Are the stack temperature and oxygen reasonable for these operating conditions? if not, what oxygen and stack temperature would
1 answer:
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Answer:
775.48 W
Explanation:
given,
diameter of disk = 0.6 cm
length of the disk = 0.4 m
T₁ = 450 K T₂ = 450 K T₃ = 300 K
= 1.33
now,
the value of view factor (F₁₂)corresponding to 1.33
F₁₂ = 0.265
F₁₃ = 1 - 0.265 = 0.735
now,
net rate of radiation heat transfer from the disk to the environment:
= 2 F₁₃ A₁ σ (T₁⁴ - T₃⁴)
= 2 x 0.735 x π x (0.3)² x (5.67 x 10⁻⁸ W/m²) (450⁴ - 300⁴)
= 775.48 W
Net radiation heat transfer from the disks to the environment = 775.48 W
Answer:
<em>Radius at liftoff 8.98 m</em>
Explanation:
At the working altitude;
maximum radius = 24 m
air pressure = 0.030 atm
air temperature = 200 K
At liftoff;
temperature = 349 K
pressure = 1 atm
radius = ?
<em>First, we assume balloon is spherical in nature,</em>
<em>and that the working gas obeys the gas laws.</em>
from the radius, we can find the volume of the balloon at working atmosphere.
Volume of a sphere =
volume of balloon = x 3.142 x
= 57913.34 m^3
using the gas equation,
=
<em>The subscript 1 indicates the properties of the gas at working altitude, and the subscript 2 indicates properties of the gas at liftoff.</em>
imputing values, we have
=
0.03 x 57913.34 x 349 = 200V2
V2 = 606352.67/200 = <em>3031.76 m^3 this is the volume occupied by the gas in the balloon at liftoff.</em>
from the formula volume of a sphere,
V = =
x 3.142 x
= 3031.76
4.19 = 3031.76
= 3031.76/4.19
radius r of the balloon on liftoff = = <em>8.98 m</em>