Question Bank • QB_FP_041 NAT • 2 Marks

Section 3: Farm Power

Farm Power

Subtopic: Lubricants and their properties

Question Statement:

A tractor crankshaft journal bearing has a journal diameter of $d = 60\text{ mm}$ ($R = 30\text{ mm}$), bearing length $L = 40\text{ mm}$, and radial clearance $c = 0.03\text{ mm}$. The journal rotates at $N = 1800\text{ rpm}$ ($\omega = 188.5\text{ rad/s}$) in lubricating oil of dynamic viscosity $\mu = 0.025\text{ Pa}\cdot\text{s}$. Using Petroff's equation for a concentric journal ($P_f = \frac{2 \pi \mu \omega^2 L R^3}{c}$), the viscous friction power lost in the bearing in Watts is ________ (round off to 1 decimal place). Take $\pi = 3.1416$.

Answer Key:

200.9

Step-by-Step Solution & Explanation

Petroff's equation for viscous friction power loss in a concentric journal bearing: $P_f = \frac{2 \pi \mu \omega^2 L R^3}{c}$ Where: • $\mu = 0.025\text{ Pa}\cdot\text{s}$ • $\omega = \frac{2 \pi \times 1800}{60} = 60\pi \approx 188.496\text{ rad/s} \implies \omega^2 \approx 35,530.6\text{ rad}^2/\text{s}^2$ • $L = 0.040\text{ m}$ • $R = 0.030\text{ m} \implies R^3 = 2.7 \times 10^{-5}\text{ m}^3$ • $c = 3.0 \times 10^{-5}\text{ m}$ $P_f = \frac{2 \times 3.1416 \times 0.025 \times 35,530.6 \times 0.040 \times (2.7 \times 10^{-5})}{3.0 \times 10^{-5}}$ $P_f = \frac{2 \times 3.1416 \times 0.025 \times 35,530.6 \times 0.040 \times 0.90}{1} \approx 200.93\text{ W} \approx 200.9\text{ W}$
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