Sliding contact Problem 2.pdf

1. Sliding Contact Bearings
Navaneeth Krishna Vernekar V

2. Problem Statement 2
The main bearing of a steam turbine runs at 1500 rpm and the
diameter of the journal is 40 mm. the load on the bearing is estimated
to be 3kN. Determine
i. Whether fluid film lubrication can be expected
ii. If artificial cooling is necessary
iii. Amount of oil flow
iv. Minimum film thickness
(Take temperature of lubricating oil as 60⁰C)
Mechanical Design 2 2

3. Given
N=1500 rpm
W=3kN
d=40mm
From T 15.7 /page 366 for steam turbine
p= 0.7 to 1.9 MPa
𝑍𝑁
𝑝 𝑚𝑖𝑛
=14.5
(C/r) = 0.001
(l/d) = 1 to 2 (Assume l/d =1.5)
Mechanical Design 2 3

4. Solution
𝑃 = Τ
𝑊
2𝑟𝑙 = Τ
3000
1.5×402 = 1.25 𝑀𝑃𝑎 < 1.9 𝑀𝑃𝑎 (E15.6 c page 355)
According to Mckee’s Investigation
𝑍𝑁
𝑃
≥ 3
𝑍𝑁
𝑃 𝑚𝑖𝑛
𝑍𝑁
𝑃
= 3 × 14.5 = 43.5
𝑍 =
43.5 × 1.25
1500
= 0.036 ൗ
𝑁𝑠
𝑚2
Mechanical Design 2 4

5. Note: Viscosity
𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒 = 𝜌 × 1000
𝑘𝑖𝑛𝑒𝑚𝑎𝑡𝑖𝑐 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦𝑍𝑘 =
𝐷𝑦𝑛𝑎𝑚𝑖𝑐 𝑉𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦
𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒
𝑘𝑖𝑛𝑒𝑚𝑎𝑡𝑖𝑐 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦𝑍𝑘′ =
𝐷𝑦𝑛𝑎𝑚𝑖𝑐 𝑉𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝑁𝑠
𝑚2 × 10−3
𝜌
𝑍𝑘
′
=
𝑍 𝐶𝑃
𝜌
; 𝑍𝑘
′
𝑖𝑠 𝑖𝑛 𝐶𝑒𝑛𝑡𝑖𝑠𝑡𝑜𝑘𝑒𝑠
Refer page 352 of DDHB
Mechanical Design 2 5

6. Assume 𝜌15 = 0.9 (Table 15.1)
𝜌60 = 0.9 − 0.00063 × 60 − 15
= 0.87 … 𝐸15.1𝑓/350
𝑍𝑘
′
=
𝑍 (𝑖𝑛 𝐶𝑒𝑛𝑡𝑖𝑝𝑜𝑖𝑠𝑒)
𝜌𝑡
=
36
0.87
= 41.4 𝐶𝑆𝑡
The point that is to be referred in fig
15.3 page 352 doesn’t lie on any curve,
in such case always chose a curve higher
than the point.
Mechanical Design 2 6

7. Curve G is selected. As per table
15.1/363 curve G is SAE 40 oil with
𝜌15 = 0.9275
𝜌60 = 0.9275 − 0.00063 × 60 − 15
= 0.89915
• from figure 15.3 page 352 for
temperature 60⁰ and curve G find
Kinematic viscosity in centistokes
𝑍𝑘
′
= 53 𝐶𝑠𝑡
Mechanical Design 2 7

8. 𝑍𝑘
′
= 53 =
𝑍 (𝑖𝑛 𝐶𝑒𝑛𝑡𝑖𝑝𝑜𝑖𝑠𝑒)
0.89915
→ 𝑍 = 47.6 𝐶𝑃
𝑍𝑁
𝑃
=
47.6 × 10−3
× 1500
1.25
= 57.184
3
𝑍𝑁
𝑃 𝑚𝑖𝑛
= 3 × 14.5 = 43.5
Since
𝑍𝑁
𝑃
> 3
𝑍𝑁
𝑃 𝑚𝑖𝑛
hence oil selected is satisfactory. Thick film lubrication exists.
Mechanical Design 2 8

9. Heat generated
𝐻𝑔 = 𝑓𝑤𝑣 ⋯ 15.6𝑗
356
𝑣 =
𝜋 × 40 × ൗ
1500
60
1000
= 3.1416 𝑚/𝑠
𝑓 = 𝐾𝑎
𝑍𝑛
𝑃
𝑟
𝑐
× 10−10
+ ∆𝑓 ⋯ 15.4𝑏
353
𝑤ℎ𝑒𝑟𝑒 𝐾𝑎 = 0.195 × 106; ∆𝑓 = 0.002
𝑓 = 0.195 × 106
57.184
60
1000 × 10−10 + 0.002 = 0.0206
∴ 𝐻𝑔= 0.0206 × 3000 × 3.1416 = 194.15 𝐽/𝑠
Mechanical Design 2 9

10. Heat Dissipated
𝐻𝑑 = 𝐾𝑝𝑙𝑑 × 10−6 ⋯ 15.11𝑐
360
𝐾𝑝 =
𝑡𝐵 − 𝑡𝐴 + 18 2
𝑘𝑝
𝑘𝑝 = 0.484 𝑠𝑚2 Τ
℃ 𝐽 𝑓𝑜𝑟 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝑜𝑟 𝑚𝑒𝑑𝑖𝑢𝑚 𝑐𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛 𝑖𝑛 𝑠𝑡𝑖𝑙𝑙 𝑎𝑖𝑟
𝐴𝑠𝑠𝑢𝑚𝑒 𝑘𝑝 𝑎𝑠 𝑎𝑏𝑜𝑣𝑒 𝑖𝑓 𝑛𝑜𝑡ℎ𝑖𝑛𝑔 𝑎𝑏𝑜𝑢𝑡 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑐𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑚𝑒𝑛𝑡𝑖𝑜𝑛𝑒𝑑.
Mechanical Design 2 10

11. Heat Dissipated
𝑡𝐵 − 𝑡𝐴 =
1
2
𝑡𝑜 − 𝑡𝐴 =
1
2
60 − 30 = 15
𝐾𝑝 =
15 + 18 2
0.484
= 2250 𝑗/𝑠𝑚2
𝐻𝑑 = 2250 × 60 × 40 × 10−6
= 5.4 𝐽/𝑠
𝑆𝑖𝑛𝑐𝑒 𝐻𝑑 < 𝐻𝑔, 𝐴𝑟𝑡𝑖𝑓𝑖𝑐𝑖𝑎𝑙 𝑐𝑜𝑜𝑙𝑖𝑛𝑔 𝑖𝑠 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑
Mechanical Design 2 11

12. Oil Flow Rate
Figure 15.13 page 358
𝑆 =
𝑟
𝑐
2 𝑍𝑁
𝑃
× 10−6 𝑠/𝑚𝑖𝑛
𝑆 = 1000 2
57.184 × 10−6
= 57.184 𝑠/𝑚𝑖𝑛
From Figure 15.13 for S=57.184 in x axis and ꞵ=360⁰
we get
𝑄
𝑟𝑐𝑁𝑙
= 3.1
𝑄 = 3.1 × 2 × 10−5 × 0.02 × 1500 × 1.5 × 0.04
= 1.116 × 10−4 ൗ
𝑚3
𝑠
Mechanical Design 2 12

13. Minimum film thickness
Figure 15.9 page 356
𝑆 = 57.184 𝑠/𝑚𝑖𝑛
for S=57.184 in x axis and ꞵ=360⁰ we
get
ℎ𝑜
𝑐
= 0.98
ℎ𝑜 = 0.98 × 0.02 = 0.0196 𝑚𝑚
Mechanical Design 2 13

14. Thank You
Mechanical Design 2 14