Why dues brainy exist as a learning platform when it is just full of answers and you won't learn anything?
1 answer:
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Answer:
Explanation:
= Energy for defect formation = 1.86 eV
T = Temperature =
k = Boltzmann constant =
The fraction of lattice sites that are Schottky defects is given by
The required ratio is .
Answer:
The NRCS triangular UH for a 10-minute rainfall is Qp = 49.84 m³/s
Peak flow of the aggregated runoff hydrograph is 420.58 m³/s
The total volume of runoff is 2125000 m³/s
Explanation:
We have
A = 25 km²
tr = 10 min = 1/6 hr
tc = 1.6 hr
lag time = 0.6 tc = 0.96 hr
Tp = tr/2 + 0.6 tc = 1/12 + 0.96 = 1.043 hr
Qp = 2.08×25/1.043 = 49.84 m³/s
Tb = 8/3×Tp = 8/3×1.043 = 2.782 hr
Since the area is
Time (min) Runoff (cm) Volume of runoff m³
0 0 0
10 4 1000000 m³
20 2.5 625000 m³
30 2 500000 m³
Total volume of runoff = 1000000 + 625000 + 500000 = 2125000 m³/s
For the 1st 10 minutes, we have
A = 25 km²
tr = 30 min = 1/2 hr
tc = 1.6 hr
lag time = 0.6 tc = 0.96 hr
Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr
Qp = 2.08×25×4/1.043 = 197.92 m³/s
Tb = 8/3×Tp = 8/3×1.21 = 3.227 hr
For the 2nd 10 minutes, we have
A = 25 km²
tr = 30 min = 1/2 hr
tc = 1.6 hr
lag time = 0.6 tc = 0.96 hr
Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr
Qp = 2.08×25×2.5/1.043 = 123.7 m³/s
Tb = 8/3×Tp = 8/3×1.21 = 3.227 hr
For the 3rd 10 minutes, we have
A = 25 km²
tr = 30 min = 1/2 hr
tc = 1.6 hr
lag time = 0.6 tc = 0.96 hr
Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr
Qp = 2.08×25×2.5/1.043 = 98.96 m³/s
Tb = 8/3×Tp = 8/3×1.21 = 3.227 hr
Peak flow of aggregate runoff is given by
Qp (total) = 98.96 + 123.7 +197.92 = 420.58 m³/s
Total volume of runoff is given by
Total volume of runoff = 1000000 + 625000 + 500000 = 2125000 m³/s
