Consider a thin circular disk that has been heated up to 400 °C and then left inside a chamber to cool down. The chamber surface
is kept at 40 °C and the air inside is maintained at 25 °C. Assume that the disk is held at the center of the chamber, and we can ignore heat conduction effects in this problem. The disk surface property is known as ε=0.65. For the diameter of D=200 mm, what is the total rate of heat transferfrom the disk?
1 answer:
Answer:
hello your question lacks the required diagram attached below is the complete question with the required diagram
answer : Qtotal = 807.4 Mw
Explanation:
Given Data :
disk properties :
∈ = 0.65
D = 200 mm
Ts = 400⁰c
attached below is the detailed solution
The total rate of Heat transferred from the disk
Qtotal = 807.4 Mw
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The time taken for the two balls to hit each other is 8 s.
The given parameters:
- <em>Acceleration of the rocket, a = 2 m/s²</em>
- <em>Length of the chamber, s = 4 m</em>
- <em>Speed of the first ball, = V1 = 0.3 m/s</em>
- <em>Speed of the second ball, V2 = 0.2 m/s</em>
The time taken for the two balls to hit each other is calculated by applying relative velocity formula as shown below;
Thus, the time taken for the two balls to hit each other is 8 s.
Learn more about relative velocity here: brainly.com/question/17228388
Refer to the diagram shown below.
After 5 minutes (300 seconds):
The man travels north by (2 ft/s)*(300 s) = 600 ft
The woman, located at q, 500 east of p, begins walking south at 4 ft/s.
The distance separating them is
d₁ = √(600² + 500²) = 781.025 ft
After 20 minutes:
The man has traveled for 20 minutes (1200 s).
The woman has traveled for 15 minutes (900 s).
The man has moved (2 ft/s)*(1200 s) = 2400 ft north of p.
The woman has moved (4 ft/s)*(900 s) = 3600 ft south of q.
The distance separating them is
d₂ = √(6000² + 500²) = 6020.8 ft
The separation from d₁ to d₂ occurs in 15 minutes (900s).
Therefore the rate of separation is
Rate = (d₂ - d₁ ft)/(900 s) = (6020.8 - 781.025)/900 = 5.822 ft/s
or
Rate = (5.822 ft/s)*(60 s/min) = 349.32 ft/min
Answer: 349.32 ft/min (or 5.82 ft/s)
After 5 minutes (300 seconds):
The man travels north by (2 ft/s)*(300 s) = 600 ft
The woman, located at q, 500 east of p, begins walking south at 4 ft/s.
The distance separating them is
d₁ = √(600² + 500²) = 781.025 ft
After 20 minutes:
The man has traveled for 20 minutes (1200 s).
The woman has traveled for 15 minutes (900 s).
The man has moved (2 ft/s)*(1200 s) = 2400 ft north of p.
The woman has moved (4 ft/s)*(900 s) = 3600 ft south of q.
The distance separating them is
d₂ = √(6000² + 500²) = 6020.8 ft
The separation from d₁ to d₂ occurs in 15 minutes (900s).
Therefore the rate of separation is
Rate = (d₂ - d₁ ft)/(900 s) = (6020.8 - 781.025)/900 = 5.822 ft/s
or
Rate = (5.822 ft/s)*(60 s/min) = 349.32 ft/min
Answer: 349.32 ft/min (or 5.82 ft/s)