Copper Tube — Free Convection and Fins: 23-Chem-A3 Heat and Mass Transfer, May 2017, Q3

Question 3: Copper Tube — Free Convection and Fins (25 points)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Water at $40\ \text{°C}$, $v = 0.10$ m/s inside a thin copper tube $D = 0.025$ m, $L = 1$ m, $k_{Cu} = 380\ \text{W/m}\cdot\text{K}$; air at $20\ \text{°C}$ outside with $h_a = 1.32\{(T_s-T_a)/D\}^{0.25}$. Water: $\rho = 965.3$, $k = 0.675$, $\mu = 3.15\times10^{-4}$ (SI); $c_p \approx 4180\ \text{J/kg}\cdot\text{K}$ assumed (not printed — see note).

Quantity	Value
Water T / velocity	40 °C / 0.10 m/s
Tube ID / length	0.025 m / 1 m
Air T	20 °C
Fins	8 × (2 mm thick, 20 mm high), copper

Find. (a) $U$, (b) wall temperature $T_s$, (c) heat loss $Q$ for the bare tube; then (d) $U$ and (e) $Q$ with the 8 fins.

Figure 3 — End view of the finned copper tube: warm water inside, eight longitudinal copper fins radiating into still air. The outside free-convection film is the controlling resistance, so extending the outer area is what raises the overall coefficient.

Check

The water $c_p$ is not printed; 4180 J/kg·K is assumed. It is not load-bearing: the inside coefficient comes out ~1000 W/m²·K against an outside free-convection coefficient of ~7 W/m²·K, so the air film controls $U$ and any reasonable $c_p$ leaves the answers unchanged to three figures. The "8 rectangular radial fins" are taken as straight fins running the full 1 m length of the tube and projecting radially outward (a star-shaped cross-section), which is the arrangement the fin count and the adiabatic-tip wording describe.

Approach. Get the inside coefficient from Dittus–Boelter, then close the wall energy balance $h_i(T_w-T_s)=h_a(T_s)(T_s-T_a)$ (with the temperature-dependent $h_a$) for $T_s$; $U$ and $Q$ follow. For the fins, compute the fin efficiency, build the effective outside area, and re-close the balance.

Inside (water) coefficient. $Re = \rho v D/\mu = (965.3)(0.10)(0.025)/(3.15\times10^{-4}) = 7660$ (turbulent), $Pr = \mu c_p/k = 1.95$. The water is being cooled, so Dittus–Boelter takes $n = 0.3$ (at $Re \approx 7700$ the flow is only just turbulent, but $h_i$ barely matters here): $$Nu = 0.023\,Re^{0.8}Pr^{0.3} = 36.0,\qquad h_i = \frac{Nu\,k}{D} = \boxed{972\ \text{W/m}^2\text{K}}.$$
Wall temperature from the coupled balance (part b). Through the thin, high-conductivity wall $T_s$ is essentially uniform, so at steady state the inside and outside fluxes match: $$h_i(T_w - T_s) = 1.32\left(\frac{T_s-T_a}{D}\right)^{0.25}(T_s-T_a).$$ Solving (the outside coefficient is tiny) gives $\boxed{T_s = 39.9\ \text{°C}}$ — the wall sits within $0.1\ \text{°C}$ of the water, because the air film chokes the heat flow.
Outside coefficient and overall $U$ (part a). At $T_s = 39.9\ \text{°C}$, $$h_a = 1.32\left(\frac{39.9-20}{0.025}\right)^{0.25} = 7.01\ \text{W/m}^2\text{K}.$$ With the thin wall ($A_i \approx A_o$) the series resistance gives $$U = \left(\frac1{h_i} + \frac1{h_a}\right)^{-1} = \left(\frac1{972}+\frac1{7.01}\right)^{-1} = \boxed{6.96\ \text{W/m}^2\text{K}}.$$
Bare-tube heat loss (part c). Outer area $A_o = \pi D L = \pi(0.025)(1) = 0.0785\ \text{m}^2$: $$Q = U A_o (T_w - T_a) = 6.96(0.0785)(20) = \boxed{10.9\ \text{W}}.$$
Fin efficiency (parts d–e). Each rectangular fin ($t = 2$ mm, height $L_f = 20$ mm, adiabatic tip) has $m = \sqrt{2h_a/(k_{Cu}t)} = \sqrt{2(7.01)/(380\times0.002)} = 4.30\ \text{m}^{-1}$, so $mL_f = 0.086$ and $$\eta_f = \frac{\tanh(mL_f)}{mL_f} = 0.998.$$ The copper fins are essentially isothermal — as expected against so weak an air film.
Effective outside area and new $U$. Eight fins add $A_{fin} = 8(2 L_f)L = 8(0.040)(1) = 0.320\ \text{m}^2$ (two faces each); the exposed base is $A_{base} = \pi D L - 8tL = 0.0785 - 0.016 = 0.0625\ \text{m}^2$, so $$A_{eff} = A_{base} + \eta_f A_{fin} = 0.0625 + 0.998(0.320) = 0.382\ \text{m}^2.$$ Combining resistances on a UA basis and referring $U$ to the bare outer area $A_o$: $$UA = \left(\frac1{h_i A_i} + \frac1{h_a A_{eff}}\right)^{-1} = 2.57\ \text{W/K}, \qquad U = \frac{UA}{A_o} = \boxed{32.7\ \text{W/m}^2\text{K}}.$$
Finned heat-transfer rate (part e). $$Q = UA\,(T_w - T_a) = 2.57(20) = \boxed{51.4\ \text{W}},$$ a 4.7-fold increase over the bare tube — the fins pay off precisely because the outside film is the bottleneck.

Quantity	Bare tube	With 8 fins
Inside coefficient $h_i$	972 W/m²·K	972 W/m²·K
Outside coefficient $h_a$	7.01 W/m²·K	7.0 W/m²·K
Wall temperature $T_s$	39.9 °C	≈ 39.3 °C
Overall $U$ (bare-area basis)	6.96 W/m²·K	32.7 W/m²·K
Heat loss $Q$	10.9 W	51.4 W

Concept & Theory

Topic. Series convective resistances with a controlling free-convection film, and longitudinal-fin enhancement.

Key relations. Dittus–Boelter $Nu = 0.023Re^{0.8}Pr^{n}$ ($n=0.4$ fluid heated, $0.3$ fluid cooled); series $1/U = 1/h_i + 1/h_a$ (thin wall); fin parameter $m = \sqrt{2h/(kt)}$; adiabatic-tip efficiency $\eta_f=\tanh(mL_f)/(mL_f)$; effective area $A_{eff}=A_{base}+\eta_f A_{fin}$.

Why this works. With $h_a \ll h_i$, the outside film owns ~99% of the resistance, so $U \approx h_a$ and the wall floats up to the water temperature. Fins help only where the finned side controls — exactly here — and because the air coefficient is small, $mL_f$ is tiny and the copper fins run at nearly the base temperature ($\eta_f\to1$). The gain is essentially the ratio of finned to bare outside area.

Common pitfall. Putting fins on the high-$h$ (water) side, or using bare-tube area for the finned rate. Also, $h_a$ depends on $T_s$, so the balance must be solved iteratively rather than assuming $T_s$.

Source. Incropera & DeWitt §3.6 (fins, efficiency of an adiabatic-tip rectangular fin) and §8.5 (internal-flow correlations).

Question 3 of 6: Copper Tube — Free Convection and Fins

Part A — Heat Transfer

Question 3: Copper Tube — Free Convection and Fins (25 points)