Chapter 7, Problem 7.56P

Interpretation Introduction

(a)

Interpretation:

To calculate the velocity at point 2 by neglecting friction.

Concept introduction:

Following assumptions are made in the derivation of Bernoulli’s equation:

• The liquid is ideal and incompressible.

The flow is at steady state and continuous.

The flow is along the streamline

The velocity is uniform over the section and is equal to the mean velocity.

The only force acting on the fluid are gravity forces and the pressure forces.

Now apply Bernoulli equation at point 1 and 2
$\frac{P_1}{\rho g}+\frac{V^2{}_1}{2g}+z_1=\frac{P_2}{\rho g}+\frac{V^2{}_2}{2g}+z_2+h_f$ ………………(1)
Where

• P₁ & P₂ are pressure at section I and 2.

• V₁ & V₂ are velocity at section 1 and 2.

• z₁ & z₂ are the elevation from the horizontal at section 1 and 2.

• h_f is the frictional losses

Interpretation Introduction

(b)

Interpretation:

To determine the diameter at point 1 if the pipe diameter at point 2 is 6.00 cm.

Concept introduction:

Mass balance for study flow processes:

The total amount of mass within the control volume does not change with time.

So that, the total amount of mass entering a control volume equal to the total amount of mass leaving it.

If the mass flow is $\stackrel{.}{m}$ kg/s and volumetric flow rate is $\stackrel{.}{v}$ m³/s
Then from conservation of mass for steady flow process will be as follows
$\sum \stackrel{\text{.}}{{\text{m}}_{\text{in}}}\text{=}\sum \stackrel{\text{.}}{{\text{m}}_{\text{out}}}$ (Kg/s)
Where $\stackrel{.}{m_{in}}$ = (? ×A ×V) _in $\stackrel{.}{m_{out}}$

• = (? A× V)_out
Conservation of mass at section 1 and section 2.

${\rho }_1{A}_1{V}_1={\rho }_2{A}_2{V}_2$

If density is constant then above equation reduce to
$A_1{V}_1=A_2{V}_2$

• A is the area at respective section of the flowing processes
• V is the velocity at respective section
• ? is the density of flowing fluid in control volume