A(n) _________ is a current greater than the equipment rated current or conductor ampacity, which is confined to the normal cond
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
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Answer:
308 acre-ft of water
Explanation:
Given:
Area of the watershed = 22 square miles
Depth of Rainfall = 0.75 in ==0.0625 ft
Percentage rainfall falling in reservoir as runoff = 35%
Now,
1 square mile = 640 acre
Thus,
22 square miles = 22 × 640 = 14,080 acres
Thus,
The total volume of rainfall = Area of watershed × Depth of the rainfall
or
The total volume of rainfall = 14,080 acres × 0.0625 ft = 880 acre-ft
also,
only 35% of the total rainfall is contributing as runoff
thus,
Runoff = 0.35 × 880 acre-ft = 308 acre-ft of water
Solution :
Given :
k = 0.5 per day
Volume, V
Now, input rate = output rate + KCV ------------- (1)
Input rate
The output rate
= ( 40 + 0.5 ) x C x 1000
Decay rate = KCV
∴
= 1.16 C mg/s
Substituting all values in (1)
C = 4.93 mg/L