Figure 8.1: Heat transfer analysis to find ∆T

This heat loss rate is due to friction between the layers of the lubricant and is calculated as 2πn′Tr =2π f W r n′ = 2πW n′c r f

c = 4πP ′r l n′c r fc . All variables can be known from the nomenclature at the

beginning of this sheet.So,

4πP ′r l n′c r fc = ρCpQ∆T

(1− 1

2QsQ

)Rearranging,

ρCp∆T

4πP ′ = r fc(

1− 12

QsQ

)(Q

r cn′l

) (8.1)

If Cp = 1764 Jkg ◦C and ρ = 861.5 kg

m3 (which are values for common petroleum based lubricant)we can write Equation 12.1 as

0.120∆T

P ′ = r fc(

1− 12

QsQ

)(Q

r cn′l

) (8.2)

Thus, from Equation. 12.2 the right hand side is known once the Somerfeld number is known for agiven l

d ratio. Basically Figures. 14.5, 14.6 and 14.7, respectively. If the mean pressure P ′ is known,the ∆T value is also known. A chart which combines all these three figures into one is presentedin Figure 14.12

9 USING RAIMONDI AND BOYD FOR ld RATIOS OTHER THAN 1

4 , 12 1

AND ∞All the Raimondi and Boyd charts are plotted for l

d ratios 14 , 1

2 1 and ∞. If we want to find forothers, we use the following expression

y

(l

d

)3

=−1

8

(1− l

d

)(1−2

l

d

)(1−4

l

d

)y∞+ 1

3

(1−2

l

d

)(1−4

l

d

)y1

− 1

4

(1− l

d

)(1−4

l

d

)y 1

2+ 1

24

(1− l

d

)(1−2

l

d

)y 1

4(9.1)

6

y is the desired variable within the interval ∞ > ld > 1

4 and y∞, y1, y 12

and y 14

are the variables

corresponding to ld ratios of ∞,1, 1

2 and 14 , respectively

9.1 EXAMPLE FOR ITERATIVE DETERMINATION OF VISCOSITY

An electrical generator rotor weighs 600,000N and rotates at 3600rpm. It is supported by twojournal bearings. The journal diameter is 400mm and the oil used in light SAE 30 oil. The oil filterallows particles of size 60µm and the peak surface roughnesses are 5µm. Design the bearings.The sump temperature is 20◦C.

• rcr

ratio is ≈ 1.00×103. Which means the crr ratio is 1.00×10−3. This givescr ≈ 0.2mm

• The hmi n ≥ 2(5+5)+60 = 80µm and hence hmi ncr

≥ 0.4

• Taking ld = 1, and hmi n

cr= 0.4 we get

(rcr

)2 µNP ′ = 0.12. With this we get a viscosity of ≈ 0.0037

Pa s (or) ≈ 3.7m Pa s [Using figure14.3]

• The average operating temperature and the viscosity needs to be found

1. First trial, from figure 14.13, Tav is about 125◦C. Now, using figure 14.14 we get a ∆T of≈ 17◦C. The ∆T for the viscosity we have taken is 2× (125−20) = 210◦C. The error in∆T (Assumed - calculated) is 210−17 = 193◦C, which is huge.

2. Trial 2: Taking a Tav ≈ 40 ≈ (20+17), we get S = 2.57. This gives a∆T of 251.7◦C.∆T as-sumed was 2×(Tav−20) = 20◦C and the difference between the assumed and obtained∆T is 20−251.7 =−231.7◦C. Given that the error is negative, we have to decrease theviscosity and increase the average working temperature.

3. Let us try Tav = 80. This gives a η of 12.5m Pa s. This gives S = 0.4028 and ∆T = 44.Assumed ∆T is (80−20)×2 = 120◦C. So the difference is 120−44 = 76◦C.

4. Trial 3: Take Tav = 20+ 40 = 60◦. This gives, S = 0.8862, ∆T = 89.2. Assumed ∆T =(60−20)×2 = 80. So the difference is 80−89.2 =−9.2 is negative.

5. Trial 4: Take now Tav = 70◦C. You get S = 0.58 and ∆T = 60.17. Assumed ∆T = (70−20)×2 = 100◦C. Difference (Assumed - obtained) is (100−60) = 40.

6. Trial 5: Take Tav = 65, S = 0.7259, ∆T = 73.92. Assume ∆T = (65−20)×2 ≈ 90. Differ-ence between the assumed and the obtained ∆T is (90−73) = 17.

7. Trial 6: Take Tav = 62.5, S = 0.7743 and ∆T ≈ 79. The assumed value of ∆T is (62.5−20)×2 = 85◦C. The difference between assumed and the obtained value of ∆T is 85−79 = 6◦C.

8. Trial7: Take Tav = 61, S = 0.8056 and the ∆T = 81.6. The assumed ∆T is (61−20)×2 =82. The difference between the assumed and the obtained value of ∆T is (82−81.6) =0.4. We can live with this error.

• So we got η = 25mPas. The S = 0.8056. For this value of S, the hmi ncr

≈ 0.83 > 0.4. So we arefine. All the parameters can be obtained using the other charts.

10 FIGURES

7

Figure 10.1: Nomenclature for the first problem

Figure 10.2: Viscosity

8

Figure 10.3: Minimum film thickness

Figure 10.4: Minimum film thickness position

9

Figure 10.5: Friction

10