Answer:
a) 18.75 ft^2
b) 2.5 ft
c) 12.07 ft
d) 10 ft
e) 1.553 ft
f) 1.875 ft
Explanation:
Given data :
5.0-ft bed width, ( b )
2.5-ft depth ( y )
1 : 1 side slope
Evaluate
a) Area of trapezoidal section
A = by + my^2
we assume m = 1
A = [5 + (1 * 2.5 ) ] *2.5
= ( 7.5 ) 2.5 = 18.75 ft^2
b) Calculate water depth
water depth = 2.5 ft
c) Calculate wetted perimeter
P = b + 2y √ 1 + m^2
= 5 + (2.5*2) √ 1 + 1 ^2 = 12.07 ft
d) calculate top width
T = b + 2my
= 5 + 2 ( 1 * 2.5 ) = 10 ft
e) calculate hydraulic radius
R = A / P = 18.75 / 12.07
= 1.553 ft
f) calculate hydraulic depth
D = A / T = 18.75 / 10 = 1.875 ft
Answer:
Free points................
Explanation:
Thanks for the points.
Answer:
331809.5gallon/hr or 92.16gallon/s
Explanation:
What is the peak runoff discharge for a 1.3 in/hr storm event from a 9.4-acre concrete-paved parking lot (C
convert 9.4 acre to inches we have=5.896*10^7
How to calculate Peak runoff discharge
1. take the dimension of the roof
2. multiply the dimension by the n umber of inches of rainfall
3. Divide by 231 to get gallon equivalence (because 1 gallon = 231 cubic inches)
5.896*10^7*1.3
7.66*10^7 cubic inches/hr
1 gallon=231 cubic inches
7.66*10^7 cubic inches=331809.5gallon/hr or 92.16gallon/s
this is gotten by converting 1 hr to seconds
331809.5gallon/hr /3600s=92.16gallon/s
Answer:
Explanation:
Hello,
In this case, we consider the reaction:
For which the law of mass action is expressed as:
Whereas the exponents are referred to the stoichiometric coefficients in the chemical reaction. Moreover, in table A-28 (Cengel's thermodynamics) the natural logarithm of the undergoing reaction at 2600 K is 2.801, thus:
In such a way, in terms of the change 
the equilibrium goes:
Hence, solving for :
Thus, the moles at equilibrium:
Finally the compositions:
Best regards.
