For turbulent liquid flow inside tubes, the tube-side heat-transfer coefficient varies approximately with mass velocity G as:
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Ah ∝ G^0.2
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Bh ∝ G^0.5
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Ch ∝ G^0.8
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Dh ∝ G^1.5
Answer
Correct Answer: h ∝ G^0.8
Explanation
Introduction / Context:Correlations like Dittus–Boelter and Sieder–Tate are used to estimate convective heat-transfer coefficients for turbulent flow in tubes. These relate Nusselt number to Reynolds and Prandtl numbers, revealing how h changes with mass velocity G (or Reynolds number) at otherwise similar conditions.
Given Data / Assumptions:
- Turbulent regime in smooth tubes (Re > ~10^4).
- Moderate Prandtl number liquids, heating or cooling without strong property variation.
- Comparable diameters and fluid properties for scaling arguments.
Concept / Approach:Dittus–Boelter: Nu = 0.023 * Re^0.8 * Pr^n (n ≈ 0.3 for heating, ≈ 0.4 for cooling). Since h = Nu * k / D and Re ∝ G * D / μ, holding D and properties fixed implies h ∝ Re^0.8 ∝ G^0.8. Hence, the strongest dependence among common exponents listed is roughly the 0.8 power for turbulent single-phase flow in clean tubes.
Step-by-Step Solution:
Start with Nu ∝ Re^0.8 Pr^n.Hold Pr and geometry approximately constant → h ∝ Nu ∝ Re^0.8.Because Re ∝ G, conclude h ∝ G^0.8.Verification / Alternative check:Other correlations (e.g., Sieder–Tate) give similar exponents near 0.8 for Re dependence in smooth tubes, validating the trend.
Why Other Options Are Wrong:
- 0.2 and 0.5 underpredict the dependence for turbulent regimes.
- 1.5 is unrealistically strong and inconsistent with established correlations.
Common Pitfalls:Extending the 0.8 exponent to laminar or transitional regimes, or to fouled tubes where h is dominated by deposits and not strongly by G.
Final Answer:h ∝ G^0.8