Consider a pillow-block bearing with a keyway sump, whose journal rotates at 900 rev/min in shaft-stirred air at 70◦F with α = 1.The lateral area of the bearing is 40 in2. The lubricant is SAE grade 20 oil.The gravity radial load is 100 lbf and the l/d ratio is unity.The bearing has a journal diam

Question

Consider a pillow-block bearing with a keyway sump, whose journal rotates at [latex] 900 \ rev/min[/latex] in shaft-stirred air at [latex] 70^\circ F [/latex] with [latex] \alpha = 1[/latex] . The lateral area of the bearing is [latex] 40 \ in^2[/latex].The lubricant is SAE grade 20 oil.The gravity radial load is[latex] 100 \ lbf \ and \ the \ {l}/{d}[/latex] ratio is unity. The bearing has a journal diameter of [latex] {2.000 + 0.000}/{−0.002} [/latex] in, a bushing bore of [latex] {2.002 + 0.004}/{−0.000} [/latex] in. For a minimum clearance assembly estimate the steady-state temperatures as well as the minimum film thickness and coefficient of friction.

Accepted Answer

The minimum radial clearance, [latex]c_{min}[/latex] , is

[latex]c_{min}= \frac{2.002 − 2.000}{2} =0.001 \ in \ P=\frac{W}{ld} =\frac{100}{2\left(2 ight) } =25 \ psi \ S =\left(\frac{r}{c} ight)^2 \frac{\mu N}{P}= \left(\frac{1}{0.001} ight)^2 \frac{\mu^{\backprime} \left(15 ight)}{10^6 \left(25 ight)}= 0.6 \mu^{\backprime}[/latex]

where [latex]\mu^{\backprime}[/latex] is viscosity in [latex]\mu reyn[/latex] . The friction horsepower loss,[latex]\left(hp ight)_f[/latex],is found as follows:

[latex]\left(hp ight)_f =\frac{f WrN}{1050} = \frac{WNc}{1050}\frac{fr}{c}=\frac{100 \left({900}/{60} ight)0.001}{1050} \frac{fr}{c}= 0.001 429 \frac{fr}{c}[/latex] hp

The heat generation rate [latex]H_{gen}[/latex], in Btu/h, is

[latex]H_{gen}= 2545\left(hp ight)_f = 2545\left(0.001 429 ight){fr}/{c} = 3.637 {fr}/{c}[/latex] Btu/h

From Eq. (12–19a)

[latex]H_{loss}=\frac{\hbar_{CR} A}{1+\alpha} \left(\bar{T}_f - T_{\infty } ight) [/latex]

with [latex]\hbar_{CR} = 2.7 \ {Btu}/{\left(h · ft^2 · ^\circ F ight)}[/latex], the rate of heat loss to the environment [latex]H_{loss}[/latex] is
[latex]H_{loss}= \frac{\hbar_{CR} A}{\alpha + 1} \left(\bar{T}_f − 70 ight) = \frac{2.7\left({40}/{144} ight)}{\left(1 + 1 ight)}\left(\bar{T}_f -70 ight) = 0.375\left(\bar{T}_f -70 ight)[/latex] Btu/h

Build a table as follows for trial values of [latex]\bar{T}_f \ of \ 190 \ and \ 195^\circ F[/latex]:

[latex] Trial \bar{T}_f[/latex]	[latex] \mu [/latex]	S	[latex]{fr}/{c} [/latex]	[latex]H_{gen} [/latex]	[latex]H_{loss}[/latex]
190	1.15	0.69	13.6	49.5	45.0
195	1.03	0.62	12.2	44.4	46.9

The temperature at which [latex]H_{gen}=H_{loss} = 46.3 \ Btu/h \ is \ 193.4^\circ F[/latex]. Rounding [latex] \bar{T}_f \ to \ 193^\circ F[/latex] we find [latex]\mu = 1.08 \ \mu reyn \ and \ S = 0.6\left(1.08 ight) = 0.65[/latex] . From Fig. 12–24,[latex]9.70 \Delta \ {TF}/{P} \ = 4.25^\circ [/latex] F/psi and thus

[latex]\Delta T_F = {4.25 P}/{9.70} = {4.25 \left(25 ight)}/{9.70} = 11.0^\circ \ F \ T_1 = T_s = \bar{T}_f − {\Delta T}/{2} = 193 − {11}/{2} = 187.5^\circ \ F \ T_{max} = T_1 +\Delta T_F = 187.5 + 11 = 198.5^\circ \ F[/latex]

From Eq. (12–19b)

[latex]T_b = \frac{\bar{T}_f + \alpha T_{\infty}}{1 + \alpha}[/latex]

[latex]T_b = \frac{T_f + \alpha T_{\infty }}{1 + \alpha}= \frac{193 + \left(1 ight)70}{1 + 1}= 131.5^\circ \ F[/latex]

with [latex]S = 0.65[/latex] , the minimum film thickness from Fig. 12–16 is
[latex]h_0 =\frac{h_0}{c}c = 0.79 \left(0.001 ight) = 0.000 79 [/latex] in
The coefficient of friction from Fig. 12–18 is

[latex]f = \frac{f r}{c}\frac{c}{r}= 12.8 \frac{0.001}{1}= 0.012 8[/latex]

The parasitic friction torque [latex]T[/latex] is

[latex]T = f Wr = 0.012 8 \left(100 ight) \left(1 ight) = 1.28 \ lbf · in[/latex]

Question 3.12.5: Consider a pillow-block bearing with a keyway sump, whose jo...

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