A flat-plate solar collector is 2 m long and 1 m wide and is inclined 60o to the horizontal. The cover plate is separated from t
he absorber plate by an air gap of 2 cm thick. If the temperatures of the cover plate and the absorber plate are 305 K and 335 K, respectively. Calculate the convective heat loss. Take the pressure at 1 atm.
1 answer:
Answer:
164.4 W
Explanation:
We can define Heat loss as a measure of the total transfer of heat through the fabric of a building from inside to the outside, either from conduction, convection, radiation, or any combination of the these.
Please kindly check attachment for the solution of the heat loss as asked in the question.
The attached file have the solved problem.
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Answer:
Maximum height=7.3535 m
Explanation:
Solution of the problem is given in the attachments.
Answer:
Explanation:
Reynolds number:
Reynolds number describe the type of flow.If Reynolds number is too high then flow is called turbulent flow and Reynolds is low then flow is called laminar flow .
Reynolds number is a dimensionless number.Reynolds number given is the ratio of inertia force to the viscous force.
For plate can be given as
Where ρ is the density of fluid , v is the average velocity of fluid and μ is the dynamic viscosity of fluid.
Flow on plate is a external flow .The values of Reynolds number for different flow given as
Answer:
Step 1 of 3
Case A:
AISI 1018 CD steel,
Fillet radius at wall=0.1 in,
Diameter of bar
From table deterministic ASTM minimum tensile and yield strengths for some hot rolled and cold drawn steels for 1018 CD steel
Tensile strength
Yield strength
The cross section at A experiences maximum bending moment at wall and constant torsion throughout the length. Due to reasonably high length to diameter ratio transverse shear will be very small compared to bending and torsion.
At the critical stress elements on the top and bottom surfaces transverse shear is zero
Explanation:
See the next steps in the attached image
