Question

A square aluminum plate 5 mm thick and 150 mm on a side is heated while vertically suspended in quiescent air at 75°C. The temperature of the plate is 15°C

a. Write the assumptions of the problem (at least two)

b. What are the properities of the problem

c. Calculate the Rayleigh number

d. What is the nature of the flow

e. Calculate the average Nusselt number

f. Determine the average heat transfer coefficient for the plate by two methods: using results from the similarity solution to the boundary layer equations, and using results from an empirical correlation

Answer 1

Solution

Known: Vertical plate is in quiescent air at a higher  temperature.

Schematic:

Plate 150 mm square & 5 mm thick Air Too 19

T = 288 K
$T_{\infty }=348 \: \: K$   
L=0.150 m

a)

Assumptions:

(1) Uniform plate surface temperature,

(2) Quiescent room air,

(3) Surface radiation exchange with surroundings negligible,

(4) Perfect gas behavior for air

(5) Constant properties

b) Properties of air

Filim temerture

$T_{f}= \frac{T_{s}+T_{\infty }}{2}=\frac{288+348}{2}$

and at 1 atm

From table A4

TP lo? Ns/m) P. 10 (m/s) k. 10 (W/ mK) a 10% (ms) Pr (K) kg/m") (kJ/kg-K) Air, M = 28.97 kg/kmol 100 3.5562 1.032 150 2.3364 1.012 200 1.7458 1.007 250 1.3947 1.006 300 1.1614 1.007 71.1 103.4 132.5 159.6 184.6 2.00 4.426 7.590 9.34 13.8 18.1 223 2.54 5.84 10.3 15.9 225 0.786 0.758 0.737 0.720 0.707 11.44 15.89 263 350 0.9950 1.009 208.2 20.92 30.0 29.9 0.700

p= 1.101496 kg/m3

$\nu =17.7\times10^{-6} \: \: m^2/s$

$k =27.63\times10^{-3} \: \: W/m-k$

$\alpha =25.16\times10^{-6} \: \: m^2/s$

$Pr=0.704$

c)

Rayleigh number:

We know that rayleigh number is given by following equation

$R_{aL}=\frac{g\beta \Delta TL^3}{\nu \alpha }$

Where

$\beta =1/T_{f}=1/318\: \: K^{-1}$

$\Delta T= T_{\infty }-T_{s}=348-288$

$\Delta T= 60\: \: K$

$R_{aL}=\frac{9.81\times 60\times 0.150^3}{318\times17.7\times 10^{-6}\times25.16\times10^{-6} }$

$R_{aL}=1.4\times 10^{7}$

d)

The nature of the flow :

we have calculated the Rayleigh number to determine the boundary layer flow conditions

Since

,

The flow is laminar.

Nature of the flow : laminar flow

e)

The average Nusselt number

Using similarity solution:

The flow is laminar and the similarity solution is given below

$g(pr)= \frac{0.75pr^{1/2}}{(0.609+1.221pr^{1/2}+1.238pr)^{1/4}}$    $for \: \: \: \: 0\leq pr \leq \infty$

$g(pr)= \frac{0.75\times 0.704^{1/2}}{(0.609+1.221\times 0.704^{1/2}+1.238\times 0.704)^{1/4}}$

$\bar{Nu_{L}}=\frac{\bar{h}L}{k}=\frac{4}{3}(\frac{Gr_{L}}{4})^{1/4}g(Pr)$

substituting numerical values with

$Gr_{L}=R_{aL}/Pr$

$\bar{Nu_{L}}=\frac{4}{3}(\frac{1.4\times 10^{7}}{0.704\times4})^{1/4}\times 0.5002$

$\bar{Nu_{L}}=31.49$

Using empirical correlation

$\bar{Nu_{L}}=0.68+\frac{0.670Ra_{L}^{1/4}}{(1+(0.492/Pr)^{9/16})^{4/9}}$$Ra_{L}\leq 10^9$

$\bar{Nu_{L}}=0.68+\frac{0.670(1.4\times10^{7})^{1/4}}{(1+(0.492/0.704)^{9/16})^{4/9}}$

$\bar{Nu_{L}}=32.106\: \: W/m^2-K$

f) The average heat transfer coefficient

1. using similarity solution

we have calculated average Nusslet number using  similarity solution in part (e)

$\bar{Nu_{L}}=31.49$

$\frac{\bar{h}\times 0.150}{27.63\times 10^{-3}}=31.49$

Average heat transfer coefficient

$\bar{h}=5.8\: \: \: W/m^2-K$

2)   using empirical correlation ( Churchill-Chu correlation)

we have calculated average Nusslet number using using empirical correlation in part (e)

$\bar{Nu_{L}}=32.106\: \: W/m^2-K$

$\frac{\bar{h}\times 0.150}{27.63\times 10^{-3}}=32.105$

The average heat transfer coefficient

$\bar{h}=5.913\: \: \: W/m^2-K$

COMMENTS: The agreement of   $\bar{h}$  calculated by these two methods is excellent.