A thick oak wall, initially at 25 C, is suddenly exposed to combustion products for which T 800 C and h 20 W/m2 K. (a) Determine the time of exposure required for the surface to reach the ignition temperature of 400 C.

The time of exposure required for the surface to reach the ignition temperature of 400 °C is 310 seconds.

Explanation:

For us to solve the question, if we consider the wall to be acting like a semi - infinite wall, the equation to solve it will be given by;

(T(o) - T)/(T(∞) - T(o)) = [erf x/√(2αt)] + [exp ((hx/k) + (h²αt/k²))] x [1 - erf {(h√(αt))/k} + {(x/(2√(αt))}]

Where;

T(o) = initial uniform temperature of wall

T = required ignition temperature of the wall,

T(∞) = temperature of combustion exhaust.

h = surface coefficient between the wall and combustion gas,

α = thermal diffusivity of oak,

k = thermal conductivity of oak

x = distance from the surface of wall.

From the image i attached, looking at oak under hardwood, we have the following properties ;

ρ = 545 kg/m³

c = 2385 J/Kg.k

k = 0.17 W/m.k

Now, thermal diffusivity of oak, α is given by; α = k/ρc = 0.17/(545 x 2385) = 1.308 x 10^(-7) m²/s

Now, from the question,

T(o) = 400 °C

T = 25 °C

T(∞) = 800 °C

Thus;

(T(o) - T)/(T(∞) - T(o)) will give;

(400 - 25)/(800 - 25) = 375/775 = 0.484

At the surface, x =0,we have;

(x/(2√(αt)) = 0 and erf(0) = 0 and

1 - erf(0) = 1

also, since h= 20 W/m²K, thus, (h√(αt))/k = [20 x √((1.308 x 10^(-7) x t))]/0.17 = [20 x √(1.308 x 10^(-7) x √t)]/0.17 = 0.0425√t

Putting x = 0 in the initial equation as well as the other relevant values and rearranging, we obtain ;

0.484 = [erf 0] + [((h√αt)/k)²)] x [1 - erf {(h√(αt))/k}]

Now, from earlier, (h√(αt))/k = 0.0425√t and also erf 0 = 0 and 1 - erf (0) = 1. So;

Thus, 0.484 = e^(0.0425√t)² [(1 - erf 0.0425√t)]

By using error function calculator and trial and error method, t is approximately equal to 310 seconds.

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm