Fluid Mechanics: Fundamentals and Applications
Fluid Mechanics: Fundamentals and Applications
4th Edition
ISBN: 9781259696534
Author: Yunus A. Cengel Dr., John M. Cimbala
Publisher: McGraw-Hill Education
bartleby

Concept explainers

bartleby

Videos

Textbook Question
Book Icon
Chapter 7, Problem 62P

An incompressible fluid of density ρ and viscosity μ flows at average speed V through a long, horizontal section of round pipe of length L, inner diameter D, and inner wall roughness height ε (Fig. P7-62). The pipe is long enough that the flow is fully developed, meaning that the velocity profile does not change down the pipe. Pressure decreases (linearly) down the pipe in order to “push” the fluid through the pipe to overcome friction. Using the method of repeating variables, develop a nondimensional relationship between pressure drop Δ P = P 1 P 2 and the other in the problem. Be sure to modify your Π groups as necessary to achieve established nondimensional parameters, and name them. (Hint: For consistency, choose D rather than L or ε as one of your repeating parameters.)
Answer: Eu = f (Re, ε /D, UD)

Expert Solution & Answer
Check Mark
To determine

The non -dimensional relationship parameters.

The non-dimensional for first pi terms.

The non-dimensional for second pi terms.

The non- dimensional for third pi terms.

The non-dimensional for fourth pi terms.

Answer to Problem 62P

The non -dimensional parameter for first pi terms is Euler number.

The non- dimensional parameter for second pi terms is Reynolds number.

The non -dimensional parameter for third pi terms is aspect ratio.

The non -dimensional parameter for fourth pi terms is roughness ratio.

The non-dimensional relationship is Eu=f(Re,LD,εD).

Explanation of Solution

Given information:

A homogenous wire with a mass per unit length is 0.056kg/m.

Write the expression for the moment of inertia of the link 3.

  (Iy)3=(Iy)3+ml2   .....(I)

Here, the moment of inertia of the link 3 is (Iy)3, the centroidal component of the moment of inertia of link 3 is (Iy)3, the mass is m and the length of link 3 is l.

Write the expression for the moment of inertia of the link 4.

  (Iy)4=(Iy)4+m(x¯2+z¯2)   .....(II)

Here, the moment of inertia of the link 4 is (Iy)4, the centroidal component of the moment of inertia of link 4 is (Iy)4, the mass is m and the length of link 4 is l.

Write the expression for the centroidal component.

  (Iy)4=ml212

Write the expression for the moment of inertia of the link 5.

  (Iy)5=(Iy)5+m(x¯2+z¯2)   .....(III)

Here, the moment of inertia of the link 5 is (Iy)5, the centroidal component of the moment of inertia of link 5 is (Iy)5, the mass is m and the length of link 5 is l.

Write the dimension of the diameter of the pipe in FLT unit system.

  D=[L]

Here, the dimension for diameter of the pipe is [L].

Write the dimension of the length of pipe in FLT unit system.

  L=[L]

Here, the dimensions for length of the pipe is [L].

Write the dimension of the height of pipe in FLT unit system.

  ε=[L]

Here, the dimension for the height of the pipe is [L].

Write the expressions for the density.

  ρ=mv   .....(I)

Here, the mass is m, the volume is v and the density is ρ.

Substitute [M] for m and [L3] for v in Equation (I).

  ρ=[M][L3]=[ML3]

Write the expression for the pressure.

  ΔP=FA   .....(II)

Here, the pressure is ΔP, the force is F and the area is A.

Substitute [MLT2] for F and [L2] for A in Equation (II).

  ΔP=[MLT 2][L2]=[ML1T2]

Write the dimension for the viscosity.

  μ=[ML1T1]

Write the dimension for the velocity.

  V=[LT1]

Write the expression for the number of pi-terms.

  n=kr   .....(III)

Here, the number of variable is k, the number of dimensions is r and the number of pi terms is n.

Write the expression for first pi terms.

  Π1=ΔPDaρbVc    .....(IV)

Here, the constant are a, c and b.

Write the dimension for pi term.

  Π=L0T0

Write the expression for second pi terms.

  Π2=μDaρbVc    .....(V)

Write the expression for third pi terms.

  Π3=LDaρbVc    .....(VI)

Write the expression for fourth pi terms.

  Π4=εDaρbVc    .....(VII)

Write the expression for relation between the pi terms.

  Π1=f(Π2,Π3,Π4)   .....(VIII)

Calculation:

The number of variables are 7 and the number of dimensions is 3.

Substitute 7 for k and 3 for r in Equation (III).

  n=73=4

Substitute [L] for D, [M0L0T0] for Π1, [ML1T2] for ΔP, [ML3] for ρ and [LT1] for V in Equation (IV).

  [M0L0T0]=[ML1T2][L]a[ML3]b[LT1]c[M0L0T0]=[ML1T2][La][MbL3b][LcTc][M0L0T0]=[M1+bL1+a3b+cT2c]    .....(IX)

Compare the coefficients of M in Equation (IX).

  1+b=0b=1

Compare the coefficients of T in Equation (IX).

  2c=0c=2

Compare the coefficients of L in Equation (IX).

  1+a3b+c=0   .....(X)

Substitute 2 for c and 1 for b in Equation (XI).

  1+a3(1)2=01+a+32=0a+22=0a=0

Substitute 2 for c and 1 for b and 0 for a in Equation (IV).

  Π1=ΔPD0ρ1V2=ΔPρ1V2=ΔPρV2

The non-dimensional for first pi terms is Euler number.

Substitute [L] for D, [M0L0T0] for Π2, [ML1T1] for μ, [ML3] for ρ and [LT1] for V in Equation (V).

  [M0L0T0]=[ML1T1][L]a[ML3]b[LT1]c[M0L0T0]=[ML1T1][La][MbL3b][LcTc][M0L0T0]=[M1+bL1+a3b+cT1c]    .....(XI)

Compare the coefficients of M in Equation (XI).

  1+b=0b=1

Compare the coefficients of T in Equation (XI).

  1c=0c=1

Compare the coefficients of L in Equation (XI).

  1+a3b+c=0   .....(XII)

Substitute 1 for c and 1 for b in Equation (XII).

  1+a3(1)1=01+a+31=0a+21=0a=1

Substitute 1 for c and 1 for b and 1 for a in Equation (V).

  Π2=μD1ρ1V1=μDρV

The non-dimensional for second pi terms is Reynolds number.

Substitute [L] for D, [M0L0T0] for Π3, [L] for L μ, [ML3] for ρ and [LT1] for V in Equation (VI).

  [M0L0T0]=[L][L]a[ML3]b[LT1]c[M0L0T0]=[L][La][MbL3b][LcTc][M0L0T0]=[MbL1+a3b+cTc]   .....(XIII)

Compare the coefficients of M in Equation (XIII).

  b=0

Compare the coefficients of T in Equation (XIII).

  c=0

Compare the coefficients of L in Equation (XIII).

  1+a3b+c=0......... (XIV)

Substitute 0 for c and 0 for b in Equation (XIV).

  1+a3(0)0=01+a=0a=1

Substitute 0 for c and 0 for b and 1 for a in Equation (V).

  Π3=LD1ρ0V0=LD

The non-dimensional for third pi terms is aspect ratio.

Substitute [L] for D, [M0L0T0] for Π4, [L] for ε, [ML3] for ρ and [LT1] for V in Equation (VII).

  [M0L0T0]=[L][L]a[ML3]b[LT1]c[M0L0T0]=[L][La][MbL3b][LcTc][M0L0T0]=[MbL1+a3b+cTc]    .....(XV)

Compare the coefficients of M in Equation (XV).

  b=0

Compare the coefficients of T in Equation (XV).

  c=0

Compare the coefficients of L in Equation (XV).

  1+a3b+c=0   .....(XVI)

Substitute 0 for c and 0 for b in Equation (XVI).

  1+a3(0)0=01+a=0a=1

Substitute 0 for c and 0 for b and 1 for a in Equation (V).

  Π4=εD1ρ0V0=εD1=εD

The non-dimensional for fourth pi terms is roughness ratio.

Substitute εD for Π4, LD for Π3, Re for Π2 and Eu for Π1 in Equation (VIII).

  Eu=f(Re,LD,εD)

Conclusion:

The non -dimensional parameter for first pi terms is Euler number.

The non- dimensional parameter for second pi terms is Reynolds number.

The non -dimensional parameter for third pi terms is aspect ratio.

The non -dimensional parameter for fourth pi terms is roughness ratio.

The non-dimensional relationship is Eu=f(Re,LD,εD).

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Hello sir Muttalibi is a step solution in detailing mathematics the same as an existing step solution EXAMPLE 6-1 Momentum-Flux Correction Factor for Laminar Pipe Flow CV Vavg Consider laminar flow through a very long straight section of round pipe. It is shown in Chap. 8 that the velocity profile through a cross-sectional area of the pipe is parabolic (Fig. 6-15), with the axial velocity component given by r4 V R V = 2V 1 avg R2 (1) where R is the radius of the inner wall of the pipe and Vavg is the average velocity. Calculate the momentum-flux correction factor through a cross sec- tion of the pipe for the case in which the pipe flow represents an outlet of the control volume, as sketched in Fig. 6-15. Assumptions 1 The flow is incompressible and steady. 2 The control volume slices through the pipe normal to the pipe axis, as sketched in Fig. 6-15. Analysis We substitute the given velocity profile for V in Eq. 6-24 and inte- grate, noting that dA, = 2ar dr, FIGURE 6–15 %3D Velocity…
Two infinite plates a distance h apart are parallel to the xzplane with the upper plate moving at speed V, as inFig.  There is a fluid of viscosity μ and constant pressurebetween the plates. Neglecting gravity and assumingincompressible turbulent flow u(y) between the plates, usethe logarithmic law and appropriate boundary conditions toderive a formula for dimensionless wall shear stress versusdimensionless plate velocity. Sketch a typical shape of theprofile u(y).
A capillary tube has an 8mm inside diameter through which liquid fluorine refrigerant R-11 flows at a rate of 0.03 cm3/s. The tube isto be used as a throttling device in an air conditioning unit. A model of this flow is constructed by using a pipe of 3cm inside diameter and water as the fluid medium. (Density of R-11 = 1.494 g/cm3 and its viscosity is 4.2 x10-4 Pa.s; Density of water is 1g/cm3 and its viscosity 8.9 x10-4 Pa.s)a) What is the required velocity in the model for dynamic similarity? Hint: For flow through a tube the Ne number can be expressed in terms of the Reynolds numberb) When dynamic similarity is achieved the pressure drop is measured at 50 Pa. What is the corresponding pressure drop in the capillary tube?Hint: In this case the Euler number defines dynamic similarity with reference to the static pressure drop

Chapter 7 Solutions

Fluid Mechanics: Fundamentals and Applications

Ch. 7 - You are probably familiar with Ohm law for...Ch. 7 - Write the primary dimensions of each of the...Ch. 7 - Prob. 13PCh. 7 - Thermal conductivity k is a measure of the ability...Ch. 7 - Write the primary dimensions of each of the...Ch. 7 - Prob. 16PCh. 7 - Explain the law of dimensional homogeneity in...Ch. 7 - Prob. 18PCh. 7 - Prob. 19PCh. 7 - An important application of fluid mechanics is the...Ch. 7 - Prob. 21PCh. 7 - Prob. 22PCh. 7 - In Chap. 4, we defined the material acceleration,...Ch. 7 - Newton's second law is the foundation for the...Ch. 7 - What is the primary reason for nondimensionalizing...Ch. 7 - Prob. 26PCh. 7 - In Chap. 9, we define the stream function for...Ch. 7 - In an oscillating incompressible flow field the...Ch. 7 - Prob. 29PCh. 7 - Consider ventilation of a well-mixed room as in...Ch. 7 - In an oscillating compressible flow field the...Ch. 7 - List the three primary purposes of dimensional...Ch. 7 - List and describe the three necessary conditions...Ch. 7 - A student team is to design a human-powered...Ch. 7 - Repeat Prob. 7-34 with all the same conditions...Ch. 7 - This is a follow-tip to Prob. 7-34. The students...Ch. 7 - The aerodynamic drag of a new sports car is lo be...Ch. 7 - This is a follow-tip to Prob. 7-37E. The...Ch. 7 - Consider the common situation in which a...Ch. 7 - Prob. 40PCh. 7 - Some students want to visualize flow over a...Ch. 7 - A lightweight parachute is being designed for...Ch. 7 - Prob. 43PCh. 7 - Prob. 44PCh. 7 - Prob. 45PCh. 7 - The Richardson number is defined as Ri=L5gV2...Ch. 7 - Prob. 47PCh. 7 - Prob. 48PCh. 7 - A stirrer is used to mix chemicals in a large tank...Ch. 7 - Prob. 50PCh. 7 - Albert Einstein is pondering how to write his...Ch. 7 - Consider filly developed Couette flow-flow between...Ch. 7 - Consider developing Couette flow-the same flow as...Ch. 7 - The speed of sound c in an ideal gas is known to...Ch. 7 - Repeat Prob. 7-54, except let the speed of sound c...Ch. 7 - Repeat Prob. 7-54, except let the speed of sound c...Ch. 7 - Prob. 57PCh. 7 - When small aerosol particles or microorganisms...Ch. 7 - Prob. 59PCh. 7 - Prob. 60PCh. 7 - Prob. 61PCh. 7 - An incompressible fluid of density and viscosity ...Ch. 7 - Prob. 63PCh. 7 - In the study of turbulent flow, turbulent viscous...Ch. 7 - Bill is working on an electrical circuit problem....Ch. 7 - A boundary layer is a thin region (usually along a...Ch. 7 - A liquid of density and viscosity is pumped at...Ch. 7 - A propeller of diameter D rotates at angular...Ch. 7 - Repeat Prob. 7-68 for the case an which the...Ch. 7 - Prob. 70PCh. 7 - Prob. 71PCh. 7 - Consider a liquid in a cylindrical container in...Ch. 7 - Prob. 73PCh. 7 - One of the first things you learn in physics class...Ch. 7 - Prob. 75CPCh. 7 - Prob. 76CPCh. 7 - Define wind tunnel blockage. What is the rule of...Ch. 7 - Prob. 78CPCh. 7 - Prob. 79CPCh. 7 - In the model truck example discussed in Section...Ch. 7 - Prob. 83PCh. 7 - A small wind tunnel in a university's...Ch. 7 - There are many established nondimensional...Ch. 7 - Prob. 86CPCh. 7 - For each statement, choose whether the statement...Ch. 7 - Prob. 88PCh. 7 - Prob. 89PCh. 7 - Prob. 90PCh. 7 - Prob. 91PCh. 7 - From fundamental electronics, the current flowing...Ch. 7 - Prob. 93PCh. 7 - Prob. 94PCh. 7 - The Archimedes number listed in Table 7-5 is...Ch. 7 - Prob. 96PCh. 7 - Prob. 97PCh. 7 - Prob. 98PCh. 7 - Prob. 99PCh. 7 - Prob. 100PCh. 7 - Repeal Prob. 7-100 except for a different...Ch. 7 - A liquid delivery system is being designed such...Ch. 7 - Prob. 103PCh. 7 - Au aerosol particle of characteristic size DPmoves...Ch. 7 - Prob. 105PCh. 7 - Prob. 106PCh. 7 - Prob. 107PCh. 7 - Prob. 108PCh. 7 - Prob. 109PCh. 7 - Prob. 110PCh. 7 - Repeat pall (a) of Prob. 7-110, except instead of...Ch. 7 - Sound intensity I is defined as the acoustic power...Ch. 7 - Repeal Prob. 7-112, but with the distance r from...Ch. 7 - Engineers at MIT have developed a mechanical model...Ch. 7 - Prob. 116PCh. 7 - Prob. 117PCh. 7 - An electrostatic precipitator (ESP) is a device...Ch. 7 - Prob. 119PCh. 7 - Prob. 120PCh. 7 - Prob. 121PCh. 7 - Prob. 122PCh. 7 - Prob. 123PCh. 7 - Prob. 124PCh. 7 - The primary dimensions of kinematic viscosity are...Ch. 7 - There at four additive terms in an equation, and...Ch. 7 - Prob. 127PCh. 7 - Prob. 128PCh. 7 - Prob. 129PCh. 7 - A one-third scale model of a car is to be tested...Ch. 7 - Prob. 131PCh. 7 - A one-third scale model of an airplane is to be...Ch. 7 - Prob. 133PCh. 7 - Prob. 134PCh. 7 - Consider a boundary layer growing along a thin...Ch. 7 - Prob. 136P
Knowledge Booster
Background pattern image
Mechanical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Text book image
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Text book image
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Text book image
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Text book image
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
Introduction to Kinematics; Author: LearnChemE;https://www.youtube.com/watch?v=bV0XPz-mg2s;License: Standard youtube license