$$CE = C + \frac{Mn}{6} + \frac{(Cr+Mo+V)}{5} + \frac{(Ni+Cu)}{15}$$
Where,
- C is the carbon content in weight percentage.
- Mn is the manganese content in weight percentage.
- Cr is the chromium content in weight percentage.
- Mo is the molybdenum content in weight percentage.
- V is the vanadium content in weight percentage.
- Ni is the nickel content in weight percentage.
- Cu is the copper content in weight percentage.
For API 5L GrB material, the typical composition is as follows:
- Carbon (C): 0.25%
- Manganese (Mn): 1.40%
- Chromium (Cr): 0.50%
- Molybdenum (Mo): 0.25%
- Vanadium (V): 0.05%
- Nickel (Ni): 0.50%
- Copper (Cu): 0.40%
Substituting these values in the CE formula, we get:
$$CE = 0.25 + \frac{1.40}{6} + \frac{(0.50+0.25+0.05)}{5} + \frac{(0.50+0.40)}{15}$$
$$CE = 0.25 + 0.23 + 0.16 + 0.06$$
$$CE = 0.70$$
