Chapter 6, Problem 6.31P

Interpretation Introduction

Interpretation:

The flow rate and pressure head at a point across a siphon is to be calculated along with the time required by water to fall from a certain height to another height.

Concept introduction:

The exit velocity of any liquid flowing in a tank and passing through a pipe is given as,

  ${\text{v}}_{\text{exit}}\text{=}\sqrt{\text{2gy}}$ ....... (1)

Here, g is the acceleration due to gravity and y is the height or the distance covered by liquid from initial to final point.

The flow rate is given as,

  $\text{Q}\text{=}\text{A×v}$ ....... (2)

Q = Flow rate of liquid

A = Area of tank/pipe

v = Velocity of liquid

The pressure head at the point B is given by the difference between the atmospheric pressure and the pressure due to the length of the tube, i.e.,

  ${\text{P}}_{\text{B}}\text{=}{\text{P}}_{\text{atm}}\text{-}\text{(ρgy)}\mathrm{.......}\text{(3)}$

P_atm is the atmospheric pressure

P_B is the pressure at point B

  $\text{ρ}$ is the density of water