Review Example 7.2 and solve this problem: A flat plate of width w = 1 m is maintained at a uniform surface temperature of Ts = 230°C by using independently controlled heated segments, each of which is Lh = 50 mm long. If atmospheric air at 25°C flows over the plate at a velocity of 50 m/s, at what heater is the electrical input a maximum? What is the value of this input?

The question in the comments section seems to contradict the main question because though it clearly states the thickness, the length and the distance of the moduled position, the value of the velocity is different. So i will use the velocity of 50 m/s in the main question. Lastly, the value of k for the module is missing and it's 5.2 W/m.k

A) Required power generation = 15.35860 x 10^(5) W/m³

B) The heat at which the electric input will be maximum is at a temperature of 244.77°C

Explanation:

From the question, we are given that;

T(s) = 230°C

T(o) = 25°C

a = 10mm = 0.01m

b = 50mm = 0.05m

L = 700mm = 0.7m

U(∞) = 50 m/s

First of all, calculation of average temperature;

T(avg) = [T(s) + T(o)]/2 = (230 + 25)/2 = 255/2 = 127.5°C

And converting to degree kelvin, to get; 273 + 127.5 ≈ 400 K

From the table i attached, the thermophysical properties are gotten as;

K = 0.0329 W/m.k

v = 25.78 x 10^(-6) m²/s

Pr = 0.703

Now,the required power generation is given by;

q' = [h(T(s) - T(o))]/a

h is unknown, so let's calculate it.

h is given by;

h = [Nu(x) • k]/x = [0.0296 • (Re)^(4/5) • (Pr)^(1/3) • k]/(L + b/2)

Where Re is reynolds number.

Now, reynolds number is given as;

Re = [U(∞) • (L + (b/2))]/v

= [50(0.7 + (0.05/2))]/25.78 x 10^(-6) = 1.41 x 10^(6)

Now, we can calculate h by plugging in the relevant values ;

[0.0296 • (1.41 x 10^(6))^(4/5) • (0.703)^(1/3) • 0.0329]/(0.7+(0.05/2)) = (2429.72 x 0.0296)/0.96 = 74.92 W/Km²

Since, we have gotten h, let's plug in the relevant values to get the required power generation;

q' = [h(T(s) - T(o))]/a

q' = [74.92(230 - 25)]/0.01 =15.35860 x 10^(5) W/m³

B) maximum temperature is given as;

Tmax = [(q'•a²)/2k] + T(s)

Now, k here will be that of the module

Tmax = [(15.35860 x 10^(5) x 0.01²)/(2 x 5.2)] + 230 = 244.77°C

i think i backed my dad

a true.

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