Mechanical shaft seals for pumps - Grundfos

16
Introduction
Calculation example, unbalanced and balanced shaft seal
A = h
In this example, we shall look at the closing force of a liquid-lubricated mechanical shaft seal.
The data below apply to an unbalanced Grundfos type A shaft seal. For more details on this
shaft seal type, see Chapter 2, type A, page 27.
π 2 2
(D — Ds
o ) = π (222 — 162 ) = 179 mm2 4 4
Do
Di
D s
Fig. 1.16: Unbalanced
Grundfos type
A shaft seal
D
D
D s
Fig. 1.17: Balanced Grundfos
type H shaft seal
i
o
Shaft diameter, D = 16 mm
s
Sliding seal face, inner diameter, D = 17 mm
i
Sliding seal face, outside diameter, D = 22 mm
o
Spring force, F = 45 N
s
This gives the following results:
Hydraulically loaded area:
k = A A = s
h 179
= = 1.17
A 153
s
π 2 2
(D — Di
o ) = π (222 — 172 ) = 153 mm2 A = h π 2 2
(D — Ds
o ) = π (222 — 162 ) = 179 mm2 4 4
Sliding face area:
4 4
k = A A = s
h 179
= = 1.17
A 153
s
π 2 2
(D — Di
o ) = π (222 — 172 ) = 153 mm2 A = h π 2 2
(D — Ds
o ) = π (222 — 162 ) = 179 mm2 4 4
4 4
Balancing ratio, according to formula 1, page 15:
k = A A = s
h 179
= = 1.17
A 153
s
π 2 2
(D — Di
o ) = π (222 — 172 ) = 153 mm2 A = h π 2 2
(D — Ds
o ) = π (222 — 162 ) = 179 mm2 4 4
k =
4 4
The closing force, F , at a 10-bar pressure
c
(P = 1 MPa) is calculated as follows:
Ah 179
= = 1.17
A 153
s
k = A A = s
h 179
= = 1.17
A 153
s
π 2 2
(D — Di
o ) = π (222 — 172 ) = 153 mm2 A = h π 2 2
(D — Ds
o ) = π (222 — 162 ) = 179 mm2 4 4
4 4
k = A A = s
h 179
= = 1.17
A 153
s
π 2 2
(D — Di
o ) = π (222 — 172 ) = 153 mm2 A = h π 2 2
(D — Ds
o ) = π (222 — 162 ) = 179 mm2 4 4
4 4
k = Ah 179
= = 1.17
A 153
s
k = A A = s
h 179
= = 1.17
153
π 2 2
(D — Di
o ) = π (222 — 172 ) = 153 mm2 A = h π 2 2
(D — Ds
o ) = π (222 — 162 ) = 179 mm2 4 4
4 4
F = A x P + F = 179 mm c h s 2 k =
x 1 MPa + 45 N = 224 N
Ah 179
= = 1.17
A 153
s
k = A A = s
h 179
= = 1.17
A 153
s
π 2 2
(D — Di
o ) = π (222 — 172 ) = 153 mm2 4 4
k = A A = s
h 179
= = 1.17
A 153
s
π 2 2
(D — Di
o ) = π (222 — 172 ) = 153 mm2 A = h π 2 2
(D — Ds
o ) = π (222 — 162 ) = 179 mm2 4 4
4 4
A s
F = A x P + F = 179 mm c h s 2 k =
x 1 MPa + 45 N = 224 N
Ah 179
= = 1.17
A 153
s
F = A x P + F = 179 mm c h s 2 k =
x 1 MPa + 45 N = 224 N
Ah 179
= = 1.17
A 153
s
F = A x P + F = 179 mm c h s 2 k =
x 1 MPa + 45 N = 224 N
A F = A x P + F = 179 mm c h s
h 179
= = 1.17
A 153
s
2 x 1 MPa + 45 N = 224 N
A = h π 2 2
(D — Ds
o ) = π (222 — 17.12 ) = 150 mm2 For a balanced Grundfos type H shaft seal for a Ø16 shaft, the
calculation is as follows:
Sleeve diameter, D = 17.1 mm
s
Sliding F = A x seal P + Fface, = 179 inner mm c h s diameter, D = 17 mm
i
Sliding 4 seal face, outside 4 diameter, D = 22 mm
o
Spring force, F = 45 N
s
Hydraulically loaded area:
2 k =
x 1 MPa + 45 N = 224 N
Ah 179
= = 1.17
A 153
s
F = A x P + F = 179 mm c h s 2 x 1 MPa + 45 N = 224 N
A = h π 2 2
(D — Ds
o ) = π (222 — 17.12 ) = 150 mm2 A = s
4 4
π 2 2
(D — Di
o ) = π (222 — 172 ) = 153 mm2 A = h π 2 2
(D — Ds
o ) = π (222 — 17.12 ) = 150 mm2 4 4
F = A x P + F = 179 mm c h s
4 4
2 k =
x 1 MPa + 45 N = 224 N
Ah 179
= = 1.17
A 153
s
A = s π 2 2
(D — Di
o ) = π (222 — 172 ) = 153 mm2 A = h π 2 2
(D — Ds
o ) = π (222 — 17.12 ) = 150 mm2 4 4
A = s
4 4
π 2 2
(D — Di
o ) = π (222 — 172 ) = 153 mm2 F = A x P + F = 150 mm c h s 2 A = h
x 1 MPa + 45 N = 195 N
π 2 2
(D — Ds
o ) = π (222 — 17.12 ) = 150 mm2 4 4
A = s
4 4
π 2 2
(D — Di
o ) = π (222 — 172 ) = 153 mm2 F = A x P + F = 150 mm c h s 2 A = h
x 1 MPa + 45 N = 195 N
π 2 2
(D — Ds
o ) = π (222 — 17.12 ) = 150 mm2 A = s
4 4
4 4
π 2 2
(D — Di
o ) = π (222 — 172 ) = 153 mm2 F = A x P + F = 150 mm c h s 2 x 1 MPa + 45 N = 195 N
k = A 4 4
A = s
h 150
= = 0.98
153
π 2 2
(D — Di
o ) = π (222 — 172 ) = 153 mm2 F = A x P + F = 150 mm c h s 2 A = h
x 1 MPa + 45 N = 195 N
π 2 2
(D — Ds
o ) = π (222 — 17.12 ) = 150 mm2 4 4
Sliding face area:
As =
4 4
π 2 2
(D — Di
o ) = π (222 — 172 ) = 153 mm2 F = A x P + F = 150 mm c h s 2 F = A x P + F = 150 mm c h s
4 4
Balancing ratio:
x 1 MPa + 45 N = 195 N
2 x 1 MPa + 45 N = 195 N
k = A A = s h 150
= = 0.98
A 153
s
The closing force, F , at a 10-bar pressure (P = 1 MPa)
c
is calculated as follows:
π 2 2
(D — Di
o ) = π (222 — 172 ) = 153 mm2 F = A x P + F = 150 mm c h s 2 A = h
x 1 MPa + 45 N = 195 N
π 2 2
(D — Ds
o ) = π (222 — 17.12 ) = 150 mm2 4 4
4 4
A s
F c = A h x P + F s = 150 mm 2 x 1 MPa + 45 N = 195 N
k = Ah 150
k = = = 0.98
153
Hydraulically loaded area = Ah Sliding face area As A s
k = A k = h 150
= = 0.98
A k =
h 150
= = 0.98
Ah 150
= = 0.98
k = A 153
s
Ah 150
= = 0.98
153
A s