13.when circuit breakers are classified as 100a through 2,000a frames, these numbers are normally referred to as the .
1 answer:
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Answer: 383.22K
Explanation:
L = 3m, w = 1.5m
Area A = 3 x 1.5 = 4.5m2
Q' = 750W/m2 (heat from sun) ,
& = 0.87
Q = &Q' = 0. 87x750 = 652.5W/m2
E = QA = 652.5 x 4.5 = 2936.25W
T(sur) = 300K, T(panel) = ?
Using E = §€A(T^4(panel) - T^4(sur))
§ = Stefan constant = 5.7x10^-8
€ = emmisivity = 0.85
2936.25 = 5.7x10^-8 x 0.85 x 4.5 x (T^4(panel) - 300^4)
T(panel) = 383.22K
See image for further details.
Answer:
48.61
Explanation:
See attached diagram.
The level rise in the tube is l sin α.
The level drop in the cylinder (let's call it y) is:
π/4 D² y = π/4 d² l
D² y = d² l
y = l (d/D)²
The elevation difference is the sum:
l sin α + l (d/D)²
l (sin α + (d/D)²)
From Bernoulli's principle:
P = ρgl (sin α + (d/D)²)
Divide both sides by density of water (ρw) and gravity:
P/(ρw g) = (ρ/ρw) l (sin α + (d/D)²)
h = S l (sin α + (d/D)²)
If we disregard the level change in the cylinder:
h = S l (sin α)
We want the percent error between these two expressions for h to be 0.1% when α = 25°.
[ S l (sin α + (d/D)²) − S l (sin α) ] / [ S l (sin α + (d/D)²) ] = 0.001
[ S l sin α + S l (d/D)² − S l sin α ] / [ S l (sin α + (d/D)²) ] = 0.001
[ S l (d/D)² ] / [ S l (sin α + (d/D)²)] = 0.001
(d/D)² / (sin α + (d/D)²) = 0.001
(d/D)² = 0.001 (sin α + (d/D)²)
(d/D)² = 0.001 sin α + 0.001 (d/D)²
0.999 (d/D)² = 0.001 sin α
d/D = √(0.001 sin α / 0.999)
When α = 25°:
d/D ≈ 0.02057
D/d ≈ 48.61
