Question

In hydrodynamic journal bearing, if the clearance ratio is halved then the Sommer field number 'S' and the coefficient of friction '$ \mu $' will change as

• A. ‘S’ becomes doubled and ‘\( \mu \)’ is halved
• B. ‘S’ becomes four times and ‘\( \mu \)’ is halved
• C. ‘S’ becomes four times and ‘\( \mu \)’ is doubled
• D. ‘S’ becomes doubled and ‘\( \mu \)’ is doubled

Hint:

For hydrodynamic journal bearings, the Sommerfeld number (S) is proportional to \( (\frac{R}{c})^2 \) or inversely proportional to \( (\text{clearance ratio})^2 \). The coefficient of friction \( \mu \) is directly proportional to the clearance ratio and also related to S. Specifically, \( \mu \propto S \cdot (\text{clearance ratio}) \). Use these proportionality relationships to quickly assess the change.

Answer 1

The Correct Option is C

Step 1: Understand the definitions of Sommerfeld number and coefficient of friction in journal bearings.
The Sommerfeld number (S) is a dimensionless parameter used in the design of hydrodynamic bearings. It is given by: \[ S = \frac{\mu_v N}{P} \left(\frac{R}{c}\right)^2 \] where:
\( \mu_v \) = absolute viscosity of the lubricant
\( N \) = journal rotational speed (rps)
\( P \) = bearing pressure = \( \frac{\text{Load}}{\text{Projected Area}} = \frac{W}{DL} \)
\( R \) = journal radius
\( c \) = radial clearance = \( R_{bearing} - R_{journal} \)
The clearance ratio is \( \frac{c}{R} \). So, \( \frac{R}{c} = \frac{1}{\text{clearance ratio}} \).
Thus, the Sommerfeld number can be written as: \[ S = \frac{\mu_v N}{P \times (\text{clearance ratio})^2} \] The coefficient of friction (\( \mu \)) in a hydrodynamic journal bearing is typically expressed as: \[ \mu = k \frac{f}{D} + \frac{c}{R} \] A more direct relationship derived from the Petroff equation (for ideal bearings) or through dimensional analysis related to the Sommerfeld number is: \[ \mu \propto \frac{1}{\sqrt{S}} \left(\frac{c}{R}\right) \text{ or } \mu \propto \frac{1}{S^{1/2}} \cdot \left(\frac{c}{R}\right) \] A common relationship from bearing charts or dimensionless groups is: \[ \frac{\mu R}{c} = \phi(S) \text{ or } \mu = \frac{c}{R} \cdot \phi(S) \] where \( \phi(S) \) is a function of the Sommerfeld number. For lightly loaded bearings, Petroff's equation gives \( \mu = 2\pi^2 \frac{\mu_v N}{P} \frac{R}{c} \).
However, a more direct relationship commonly used for evaluating changes is derived from the fact that \( \mu S \) is a characteristic parameter. For example, for a full journal bearing, the friction variable \( \frac{\mu R}{c} \) is related to the Sommerfeld number.
From basic friction considerations, \( \mu = \frac{\text{friction force}}{\text{normal load}} \).
The friction force in a hydrodynamic bearing is often proportional to \( \frac{\mu_v N L R^2}{c} \).
So, \( \mu \propto \frac{\mu_v N R^2}{c \cdot P \cdot D L} \propto \frac{\mu_v N R^2}{c P (2R) L} \propto \frac{\mu_v N R}{c P} \).
Combining this with \( S \propto \frac{\mu_v N}{P} (\frac{R}{c})^2 \), we get:
\( \mu \propto \frac{S}{(R/c)} = S \left(\frac{c}{R}\right) \).
Step 2: Analyze the effect of halving the clearance ratio.
Let the original clearance ratio be \( CR = \frac{c}{R} \).
The new clearance ratio \( CR' = \frac{1}{2} CR = \frac{1}{2} \left(\frac{c}{R}\right) \).
New Sommerfeld number \( S' \): \[ S' = \frac{\mu_v N}{P} \left(\frac{R}{c'}\right)^2 = \frac{\mu_v N}{P} \left(\frac{1}{CR'}\right)^2 \] Substitute \( CR' = \frac{1}{2} CR \): \[ S' = \frac{\mu_v N}{P} \left(\frac{1}{\frac{1}{2}CR}\right)^2 = \frac{\mu_v N}{P} \left(\frac{2}{CR}\right)^2 \] \[ S' = \frac{\mu_v N}{P} \frac{4}{(CR)^2} = 4 \left[ \frac{\mu_v N}{P} \left(\frac{R}{c}\right)^2 \right] = 4S \] So, the Sommerfeld number becomes four times its original value. New coefficient of friction \( \mu' \): Using the relationship \( \mu \propto S \left(\frac{c}{R}\right) \): \[ \mu' \propto S' \left(\frac{c'}{R}\right) \] Substitute \( S' = 4S \) and \( \frac{c'}{R} = \frac{1}{2} \left(\frac{c}{R}\right) \): \[ \mu' \propto (4S) \left(\frac{1}{2} \frac{c}{R}\right) \] \[ \mu' \propto 2 S \left(\frac{c}{R}\right) \] Since \( \mu \propto S \left(\frac{c}{R}\right) \), we have \( \mu' = 2\mu \). So, the coefficient of friction becomes doubled. Therefore, 'S' becomes four times and '\( \mu \)' is doubled. The final answer is $\boxed{\text{3}}$.