Question 3.12.2: Using the parameters given in Ex. 12–1, determine the coeffi...

We enter Fig. 12–18 with S = 0.135 \ and \ {l}/{d} = 1 \ and \ find \left( {r}/{c} \right) f = 3.50 . The coefficient of friction f is

f = 3.50 {c}/{r} = 3.50 \left({0.0015}/{0.75}\right) = 0.0070

The friction torque on the journal is

T = f W r = 0.007 \left(500\right)0.75 = 2.62 lbf · in

The power loss in horsepower is

\left(hp\right)_{loss} = \frac{T N}{1050}= \frac{2.62 \left(30\right)}{1050}= 0.075 hp

or, expressed in Btu/s,

H = \frac{2 \pi T N}{778 \left(12\right)}= \frac{2 \pi \left(2.62\right)30}{778 \left(12\right)}= 0.0529 Btu/s