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.
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
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.