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Sagot :
Answer:
a) P = 118.4 W, b) t = 7.9 10⁻⁵ s
Explanation:
a) Let's analyze this interesting exercise a bit, we suppose that all the laser
energy is used to heat the aluminum, we should calculate the energy necessary to bring the solid aluminum to the melting temperature and add the energy to carry out the change of solid state to liquid,
let's use the calorimeter equation
Q₁ = m c_e ΔT
and the energy of change of these solid to liquid (fusion process)
Q₂ = m L
the energy required to create the hole is
Q_ {total} = Q₁ + Q₂
if there are no losses this is the laser energy
E = Q_ {total}
The aluminun data c_e =9000 J/kgC, L = 322 103 J/kg, ρ = 2.7 103 kg/m3 , T₂ = 660C, T₀= 25C
Let's find the mass of the hole, which we approximate by a cylinder of diameter d = 0.25 mm = 0.25 10⁻³ m and a thickness of e = 1 mm = 1 10⁻³ m
let's use the concept of density
ρ = m / V
the volume of a cylinder is
V = π r² e = π (d²/4) e
we substitute
m = [tex]\rho \pi \frac{d^2 e}{4}[/tex]
let's calculate
m = π/4 2.7 10³ (0.25 10⁻³)² 1 10⁻³
m = 1,325 10⁻⁷ kg
we calculate the energy
E = 1,325 10⁻⁷ 900 (660 - 25) + 1,325 10⁻⁷ 322 10³
E = 7.57 10⁻² + 4.27 10⁻²
E = 1.184 10⁻¹ J
Let's use the power ratio
P = E / t
P = 0.1184 /1 10⁻³
P = 118.4 W
b) In this part they indicate that the laser power is P = 1500 W, find the time to deposit the energy to melt the aluminum
P = E / t
t = E / P
t = 0.1184 / 1500
t = 7.9 10⁻⁵ s
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