If 'x' is the depth of flow in an open channel of large width, then the hydraulic radius is equal to
A. x
B. $$\frac{{\text{x}}}{2}$$
C. $$\frac{{\text{x}}}{3}$$
D. $$\frac{{2{\text{x}}}}{3}$$
Answer: Option A
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Comments ( 2 )
A. Thermal conductivity
B. Electrical conductivity
C. Specific gravity
D. Electrical resistivity
A. $$\frac{{\text{V}}}{{{{\text{V}}_{\max }}}} = {\left( {\frac{{\text{x}}}{{\text{r}}}} \right)^{\frac{1}{7}}}$$
B. $$\frac{{\text{V}}}{{{{\text{V}}_{\max }}}} = {\left( {\frac{{\text{r}}}{{\text{x}}}} \right)^{\frac{1}{7}}}$$
C. $$\frac{{\text{V}}}{{{{\text{V}}_{\max }}}} = {\left( {{\text{x}} \times {\text{r}}} \right)^{\frac{1}{7}}}$$
D. None of these
A. d
B. $$\frac{1}{{\text{d}}}$$
C. $$\sigma $$
D. $$\frac{l}{\sigma }$$
A. $$\frac{{4\pi {\text{g}}}}{3}$$
B. $$\frac{{0.01\pi {\text{gH}}}}{4}$$
C. $$\frac{{0.01\pi {\text{gH}}}}{8}$$
D. $$\frac{{0.04\pi {\text{gH}}}}{3}$$
Rh = (x*W)/((2*x)+W)
Since W>>x neglect x in denominator
Rh = (x*W)/W
Rh = x
Rh=flow area / wetted perimeter = (W * X) / (2 * (X+W)).
Since W (width)>>>X , Rh= (W * X)/(2 * W)= X/2.
So, I think the answer is B.