Consider an opaque horizontal plate that is well insulated on its back side. The irradiation on the plate is 2500 W/m2, of which
500 W/m2 is reflected. The plate is at 227°C and has an emissive power of 1200 W/m2. Air at 127°C flows over the plate with a convection heat transfer coefficient of 15 W/m2 ⋅ K. Determine the emissivity, absorptivity, and radiosity of the plate. What is the net heat transfer rate per unit area?
1 answer:
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Answer:
Diameter decreases by the diameter of 0.0312 m.
Explanation:
Given that,
Bulk modulus = 14.0 × 10¹⁰ N/m²
Diameter d = 2.20 m
Depth = 2.40 km
Pressure = ρ g h = 1030 × 9.81 × 2.4 × 1000
= 24.25 × 10⁶ N/m²
Volume =
Bulk modulus is equal to
now
Δ r = -0.0156 m
change in diameter
Δ d = -2 × 0.0156
Δ d = -0.0312 m
Diameter decreases by the diameter of 0.0312 m.
Answer:
<em>a) 318.2 W/m^2</em>
<em>b) 2.5 x 10^-4 J</em>
<em>c) 1.55 x 10^-8 v/m</em>
<em></em>
Explanation:
Power of laser P = 1 mW = 1 x 10^-3 W
exposure time t = 250 ms = 250 x 10^-3 s
If beam diameter = 2 mm = 2 x 10^-3 m
then
cross-sectional area of beam A = = (3.142 x
)/4
A = 3.142 x 10^-6 m^2
a) Intensity I = P/A
where P is the power of the laser
A is the cros-sectional area of the beam
I = ( 1 x 10^-3)/(3.142 x 10^-6) = <em>318.2 W/m^2</em>
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b) Total energy delivered E = Pt
where P is the power of the beam
t is the exposure time
E = 1 x 10^-3 x 250 x 10^-3 = <em>2.5 x 10^-4 J</em>
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c) The peak electric field is given as
E =
where I is the intensity of the beam
E is the electric field
c is the speed of light = 3 x 10^8 m/s
= 8.85 x 10^9 m kg s^-2 A^-2
E = = <em>1.55 x 10^-8 v/m</em>