Answer:
www stands for world wide web
Explanation:
It will really help you thank you.
The complete Question is:
Airflow through a long, 0.15-m-square air conditioning duct maintains the outer duct surface temperature at 10°C. If the horizontal duct is uninsulated and exposed to air at 35°C in the crawlspace beneath a home, what is the heat gain per unit length of the duct? Evaluate the properties of air at 300 K. For the sides of the duct, use the more accurate Churchill and Chu correlations for laminar flow on vertical plates.
What is the Rayleigh number for free convection on the outer sides of the duct?
What is the free convection heat transfer coefficient on the outer sides of the duct, in W/m2·K?
What is the Rayleigh number for free convection on the top of the duct?
What is the free convection heat transfer coefficient on the top of the duct, in W/m2·K?
What is the free convection heat transfer coefficient on the bottom of the duct, in W/m2·K?
What is the total heat gain to the duct per unit length, in W/m?
Answers:
- 7709251 or 7.709 ×10⁶
- 4.87
- 965073
- 5.931 W/m² K
- 2.868 W/m² K
- 69.498 W/m
Explanation:
Find the given attachments for complete explanation
Seriously? Ok
first, get some good quality metal, preferably aluminum, if you want to avoid rust.
build in the the shape of a robot, you can use a doll to help you if you are a beginner, but feel free to shape it the a Star Wars Character!
create an AI (now you said robot not AI) and fix everything up.
Answer:
T1 = 625.54 K
Explanation:
We are given;
T_α = Tsur = 25°C = 298K
h = 20 W/m².K,
L = 0.15 m
K = 1.2 W/m.K
ε = 0.8
Ts = T2 = 100°C = 373K
T1 = ?
Assumption:
-Steady- state condition
-One- dimensional conduction
-No uniform heat generation
-Constant properties
From Energy balance equation;
E°in - E°out = 0
Thus,
q"cond – q"conv – q"rad = 0
K[(T1 - T2)/L] - h(Ts-T_α) - εσ (Ts⁴ – Tsur⁴)
Where σ is Stephan-Boltzmann constant and has a value of 5.67 x 10^(-8)
Thus;
K[(T1 - T2)/L] - h(Ts-T_α) - εσ (Ts⁴ – Tsur⁴) = 1.2[(T1 - 373)/0.15] - 20(373 - 298] - 0.8x5.67x10^(-8)[373⁴ - 298⁴] = 0
This gives;
(8T1 - 2984) - (1500) - 520.31 = 0
8T1 = 2984 + 1500 + 520.31
8T1 = 5004.31
T1 = 5004.31/8
T1 = 625.54 K