControlLiquidHeatControl
LiquidHeatControlMohammad Abu Sayed
107 views17 Folien
Flash Steam and Steam Condensates in Return Lines von
Flash Steam and Steam Condensates in Return LinesFlash Steam and Steam Condensates in Return Lines
Flash Steam and Steam Condensates in Return LinesVijay Sarathy
756 views5 Folien

Similar a T3a - Finding the operating point of a pumping system 2023.pptx(20)

Episode 40 : DESIGN EXAMPLE – DILUTE PHASE PNEUMATIC CONVEYING (Part 2) von SAJJAD KHUDHUR ABBAS
Episode 40 : DESIGN EXAMPLE – DILUTE PHASE PNEUMATIC CONVEYING (Part 2)Episode 40 : DESIGN EXAMPLE – DILUTE PHASE PNEUMATIC CONVEYING (Part 2)
Episode 40 : DESIGN EXAMPLE – DILUTE PHASE PNEUMATIC CONVEYING (Part 2)
Episode 40 : DESIGN EXAMPLE – DILUTE PHASE PNEUMATIC CONVEYING von SAJJAD KHUDHUR ABBAS
Episode 40 :  DESIGN EXAMPLE – DILUTE PHASE PNEUMATIC CONVEYINGEpisode 40 :  DESIGN EXAMPLE – DILUTE PHASE PNEUMATIC CONVEYING
Episode 40 : DESIGN EXAMPLE – DILUTE PHASE PNEUMATIC CONVEYING
Flash Steam and Steam Condensates in Return Lines von Vijay Sarathy
Flash Steam and Steam Condensates in Return LinesFlash Steam and Steam Condensates in Return Lines
Flash Steam and Steam Condensates in Return Lines
Vijay Sarathy756 views
Numerical Calculation of Solid-Liquid two-Phase Flow Inside a Small Sewage Pump von theijes
Numerical Calculation of Solid-Liquid two-Phase Flow Inside a Small Sewage PumpNumerical Calculation of Solid-Liquid two-Phase Flow Inside a Small Sewage Pump
Numerical Calculation of Solid-Liquid two-Phase Flow Inside a Small Sewage Pump
theijes32 views
Flow measurement basics von Salman1011
Flow measurement basicsFlow measurement basics
Flow measurement basics
Salman10111.9K views
Metal cutting tool position control using static output feedback and full sta... von Mustefa Jibril
Metal cutting tool position control using static output feedback and full sta...Metal cutting tool position control using static output feedback and full sta...
Metal cutting tool position control using static output feedback and full sta...
Mustefa Jibril66 views
Performance of a_centrifugal_pump_autosaved von Dickens Mimisa
Performance of a_centrifugal_pump_autosavedPerformance of a_centrifugal_pump_autosaved
Performance of a_centrifugal_pump_autosaved
Dickens Mimisa1.8K views
Optimum overhaul of pumps 2014 von Ray Beebe
Optimum overhaul of pumps 2014Optimum overhaul of pumps 2014
Optimum overhaul of pumps 2014
Ray Beebe1.2K views
Energy efficiency in pumps and fans ppt von D.Pawan Kumar
Energy efficiency in pumps and fans pptEnergy efficiency in pumps and fans ppt
Energy efficiency in pumps and fans ppt
D.Pawan Kumar6.8K views
5_2018_05_15!02_24_23_AM.pptx von PrakashGohil5
5_2018_05_15!02_24_23_AM.pptx5_2018_05_15!02_24_23_AM.pptx
5_2018_05_15!02_24_23_AM.pptx
PrakashGohil526 views

Más de Keith Vaugh

T2c - Centrifugal Pumps, turbines and Impeller calculations 2023.pptx von
T2c - Centrifugal Pumps, turbines and Impeller calculations 2023.pptxT2c - Centrifugal Pumps, turbines and Impeller calculations 2023.pptx
T2c - Centrifugal Pumps, turbines and Impeller calculations 2023.pptxKeith Vaugh
1.5K views76 Folien
T2b - Momentum of Fluids 2023.pptx von
T2b - Momentum of Fluids 2023.pptxT2b - Momentum of Fluids 2023.pptx
T2b - Momentum of Fluids 2023.pptxKeith Vaugh
1.5K views18 Folien
T2a - Fluid Discharge 2023.pptx von
T2a - Fluid Discharge 2023.pptxT2a - Fluid Discharge 2023.pptx
T2a - Fluid Discharge 2023.pptxKeith Vaugh
1.5K views27 Folien
T1 - Essential Fluids - 2023.pptx von
T1 - Essential Fluids - 2023.pptxT1 - Essential Fluids - 2023.pptx
T1 - Essential Fluids - 2023.pptxKeith Vaugh
1.5K views44 Folien
L7 - SecondLawThermo 2023.pptx von
L7 - SecondLawThermo 2023.pptxL7 - SecondLawThermo 2023.pptx
L7 - SecondLawThermo 2023.pptxKeith Vaugh
1.5K views44 Folien
L6 - Mass&EnergyClosedVol 2023.pptx von
L6 - Mass&EnergyClosedVol 2023.pptxL6 - Mass&EnergyClosedVol 2023.pptx
L6 - Mass&EnergyClosedVol 2023.pptxKeith Vaugh
1.5K views35 Folien

Más de Keith Vaugh(20)

T2c - Centrifugal Pumps, turbines and Impeller calculations 2023.pptx von Keith Vaugh
T2c - Centrifugal Pumps, turbines and Impeller calculations 2023.pptxT2c - Centrifugal Pumps, turbines and Impeller calculations 2023.pptx
T2c - Centrifugal Pumps, turbines and Impeller calculations 2023.pptx
Keith Vaugh1.5K views
T2b - Momentum of Fluids 2023.pptx von Keith Vaugh
T2b - Momentum of Fluids 2023.pptxT2b - Momentum of Fluids 2023.pptx
T2b - Momentum of Fluids 2023.pptx
Keith Vaugh1.5K views
T2a - Fluid Discharge 2023.pptx von Keith Vaugh
T2a - Fluid Discharge 2023.pptxT2a - Fluid Discharge 2023.pptx
T2a - Fluid Discharge 2023.pptx
Keith Vaugh1.5K views
T1 - Essential Fluids - 2023.pptx von Keith Vaugh
T1 - Essential Fluids - 2023.pptxT1 - Essential Fluids - 2023.pptx
T1 - Essential Fluids - 2023.pptx
Keith Vaugh1.5K views
L7 - SecondLawThermo 2023.pptx von Keith Vaugh
L7 - SecondLawThermo 2023.pptxL7 - SecondLawThermo 2023.pptx
L7 - SecondLawThermo 2023.pptx
Keith Vaugh1.5K views
L6 - Mass&EnergyClosedVol 2023.pptx von Keith Vaugh
L6 - Mass&EnergyClosedVol 2023.pptxL6 - Mass&EnergyClosedVol 2023.pptx
L6 - Mass&EnergyClosedVol 2023.pptx
Keith Vaugh1.5K views
L5 - EnergyAnalysisClosedSys 2023.pptx von Keith Vaugh
L5 - EnergyAnalysisClosedSys 2023.pptxL5 - EnergyAnalysisClosedSys 2023.pptx
L5 - EnergyAnalysisClosedSys 2023.pptx
Keith Vaugh1.5K views
L4 - PropertiesPureSubstances 2023.pptx von Keith Vaugh
L4 - PropertiesPureSubstances 2023.pptxL4 - PropertiesPureSubstances 2023.pptx
L4 - PropertiesPureSubstances 2023.pptx
Keith Vaugh1.5K views
L2 - Basic Concepts 2023 UD.pptx von Keith Vaugh
L2 - Basic Concepts 2023 UD.pptxL2 - Basic Concepts 2023 UD.pptx
L2 - Basic Concepts 2023 UD.pptx
Keith Vaugh1.5K views
L1 - ES & Thermofluids 2023 Master SS.pptx von Keith Vaugh
L1 - ES & Thermofluids 2023 Master SS.pptxL1 - ES & Thermofluids 2023 Master SS.pptx
L1 - ES & Thermofluids 2023 Master SS.pptx
Keith Vaugh1.5K views
L1 - Energy Systems and Thermofluids 2021-22 von Keith Vaugh
L1 - Energy Systems and Thermofluids 2021-22L1 - Energy Systems and Thermofluids 2021-22
L1 - Energy Systems and Thermofluids 2021-22
Keith Vaugh3.1K views
CAD & Analysis Introduction von Keith Vaugh
CAD & Analysis IntroductionCAD & Analysis Introduction
CAD & Analysis Introduction
Keith Vaugh1.7K views
Wind Energy Lecture slides von Keith Vaugh
Wind Energy Lecture slidesWind Energy Lecture slides
Wind Energy Lecture slides
Keith Vaugh19.3K views
Fluid discharge von Keith Vaugh
Fluid dischargeFluid discharge
Fluid discharge
Keith Vaugh49.7K views
Essential fluids von Keith Vaugh
Essential fluids Essential fluids
Essential fluids
Keith Vaugh29.6K views
Essential fluid mechanics von Keith Vaugh
Essential fluid mechanicsEssential fluid mechanics
Essential fluid mechanics
Keith Vaugh3K views

Último

Payment Integration using Braintree Connector | MuleSoft Mysore Meetup #37 von
Payment Integration using Braintree Connector | MuleSoft Mysore Meetup #37Payment Integration using Braintree Connector | MuleSoft Mysore Meetup #37
Payment Integration using Braintree Connector | MuleSoft Mysore Meetup #37MysoreMuleSoftMeetup
50 views17 Folien
12.5.23 Poverty and Precarity.pptx von
12.5.23 Poverty and Precarity.pptx12.5.23 Poverty and Precarity.pptx
12.5.23 Poverty and Precarity.pptxmary850239
381 views30 Folien
Retail Store Scavenger Hunt.pptx von
Retail Store Scavenger Hunt.pptxRetail Store Scavenger Hunt.pptx
Retail Store Scavenger Hunt.pptxjmurphy154
52 views10 Folien
Creative Restart 2023: Leonard Savage - The Permanent Brief: Unearthing unobv... von
Creative Restart 2023: Leonard Savage - The Permanent Brief: Unearthing unobv...Creative Restart 2023: Leonard Savage - The Permanent Brief: Unearthing unobv...
Creative Restart 2023: Leonard Savage - The Permanent Brief: Unearthing unobv...Taste
55 views21 Folien
The Future of Micro-credentials: Is Small Really Beautiful? von
The Future of Micro-credentials:  Is Small Really Beautiful?The Future of Micro-credentials:  Is Small Really Beautiful?
The Future of Micro-credentials: Is Small Really Beautiful?Mark Brown
75 views35 Folien
Monthly Information Session for MV Asterix (November) von
Monthly Information Session for MV Asterix (November)Monthly Information Session for MV Asterix (November)
Monthly Information Session for MV Asterix (November)Esquimalt MFRC
107 views26 Folien

Último(20)

Payment Integration using Braintree Connector | MuleSoft Mysore Meetup #37 von MysoreMuleSoftMeetup
Payment Integration using Braintree Connector | MuleSoft Mysore Meetup #37Payment Integration using Braintree Connector | MuleSoft Mysore Meetup #37
Payment Integration using Braintree Connector | MuleSoft Mysore Meetup #37
12.5.23 Poverty and Precarity.pptx von mary850239
12.5.23 Poverty and Precarity.pptx12.5.23 Poverty and Precarity.pptx
12.5.23 Poverty and Precarity.pptx
mary850239381 views
Retail Store Scavenger Hunt.pptx von jmurphy154
Retail Store Scavenger Hunt.pptxRetail Store Scavenger Hunt.pptx
Retail Store Scavenger Hunt.pptx
jmurphy15452 views
Creative Restart 2023: Leonard Savage - The Permanent Brief: Unearthing unobv... von Taste
Creative Restart 2023: Leonard Savage - The Permanent Brief: Unearthing unobv...Creative Restart 2023: Leonard Savage - The Permanent Brief: Unearthing unobv...
Creative Restart 2023: Leonard Savage - The Permanent Brief: Unearthing unobv...
Taste55 views
The Future of Micro-credentials: Is Small Really Beautiful? von Mark Brown
The Future of Micro-credentials:  Is Small Really Beautiful?The Future of Micro-credentials:  Is Small Really Beautiful?
The Future of Micro-credentials: Is Small Really Beautiful?
Mark Brown75 views
Monthly Information Session for MV Asterix (November) von Esquimalt MFRC
Monthly Information Session for MV Asterix (November)Monthly Information Session for MV Asterix (November)
Monthly Information Session for MV Asterix (November)
Esquimalt MFRC107 views
CUNY IT Picciano.pptx von apicciano
CUNY IT Picciano.pptxCUNY IT Picciano.pptx
CUNY IT Picciano.pptx
apicciano64 views
Nelson_RecordStore.pdf von BrynNelson5
Nelson_RecordStore.pdfNelson_RecordStore.pdf
Nelson_RecordStore.pdf
BrynNelson546 views
INT-244 Topic 6b Confucianism von S Meyer
INT-244 Topic 6b ConfucianismINT-244 Topic 6b Confucianism
INT-244 Topic 6b Confucianism
S Meyer45 views
Six Sigma Concept by Sahil Srivastava.pptx von Sahil Srivastava
Six Sigma Concept by Sahil Srivastava.pptxSix Sigma Concept by Sahil Srivastava.pptx
Six Sigma Concept by Sahil Srivastava.pptx
Sahil Srivastava44 views
NodeJS and ExpressJS.pdf von ArthyR3
NodeJS and ExpressJS.pdfNodeJS and ExpressJS.pdf
NodeJS and ExpressJS.pdf
ArthyR348 views
Parts of Speech (1).pptx von mhkpreet001
Parts of Speech (1).pptxParts of Speech (1).pptx
Parts of Speech (1).pptx
mhkpreet00146 views

T3a - Finding the operating point of a pumping system 2023.pptx

  • 1. KV WORKED EXAMPLES {Pumping Systems Example 2} Keith Vaugh BEng (AERO) MEng
  • 2. PUMPING SYSTEM OPERATING POINT A centrifugal pump pumps water at 25 °C through a cast iron pipes in the system as illustrated. The pump has an impeller of 200 mm diameter and a shutoff head H0 = 7.6 m off water when operated at 1170 rpm. The best efficiency occurs at a volumetric flow rate of 68m3/h where the head H is 6.7m for this speed. Given these conditions it can be shown that the parabolic equation representing this pump system is given by; 𝐻 = 7.6 − 1.95 × 10−4 𝑉 · 2 If the pump is scaled to 1750 rpm, the parabolic equation can be shown to be 𝐻 = 17 − 1.95 × 10−4 𝑉 · 2 For this case; • Develop an algebraic expression for the general shape of the system resistance curve. • Calculate and plot the system resistance curve. • Solve graphically for the system operating point.
  • 3. PUMPING SYSTEM OPERATING POINT 0.6 m 900 m 250 mm diameter 200 mm diameter
  • 4. Given Pump operating at 1750 rpm with a 𝐻 = 𝐻0 − 𝐴𝑉 · 2 where H0 = 17 m and 𝐴 = 1.95 × 10−4 𝑚/ (𝑚3 /ℎ)2 . The pipe from the first reservoir to the pump has a length of 0.6 m and a diameter of 250 mm, and the pipe from the pump to the second reservoir has a length of 900 m and a diameter of 200 mm. Water at 25 degrees celsius is transferred horizontally between the reservoirs and the level of water in each is at the same height. Find: (a) A general algebraic expression for the system head curve (b) The system head curve by direct calculation (c) The system operating point using a graphical solution Solution Apply the energy equation to the flow system
  • 5. Total head loss is the summation of the major and minor losses in the system 𝑓 = −1.8𝑙𝑜𝑔10 𝜖 𝐷 3.7 1.11 − 6.9 𝑅𝑒 −2 𝑃𝑖𝑛 𝜌𝑔 + 𝑈𝑖𝑛 2 2𝑔 + 𝑧𝑖𝑛 = 𝑃𝑜𝑢𝑡 𝜌𝑔 + 𝑈𝑜𝑢𝑡 2 2𝑔 + 𝑧𝑜𝑢𝑡 + 𝑓 𝐿 𝐷 𝑈𝑝𝑖𝑝𝑒 2 2𝑔 + ∑𝑓 𝐿𝑒 𝐷 𝑈𝑝𝑖𝑝𝑒 2 2𝑔 + ∑𝐾 𝑈𝑝𝑖𝑝𝑒 2 2𝑔 − 𝐻 𝐻 = ℎ𝑝 𝑔 ℎ𝐿 = 𝑓 𝐿 𝐷 𝑈𝑝𝑖𝑝𝑒 2 2𝑔 + ∑𝑓 𝐿𝑒 𝐷 𝑈𝑝𝑖𝑝𝑒 2 2𝑔 + ∑𝐾 𝑈𝑝𝑖𝑝𝑒 2 2𝑔 𝑃𝑖𝑛 𝜌𝑔 + 𝑈𝑖𝑛 2 2𝑔 + 𝑧𝑖𝑛 = 𝑃𝑜𝑢𝑡 𝜌𝑔 + 𝑈𝑜𝑢𝑡 2 2𝑔 + 𝑧𝑜𝑢𝑡 + ℎ𝐿 − 𝐻 Friction factor Total head loss is the summation of the major and minor losses in the system The energy equation for steady incompressible pipe flow can be written as; 𝑁𝑃𝑆𝐻𝐴 = 𝑃𝑝𝑢𝑚𝑝 + 𝑃𝑎𝑡𝑚 − 𝑃𝑣𝑎𝑝𝑜𝑢𝑟 𝜌𝑔 Net Positive Suction Head Available GOVERNING EQUATIONS
  • 6. GOVERNING EQUATIONS Piping system - 𝑃0 𝜌𝑔 + 𝑈0 2 2𝑔 + 𝑧0 + 𝐻𝑎 = 𝑃3 𝜌𝑔 + 𝑈3 2 2𝑔 + 𝑧3 + ℎ𝑙𝑡 𝑔 Friction Factor - 𝑓 = −1.8𝑙𝑜𝑔10 𝜖 𝐷 3.7 1.11 + 6.9 𝑅𝑒 −2 where z0 and z3 are the surface levels for the supply and discharge reservoirs respectively ASSUMPTIONS P0 = P3 =Patm U0 = U3 = 0 z0 =z3
  • 7. 𝐻𝑎 = ℎ𝑙𝑡 𝑔 = ℎ𝑙𝑇01 𝑔 + ℎ𝑙𝑇23 𝑔 = 𝐻𝑙𝑇 Simplifying the Governing Equation where section ① and ② are located just upstream and downstream from the pump, respectively. ℎ𝑙𝑇23 = 𝑓2 𝐿2 𝐷2 𝑈2 2 2 + 𝐾𝑒𝑥𝑖𝑡 𝑈2 2 2 = 𝑓2 𝐿2 𝐷2 + 𝐾𝑒𝑥𝑖𝑡 𝑈2 2 2 ℎ𝑙𝑇01 = 𝐾𝑒𝑛𝑡 𝑈1 2 2 + 𝑓1 𝐿1 𝐷1 𝑈1 2 2 = 𝐾𝑒𝑛𝑡 + 𝑓1 𝐿1 𝐷1 𝑈1 2 2 ① ② ③ ⓪ The total heads losses are the sum of the major and minor losses, so
  • 8. From continuity, 𝑈1𝐴1 = 𝑈2𝐴2 therefore 𝑈1 = 𝑈2 𝐴2 𝐴1 = 𝑈2 𝐷2 𝐷1 2 Hence 𝐻𝑙𝑇 = 𝐾𝑒𝑛𝑡 + 𝑓1 𝐿1 𝐷1 𝐷2 𝐷1 4 + 𝑓2 𝐿2 𝐷2 + 𝐾𝑒𝑥𝑖𝑡 𝑈2 2 2𝑔 𝐻𝑙𝑇 = ℎ𝑙𝑡 𝑔 = 𝐾𝑒𝑛𝑡 + 𝑓1 𝐿1 𝐷1 𝑈2 2 2𝑔 𝐷2 𝐷1 4 + 𝑓2 𝐿2 𝐷2 + 𝐾𝑒𝑥𝑖𝑡 𝑈2 2 2𝑔 Simplifying this equation results in an equation representative of the Total Head Loss in the pipes as consequence of the Major and Minor Head Losses At the operating point as per the simplified governing equation 𝐻𝑎 = ℎ𝑙𝑡 𝑔 = ℎ𝑙𝑇01 𝑔 + ℎ𝑙𝑇23 𝑔 = 𝐻𝑙𝑇 the head loss is equal to the head produced nay the pump given by 𝐻 = 𝐻0 − 𝐴𝑉 · 2 where 𝐻0 = 17𝑚 and 𝐴 = 1.95 × 10−4 𝑚/ (𝑚3 /ℎ)2
  • 9. 𝐻𝑙𝑇 = ℎ𝑙𝑡 𝑔 = 𝐾𝑒𝑛𝑡 + 𝑓1 𝐿1 𝐷1 𝑈1 2 2𝑔 + 𝑓2 𝐿2 𝐷2 + 𝐾𝑒𝑥𝑖𝑡 𝑈2 2 2𝑔 𝑈1 = 𝑈2 𝐴2 𝐴1 = 𝑈2 𝐷2 𝐷1 2 ℎ𝑙𝑇23 = 𝑓2 𝐿2 𝐷2 𝑈2 2 2 + 𝐾𝑒𝑥𝑖𝑡 𝑈2 2 2 = 𝑓2 𝐿2 𝐷2 + 𝐾𝑒𝑥𝑖𝑡 𝑈2 2 2 ℎ𝑙𝑇01 = 𝐾𝑒𝑛𝑡 𝑈1 2 2 + 𝑓1 𝐿1 𝐷1 𝑈1 2 2 = 𝐾𝑒𝑛𝑡 + 𝑓1 𝐿1 𝐷1 𝑈1 2 2 Pipe on the Left side of the pump Pipe on the Right side of the pump 𝐻𝑙𝑇 = ℎ𝑙𝑇01 + ℎ𝑙𝑇23 Add these two equations to get the total head loss in the system But from continuity Gives 𝑈1 2 = 𝑈2 𝐷2 𝐷1 2 2 = 𝑈2 2 𝐷2 𝐷1 4 𝐻𝑙𝑇 = ℎ𝑙𝑡 𝑔 = 𝐾𝑒𝑛𝑡 + 𝑓1 𝐿1 𝐷1 1 2𝑔 𝑈2 2 × 𝐷2 𝐷1 4 + 𝑓2 𝐿2 𝐷2 + 𝐾𝑒𝑥𝑖𝑡 𝑈2 2 2𝑔
  • 10. Table 1 - Data given in question or sourced from fluids tables Given Data Value Units Source Water at 25 Degrees Pipe Diameter D1 25 cm Pipe Diameter D2 20 cm ε 2.6E-04 m Tables Patm 101.3 kPa Kinematic Viscosity 8.96E-07 m2/s Tables Density 997 kg/m3 z1 0 m z2 0 m Lsuction 0.6 m Side of pump Ldelivery 900 m Side of pump LT 900.6 m For minor losses, K Reentrant 0.5 Tables Sudden Expansion 1 Tables KT 1.5 Compile all available data into a table Some data needs to be sourced from reference tables, databases etc… EXAMPLE Table: Roughness for pipes of common Engineering Materials Pipe Roughness, ε (mm) Riveted steel 0.9 - 9 Concrete 0.3 - 3 Wood Stave 0.2 - 0.9 Cast Iron 0.26 Glavanised Iron 0.15 Asphalted Cast Iron 0.12 Commercial Steel or Wrought Iron 0.046 Dran Tubing 0.0015
  • 11. Generate a data table populated with the calculations distilled from the formulas developed Table 2 - Data Tabulation and Analysis to determine the operating point of the system Volumetric Flow Rate (m3/hr) U1 (m/s) Reynolds Number, Re Friction Factor, f1 U2 (m/s) Reynolds Number, Re Friction Factor, f2 New Pipes (m) Pump Curve (m) 0 0.00 0.00 0.0000 0.00 0.00 0.00 0.00 17.00 25 0.14 39457.07 0.0246 0.22 49321.34 0.0246 0.28 16.88 50 0.28 78914.14 0.0226 0.44 98642.68 0.0230 1.04 16.51 75 0.42 118371.21 0.0218 0.66 147964.02 0.0224 2.28 15.90 100 0.57 157828.28 0.0214 0.88 197285.35 0.0221 4.00 15.05 125 0.71 197285.35 0.0211 1.10 246606.69 0.0219 6.19 13.95 150 0.85 236742.42 0.0209 1.33 295928.03 0.0217 8.86 12.61 175 0.99 276199.49 0.0208 1.55 345249.37 0.0216 12.01 11.03 200 1.13 315656.57 0.0207 1.77 394570.71 0.0215 15.63 9.20 225 1.27 355113.64 0.0206 1.99 443892.05 0.0215 19.73 7.13 250 1.41 394570.71 0.0205 2.21 493213.38 0.0214 24.31 4.81
  • 12. Table 2 - Data Tabulation and Analysis to determine the operating point of the system Volumetric Flow Rate (m3/hr) U1 (m/s) Reynolds Number, Re Friction Factor, f1 U2 (m/s) Reynolds Number, Re Friction Factor, f2 New Pipes (m) Pump Curve (m) 0 0.00 0.00 0.0000 0.00 0.00 0.00 0.00 17.00 25 0.14 39457.07 0.0246 0.22 49321.34 0.0246 0.28 16.88 50 0.28 78914.14 0.0226 0.44 98642.68 0.0230 1.04 16.51 75 0.42 118371.21 0.0218 0.66 147964.02 0.0224 2.28 15.90 100 0.57 157828.28 0.0214 0.88 197285.35 0.0221 4.00 15.05 125 0.71 197285.35 0.0211 1.10 246606.69 0.0219 6.19 13.95 150 0.85 236742.42 0.0209 1.33 295928.03 0.0217 8.86 12.61 175 0.99 276199.49 0.0208 1.55 345249.37 0.0216 12.01 11.03 200 1.13 315656.57 0.0207 1.77 394570.71 0.0215 15.63 9.20 225 1.27 355113.64 0.0206 1.99 443892.05 0.0215 19.73 7.13 250 1.41 394570.71 0.0205 2.21 493213.38 0.0214 24.31 4.81 𝐻𝑙𝑇 = 𝐾𝑒𝑛𝑡 + 𝑓1 𝐿1 𝐷1 𝐷2 𝐷1 4 + 𝑓2 𝐿2 𝐷2 + 𝐾𝑒𝑥𝑖𝑡 𝑈2 2 2𝑔 𝐻 = 17 − 1.95 × 10−4 𝑉 · 2
  • 13. Plot the relevant curves and identify the operating point for the pumping system. 0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 16.00 18.00 20.00 0 50 100 150 200 250 Pump Head (m) Volumetric Flow Rate in m^3/hr New Pipes (m) Where curves cross is the optimum operating point for the system. The graphical solution is shown on the plot. At the operating point H ≈ 11.4 m and the volumetric flow rate, 170 m3/h.
  • 14. As pipes age with time, build up forms on the inner walls. This must be taken into account when designing the system. Typical multipliers are available that can be applied to the calculations Table 2 - Data Tabulation and Analysis to determine the operating point of the system Volumetric Flow Rate (m3/hr) U1 (m/s) Reynolds Number, Re Friction Factor, f1 U2 (m/s) Reynolds Number, Re Friction Factor, f2 New Pipes (m) Age pipes 10 years Age pipes 20 years Pump Curve (m) 0 0.00 0.00 0.0000 0.00 0.00 0.00 0.00 17.00 25 0.14 39457.07 0.0246 0.22 49321.34 0.0246 0.28 16.88 50 0.28 78914.14 0.0226 0.44 98642.68 0.0230 1.04 16.51 75 0.42 118371.21 0.0218 0.66 147964.02 0.0224 2.28 15.90 100 0.57 157828.28 0.0214 0.88 197285.35 0.0221 4.00 15.05 125 0.71 197285.35 0.0211 1.10 246606.69 0.0219 6.19 13.95 150 0.85 236742.42 0.0209 1.33 295928.03 0.0217 8.86 12.61 175 0.99 276199.49 0.0208 1.55 345249.37 0.0216 12.01 11.03 200 1.13 315656.57 0.0207 1.77 394570.71 0.0215 15.63 9.20 225 1.27 355113.64 0.0206 1.99 443892.05 0.0215 19.73 7.13 250 1.41 394570.71 0.0205 2.21 493213.38 0.0214 24.31 4.81 Add two new columns to the table Table 2: Typical Multipliers applied to friction factors, with ageing pipes Pipe Age (Years) Small Pipes, 100 - 250 mm Large Pipes, 300 - 1500 mm New 1.00 1.00 10 2.2 1.60 20 5.00 2.00 30 7.25 2.20 40 8.75 2.40 50 9.6 2.86 60 10.0 3.70 70 10.1 4.70 From the table locate the multipliers for the ageing pipe Multiply these values by the fraction factor for that condition
  • 15. Given that each pipe is ≤ 250 mm diameter the multiplier for 10 years is 2.2 and for 20 years is 5 Table 3 - Data Tabulation and Analysis to determine the operating point of the system Volumetric Flow Rate (m3/hr) U1 (m/s) Reynolds Number, Re Friction Factor, f1 U2 (m/s) Reynolds Number, Re Friction Factor, f2 New Pipes (m) Ageing pipes 10 years Ageing pipes 20 years 0 0.00 0.00 0.0000 0.00 0.00 0.00 0.00 0.00 0.00 17.00 25 0.14 39457.07 0.0246 0.22 49321.34 0.0246 0.28 0.61 2.41 16.88 50 0.28 78914.14 0.0226 0.44 98642.68 0.0230 1.04 2.28 9.02 16.51 75 0.42 118371.21 0.0218 0.66 147964.0 2 0.0224 2.28 4.99 19.76 15.90 100 0.57 157828.28 0.0214 0.88 197285.3 5 0.0221 4.00 8.74 34.62 15.05 125 0.71 197285.35 0.0211 1.10 246606.6 9 0.0219 6.19 13.54 53.61 13.95 150 0.85 236742.42 0.0209 1.33 295928.0 3 0.0217 8.86 19.37 76.72 12.61 175 0.99 276199.49 0.0208 1.55 345249.3 7 0.0216 12.01 26.25 103.95 11.03 200 1.13 315656.57 0.0207 1.77 394570.7 1 0.0215 15.63 34.16 135.30 9.20 225 1.27 355113.64 0.0206 1.99 443892.0 5 0.0215 19.73 43.12 170.77 7.13
  • 16. 0.00 2.50 5.00 7.50 10.00 12.50 15.00 17.50 20.00 0 75 150 225 300 Pump Head (m) Volumetric Flow Rate (m^3/hr) New Pipes (m) Ageing pipes 10 years Ageing pipes 20 years Plot the relevant curves and identify the operating point for the ageing pipes in the pumping system. The graphical solution is shown on the plot. At the operating point for new pipes H ≈ 11.4 m and the volumetric flow rate, 170 m3/h, for 10 year old pipes H ≈ 13.75 m and the volumetric flow rate, 127 m3/h, and for 10 year old pipes H ≈ 16 m and the volumetric flow rate, 67 m3/h.