Determine the discharge through the following sections for a normal depth of 4.1 ft; n=0.013, and blttom slope S= 0.20%. Solution: a. A rectangular section 16.4 ft wide. 1). Section area A = 2). Wetted perimeter Pw = 3). Hydraulic radius Rh = 4). Discharge/flow rate Q = 2). Section area A = 3). Hydraulic radius Rh = b. A circular section 16.4 ft in diameter. 1). Inner angle 0 = 2). Section area A = ft²: 3). Hydraulic radius Rh = I 4) Discharge/flow rate Q= 4). Discharge/flow rate Q = c. A right angled triangular section. 1). Side slope m = rad = ft²; ft; ft; ft; ft²; ft; cfs; 0. cfs; cfs:
Determine the discharge through the following sections for a normal depth of 4.1 ft; n=0.013, and blttom slope S= 0.20%. Solution: a. A rectangular section 16.4 ft wide. 1). Section area A = 2). Wetted perimeter Pw = 3). Hydraulic radius Rh = 4). Discharge/flow rate Q = 2). Section area A = 3). Hydraulic radius Rh = b. A circular section 16.4 ft in diameter. 1). Inner angle 0 = 2). Section area A = ft²: 3). Hydraulic radius Rh = I 4) Discharge/flow rate Q= 4). Discharge/flow rate Q = c. A right angled triangular section. 1). Side slope m = rad = ft²; ft; ft; ft; ft²; ft; cfs; 0. cfs; cfs:
Elements Of Electromagnetics
7th Edition
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Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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![Determine the discharge through the following sections for a normal depth of 4.1 ft; n=0.013, and blttom slope S=
0.20%.
Solution:
a. A rectangular section 16.4 ft wide.
1). Section area A =
2). Wetted perimeter Pw =
3). Hydraulic radius Rh =
4). Discharge/flow rate Q =
2). Section area A =
3). Hydraulic radius Rh =
b. A circular section 16.4 ft in diameter.
1). Inner angle 0 =
2). Section area A =
ft²:
3). Hydraulic radius Rh =
I
4) Discharge/flow rate Q=
4). Discharge/flow rate Q =
c. A right angled triangular section.
1). Side slope m =
rad =
ft²;
ft;
ft;
ft;
ft²;
ft;
cfs;
0.
cfs;
cfs:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9a42e238-4116-4e80-81a7-f4d567b4bd0e%2Faf3928dd-27ea-4bf0-9b83-5abcaba999d2%2Fb0zr219_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Determine the discharge through the following sections for a normal depth of 4.1 ft; n=0.013, and blttom slope S=
0.20%.
Solution:
a. A rectangular section 16.4 ft wide.
1). Section area A =
2). Wetted perimeter Pw =
3). Hydraulic radius Rh =
4). Discharge/flow rate Q =
2). Section area A =
3). Hydraulic radius Rh =
b. A circular section 16.4 ft in diameter.
1). Inner angle 0 =
2). Section area A =
ft²:
3). Hydraulic radius Rh =
I
4) Discharge/flow rate Q=
4). Discharge/flow rate Q =
c. A right angled triangular section.
1). Side slope m =
rad =
ft²;
ft;
ft;
ft;
ft²;
ft;
cfs;
0.
cfs;
cfs:
![4). Discharge/flow rate Q =
c. A right angled triangular section.
1). Side slope m =
2). Section area A =
3). Hydraulic radius Rh =
4). Discharge/flow rate Q =
2). Wetted perimeter Pw =
3). Hydraulic radius Rh =
ft²;
4). Discharge/flow rate Q =
I
ft;
d. A trapezoidal section with a bottom width of 16.4 ft and side slope of V:H=1:2.
1). Section area A =
ft²;
cfs;
ft;
cfs;
ft;
cfs;](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9a42e238-4116-4e80-81a7-f4d567b4bd0e%2Faf3928dd-27ea-4bf0-9b83-5abcaba999d2%2Fgz4mm3_processed.jpeg&w=3840&q=75)
Transcribed Image Text:4). Discharge/flow rate Q =
c. A right angled triangular section.
1). Side slope m =
2). Section area A =
3). Hydraulic radius Rh =
4). Discharge/flow rate Q =
2). Wetted perimeter Pw =
3). Hydraulic radius Rh =
ft²;
4). Discharge/flow rate Q =
I
ft;
d. A trapezoidal section with a bottom width of 16.4 ft and side slope of V:H=1:2.
1). Section area A =
ft²;
cfs;
ft;
cfs;
ft;
cfs;
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