Transport analogies — For which case are mass diffusivity, thermal diffusivity, and eddy momentum diffusivity effectively equal (i.e., NPr = NSc = ______ )?
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A1
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B0.5
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C10
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D0
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E2
Answer
Correct Answer: 1
Explanation
Introduction:Transport property analogies (Reynolds, Chilton–Colburn) connect momentum, heat, and mass transfer. When dimensionless groups align, similar boundary-layer behavior allows property analogies to be applied cleanly. This question asks the special condition under which momentum, heat, and mass diffusivities are equal in a normalized sense.
Given Data / Assumptions:
- Prandtl number NPr = ν/α (momentum diffusivity/thermal diffusivity).
- Schmidt number NSc = ν/D (momentum diffusivity/mass diffusivity).
- Fully turbulent regime with similar eddy diffusivities for momentum, heat, and mass.
Concept / Approach:
When NPr = 1 and NSc = 1, momentum, heat, and mass boundary layers have similar thicknesses because ν ≈ α ≈ D (or, in turbulent flow, their eddy counterparts are comparable). Under this condition, the Chilton–Colburn j-factor analogy reduces to direct equivalence, simplifying transfer coefficient estimation and permitting the use of a single Stanton number relation for all three modes.
Step-by-Step Solution:
Write definitions: NPr = ν/α; NSc = ν/D.Set NPr = NSc = 1 → ν = α = D (effective diffusivities).Conclude that the three transport processes behave analogously.Therefore fill the blank with 1.Verification / Alternative check:
Textbook analogies yield j_H ≈ j_D when Pr = Sc, with best agreement near unity; many gases at moderate temperatures have Pr of order 1, supporting the approximation.
Why Other Options Are Wrong:
B, C, E: Values other than 1 imply unequal diffusivities and dissimilar boundary layers. D: NPr or NSc = 0 is non-physical for real fluids.
Common Pitfalls:
Assuming perfect equality is required; in practice, analogies work reasonably well for Pr and Sc not too far from unity.
Final Answer:
1