Question

For a fully developed flow through a pipe, how will the heat transfer coefficient (h) change, when the mean flow rate is doubled with all other conditions remaining the same?

• A. h increases by 50% (approx)
• B. h increases by 74% (approx)
• C. h decreases by 87% (approx)
• D. h increases by 20% (approx)

Hint:

The exponent 0.8 in the Dittus-Boelter equation is key for turbulent flow problems: - If flow velocity doubles ($2\times$), then $h$ scales by $2^{0.8} \approx 1.74 \implies 74\%$ increase. - Note: If the flow field were fully developed *laminar*, the Nusselt number would be constant ($Nu = 3.66$ or $4.36$), meaning $h$ would not change with flow rate. The options here clearly indicate a turbulent flow context.

Answer 1

The Correct Option is B

Step 1: Write the governing proportionality.
For fully developed turbulent flow, the Dittus-Boelter correlation gives \(Nu \propto Re^{0.8}\), and since \(Nu = hd/k\) with \(d\) and \(k\) fixed, this means \(h \propto v^{0.8}\). Doubling the flow rate doubles the mean velocity, so: \[ \frac{h_2}{h_1} = 2^{0.8} \]

Step 2: Evaluate \(2^{0.8}\) using logarithms instead of roots.
Taking \(\log_{10}\) of both sides: \[ \log_{10}\left(\frac{h_2}{h_1}\right) = 0.8 \times \log_{10}(2) = 0.8 \times 0.301 = 0.2408 \] Converting back from the log value: \[ \frac{h_2}{h_1} = 10^{0.2408} \approx 1.741 \]

Step 3: Convert to a percentage change.
\[ \text{Increase} = (1.741 - 1) \times 100\% \approx 74\% \] \[ \boxed{h \text{ increases by about } 74\%} \] This matches option 2.