Question 1 of 6: Part A: Free-Convection Heat Loss from a Pipe, Horizontal vs. Vertical
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2017 — 16-Chem-A3 Heat and Mass Transfer. Three-hour, open-book exam (one textbook of the candidate’s choice; any non-communicating calculator). Format: two parts — Part A — Heat Transfer (Q1–Q3) and Part B — Mass Transfer (Q4–Q6). All problems are worth 25 points; the rubric requires at least two problems attempted from each part and marks only the first two per part, so a complete paper is four questions (100 points). All six problems are solved below for completeness. Property data are quoted as stated in each problem; where a value applies to a film condition it is used at the stated film temperature.
Reference texts: F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — free/forced convection correlations, conduction shape factors, the heat–mass-transfer analogy; J. P. Holman, Heat Transfer (McGraw-Hill) — natural-convection constants $C,m$ and cylinder conduction; R. E. Treybal, Mass-Transfer Operations (McGraw-Hill) — wetted-wall $k_G$, gas absorption and the Chilton–Colburn analogy; R. B. Bird, W. E. Stewart & E. N. Lightfoot, Transport Phenomena (Wiley); supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.).
Question 1 — Part A: Free-Convection Heat Loss from a Pipe, Horizontal vs. Vertical (25 points)
Given. A pipe of diameter $D=0.040$ m and length $L=0.60$ m at surface temperature $T_s=85$ °C loses heat by free convection to still water at $T_\infty=15$ °C, so $\Delta T = 70$ K. Water properties (at the film temperature $T_f=50$ °C) are tabulated below.
Property
Value
Density $\rho$
988 kg/m³
Thermal conductivity $k$
0.647 W/m·K
Dynamic viscosity $\mu$
$5.493\times10^{-4}$ kg/m·s
Specific heat $c_p$
4178 J/kg·K
Volume-expansion coeff. $\beta$
$5.1\times10^{-4}$ K$^{-1}$
Find. The convective heat-loss rate $Q$ (a) with the pipe horizontal and (b) with the pipe vertical.
Figure 1 — The same pipe loses heat to the surrounding water in two orientations. The characteristic length in $Gr$ is the diameter $D$ when horizontal and the height $L$ when vertical, which is what makes the two answers differ.
Approach. Form the Prandtl number, then the Grashof number on the correct characteristic length for each orientation, select the matching $(C,m)$ from the $Gr\,Pr$ band, evaluate $Nu$ and hence the film coefficient $h=Nu\,k/L_c$, and finally $Q=hA\,\Delta T$ over the same lateral area $A=\pi D L$.
Prandtl number. This is a fluid property, common to both orientations:
$$Pr = \frac{c_p\,\mu}{k} = \frac{(4178)(5.493\times10^{-4})}{0.647} = \boxed{3.55}.$$
Horizontal pipe — Grashof number on $L_c=D$. The free-convection driving group uses the diameter for a horizontal cylinder:
$$Gr_D = \frac{g\,\beta\,\Delta T\,D^{3}\rho^{2}}{\mu^{2}} = \frac{(9.81)(5.1\times10^{-4})(70)(0.040)^3(988)^2}{(5.493\times10^{-4})^2} = 7.25\times10^{7}.$$
Then $Gr_D\,Pr = (7.25\times10^{7})(3.55) = 2.57\times10^{8}$, which lies in the band $10^4$–$10^9$, so $C=0.53,\ m=0.25$.
Horizontal film coefficient and heat loss. With $Nu = 0.53\,(Gr_D Pr)^{0.25}$,
$$Nu = 0.53\,(2.57\times10^{8})^{0.25} = 67.1,\qquad h = \frac{Nu\,k}{D} = \frac{(67.1)(0.647)}{0.040} = 1086\ \text{W/m}^2\text{K}.$$
The lateral area is $A=\pi D L = \pi(0.040)(0.60)=0.0754\ \text{m}^2$, so
$$Q_\text{horiz} = hA\,\Delta T = (1086)(0.0754)(70) = \boxed{5.73\times10^{3}\ \text{W} \approx 5.73\ \text{kW}}.$$
That is part (a).
Vertical pipe — Grashof number on $L_c=L$. For a vertical cylinder the height replaces the diameter, and because $Gr\propto L_c^{3}$ the group jumps by $(L/D)^3=3375$:
$$Gr_L = \frac{g\,\beta\,\Delta T\,L^{3}\rho^{2}}{\mu^{2}} = 2.45\times10^{11},\qquad Gr_L\,Pr = 8.68\times10^{11} > 10^{9},$$
so now $C=0.13,\ m=0.33$.
Vertical film coefficient and heat loss. With $Nu = 0.13\,(Gr_L Pr)^{0.33}$,
$$Nu = 0.13\,(8.68\times10^{11})^{0.33} = 1132,\qquad h = \frac{Nu\,k}{L} = \frac{(1132)(0.647)}{0.60} = 1220\ \text{W/m}^2\text{K}.$$
Over the same area $A=0.0754\ \text{m}^2$,
$$Q_\text{vert} = (1220)(0.0754)(70) = \boxed{6.44\times10^{3}\ \text{W} \approx 6.44\ \text{kW}}.$$
That is part (b).
