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

Keith Vaugh
Keith VaughSTEM Education & Design
โ€ขโ€ข
KV
WORKED EXAMPLES
{Pumping Systems Example 2}
Keith Vaugh BEng (AERO) MEng
PUMPING SYSTEM OPERATING POINT
A centrifugal pump pumps water at 25 ยฐC 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; ๐ป = 7.6 โˆ’ 1.95 ร— 10โˆ’4
๐‘‰
ยท
2
If the pump is scaled to 1750 rpm, the parabolic equation can be shown to be ๐ป = 17 โˆ’ 1.95 ร—
10โˆ’4 ๐‘‰
ยท
2
For this case;
โ€ข Develop an algebraic expression for the general shape of the system resistance curve.
โ€ข Calculate and plot the system resistance curve.
โ€ข 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 ๐ป = ๐ป0 โˆ’ ๐ด๐‘‰
ยท
2
where H0 = 17 m and ๐ด =
1.95 ร— 10โˆ’4
๐‘š/ (๐‘š3
/โ„Ž)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
๐‘“ = โˆ’1.8๐‘™๐‘œ๐‘”10
๐œ–
๐ท
3.7
1.11
โˆ’
6.9
๐‘…๐‘’
โˆ’2
๐‘ƒ๐‘–๐‘›
๐œŒ๐‘”
+
๐‘ˆ๐‘–๐‘›
2
2๐‘”
+ ๐‘ง๐‘–๐‘› =
๐‘ƒ๐‘œ๐‘ข๐‘ก
๐œŒ๐‘”
+
๐‘ˆ๐‘œ๐‘ข๐‘ก
2
2๐‘”
+ ๐‘ง๐‘œ๐‘ข๐‘ก + ๐‘“
๐ฟ
๐ท
๐‘ˆ๐‘๐‘–๐‘๐‘’
2
2๐‘”
+ โˆ‘๐‘“
๐ฟ๐‘’
๐ท
๐‘ˆ๐‘๐‘–๐‘๐‘’
2
2๐‘”
+ โˆ‘๐พ
๐‘ˆ๐‘๐‘–๐‘๐‘’
2
2๐‘”
โˆ’ ๐ป
๐ป =
โ„Ž๐‘
๐‘”
โ„Ž๐ฟ = ๐‘“
๐ฟ
๐ท
๐‘ˆ๐‘๐‘–๐‘๐‘’
2
2๐‘”
+ โˆ‘๐‘“
๐ฟ๐‘’
๐ท
๐‘ˆ๐‘๐‘–๐‘๐‘’
2
2๐‘”
+ โˆ‘๐พ
๐‘ˆ๐‘๐‘–๐‘๐‘’
2
2๐‘”
๐‘ƒ๐‘–๐‘›
๐œŒ๐‘”
+
๐‘ˆ๐‘–๐‘›
2
2๐‘”
+ ๐‘ง๐‘–๐‘› =
๐‘ƒ๐‘œ๐‘ข๐‘ก
๐œŒ๐‘”
+
๐‘ˆ๐‘œ๐‘ข๐‘ก
2
2๐‘”
+ ๐‘ง๐‘œ๐‘ข๐‘ก + โ„Ž๐ฟ โˆ’ ๐ป
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;
๐‘๐‘ƒ๐‘†๐ป๐ด =
๐‘ƒ๐‘๐‘ข๐‘š๐‘ + ๐‘ƒ๐‘Ž๐‘ก๐‘š โˆ’ ๐‘ƒ๐‘ฃ๐‘Ž๐‘๐‘œ๐‘ข๐‘Ÿ
๐œŒ๐‘”
Net Positive Suction Head Available
GOVERNING EQUATIONS
GOVERNING EQUATIONS
Piping system -
๐‘ƒ0
๐œŒ๐‘”
+
๐‘ˆ0
2
2๐‘”
+ ๐‘ง0 + ๐ป๐‘Ž =
๐‘ƒ3
๐œŒ๐‘”
+
๐‘ˆ3
2
2๐‘”
+ ๐‘ง3 +
โ„Ž๐‘™๐‘ก
๐‘”
Friction Factor - ๐‘“ = โˆ’1.8๐‘™๐‘œ๐‘”10
๐œ–
๐ท
3.7
1.11
+
6.9
๐‘…๐‘’
โˆ’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
๐ป๐‘Ž =
โ„Ž๐‘™๐‘ก
๐‘”
=
โ„Ž๐‘™๐‘‡01
๐‘”
+
โ„Ž๐‘™๐‘‡23
๐‘”
= ๐ป๐‘™๐‘‡
Simplifying the Governing Equation
where section โ‘  and โ‘ก are located just upstream and downstream from the pump, respectively.
โ„Ž๐‘™๐‘‡23
= ๐‘“2
๐ฟ2
๐ท2
๐‘ˆ2
2
2
+ ๐พ๐‘’๐‘ฅ๐‘–๐‘ก
๐‘ˆ2
2
2
= ๐‘“2
๐ฟ2
๐ท2
+ ๐พ๐‘’๐‘ฅ๐‘–๐‘ก
๐‘ˆ2
2
2
โ„Ž๐‘™๐‘‡01
= ๐พ๐‘’๐‘›๐‘ก
๐‘ˆ1
2
2
+ ๐‘“1
๐ฟ1
๐ท1
๐‘ˆ1
2
2
= ๐พ๐‘’๐‘›๐‘ก + ๐‘“1
๐ฟ1
๐ท1
๐‘ˆ1
2
2
โ‘  โ‘ก โ‘ข
โ“ช
The total heads losses are the sum of the
major and minor losses, so
From continuity, ๐‘ˆ1๐ด1 = ๐‘ˆ2๐ด2 therefore ๐‘ˆ1 = ๐‘ˆ2
๐ด2
๐ด1
= ๐‘ˆ2
๐ท2
๐ท1
2
Hence
๐ป๐‘™๐‘‡
= ๐พ๐‘’๐‘›๐‘ก + ๐‘“1
๐ฟ1
๐ท1
๐ท2
๐ท1
4
+ ๐‘“2
๐ฟ2
๐ท2
+ ๐พ๐‘’๐‘ฅ๐‘–๐‘ก
๐‘ˆ2
2
2๐‘”
๐ป๐‘™๐‘‡
=
โ„Ž๐‘™๐‘ก
๐‘”
= ๐พ๐‘’๐‘›๐‘ก + ๐‘“1
๐ฟ1
๐ท1
๐‘ˆ2
2
2๐‘”
๐ท2
๐ท1
4
+ ๐‘“2
๐ฟ2
๐ท2
+ ๐พ๐‘’๐‘ฅ๐‘–๐‘ก
๐‘ˆ2
2
2๐‘”
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 ๐ป๐‘Ž =
โ„Ž๐‘™๐‘ก
๐‘”
=
โ„Ž๐‘™๐‘‡01
๐‘”
+
โ„Ž๐‘™๐‘‡23
๐‘”
= ๐ป๐‘™๐‘‡
the head loss is equal to
the head produced nay the pump given by ๐ป = ๐ป0 โˆ’ ๐ด๐‘‰
ยท
2
where ๐ป0 = 17๐‘š and ๐ด = 1.95 ร— 10โˆ’4
๐‘š/ (๐‘š3
/โ„Ž)2
๐ป๐‘™๐‘‡
=
โ„Ž๐‘™๐‘ก
๐‘”
= ๐พ๐‘’๐‘›๐‘ก + ๐‘“1
๐ฟ1
๐ท1
๐‘ˆ1
2
2๐‘”
+ ๐‘“2
๐ฟ2
๐ท2
+ ๐พ๐‘’๐‘ฅ๐‘–๐‘ก
๐‘ˆ2
2
2๐‘”
๐‘ˆ1 = ๐‘ˆ2
๐ด2
๐ด1
= ๐‘ˆ2
๐ท2
๐ท1
2
โ„Ž๐‘™๐‘‡23
= ๐‘“2
๐ฟ2
๐ท2
๐‘ˆ2
2
2
+ ๐พ๐‘’๐‘ฅ๐‘–๐‘ก
๐‘ˆ2
2
2
= ๐‘“2
๐ฟ2
๐ท2
+ ๐พ๐‘’๐‘ฅ๐‘–๐‘ก
๐‘ˆ2
2
2
โ„Ž๐‘™๐‘‡01
= ๐พ๐‘’๐‘›๐‘ก
๐‘ˆ1
2
2
+ ๐‘“1
๐ฟ1
๐ท1
๐‘ˆ1
2
2
= ๐พ๐‘’๐‘›๐‘ก + ๐‘“1
๐ฟ1
๐ท1
๐‘ˆ1
2
2
Pipe on the Left side of the pump
Pipe on the Right side of the pump
๐ป๐‘™๐‘‡
= โ„Ž๐‘™๐‘‡01
+ โ„Ž๐‘™๐‘‡23
Add these two equations to get the total head loss in the system
But from continuity
Gives
๐‘ˆ1
2
= ๐‘ˆ2
๐ท2
๐ท1
2 2
= ๐‘ˆ2
2
๐ท2
๐ท1
4
๐ป๐‘™๐‘‡
=
โ„Ž๐‘™๐‘ก
๐‘”
= ๐พ๐‘’๐‘›๐‘ก + ๐‘“1
๐ฟ1
๐ท1
1
2๐‘”
๐‘ˆ2
2
ร—
๐ท2
๐ท1
4
+ ๐‘“2
๐ฟ2
๐ท2
+ ๐พ๐‘’๐‘ฅ๐‘–๐‘ก
๐‘ˆ2
2
2๐‘”
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
ฮต 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 etcโ€ฆ
EXAMPLE Table: Roughness for pipes of common Engineering Materials
Pipe Roughness, ฮต (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
๐ป๐‘™๐‘‡
= ๐พ๐‘’๐‘›๐‘ก + ๐‘“1
๐ฟ1
๐ท1
๐ท2
๐ท1
4
+ ๐‘“2
๐ฟ2
๐ท2
+ ๐พ๐‘’๐‘ฅ๐‘–๐‘ก
๐‘ˆ2
2
2๐‘”
๐ป = 17 โˆ’ 1.95 ร— 10โˆ’4 ๐‘‰
ยท
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 โ‰ˆ 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 โ‰ค 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 โ‰ˆ 11.4 m and the
volumetric flow rate, 170
m3/h, for 10 year old pipes
H โ‰ˆ 13.75 m and the
volumetric flow rate, 127
m3/h, and for 10 year old
pipes H โ‰ˆ 16 m and the
volumetric flow rate, 67
m3/h.
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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 ยฐC 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; ๐ป = 7.6 โˆ’ 1.95 ร— 10โˆ’4 ๐‘‰ ยท 2 If the pump is scaled to 1750 rpm, the parabolic equation can be shown to be ๐ป = 17 โˆ’ 1.95 ร— 10โˆ’4 ๐‘‰ ยท 2 For this case; โ€ข Develop an algebraic expression for the general shape of the system resistance curve. โ€ข Calculate and plot the system resistance curve. โ€ข 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 ๐ป = ๐ป0 โˆ’ ๐ด๐‘‰ ยท 2 where H0 = 17 m and ๐ด = 1.95 ร— 10โˆ’4 ๐‘š/ (๐‘š3 /โ„Ž)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 ๐‘“ = โˆ’1.8๐‘™๐‘œ๐‘”10 ๐œ– ๐ท 3.7 1.11 โˆ’ 6.9 ๐‘…๐‘’ โˆ’2 ๐‘ƒ๐‘–๐‘› ๐œŒ๐‘” + ๐‘ˆ๐‘–๐‘› 2 2๐‘” + ๐‘ง๐‘–๐‘› = ๐‘ƒ๐‘œ๐‘ข๐‘ก ๐œŒ๐‘” + ๐‘ˆ๐‘œ๐‘ข๐‘ก 2 2๐‘” + ๐‘ง๐‘œ๐‘ข๐‘ก + ๐‘“ ๐ฟ ๐ท ๐‘ˆ๐‘๐‘–๐‘๐‘’ 2 2๐‘” + โˆ‘๐‘“ ๐ฟ๐‘’ ๐ท ๐‘ˆ๐‘๐‘–๐‘๐‘’ 2 2๐‘” + โˆ‘๐พ ๐‘ˆ๐‘๐‘–๐‘๐‘’ 2 2๐‘” โˆ’ ๐ป ๐ป = โ„Ž๐‘ ๐‘” โ„Ž๐ฟ = ๐‘“ ๐ฟ ๐ท ๐‘ˆ๐‘๐‘–๐‘๐‘’ 2 2๐‘” + โˆ‘๐‘“ ๐ฟ๐‘’ ๐ท ๐‘ˆ๐‘๐‘–๐‘๐‘’ 2 2๐‘” + โˆ‘๐พ ๐‘ˆ๐‘๐‘–๐‘๐‘’ 2 2๐‘” ๐‘ƒ๐‘–๐‘› ๐œŒ๐‘” + ๐‘ˆ๐‘–๐‘› 2 2๐‘” + ๐‘ง๐‘–๐‘› = ๐‘ƒ๐‘œ๐‘ข๐‘ก ๐œŒ๐‘” + ๐‘ˆ๐‘œ๐‘ข๐‘ก 2 2๐‘” + ๐‘ง๐‘œ๐‘ข๐‘ก + โ„Ž๐ฟ โˆ’ ๐ป 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; ๐‘๐‘ƒ๐‘†๐ป๐ด = ๐‘ƒ๐‘๐‘ข๐‘š๐‘ + ๐‘ƒ๐‘Ž๐‘ก๐‘š โˆ’ ๐‘ƒ๐‘ฃ๐‘Ž๐‘๐‘œ๐‘ข๐‘Ÿ ๐œŒ๐‘” Net Positive Suction Head Available GOVERNING EQUATIONS
  • 6. GOVERNING EQUATIONS Piping system - ๐‘ƒ0 ๐œŒ๐‘” + ๐‘ˆ0 2 2๐‘” + ๐‘ง0 + ๐ป๐‘Ž = ๐‘ƒ3 ๐œŒ๐‘” + ๐‘ˆ3 2 2๐‘” + ๐‘ง3 + โ„Ž๐‘™๐‘ก ๐‘” Friction Factor - ๐‘“ = โˆ’1.8๐‘™๐‘œ๐‘”10 ๐œ– ๐ท 3.7 1.11 + 6.9 ๐‘…๐‘’ โˆ’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. ๐ป๐‘Ž = โ„Ž๐‘™๐‘ก ๐‘” = โ„Ž๐‘™๐‘‡01 ๐‘” + โ„Ž๐‘™๐‘‡23 ๐‘” = ๐ป๐‘™๐‘‡ Simplifying the Governing Equation where section โ‘  and โ‘ก are located just upstream and downstream from the pump, respectively. โ„Ž๐‘™๐‘‡23 = ๐‘“2 ๐ฟ2 ๐ท2 ๐‘ˆ2 2 2 + ๐พ๐‘’๐‘ฅ๐‘–๐‘ก ๐‘ˆ2 2 2 = ๐‘“2 ๐ฟ2 ๐ท2 + ๐พ๐‘’๐‘ฅ๐‘–๐‘ก ๐‘ˆ2 2 2 โ„Ž๐‘™๐‘‡01 = ๐พ๐‘’๐‘›๐‘ก ๐‘ˆ1 2 2 + ๐‘“1 ๐ฟ1 ๐ท1 ๐‘ˆ1 2 2 = ๐พ๐‘’๐‘›๐‘ก + ๐‘“1 ๐ฟ1 ๐ท1 ๐‘ˆ1 2 2 โ‘  โ‘ก โ‘ข โ“ช The total heads losses are the sum of the major and minor losses, so
  • 8. From continuity, ๐‘ˆ1๐ด1 = ๐‘ˆ2๐ด2 therefore ๐‘ˆ1 = ๐‘ˆ2 ๐ด2 ๐ด1 = ๐‘ˆ2 ๐ท2 ๐ท1 2 Hence ๐ป๐‘™๐‘‡ = ๐พ๐‘’๐‘›๐‘ก + ๐‘“1 ๐ฟ1 ๐ท1 ๐ท2 ๐ท1 4 + ๐‘“2 ๐ฟ2 ๐ท2 + ๐พ๐‘’๐‘ฅ๐‘–๐‘ก ๐‘ˆ2 2 2๐‘” ๐ป๐‘™๐‘‡ = โ„Ž๐‘™๐‘ก ๐‘” = ๐พ๐‘’๐‘›๐‘ก + ๐‘“1 ๐ฟ1 ๐ท1 ๐‘ˆ2 2 2๐‘” ๐ท2 ๐ท1 4 + ๐‘“2 ๐ฟ2 ๐ท2 + ๐พ๐‘’๐‘ฅ๐‘–๐‘ก ๐‘ˆ2 2 2๐‘” 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 ๐ป๐‘Ž = โ„Ž๐‘™๐‘ก ๐‘” = โ„Ž๐‘™๐‘‡01 ๐‘” + โ„Ž๐‘™๐‘‡23 ๐‘” = ๐ป๐‘™๐‘‡ the head loss is equal to the head produced nay the pump given by ๐ป = ๐ป0 โˆ’ ๐ด๐‘‰ ยท 2 where ๐ป0 = 17๐‘š and ๐ด = 1.95 ร— 10โˆ’4 ๐‘š/ (๐‘š3 /โ„Ž)2
  • 9. ๐ป๐‘™๐‘‡ = โ„Ž๐‘™๐‘ก ๐‘” = ๐พ๐‘’๐‘›๐‘ก + ๐‘“1 ๐ฟ1 ๐ท1 ๐‘ˆ1 2 2๐‘” + ๐‘“2 ๐ฟ2 ๐ท2 + ๐พ๐‘’๐‘ฅ๐‘–๐‘ก ๐‘ˆ2 2 2๐‘” ๐‘ˆ1 = ๐‘ˆ2 ๐ด2 ๐ด1 = ๐‘ˆ2 ๐ท2 ๐ท1 2 โ„Ž๐‘™๐‘‡23 = ๐‘“2 ๐ฟ2 ๐ท2 ๐‘ˆ2 2 2 + ๐พ๐‘’๐‘ฅ๐‘–๐‘ก ๐‘ˆ2 2 2 = ๐‘“2 ๐ฟ2 ๐ท2 + ๐พ๐‘’๐‘ฅ๐‘–๐‘ก ๐‘ˆ2 2 2 โ„Ž๐‘™๐‘‡01 = ๐พ๐‘’๐‘›๐‘ก ๐‘ˆ1 2 2 + ๐‘“1 ๐ฟ1 ๐ท1 ๐‘ˆ1 2 2 = ๐พ๐‘’๐‘›๐‘ก + ๐‘“1 ๐ฟ1 ๐ท1 ๐‘ˆ1 2 2 Pipe on the Left side of the pump Pipe on the Right side of the pump ๐ป๐‘™๐‘‡ = โ„Ž๐‘™๐‘‡01 + โ„Ž๐‘™๐‘‡23 Add these two equations to get the total head loss in the system But from continuity Gives ๐‘ˆ1 2 = ๐‘ˆ2 ๐ท2 ๐ท1 2 2 = ๐‘ˆ2 2 ๐ท2 ๐ท1 4 ๐ป๐‘™๐‘‡ = โ„Ž๐‘™๐‘ก ๐‘” = ๐พ๐‘’๐‘›๐‘ก + ๐‘“1 ๐ฟ1 ๐ท1 1 2๐‘” ๐‘ˆ2 2 ร— ๐ท2 ๐ท1 4 + ๐‘“2 ๐ฟ2 ๐ท2 + ๐พ๐‘’๐‘ฅ๐‘–๐‘ก ๐‘ˆ2 2 2๐‘”
  • 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 ฮต 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 etcโ€ฆ EXAMPLE Table: Roughness for pipes of common Engineering Materials Pipe Roughness, ฮต (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 ๐ป๐‘™๐‘‡ = ๐พ๐‘’๐‘›๐‘ก + ๐‘“1 ๐ฟ1 ๐ท1 ๐ท2 ๐ท1 4 + ๐‘“2 ๐ฟ2 ๐ท2 + ๐พ๐‘’๐‘ฅ๐‘–๐‘ก ๐‘ˆ2 2 2๐‘” ๐ป = 17 โˆ’ 1.95 ร— 10โˆ’4 ๐‘‰ ยท 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 โ‰ˆ 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 โ‰ค 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 โ‰ˆ 11.4 m and the volumetric flow rate, 170 m3/h, for 10 year old pipes H โ‰ˆ 13.75 m and the volumetric flow rate, 127 m3/h, and for 10 year old pipes H โ‰ˆ 16 m and the volumetric flow rate, 67 m3/h.