22-Mec-A1 Applied Thermodynamics and Heat Transfer · May 2014
Question 6 of 8: Water-Cooled Tube in Cross-Flow Air
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Examinations — 07-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2014. Open-book, 3-hour paper. Part A (Thermodynamics, Q1–Q4) and Part B (Heat Transfer, Q5–Q8); a complete paper is any five questions — three from one part and two from the other, all of equal value. Full worked solutions to all eight questions are given below.
Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed., McGraw-Hill) — closed- and open-system energy balances, steam tables, gas power cycles, reciprocating compressors and vapour-compression refrigeration; Çengel & Ghajar, Heat and Mass Transfer (6th ed.) and Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — composite-wall conduction, internal-flow and cross-flow convection correlations, natural convection with radiation, and the ε–NTU heat-exchanger method. Freon-12 property data are taken from the appendix supplied with the exam; steam and air data from standard tables.
Question 6: Water-Cooled Tube in Cross-Flow Air (equal value)
Given. Tube diameter $D=0.02\ \text{m}$; water in at $77\ ^\circ\text{C}$, out at $55\ ^\circ\text{C}$, $V=6\ \text{m/s}$; cross-flow air at $27\ ^\circ\text{C}$, $V_\infty=30\ \text{m/s}$. Water properties at the bulk mean 66 °C: $\rho=980$, $c_p=4187$, $k=0.659$, $\mu=4.34\times10^{-4}$, $\text{Pr}=2.75$. Air at film ≈ 46 °C: $k=0.0278$, $\nu=1.80\times10^{-5}$, $\text{Pr}=0.704$.
Find. The tube length $L$ for the water to leave at 55 °C.
Figure 5 — Water flows inside the horizontal tube (forced convection, inside coefficient $h_i$) while ambient air blows across it (cross-flow cylinder, outside coefficient $h_o$). The two coefficients combine into an overall $U$ that governs the axial cooling of the water.
Approach. Compute the inside (tube-flow) and outside (cross-flow cylinder) convection coefficients, combine into an overall $U$, then apply the single-stream exponential decay of water temperature toward the constant air temperature to solve for the length.
Overall coefficient. With a thin tube wall (equal inner/outer area),$$\frac1U=\frac1{h_i}+\frac1{h_o}=\frac1{22800}+\frac1{148.5}\;\Rightarrow\;U=147.5\ \text{W/m}^2\text{K}\ (\text{air-side controlled}).$$
Length from the temperature decay. The water mass flow is $\dot m=\rho V(\pi D^2/4)=1.847\ \text{kg/s}$. For a stream cooling toward the constant air temperature,$$\frac{T_\infty-T_{bo}}{T_\infty-T_{bi}}=\exp\!\left(-\frac{UA_s}{\dot m c_p}\right),\quad A_s=\pi D L.$$ Solving, $A_s=30.4\ \text{m}^2$ and$$L=\frac{A_s}{\pi D}=\frac{30.4}{\pi(0.02)}=\boxed{\approx 484\ \text{m}}.$$
Check
The air-side coefficient (≈149 W/m²K) is roughly 150× smaller than the water-side, so it alone controls $U$ and the enormous length reflects the poor external cooling of a bare small-diameter tube by air, plus the large water mass flow (1.85 kg/s carrying ≈170 kW). The length is sensitive to the air-property choice: a ±5 % shift in $h_o$ moves $L$ by ±5 %. In practice one would add fins or use many parallel tubes; the single bare tube is what the question specifies.
