CALCULATIONS OF A SINGLE STAGE CENTRIFUGAL
PUMP DESIGN
(model for case study)
By
Eng. Mohamed Khalil
1
1. Specification:
𝐻𝑚 = 16 (m) = 59.5 (ft)
Q = 13.5 (m3/hr) = 60 (Gpm)
N = 1450 (rev/min)
𝐻𝑠 = 13.2 (m)
2. General dimensions:
2.1 Specific Speed:
13.5
3.65×N×√Q 3.65×1450×√
3600
𝑁𝑠 = 3 = 3 = 40.512
(𝐻𝑚 )4 (16)4
It is a radial type centrifugal pump.
2.2 Nominal diameter:
3 13.5
3 Q
𝐷1𝑛𝑜𝑚 = 4.5 × 10 × √ = 4.5 × 10 × √
3 3 3600
= 62 ≅ 65 (mm)
N 1450
2.3 Hydraulic Efficiency:
0.42
𝜂ℎ𝑦 = 1 − = 0.84
[log (𝐷1𝑛𝑜𝑚 )−0.172]2
2.4 Volumetric Efficiency:
1 1
𝜂𝑣 = −
2 = −
2 = 0.9454
1+0.68×(𝑁𝑠 ) 3 1+0.68×(40.512) 3
2.5 Overall Efficiency:
From Chart:
𝜂𝑜 = 0.51
2.6 Mechanical Efficiency:
𝜂𝑜 0.51
𝜂𝑚 = = = 0.642
𝜂𝑣 ×𝜂ℎ𝑦 0.9454×0.84
2.7 Output Power:
13.5
𝑃𝑜 = ρg𝐻𝑚 Q = 1000 × 9.8 × 16 × = 588 (watt)
3600
2
2.8 Input Power:
𝑃𝑜 588
𝑃𝑠ℎ = = = 1153 (watt) = 1.55 (hp)
𝜂𝑜 0.51
Assuming an overload of 15% ,
𝑃𝑖𝑛 = 0.15 × 𝑃𝑠ℎ + 𝑃𝑠ℎ = 1.15 × 1153 = 1326 (watt) = 1.326 (kw)
= 1.78 (hp)
2.9 Torque:
𝑃𝑖𝑛 1.326
T= = 2×π×1450 = 0.008732 (KN. m)
ω
60
Taking the shaft material as ‘En8’ (Mild Steel), Ultimate Stress (𝑓𝑚 ) = 35
N/mm2 and taking factor of safety (FS) as 2 for uniform speed of
rotation.
2.10 Working Stress:
𝑓𝑚 35 N KN
𝑓𝑠 = = = 17.5 ( 2) = 1.75 × 104 ( 2)
FS 2 𝑚𝑚 𝑚
2.11 Shaft diameter:
3 16T 3 16×0.008732
𝑑𝑠 = √ =√ = 0.0136 (m) ≅ 14 (mm)
π𝑓 π×1.75×104
𝑠
Taking fatigue stress (bending and shear) into account, minimum shaft
diameter 𝑑𝑠 is taken as 25 (mm), 𝑑𝑠 = 25 (mm)
2.12 Hub diameter:
𝑑ℎ =1.2 × 𝑑𝑠 = 1.2× 25 = 30 (mm)
3. Inlet dimensions:
3.1 Theoretical discharge:
13.5
Q 3600 𝑚3
𝑄𝑡ℎ = = = 0.00397 ( )
𝜂𝑣 0.9454 sec
3.2 The suction pipe diameter (𝑫𝒔 )=(𝑫𝑶 ) [eye diameter]:
𝐷1𝑛𝑜𝑚 62
𝐷𝑠 = = = 55 (mm)
1.14 1.14
3
3.3 The axial velocity at impeller eye:
4𝑄𝑡ℎ 4×0.00397 m
𝐶𝑜 = = = 1.67 ( )
π(𝐷𝑠 )2 π×0.0552 sec
3.4 The diameter of the inlet edge of the impeller blade:
taken as 𝐷1 = 𝐷1𝑛𝑜𝑚 ≅ 65 (mm)
3.5 The flow velocity before the inlet edge of impeller blade:
3 3 m
𝐶𝑚𝑜 = 0.06√𝑄𝑡ℎ (𝑁)2 = 0.06√0.00397 × (1450)2 = 1.22 ( )
sec
m
𝐶𝑚1 = 𝐾1 × 𝐶𝑚𝑜 = 1.4 × 1.22 = 1.708 = ( )
sec
3.6 Inlet breadth:
𝑄𝑡ℎ 0.00397
b= = = 0.0114 (m) = 11.4 (mm)
𝜋𝐷1 𝐶𝑚1 π×0.065×1.708
π𝐷1 N π × 0.065 × 1450 m
𝑢1 = = = 4.935 ( )
60 60 sec
Assuming normal entry, (𝐶𝑢1 = 0), Inlet blade angle ‘𝛽1 ’ will be
𝐶𝑚1 1.708
𝛽10 = tan−1 ( ) = tan−1 ( ) = 19°
𝑢1 4.935
Allowing angle of attack δ ≈ 4.5°
𝛽1 = 𝛽10 + δ = 19 + 4.5 = 23.5°
4. Outlet dimension:
4.1 Overall head:
𝐻𝑚 16
𝐻𝑜 = = = 19.047 (𝑚)
𝜂ℎ𝑦 0.84
𝐶𝑢2
̅ =
Taking 𝐶𝑢2 = 0.5
𝑢2
̅ × 𝑢2 2
𝐻𝑚 𝐶𝑢2 × 𝑢2 𝐶𝑢2
𝐻𝑜 = = =
𝜂ℎ𝑦 𝑔 𝑔
4.2 First Approximation:
4.2.1 Outlet blade velocity:
4
𝑔𝐻𝑚 9.81 × 19.047 𝑚
𝑢2 = √ =√ = 19.33 ( )
̅
𝐶𝑢2 0.5 𝑠𝑒𝑐
4.2.2 Outlet diameter:
60𝑢2 60 × 19.33
𝐷2 = = = 0.2546 (𝑚) = 255 (𝑚𝑚)
𝜋𝑁 𝜋 × 1450
4.2.3 Outlet blade angle:
Taking,
𝑚
𝐶𝑚3 = 0.8𝐶𝑚𝑜 = 0.8 × 1.22 = 0.976 ( )
𝑠𝑒𝑐
𝐾2 = 1.2
𝐾1 = 1.4
𝑤1
= 1.18
𝑤2
𝐾2 𝑤1 𝐶𝑚3
sin 𝛽2 = sin 𝛽1 ∙ ∙ ∙
𝐾1 𝑤2 𝐶𝑚𝑜
1.2
= sin 23.5 × × 11.8 × 0.8 = 0.324
1.4
𝛽2 = 19°
4.2.4 Outlet flow velocity:
𝐶𝑚2 𝐾2 𝐶𝑚3 1.2
= = × 0.8 = 0.687
𝐶𝑚1 𝐾1 𝐶𝑚𝑜 1.4
𝑚
𝐶𝑚2 = 0.687𝐶𝑚1 = 0.687 × 1.708 = 1.1734 ( )
𝑠𝑒𝑐
4.2.5 No. of blades:
𝐷2 +𝐷1 𝛽1 +𝛽2 255+65 23.5+19
Z = 6.5 ∙ ∙ sin ( ) = 6.5 × 225−65 × sin ( )=
𝐷2 −𝐷1 2 2
3.9677 ≅ 4
𝜓 = 0.6(1 + sin 𝛽2 ) = 0.6(1 + sin 19) = 0.795
2𝜓 1 2×0.795 1
P= ∙ 𝑟 = × 65 2 = 0.409
𝑍 1−(𝑟1 )2 4 2 )
2 1−( 225
2
H∞ = (1 + 𝑃)𝐻𝑜 = 1.409 × 19.047 = 26.84
4.3 Second Approximation:
4.3.1 Outlet blade velocity:
𝐶𝑚2 𝐶𝑚2 2 1.1734 1.734 2
𝑢2 =
2 tan 𝛽2
+ √(
2 tan 𝛽2
) + 𝑔𝐻∞ = 2 tan 19 + √(2 tan 19) +
𝑚
9.81 × 26.84 = 18.019 ( )
𝑠𝑒𝑐
5
4.3.2 Outlet diameter:
60𝑢2 60 × 18.019
𝐷2 = = = 0.23734 (𝑚) = 238 (𝑚𝑚)
𝜋𝑁 𝜋 × 1450
D2 from first approximation (D2I = 255 mm) and D2 from second
approximation (D2II = 238 mm). Closely agrees, Final value of outer
diameter D2 is taken as (D2 = 250 mm)
4.3.3 Outlet breadth:
𝐶𝑚2 1.1734
𝐶𝑚3 = = = 0.9778
𝐾2 1.2
𝑄𝑡ℎ 0.00397
𝑏2 = = = 0.00517 (𝑚) = 5.17(𝑚𝑚)
𝜋𝐷2 𝐶𝑚3 𝜋×0.25×0.9778
4.3.4 Verification for flow coefficients:
1 1
𝐾1 = 𝑍𝛿1 = 4×0.005 = 1.33
1− 1−
𝜋𝐷1 sin 𝛽1 𝜋×0.065×sin 23.5
1 1
𝐾2 = 𝑍𝛿2 = 4×0.005 = 1.1
1− 1−
𝜋𝐷2 sin 𝛽2 𝜋×0.25×sin 19
𝐶𝑚1 1.708 𝑚
𝑤1 = = = 4.3 ( )
sin 𝛽1 sin 23.5 𝑠𝑒𝑐
𝐶𝑚2 1.1734 𝑚
𝑤2 = = = 3.6 ( )
sin 𝛽2 sin 19 𝑠𝑒𝑐
𝐾2 1.1
= = 0.827
𝐾1 1.33
𝑤1 4.3
= = 1.194
𝑤2 3.6
𝑄𝑝 = 𝜋𝑏1 𝐷1 (1 − 𝜀1 )𝐶𝑟1
0.00375 = 𝜋 × 0.0114 × 0.065 × (1 − 𝜀1 ) × 1.708
𝜀1 = 0.06 = 6%
𝑄𝑝 = 𝜋𝑏2 𝐷2 (1 − 𝜀2 )𝐶𝑟2
0.00375 = 𝜋 × 0.00517 × .250 × (1 − 𝜀2 ) × 1.1734
𝜀2 = 0.21 = 21%
5. Volute Design:
Calculation of circular volute with 𝑪𝒖 r = constant:
6
720𝜋𝑔 𝐻𝑚
𝐾= ∙
𝜔 𝑄360
2𝜋𝑁 2 × 𝜋 × 1450
𝜔= = = 151.844
60 60
720 × 𝜋 × 9.81 × 16
𝐾= = 623508.61
151.844 × 0.00375
𝑏3 = 𝑏2 + 0.05𝐷2 = 5.17 + 0.05 × 250 = 17.67 (𝑚𝑚)
𝐷3 = 1.04𝐷2 = 1.04 × 250 = 260 (𝑚𝑚)
𝑟3 = 130 (𝑚𝑚)
𝑎3 = 𝑟3 + 𝜌
𝜃° 𝜃°
𝜌 = + √2 𝑟3
𝐾 𝐾
Where:
a: It is the distance from the volute center to the discharge
diameter center.
S. No. 𝜃° 𝜃° 𝜃° 𝜌 (𝑚𝑚) 𝑎3 (𝑚𝑚)
(𝑚) √2 𝑟3 (𝑚)
𝐾 𝐾
1 45 7.22 × 10−5 4.33 × 10−3 4.4022 134.3
2 90 1.44 × 10−4 6.126 × 10−3 6.27 136.27
3 135 2.165 × 10−4 7.502 × 10−3 7.7185 137.7185
4 180 2.886 × 10−4 8.66 × 10−3 8.95 138.95
5 225 3.61 × 10−4 9.7 × 10−3 10.061 140.061
6 270 4.33 × 10−4 0.0106 11.033 141.033
7 315 5.052 × 10−4 0.0115 12.0052 142.0052
8 360 5.774 × 10−4 0.01225 12.83 142.83
6. Total Equivalent Length:
𝐻𝑠 = 13.2 (𝑚)
7
𝐻𝑠𝑠 = 1 (𝑚)
𝐻𝑠𝑑 = 12.2 (𝑚)
𝑚
𝑉𝑠 = 1.2 ( )
𝑠𝑒𝑐
𝑚
𝑉𝑑 = 1.6 ( )
𝑠𝑒𝑐
Just Major Losses:
𝐻𝑚 = 𝐻𝑠 + 𝐻𝑙𝑜𝑠𝑠𝑒𝑠 𝑖𝑛 𝑝𝑖𝑝𝑒𝑠
𝐻𝑙𝑜𝑠𝑠𝑒𝑠 = 𝐻𝑚 − 𝐻𝑠 = 16 − 13.2 = 2.8 (𝑚)
𝑉𝑠 2 𝐿𝑠 𝑉𝑑 2 𝐿𝑑
𝐻𝑙𝑜𝑠𝑠𝑒𝑠 = [𝑓𝑠 ∙ ∙ + 𝑓𝑑 ∙ ∙ ]
2𝑔 𝑑𝑠 2𝑔 𝑑𝑑
𝑄𝑝 = 𝑉𝑠 ∙ 𝐴𝑠 = 𝑉𝑑 ∙ 𝐴𝑑 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
13.5 𝜋
= 1.2 × × 𝑑𝑠 2
3600 4
𝑑𝑠 = 0.063 (𝑚)
13.5 𝜋
= 1.6 × × 𝑑𝑑 2
3600 4
𝑑𝑑 = 0.055 (𝑚)
𝑉∙𝐷
𝑅𝑒 =
𝜈
At temperature = 20° :
−5
𝑚2
𝜈 = 4.5 × 10 ( )
𝑠𝑒𝑐
1.2 × 0.063
𝑅𝑒,𝑠 = = 1680 < 2100 (𝑙𝑎𝑚𝑖𝑛𝑎𝑟)
4.5 × 10−5
1.6 × 0.055
𝑅𝑒,𝑑 = = 1955 < 2100 (𝑙𝑎𝑚𝑖𝑛𝑎𝑟)
4.5 × 10−5
64
𝑓=
𝑅𝑒
𝑓𝑠 = 0.038
𝑓𝑑 = 0.033
(1.2)2 𝐿𝑠 (1.6)2 𝐿𝑑
2.8 = [0.038 × × + 0.033 × × ]
2 × 9.8 0.063 2 × 9.8 0.055
2.8 = 0.044𝐿𝑠 + 0.0783𝐿𝑑
𝐿𝑠 𝐻𝑠𝑠 1
= = = 0.082
𝐿𝑑 𝐻𝑠𝑑 12.2
𝐿𝑠 = 0.082𝐿𝑑
2.8 = 0.044(0.082𝐿𝑑 ) + 0.0783𝐿𝑑
𝐿𝑑 = 34.2 (𝑚)
8
𝐿𝑠 = 2.803 (𝑚)
𝐿𝑠 + 𝐿𝑑 = 37 (𝑚)
ℎ𝑓𝑠 = 0.123 (𝑚)
ℎ𝑓𝑑 = 2.6778 (𝑚)
7. Pressures:
7.1 Suction Pressure:
𝑃𝑠 𝑃𝑎𝑡𝑚 𝑉𝑠 2
= − [ + 𝐻𝑠𝑠 + ℎ𝑓𝑠 ]
𝜌𝑔 𝜌𝑔 2𝑔
𝑃𝑠 100 × 103 1.22
= −[ + 1 + 0.123]
1000 × 9.81 1000 × 9.81 2 × 9.81
𝑃𝑠 = 88263.37 (𝑝𝑎)
7.2 Discharge Pressure:
𝑃𝑑 𝑃𝑎𝑡𝑚 𝑉𝑑 2
= + 𝐻𝑠𝑑 + ℎ𝑓𝑑 −
𝜌𝑔 𝜌𝑔 2𝑔
3
𝑃𝑑 100 × 10 1.62
= + 12.2 + 2.6778 −
1000 × 9.81 1000 × 9.81 2 × 9.81
𝑃𝑑 = 244671.218 (𝑝𝑎)
8. Net Positive Suction Head:
𝑃1 − 𝑃𝑣
𝑁𝑃𝑆𝐻𝑎 = [ ] − 𝐻𝑠𝑠 − ℎ𝑓𝑠
𝜌𝑔
At temperature = 20° :
𝑃𝑣 = 2.33 × 103 (𝑝𝑎)
105 − 2.33 × 103
𝑁𝑃𝑆𝐻𝑎 = − 1 − 0.123 = 8.833
1000 × 9.81
9. Pump Curve:
𝜂𝑜 = 0.51
Diameter of pump is constant
𝜌𝑔𝑄𝑝 𝐻𝑚
𝜂𝑜 = = 0.51
𝑃𝑠ℎ
𝜂𝑜 𝑃𝑠ℎ
𝐻𝑚 =
𝜌𝑔𝑄𝑝
10. System Curve:
𝐻𝑚 = 𝐻𝑠 + 𝐻𝑙𝑜𝑠𝑠𝑒𝑠 𝑖𝑛 𝑝𝑖𝑝𝑒𝑠
9
1 𝐿𝑠 1 1 𝐿𝑑 1
𝐻𝑚 = 𝐻𝑠 + [𝑓𝑠 ∙ ∙ ∙ 2 + 𝑓𝑑 ∙ ∙ ∙ 2 ] 𝑄𝑝 2
2𝑔 𝑑𝑠 𝐴𝑠 2𝑔 𝑑𝑑 𝐴𝑑
𝐻𝑚 = 𝐻𝑠 + 𝐾𝑄𝑝 2
𝐻𝑚 = 13.2 + (199111.111)𝑄𝑝 2
11. Final Values:
Variable Value
𝐻𝑚 16 m = 52.5 ft
Q 13.5 𝑚3 /hr = Gpm
N 1450 rpm
𝐻𝑠 13.2 m
𝑁𝑠 576 in chart - 40.512 rpm
𝜂𝑜 0.51
𝐷1 65 mm
𝜂ℎ𝑦 0.84
𝜂𝑣 0.9454
𝜂𝑚𝑒𝑐ℎ 0.642
𝑃𝑜 588 watt
𝑃𝑖𝑛 1326 watt
T 0.008732 KN.m
𝑑𝑠 25 mm
𝑑ℎ 30 mm
𝐶𝑟1 = 𝐶𝑚1 1.708 m/sec
𝑏1 = 𝐵1 11.4 mm
𝑢1 4.935 m/sec
𝛽1 23.5°
𝑢2 18 m/sec
𝐷2 250 mm
𝛽2 19°
𝐶𝑟2 = 𝐶𝑚2 1.1734 m/sec
P 0.409
𝐻𝑜 19.047 m
H∞ 26.84 m
Z 4
𝑤1 4.3 m/sec
𝑤2 3.6 m/sec
𝑏2 = 𝐵2 5.17 mm
10
𝜔 151.844 rad/sec
𝑄𝑝=𝑄 13.5
360° 𝑚3 /𝑠𝑒𝑐
3600
𝑏3 17.67 mm
𝐷3 260 mm
K 623508.61
𝐻𝑠 13.2 m
𝐻𝑚 16 m
𝐻𝑠𝑠 1m
𝐻𝑠𝑑 12.2 m
𝑉𝑠 1.2 m/sec
𝑉𝑑 1.6 m/sec
𝑑𝑠 63 mm
𝑑𝑑 55 mm
Total Equivalent Length 37 m
𝐿𝑠 2.803 m
𝐿𝑑 34.2 m
ℎ𝑓𝑠 0.123 m
ℎ𝑓𝑑 2.6778 m
𝑃𝑠 88263.37 pa
𝑃𝑑 244671.218 pa
𝑃𝑣𝑎𝑝 2.33× 103 pa
𝑁𝑃𝑆𝐻𝑎 8.833
System losses factor (K) 199111.111
𝜀1 6 %
𝜀2 21 %
11
