Determine h0 and e using the following given parameters: μ = 4 μreyn, N = 30 rev/s,W = 500 lbf (bearing load), r = 0.75 in, c = 0.0015 in, and l = 1.5 in

Question

Determine [latex]h_0[/latex] and e using the following given parameters: [latex]\mu = 4 \ μreyn \ , \ N = 30 \ rev/s \ , \ W = 500 \ lbf \ (bearing \ load) \ , \ r = 0.75 \ in \ , c = 0.0015 \ in \ , \ and \ l = 1.5 \ in[/latex]

Accepted Answer

The nominal bearing pressure (in projected area of the journal) is

[latex]P =\frac{W}{2rl}=\frac{500}{2\left(0.75\right) 1.5}= 222 [/latex] psi

The Sommerfeld number is, from Eq. (12–7),

[latex] S =\left(\frac{r}{c} \right)^2 [/latex]

where [latex] N = N_j = 30[/latex] rev/s,

[latex]S =\left(\frac{r}{c} \right)^2 \left( \frac{\mu N}{P} \right)=\left(\frac{0.75}{0.0015}\right)^2 \left[ \frac{4\left(10^{−6}\right) 30}{222}\right]= 0.135[/latex]

Also, [latex]{l}/{d} = {1.50}/{\left[2\left(0.75\right)\right]} = 1[/latex] . Entering Fig. 12–16 with [latex]S = 0.135 \ and \ {l}/{d} = 1[/latex] gives [latex]{h_0}/{c} = 0.42 \ and \ \epsilon = 0.58[/latex] . The quantity [latex]{h_0}/{c}[/latex] is called the [latex]minimum \ film \ thickness \ variable[/latex] . Since [latex]c = 0.0015[/latex] in , the minimum film thickness
[latex]h_0[/latex] is [latex]h_0 = 0.42 \left(0.0015\right) = 0.000 63[/latex] in

We can find the angular location [latex] \phi[/latex] of the minimum film thickness from the chart of Fig. 12–17. Entering with [latex]S = 0.135 \ and \ {l}/{d} = 1 \ gives \ \phi = 53^{\circ} [/latex] .

The eccentricity ratio is [latex]\epsilon = {e}/{c} = 0.58[/latex] . This means the eccentricity [latex] e[/latex] is [latex] e = 0.58 \left(0.0015\right) = 0.000 87[/latex] in

Question 3.12.1: Determine h0 and e using the following given parameters: μ =...

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