Question 21.11: A full journal bearing of 100 mm length and 100 mm diameter ......

(i) From the McKee equation, the coefficient of friction is given as

f=2 \pi^2\left(\frac{\mu N^*}{p}\right)\left(\frac{r}{C_r}\right)+0.002=2 \pi^2\left(\frac{12.5 \mu}{0.6 \times 10^6}\right)(1000)+0.002=0.411 \mu+0.002

where

p=\frac{W}{L D}=\frac{6000}{100 \times 100}=0.6 \mathrm{~N} / \mathrm{mm}^2

N^*=750 / 60=12.5 \mathrm{~rps}

Heat generated is given by

H_g=\frac{(f W)(2 \pi N r)}{60}=\frac{(f W)(\pi N D)}{60}=\frac{(0.411 \mu+0.002) \times 6000 \times\left(\pi \times 750 \times \frac{100}{1000}\right)}{60}

=9683.95 \mu+47.12 (21.75)

Heat dissipated is given by

H_d=\rho Q C_p \Delta T=0.86 Q \times 1.76 \times(62-27)=52.976 Q (21.76)

\left(\frac{Q}{r C_r N^* L}\right) is the flow variable (FV)

\left(\frac{Q}{r C_r N^* L}\right)=\mathrm{FV}

Q=r C_r N^* L(\mathrm{FV}) and from Eq. (21.76) we can write

H_d=52.976 \times r C_r N^* L(\mathrm{FV})=52.976 \times\left(\frac{50}{1000}\right)\left(\frac{50}{1000}\right)\left(\frac{750}{60}\right)\left(\frac{100}{1000}\right)(\mathrm{FV})=0.166(\mathrm{FV}) (21.77)

Equating Eqs. (21.75) and (21.77)

9683.95 \mu+47.12=0.166(\mathrm{FV})

\mathrm{FV}=58337.1 \mu+283.9 (21.78)

Assuming density of oil as 860 \mathrm{kg} / \mathrm{m}^3 \text { and } C_p=1760 \mathrm{~J} / \mathrm{kg}{ }^{\circ} \mathrm{C}

\frac{\rho C_p \Delta T}{p}=\frac{\mathrm{CFV}}{\mathrm{FV}} \times 4 \pi=\frac{0.86 \times 1.76 \times(62-27)}{0.6}=88.29

\frac{\mathrm{CFV}}{\mathrm{FV}}=\frac{88.29}{4 \pi}=7.026 (21.79)

f\left(\frac{r}{C_r}\right)=\mathrm{CFV}=(0.414 \mu+0.002)\left(\frac{50}{0.05}\right) \text { taking } \frac{r}{C_r}=1000

\text { CFV }=1000 \times(0.414 \mu+0.002)

From Eq. (21.79) we can have

\mathrm{FV}=\frac{\mathrm{CFV}}{7.026}=\frac{1000 \times(0.414 \mu+0.002)}{7.026}

\mathrm{FV}=142.33 \times(0.414 \mu+0.002) (21.80)

Equating Eqs. (21.78) and (21.80)

142.33 \times(0.414 \mu+0.002)=58337.1 \mu+283.9

Solving, we get

58337.1 \mu-58.925 \mu=283.9-0.28466

\mu=\frac{283.615}{58278.175}=4.867 \times 10^{-3} \mathrm{~kg} / \mathrm{ms}

Alternate method

(i) Without ventilation condition

The rough estimate of heat dissipation can be obtained as

H_d=k A \Delta T

where k is the overall heat transfer coefficient

k = 140 to 420 W/m²/°C without ventilation

= 490 to 1400 W/m²/°C with ventilation.

Taking k = 200 W /m²/°C

H_d=k A \Delta T=200 \times 0.1 \times 0.1 \times 35=70 \mathrm{~W}

Equating with Eq. (21.75)

9683.95 \mu+47.12=70

\mu=\frac{22.88}{9683.95}=2.36 \times 10^{-3} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}

(ii) with well ventilation

Assume k = 600 W/m²/°C

H_d=k A \Delta T=600 \times 0.1 \times 0.1 \times 35=210 \mathrm{~W}

Equating with Eq. (21.75)

9683.95 \mu+47.12=210

\mu=\frac{162.88}{9683.95}=0.0168 \mathrm{~kg} / \mathrm{ms}

These are approximate methods as the exact heat transfer coefficient is unknown and assumed from a range of the values.