Question

The relation between Tractive Efficiency (TE), Net Traction Coefficient (\(\mu\)), Gross Traction Coefficient (\(\mu_g\)) and Slip (\(s\)) is

• A. \(\text{TE} = \mu / \mu_g\)
• B. \(\text{TE} = (\mu / \mu_g)(1 - s)\)
• C. \(\text{TE} = \mu(1 - s)\mu_g\)
• D. \(\text{TE} = (\mu / \mu_g) / (1 - s)\)

Hint:

Tractive efficiency is the product of travel reduction efficiency $(1-s)$ and motion resistance efficiency $(\mu / \mu_g)$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Tractive efficiency is the ratio of drawbar power developed to the axle power supplied to the traction wheels.
Key Formula or Approach:
\[ \text{TE} = \frac{P_{\text{drawbar}}}{P_{\text{axle}}} = \frac{P \cdot V_a}{T \cdot \omega} \]
\[ \mu = \frac{P}{W}, \quad \mu_g = \frac{T/r}{W}, \quad (1 - s) = \frac{V_a}{V_t} \]

Step 2: Detailed Explanation:
Let:
- \(P = \mu W\) (Drawbar pull)
- \(T/r = \mu_g W\) (Gross tractive thrust)
- \(V_a = V_t(1 - s)\) (Actual travel speed vs theoretical speed)
The power delivered at the drawbar is:
\[ P_{\text{db}} = P \times V_a = (\mu W) \times V_t(1 - s) \]
The input axle power is:
\[ P_{\text{axle}} = (\mu_g W) \times V_t \]
Taking the ratio:
\[ \text{TE} = \frac{(\mu W) V_t(1 - s)}{(\mu_g W) V_t} = \left( \frac{\mu}{\mu_g} \right) (1 - s) \]

Step 3: Final Answer:
Therefore, the correct relation is \(\text{TE} = (\mu / \mu_g)(1 - s)\), corresponding to option (B).