The critical depth of flow in a most economical triangular channel section for a discharge of $1 m^{3} / s e c$ is given by

- 1
  
  ${(\frac{1}{9.8})}^{1 / 5}$ metre
- 2
  
  ${(\frac{1}{9.8})}^{1 / 3}$ metre
- 3
  
  ${(\frac{1}{4.9})}^{1 /} 5$ metre
- 4
  
  ${(\frac{1}{4.9})}^{1 /} 3$ metre

Display Comments

Leave a comment

Login at first then you can submit comment. Click here to login.