Determine the discharge through the following sections for a normal depth of 4.4 ft; n=0.013, and blttom slope S = 0.20%. Solution: a. A rectangular section 17.6 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 17.6 ft in diameter. 1). Inner angle 0 = 4). Discharge/flow rate Q = ft²; 2). Section area A = rad = ft²; c. A right angled triangular section. 1). Side slope m = ft; ft²; ft; ft; cfs; 0. cfs;

Elements Of Electromagnetics
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Please answer Fluid Mechanics Hw problem #8
Determine the discharge through the following sections for a normal depth of 4.4 ft; n=0.013, and blttom slope S =
0.20%.
Solution:
a. A rectangular section 17.6 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 17.6 ft in diameter.
1). Inner angle 0 =
4). Discharge/flow rate Q =
ft²;
2). Section area A =
rad =
ft²;
c. A right angled triangular section.
1). Side slope m =
ft;
ft²;
ft;
ft;
cfs;
0.
cfs;
Transcribed Image Text:Determine the discharge through the following sections for a normal depth of 4.4 ft; n=0.013, and blttom slope S = 0.20%. Solution: a. A rectangular section 17.6 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 17.6 ft in diameter. 1). Inner angle 0 = 4). Discharge/flow rate Q = ft²; 2). Section area A = rad = ft²; c. A right angled triangular section. 1). Side slope m = ft; ft²; ft; ft; cfs; 0. cfs;
b. A circular section 17.6 ft in diameter.
1). Inner angle 0 =
2). Section area A =
3). Hydraulic radius Rh =
4). Discharge/flow rate Q =
2). Section area A =
c. A right angled triangular section.
1). Side slope m =
3). Hydraulic radius Rh =
2). Wetted perimeter Pw =
rad =
3), Hydraulic radius Rh =
ft²:
4). Discharge/flow rate Q =
ft²;
ft;
ft;
4). Discharge/flow rate Q =
d. A trapezoidal section with a bottom width of 17.6 ft and side slope of V:H=1:2.
1). Section area A =
ft²;
cfs;
ft;
cfs;
ft;
cfs;
Transcribed Image Text:b. A circular section 17.6 ft in diameter. 1). Inner angle 0 = 2). Section area A = 3). Hydraulic radius Rh = 4). Discharge/flow rate Q = 2). Section area A = c. A right angled triangular section. 1). Side slope m = 3). Hydraulic radius Rh = 2). Wetted perimeter Pw = rad = 3), Hydraulic radius Rh = ft²: 4). Discharge/flow rate Q = ft²; ft; ft; 4). Discharge/flow rate Q = d. A trapezoidal section with a bottom width of 17.6 ft and side slope of V:H=1:2. 1). Section area A = ft²; cfs; ft; cfs; ft; cfs;
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