Chapter 6, Problem 6.9P

Interpretation Introduction

Interpretation: Pressure head at a point is to be calculated with the known pressure head at other point by the use of Energy balance equation.

Concept introduction: The Energy balance across the pipe is given by the Bernoulli’s equation in which the sum of Potential head, Kinetic head and Pressure head is constant. The Bernoulli equation across 2 points of any system is given by:-

  $\frac{{\text{P}}_{\text{A}}}{\text{ρg}}\text{+}\frac{{\text{v}}_{\text{A}}{}^{\text{2}}}{\text{2g}}{\text{+ Z}}_{\text{A}}\text{=}\frac{{\text{P}}_{\text{B}}}{\text{ρg}}\text{+}\frac{{\text{v}}_{\text{B}}{}^{\text{2}}}{\text{2g}}{\text{+ Z}}_{\text{B}}\text{+}{\text{h}}_{\text{f}}$ ........ (1)

P_A and P_B are pressure at 2 points of the pipe

v_A and v_B are velocity at 2 points of the pipe

Z_A and Z_B are the locations of the inlet and outlet of the liquid.

  $\begin{array}{l}\text{ρ}\text{=}\text{Density}\text{of}\text{liquid}\\ \text{g}\text{=}\text{Acceleration}\text{due}\text{to}\text{gravity}\end{array}$

h_f= Friction loss

The volumetric flow rate in the air duct is given as,

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

Q = Volumetric flow rate of liquid

A = Area of pipe

v = Velocity of liquid