Reynolds analogy (simplest form) for turbulent forced convection over surfaces relates momentum and heat transfer.\r Which proportionality captures the essence of the analogy under its ideal assumptions?

Introduction:
Reynolds analogy provides a bridge between momentum and heat transfer in turbulent boundary layers. Under simplifying assumptions (notably Prandtl number near unity and negligible pressure gradients), the analogy connects friction factor with Stanton number, enabling heat-transfer predictions from friction data.

Given Data / Assumptions:

• High-Reynolds-number turbulent flow.
• Prandtl number approximately 1.
• No form drag and negligible pressure gradients.

Concept / Approach:
The simplest Reynolds analogy states: St = f/2, where St is the Stanton number (dimensionless heat transfer coefficient) and f is the Fanning friction factor. In proportional terms, St ∝ f, matching option (a). Extended analogies (Chilton–Colburn) generalize this to St * Pr^(2/3) = f/2 for a wider range of Pr.

Step-by-Step Solution:
Recall definition: St = h/(ρ * u * Cp).Apply the simple analogy: St = f/2 ⇒ St ∝ f.Map to offered statements: Nst α f aligns; other pairings are incorrect.

Verification / Alternative check:
Using friction-factor correlations, one can estimate heat transfer where direct measurement is hard, provided Pr ~ 1 and surface conditions are comparable.

Why Other Options Are Wrong:

• Nst α NRe or NNu α f: These do not reflect the canonical simple-analogy relationship.
• NRe α f: Friction factor scales differently with Re; they are not simply proportional.

Common Pitfalls:
Applying the analogy far from its validity (very high or low Pr). Use Chilton–Colburn j-factors for broader applicability.

Final Answer:
Nst α f