Question 2 of 6: Part A: Local Free-Convection Coefficient on a Vertical Plate
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: open-book, three-hour exam; Part A (Heat Transfer, Q1–Q3) and Part B (Mass Transfer, Q4–Q6); all problems 25 points, at least two attempted per part. Every problem is worked in full below.
Reference texts: J. M. Coulson & J. F. Richardson, Chemical Engineering Vols. 1&2 (Coulson & Richardson) — boiling curves, tube-bank convection, gas absorption and column sizing; F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — free-convection correlations and the gas-property tables reproduced in the paper; J. R. Welty, C. E. Wicks & R. E. Wilson, Fundamentals of Momentum, Heat and Mass Transfer — the laminar vertical-plate local Nusselt relation; R. E. Treybal, Mass-Transfer Operations (McGraw-Hill) — stagnant-film diffusion, differential (Rayleigh) distillation and packed-tower flooding; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.).
Question 2 — Part A: Local Free-Convection Coefficient on a Vertical Plate (25 points)
Given. Vertical plate at $T_s=390$ K in quiescent gas at $T_\infty=290$ K, so $\Delta T=100$ K; position $x=0.45$ m up the plate. Properties are read at the film temperature $T_f=\tfrac12(T_s+T_\infty)=340$ K by linear interpolation of the supplied 300 K / 400 K rows ($T_f$ is 40 % of the way):
Gas @ $T_f=340$ K
$\nu$ (m²/s)
$k$ (W/m·K)
$Pr$
Oxygen (O₂)
$20.4\times10^{-6}$
0.0293
0.721
Hydrogen (H₂)
$138.2\times10^{-6}$
0.200
0.699
Find. The local natural-convection coefficient $h_x$ at $x=0.45$ m for (a) O₂ and (b) H₂.
Figure 2 — Laminar free-convection layer rising along the heated vertical plate. The local coefficient is evaluated at height $x=0.45$ m, where $h_x=Nu_x\,k/x$ with $Nu_x$ from the Ostrach similarity solution.
Approach. Evaluate the local Grashof number $Gr_x$ at $x=0.45$ m, confirm laminar flow ($Gr_x Pr<10^9$), apply the laminar vertical-plate similarity relation $Nu_x=0.508\,Pr^{1/2}(0.952+Pr)^{-1/4}Gr_x^{1/4}$, and convert to $h_x=Nu_x\,k/x$ for each gas.
Local Grashof number, Oxygen. With $\beta=1/T_f=1/340=2.94\times10^{-3}$ K$^{-1}$,
$$Gr_x=\frac{g\,\beta\,\Delta T\,x^{3}}{\nu^{2}}=\frac{(9.81)(2.94\times10^{-3})(100)(0.45)^3}{(20.4\times10^{-6})^2}=6.31\times10^{8}.$$
Then $Gr_x Pr=(6.31\times10^{8})(0.721)=4.55\times10^{8}<10^{9}$, so the layer is laminar.
Local Nusselt and coefficient, Oxygen. The Ostrach laminar relation gives
$$Nu_x=0.508\,(0.721)^{1/2}(0.952+0.721)^{-1/4}(6.31\times10^{8})^{1/4}=60.1,$$
$$h_x=\frac{Nu_x\,k}{x}=\frac{(60.1)(0.0293)}{0.45}=\boxed{3.9\ \text{W/m}^2\text{K}}\quad\text{(part a).}$$
Local Grashof number, Hydrogen. Hydrogen’s far larger $\nu$ shrinks $Gr_x$ by $(\nu_{H_2}/\nu_{O_2})^2\approx46$:
$$Gr_x=\frac{(9.81)(2.94\times10^{-3})(100)(0.45)^3}{(138.2\times10^{-6})^2}=1.38\times10^{7},\qquad Gr_x Pr=9.6\times10^{6}<10^{9}.$$
Local Nusselt and coefficient, Hydrogen. With $Pr=0.699$,
$$Nu_x=0.508\,(0.699)^{1/2}(0.952+0.699)^{-1/4}(1.38\times10^{7})^{1/4}=22.8,$$
$$h_x=\frac{(22.8)(0.200)}{0.45}=\boxed{10.2\ \text{W/m}^2\text{K}}\quad\text{(part b).}$$
Hydrogen has the smaller $Gr$ yet the larger $h$: its thermal conductivity is about seven times that of oxygen, which dominates once $Nu$ is only modestly lower.
Quantity
Oxygen
Hydrogen
$Gr_x$ at $x=0.45$ m
$6.31\times10^{8}$
$1.38\times10^{7}$
$Nu_x$
60.1
22.8
Local coefficient $h_x$
≈ 3.9 W/m²K
≈ 10.2 W/m²K
Check
The question asks for the coefficient “at a distance of 45 cm”, which is the local value $h_x$. Had it asked for the average over a plate 0.45 m tall, the laminar mean is $\bar h=\tfrac43 h_x$ (i.e. ≈ 5.2 and 13.5 W/m²K). Both interpretations use the same $Gr_x$; we report the local value as literally requested.
