Chapter 7, Problem 7.58P

Interpretation Introduction

(a)

Interpretation:

Derive the relation between the velocity u₁ and u₂ at point 1 and point 2.

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

$\sum \stackrel{.}{m_{in}}=\sum \stackrel{.}{m_{out}}$ (Kg/s)

Where

• $\stackrel{.}{m_{in}}$ = (? ×A ×u) _in $\stackrel{.}{m_{out}}$ = (?× A× u)_out
• A is the area at respective section of the flowing processes
• u is the velocity at respective section
• ? is the density of flowing fluid in control volume

Interpretation Introduction

(b)

Interpretation:

The Bernoulli equation for the system between point 1 and 2 should be used to prove the following equation:

$P_1-P_2=\frac{15\rho \stackrel{.}{V^2}}{2A_1{}^2}$

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)

Given

$A_1=4A_2$

And from assumption density is constant and process is at steady state

Now from conservation of mass ${\rho }_1{A}_1{V}_1={\rho }_2{A}_2{V}_2$

It gives $4V_1=V_2$

Now from equation (1) $\frac{P_1}{\rho g}-\frac{P_2}{\rho g}=\frac{V_1{}^2}{2g}-\frac{V_2{}^2}{2g}+z_2-z_1+h_f$

Interpretation Introduction

(c)

Interpretation:

The volumetric flow rate of the acetone should be calculated.

Concept introduction:

If manometer reading is given, then to calculate the volumetric flow rate we can use the following formula, which is a function of area at both the section, and manometer reading.

$\stackrel{.}{V}=\frac{A_1{A}_2\sqrt{2gh}}{\sqrt{A_1{}^2-A_2{}^2}}$

$\stackrel{.}{V}=A_1{V}_1$