Fluid Mechanics: Fundamentals and Applications
4th Edition
ISBN: 9781259696534
Author: Yunus A. Cengel Dr., John M. Cimbala
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 10, Problem 1CP
Discuss how nondimensalizsionalization of the Navier-Stokes equation is helpful in obtaining approximate solutions. Give an example.
Expert Solution & Answer
To determine
The advantages of solving Navier Strokes equation using non-dimensionalization.
Explanation of Solution
Non-dimensionalization of Navier-Stokes equation helps in obtaining approximate solutions. This can be explained from below mentioned points:
- Non-dimensionalization helps in reducing the complexity of any equation.
- It helps in removing the dimensions of the all the quantities present.
- The primary function is to make the dimension of the quantities present in Navier-Stokes equation unity.
- It calculates small quantities with respect to a large quantity.
- For example- If the value of Strouhal number is less with respect to Reynolds number, we can ignore the term containing Strouhal number whereas the respective value of Reynolds number must retain.
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Two large tanks, each holding 100 L of liquid, are interconnected by pipes, with the liquid flowing from tank
A into tank B at a rate of 3 L/min and from B into A at a rate of 1 L/min (see Figure Q1). The liquid inside each
tank is kept well stirred. A brine solution with a concentration of 0.2 kg/L of salt flows into tank A at a rate of
6 L/min. The diluted solution flows out of the system from tank A at 4 L/min and from tank B at 2 L/min. If,
initially, tank A contains pure water and tank B contains 20 kg of salt.
A
6 L/min
0.2 kg/L
x(t)
100 L
4 L/min
x(0) = 0 kg
3 L/min
B
y(t)
100 L
y(0) = 20 kg
2 L/min
1 L/min
Figure Q1 - Mixing problem for interconnected tanks
Determine the mass of salt in each tank at time t > 0:
Analytically (hand calculations)
Two springs and two masses are attached in a straight vertical line as shown in Figure Q3. The system is set
in motion by holding the mass m₂ at its equilibrium position and pushing the mass m₁ downwards of its
equilibrium position a distance 2 m and then releasing both masses. if m₁ = m₂ = 1 kg, k₁ = 3 N/m and
k₂ = 2 N/m.
www.m
k₁ = 3
(y₁ = 0).
m₁ = 1
k2=2
(y₂ = 0)
|m₂ = 1
Y2
y 2
System in
static
equilibrium
(Net change in
spring length
=32-31)
System in
motion
Figure Q3 - Coupled mass-spring system
Determine the equations of motion y₁(t) and y₂(t) for the two masses m₁ and m₂ respectively:
Analytically (hand calculations)
Chapter 10 Solutions
Fluid Mechanics: Fundamentals and Applications
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