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?
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?
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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.
Updated On: Jul 4, 2026
- h increases by 50% (approx)
- h increases by 74% (approx)
- h decreases by 87% (approx)
- h increases by 20% (approx)
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The Correct Option is B
Solution and Explanation
Concept:
For fully developed turbulent flow through a smooth pipe, convective heat transfer is modeled using the empirical Dittus-Boelter Equation. This relation correlates the Nusselt number (\(Nu\)) with the Reynolds number (\(Re\)) and Prandtl number (\(Pr\)):
\[
Nu = \frac{h \cdot d}{k} = 0.023 \cdot Re^{0.8} \cdot Pr^n
\]
Where:
• \(h\) = Convective heat transfer coefficient.
• \(d\) = Internal pipe diameter.
• \(k\) = Fluid thermal conductivity.
• \(Re = \frac{\rho \cdot v \cdot d}{\mu}\) = Reynolds number, which is directly proportional to the fluid mean velocity or flow rate (\(Re \propto v\)).
• \(Pr\) = Prandtl number, a fluid property constant.
Step 1: Finding the proportionality relation for \(h\).
Since all fluid properties (\(\rho, \mu, k, Pr\)) and geometric dimensions (\(d\)) remain constant, we can isolate the heat transfer coefficient \(h\): \[ h \propto Re^{0.8} \quad \Rightarrow \quad h \propto v^{0.8} \] This shows that the convective heat transfer coefficient scales with velocity to the power of 0.8.
Step 2: Calculating the change when flow rate is doubled.
Let the initial velocity be \(v_1\) and the initial heat transfer coefficient be \(h_1\). The problem states that the mean flow rate is doubled, meaning the new velocity is: \[ v_2 = 2 \cdot v_1 \] Using our proportionality relation, the new heat transfer coefficient \(h_2\) is: \[ \frac{h_2}{h_1} = \left(\frac{v_2}{v_1}\right)^{0.8} = (2)^{0.8} \] Let us compute the value of \(2^{0.8}\): \[ 2^{0.8} = 2^{4/5} = \sqrt[5]{2^4} = \sqrt[5]{16} \approx 1.7411 \] Therefore, the new heat transfer coefficient is: \[ h_2 \approx 1.7411 \cdot h_1 \]
Step 3: Calculating the percentage increase in \(h\).
The fractional change in the heat transfer coefficient is given by: \[ \text{Percentage Increase} = \left(\frac{h_2 - h_1}{h_1}\right) \times 100\% = \left(\frac{1.7411 \cdot h_1 - h_1}{h_1}\right) \times 100\% \] \[ \text{Percentage Increase} = (1.7411 - 1) \times 100\% = 0.7411 \times 100\% \approx 74.11\% \] Thus, the convective heat transfer coefficient increases by approximately 74%, which matches Option (2).
• \(h\) = Convective heat transfer coefficient.
• \(d\) = Internal pipe diameter.
• \(k\) = Fluid thermal conductivity.
• \(Re = \frac{\rho \cdot v \cdot d}{\mu}\) = Reynolds number, which is directly proportional to the fluid mean velocity or flow rate (\(Re \propto v\)).
• \(Pr\) = Prandtl number, a fluid property constant.
Step 1: Finding the proportionality relation for \(h\).
Since all fluid properties (\(\rho, \mu, k, Pr\)) and geometric dimensions (\(d\)) remain constant, we can isolate the heat transfer coefficient \(h\): \[ h \propto Re^{0.8} \quad \Rightarrow \quad h \propto v^{0.8} \] This shows that the convective heat transfer coefficient scales with velocity to the power of 0.8.
Step 2: Calculating the change when flow rate is doubled.
Let the initial velocity be \(v_1\) and the initial heat transfer coefficient be \(h_1\). The problem states that the mean flow rate is doubled, meaning the new velocity is: \[ v_2 = 2 \cdot v_1 \] Using our proportionality relation, the new heat transfer coefficient \(h_2\) is: \[ \frac{h_2}{h_1} = \left(\frac{v_2}{v_1}\right)^{0.8} = (2)^{0.8} \] Let us compute the value of \(2^{0.8}\): \[ 2^{0.8} = 2^{4/5} = \sqrt[5]{2^4} = \sqrt[5]{16} \approx 1.7411 \] Therefore, the new heat transfer coefficient is: \[ h_2 \approx 1.7411 \cdot h_1 \]
Step 3: Calculating the percentage increase in \(h\).
The fractional change in the heat transfer coefficient is given by: \[ \text{Percentage Increase} = \left(\frac{h_2 - h_1}{h_1}\right) \times 100\% = \left(\frac{1.7411 \cdot h_1 - h_1}{h_1}\right) \times 100\% \] \[ \text{Percentage Increase} = (1.7411 - 1) \times 100\% = 0.7411 \times 100\% \approx 74.11\% \] Thus, the convective heat transfer coefficient increases by approximately 74%, which matches Option (2).
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