CENTRIFUGAL PUMP DESIGN 1
The client will usually specify specify the desired head and pump capacity. The type and speed of the driver may also be specified. Speed is governed by considerations considerations of cost and efficiency as well as drivers available to the client. client. Given these parameters, the task of the engineer is to minimize cost. Which cost to minimize, first cost or life-cycle cost, however, is an important consideration. From From a life life cycl cyclee viewp viewpoin oint, t, we must take take into into accoun accountt power power consu consumpt mption ion and operat operation ion and maintenance costs. These considerations call for optimizing optimizing efficiency, reliability (the mean time between between failure) and maintainability (the mean time to repair). In general, designing to optimize optimize these categories categories results results in increased increased costs. costs. Often, Often, these considerati considerations ons are not very important important and we can design for minimum first cost. In appropriate cases, the engineer should initiate a dialog with with the client concerning available options. For example, designing a boiler feed feed pump that operates continuously would would probably probably call for maximizing efficienc efficiency. y. Efficie Efficiency ncy considerat considerations ions would not be so important, important, however, however, for a drainage pu mp that is only required to operate occasionally.
PIPE CONNECTIONS AND VELOCITIES
The diameter of the suction pipe is usually made larger that the pump suction flange and both are made larger than the discharge flange and pipe. Church recommends keeping the velocity velocity at the suction suction flange about 9 or 10 ft/s and that at the discharge flange between 18 and 25 ft/s.
LEAKAGE LOSSES
To design the impeller, account must be taken of leakage from the discharge side back to the suction side. To reduce reduce the leakage, leakage, wearing rings are fitted fitted to the impeller and casing. These rings are designed designed with specified specified clearances. clearances. The leakage across across each ring can be calculated calculated from the follow following ing formula:
Q L = CA 2 gH L where: C = flow coefficient 2 A = leakage area = π Ds / 2 D = mean clearance diameter s = diametrical clearance
= 0.010 + ( D − 6)(0.001)in3
For small wearing rings with precise machining and ball bearings, the minimum clearance may be reduced to 0.008 in.
H L
=
3
[ (U 4
2 2
2
−U1
) / 2 g ]
3
IMPELLER INLET DIMENSIONS AND VANE ANGLE
The diameter of the impeller eye, Do, eye, Do, is dependent on the shaft diameter, Ds diameter, Ds,, which must initially be approximated. approximated. The hub diameter, 1
This section is based on Church, A.H., Centrifugal Pumps and Blowers, Ch. 6, John Wiley & Sons, Sons, 1950. 2 Id. Fig. 6-1, p. 92. 3 Attributed by Church to Stepanoff, A.J., Trans. A.S.M.E., HYD-54-5, 1932.
1
D H , is made 5/16 to ½ inch larger than Ds. After estimating Ds and D H , Do is based on the known flowrate. The inlet vane edge diameter, D1, is made about the same as Do to ensure smooth flow. EXAMPLE OF IMPELLER DESIGN4
Specified conditions:
1.
Quantity flowrate :
Q= 2.
( 2500 ) gal
ft 3
min
( 60 ) s ( 7.48) gal
min
= 5.57 ft 3 / s
Mass flowrate:
( 5.57)
&= m
3.
Required head: h P = 150ft Required flowrate: Q = 2500 gpm Required speed N = 1760 rpm
ft 3 ( 62.4) lbm ft 3
s
= 348lbm / s
Specific speed: Assume a double suction impeller; then, Q = 2500/2 = 1250gpm, and:
N sd
ω ( rpm) =
[
Q( gpm)
h p ( ft )
]
3/ 4
(1760) 1250 =
(150) 3/ 4
=
1450rpm
For this specific speed, a radial flow pump is indicated. 5 4.
WHP
Water horsepower .
& mgh =
550
(348)lbm (32.2) ft (150) ft =
s
s2
s
hp
⋅
s2
(550) ft lbf 32.2 ft ⋅
WHP = 94.6hp 5.
4 5
Shaft diameter. Calculate shaft diameter based on torque. Increase the calculated value somewhat to allow for bending moment which is unknown at this point and to ensure that the critical speed exceeds the operational speed by a reasonable margin. The bending moment will depend on the weight distribution of the shaft and any unbalanced radial thrust acting on the impeller. From the figure shown below, with the given flow of 2500 gpm and calculated value of specific speed of 1450, we select a tentative value of efficiency of 80%.
See, Church, p. 107-117. See,Munson, Fig. 12.18, p. 812.
2
WHP 94.6
BHP =
Thus:
=
η
0.8
= 118hp
The required shaft torque then is:
T=
W& (118) hp (550) ft ⋅ lbf
ω
=
s ⋅ hp
min
(60) s
rev
(12)in
(1760)rev min (2)(π ) rad
ft
= 4230lbf ⋅ in
Assuming a shear stress of 4000 psi:
D s =
3
16T
π s s
=
3
(16)(4230)lbf ⋅ in ⋅ in 2 (π )(4000)lbf
= 1.75in
To account for the unknown bending moment and critical speed, increase the shaft diameter to 2 1/8 in. Church states that the hub diameter, D H , is made from 5/16 to ½ in. larger than D s: Let D H
6.
= 2 ½ in.
Suction line velocity and diameter of suction flange.
Assume a velocity of 10 ft/s at the suction flange; thus:
Q=
2 V SU (π ) D SU
4
; or, DSU
=
VSU =
4Q (π )VSU
=
(4)(5.57) ft 3 ⋅ s (144)in 2 (π )(10) ft ⋅ s ⋅ ft 2
(4)(5.57) ft 3 (144)in 2 (π ) s(10) 2 in 2 ⋅ ft 2
3
= 10.2 ft / s
= 10.1, say ,10in ;thus,
Assume the velocity at the eye of the impeller is 11 ft/s. For a double suction pump, assume that the leakage will not exceed 2%. Dividing the total flow by 2 gives:
Q (4)(1.02)Q
D0 = 7.
(π )(2)V 0
= V 0 A0 = V0 (
π D02
4
−
(4)(1.02)(5.57)(144)
2 + D H =
(π )(2)(11)
2 π D H
4
)
+ (2.5)2 = 7.33in, say, 7
5 16
in
Wheel inlet dimensions and angle.
Assume an inlet diameter, D1 , of 7 5/16 in.
U1
= ω r =
(1760)( 2)(π )( 7.315) (60)( 2)(12)
= 56.2
ft / s
The radial velocity should be slightly higher than V 0 because a converging shape is more efficient than a divergent one. Let V r be 12 ft/s. The inlet area will be decreased by the vane thickness. Assume a contraction factor, ε16 , of 0.85; the entering width then is:
b1
(1.02)(5.57 )(144 )
Q =
π D1V r 1ε1
=
(π )(2 )(7.31)(12 )(0.85)
Inlet angle: Assume that water enters vanes radially.
β 1 = tan −1
V r 1 U 1
= tan −1
=
1.75in Vr1
12 = 12.10 56.2
W1
β1 U1
β 1 is usually increased slightly to account for contraction of the stream as it passes the inlet edges as well as prerotation. The inlet angle is usually between 10 and 25 degrees7. Let β 1 be 130. 8.
Impeller diameter, D2.
The theoretical head can be found from integrating the force on a differential mass:
dF = dmrω 2 2
2
and
dP =
dF A
;
dm = ρd∀ = ρbrd φ dr
2
ρbrdφdr ⋅ r ω 2 ρω 2 2 2 = ρω ∫ rdr = (r2 − r12 ) ∫1 dP = ∫ 1 brd φ 2 1
U
dφ
6 7
ε1 is generally between 0.8 and 0.9, Church, p. 95. Church, p. 95.
4
but
U
=
r ω
H
and
P =
ρ g
H 2
;hence,
−
H 1
=
P2
−
P 1
ρ g
=
U 22
2
− U 1
2 g
For a closed rotating cylinder containing a fluid, the pressure head developed at the outer rim is:
H 2 =
U 22 2 g
Substituting D2 /2( ω ) for U 2 and solving for D2 :
D2 =
2 2 gH 2
ω
=
2 (2)(32.2) H 2 (60) ( 12 ) (2)(π ) N
=
1840 H 2 N
(12)
Where: H 2 is in feet; N is in rpm; D2 is in inches. Tests have shown that the required impeller diameter can be calculated from this expression by substituting the head corresponding to the best efficiency point for H 2 and then multiplying the right side by an experimentally determined coefficient Φ:
D2
=
1840Φ H
(13)
N
Church 8 gives several charts for Φ which have been based on a large number of tests. Most of the plotted points fall within a range of 0.9 to 1.1. Noting that if the head on test is found to be too high, the impeller diameter can be machined to an appropriate diameter, select 1.05 for Φ ; then:
D2
9.
Outlet vane angle,
(1840)(1.05) 150 =
=
(1760)
13.4in ; say, 13 ½ in.
, and impeller width.
2
The normal range for discharge angles is between 20 and 25 degrees 9. Furthermore, β2 is usually made larger than the inlet angle. Assume β2 = 200. The radial outlet velocity, V r2 , is made the same as, or slightly less than, the radial inlet velocity, V r1. Assume V r2 = 11 ft/s10. Outlet area (based on required flow plus leakage).
A2 =
Q Vr 2
=
(1.02)(5.57) ft 3 ⋅ s (144)in 2 s (11) ft ⋅ ft
2
Assume a contraction width, ε 2 , (based on experience) of 0.925:
8
Church, pp. 199-104. Id., p. 35. 10 Id., p. 110. 9
5
= 74.4in 2
b2 =
Q Vr 2π D2ε 2
(1.02)(5.57) ft 3 ⋅ s ⋅ (144)in 2
=
= 1.896in
s ⋅ (11) ft (π )(13.5)in ⋅ ft 2 (0.925)
10. Outlet velocity diagram. The absolute outlet velocity, V 2 , is used in the design of the volute. We proceed as follows:
U2
=
ω r2
=
(1760) rev ⋅ min(2)(π ) rad (13.5) in ⋅ ft
Theoretical tangential outlet velocity, V
Vθ 2
=
min(60) s(2)(12) in ⋅ rev
= U 2 −
103.7 ft / s
2.
V r 2 tan β 2
=
1037 . −
11 tan 20
0
=
73.5 ft / s
Actual tangential outlet velocity, V 2’.11
The inertia of the rotating fluid causes a circulatory flow opposite to the direction of rotation of the impeller. This flow, superimposed on the outward flow, results in the fluid leaving the impeller at a n angle less than that calculated from angular momentum theory. Thus β 2 must be decreased and , therefore, the absolute angle, α 2 , increased. The effect of circulatory flow is to reduce V 2 and the theoretical head. Church defines a circulatory flow coefficient, η θ , as:
η θ =
V θ ' 2 V θ 2
Church assumes a value of η ∞ of 0.7. This coefficient can be calculated from tests. Pump manufacturers will maintain records from which a reasonable value might be estimated for a given design.
Vθ ' 2
. ft = (0.7)( 73.5) = 515
/s
The outlet vector diagram can now be drawn:
α 2' '
V2
=
=
tan 2
Vr 2
−1
11 515 .
0
= 12.1
'2
+Vθ 2 =
2
11
, say, 13 0
2
. + 515
= 52.7 ft
/s
V2’
V2
α'2
Vr2
α2 V’θ2 Vθ2 U2 11. Cross-section of impeller.
11
See, Church, p. 28 for a discussion of circulatory flow.
6
β2 Vr2
β2’
Wall and vane thicknesses are usually made a minimum consistent with good foundry practice. The stresses due to centrifugal force and fluid pressure are relatively low for average applications; otherwise, they need to be taken into account 12.
Table of Calculated or Assumed Dimensions
b1 = 1.75 in per side b2 = 1.90 in D2 = 13 ½ in D0 = 7 5/16 in Dr = 8 ½ in (to outside of impeller wearing ring) Impeller shroud tip thickness - 3/16 in Connect outlet to inlet by a straight line faired into entrance to provide a smooth transition. Make tip of hub core 3/16 in and fair into hub diameter. The drawing is shown in the figure on the following page. 12. Check leakage loss. From the figure on page 8, the mean diameter of the clearance is 8 ½ in. Let s be the diametral clearance. Church states that the wearing ring clearance for good practice is 0.01 in for rings of 6 in diameter and less. For rings greater than 6 in, increase the clearance by 0.001 in for every inch of ring diameter greater than 6 in:
s
=
0.010 + ( D − 6)( 0.001)
=
0.010 + ( 8.5 − 6)( 0. 001)
=
0. 0125, say, 0. 013in
The clearance area is:
A = π Ds / 2 = (π / 2)(8.5)(0.013) = 0.174in 2 = 0.00121 ft 2 Head across the rings 13: 2 2 3 U 2 − U 1 (3)(103.7 2 − 56.2 2 ) H L = = = 88.5 ft 4 2 g (4 )( 2 )( 32.2 )
From Figure 6-1, p 92, Church, the flow coefficient for 1760 rpm and a 0.013 in clearance is 0.410. Thus, the leakage is:
Q L
= CA
The per cent leakage is
2 gH L
0.075 5.58
= ( 0.410)( 0.00121)
( 2)( 32.2)( 88.5)
=
0. 0375 ft 3 / s
(100) ; or 1.35 %, which is, close enough to the assumed value of 2 %.
12
Id,, p. 152. Church attributes this equation to A.J. Stepanoff: “Leakage Loss and Axial Thrust in Centrifugal Pumps,” A.S.M.E. Trans., HYD-54-5, 1932. 13
7
8
DESIGN OF VANES
0 0 The entrance vane angle, β 1 , has been found to be 13 ; that at the exit, 20 . For smooth flow, we must design the vane such t hat this angle increases smoothly from 13 0 to 200 . We note also that the radial components of velocity to these two angles are 12 and 11 ft/s, respectively. We also see from the vector diagram that W = V r / sin β . The relative velocities corresponding to the entrance and outlet
stations are then: 12 / sin 130
= 53.3 ft
/ s and 11 / sin 20 0
= 32.2
ft / s . To obtain intermediate
values of radii corresponding to intermediate values of the position angle, θ , we proceed as follows (see Fig. 3): 1) Plot β, V r , and W against vane radius, r , for the entrance and outlet stations and connect by a straight line (or a smooth curve). 2) The corresponding values for vane angle, β , are computed from sin β = Vr / W . These values are also plotted against their radii.
Alternatively, write a computer program to perform the above functions. Referring to the figure below:
tan β =
dr rd θ
or
d θ =
dr
dr
r tan β
rdθ
r
180 dr 180 r ∇r θ = = ∑ π ∫ π r 1 r tan β r tan β r 1 0
Note: Use MATLAB or other computer system to perform the integration. Use a sufficiently close spacing of r to obtain a smooth vane shape. 3) Plot the radii against θ to give the shape of the trailing edge of the vane.
9
dθ
r
Draw the front edge of the vane with the same curvature as the back edge with a thickness of about 1/8 in 14.
NUMBER OF VANES
The number of vanes is given by the Pfleiderer equation 15. First, calculate the average vane angle:
β m z = no . vanes = 6.5
=
D2 + D1 D2 − D1
β1
+
β 2
2
=
13 + 20 2
sin β m = ( 6.5)
The circumferential pitch of the vanes is:
=
(13.5 + 7.312 ) sin 16.5 0 = 6.21, say ,6. (13.5 − 7.312 )
(π )( 7.312) =
(6)
0 16.5 ; then,
3.83in
Check the contraction factor: D − π
ε =
ε 1 = 1 − ε 2 = 1 −
zt sin β
π D
=1−
zt
π D sin β
( 6)( 0.125) = 0.855(0.85assumed ) π ( 7.31) sin 13 0
( 6)( 0.125) = 0.948(0.925assumed ) π (13.50) sin 20 0
The assumed values agree reasonably with those calculated.
SUMMARY
Diameter of suction flange, D su ------------------------------- 10in Velocity in suction flange, V su ---------------------------------10.22 ft/s Shaft diameter, D s -----------------------------------------------------------------------2 1/8 in Impeller hub diameter, D H --------------------------------------2 ½ in Impeller eye diameter, D0 --------------------------------------7 5/16 in Velocity through impeller eye,V 0-------------------------------11 ft/s Diameter of inlet vane edge, D1--------------------------------7 5/16 in Velocity at inlet vane edge, V 1 = V r1 ---------------------------12 ft/s 14 15
Church, p. 115. Id.
10
Passage width at inlet, b1 -----------------------------------------1.75 in per side Tangential velocity of inlet vane edge, U 1 -------------------56.2 ft/s Vane angle at inlet, β 1 --------------------------------------------130 Impeller outlet diameter, D2 -------------------------------------13 ½ in Radial component of outlet velocity, V r2 ----------------------11 ft/s Vane angle at outlet, β 2 -------------------------------------------200 Total passage width at outlet, b2 ---------------------------------1.98 in Tangential velocity of outlet vane edge, U 2 -------------------103.7 ft/s '
Absolute velocity leaving impeller, V 2 -------------------------52.5 ft/s '
Tangential component of absolute leaving velocity, V θ 2 ---51.5 ft/s Angle of water leaving impeller,
α 2' ----------------------------130
Number of impeller vanes, z ------------------------------------6
11
