Concept explainers
An incompressible fluid of density
Answer: Eu = f (Re,
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
Explanation of Solution
Given information:
A homogenous wire with a mass per unit length is
Write the expression for the moment of inertia of the link 3.
Here, the moment of inertia of the link 3 is
Write the expression for the moment of inertia of the link 4.
Here, the moment of inertia of the link 4 is
Write the expression for the centroidal component.
Write the expression for the moment of inertia of the link 5.
Here, the moment of inertia of the link 5 is
Write the dimension of the diameter of the pipe in
Here, the dimension for diameter of the pipe is
Write the dimension of the length of pipe in
Here, the dimensions for length of the pipe is
Write the dimension of the height of pipe in
Here, the dimension for the height of the pipe is
Write the expressions for the density.
Here, the mass is
Substitute
Write the expression for the pressure.
Here, the pressure is
Substitute
Write the dimension for the viscosity.
Write the dimension for the velocity.
Write the expression for the number of pi-terms.
Here, the number of variable is
Write the expression for first pi terms.
Here, the constant are
Write the dimension for pi term.
Write the expression for second pi terms.
Write the expression for third pi terms.
Write the expression for fourth pi terms.
Write the expression for relation between the pi terms.
Calculation:
The number of variables are
Substitute
Substitute
Compare the coefficients of
Compare the coefficients of
Compare the coefficients of
Substitute
Substitute
The non-dimensional for first pi terms is Euler number.
Substitute
Compare the coefficients of
Compare the coefficients of
Compare the coefficients of
Substitute
Substitute
The non-dimensional for second pi terms is Reynolds number.
Substitute
Compare the coefficients of
Compare the coefficients of
Compare the coefficients of
Substitute
Substitute
The non-dimensional for third pi terms is aspect ratio.
Substitute
Compare the coefficients of
Compare the coefficients of
Compare the coefficients of
Substitute
Substitute
The non-dimensional for fourth pi terms is roughness ratio.
Substitute
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
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Chapter 7 Solutions
FLUID MECHANICS FUNDAMENTALS+APPS
- reading is 0.4 mas SHOWN. Assume h₁ = 0.4 m, h₂ = 0.5 m. (a) Do you know the specific weight of mercury? (b) Do you know the specific weight of gasoline? (c) Do you know the specific weight of oil? (a) YHg = 133,000 (b) Ygas = 6867 (c) Yoil = 8829 eTextbook and Media Part 2 N/m³ N/m³ N/m³ A+ Gasoline t +B Oil -Mercury Attempts: unlimited Did you calculate the pressure difference between two locations using the correct specific weight? Did you assume that the pressures in fluid are the same in a horizontal plane even though they are in different tubes? Are the calculated pressures in a column of fluid always higher at lower elevations? Did you account for the fact that the two horizontal tubes of the U-tube are above the ground? Concepts: The pressure in a fluid is a function of the specific weight of the fluid and the height relative to a reference. Pressure is constant in a horizontal plane of a continuous mass of fluid. (a) What is the initial pressure difference? (PA-PB) (b) What is…arrow_forwardFind the solution of the following Differential Equations 1) "-4y+3y=0 3) "+16y=0 2) y"-16y=0 4) y"-y-6y=0 5) y"+2y=0 7) y"+y=0, (#0) 9) y"-y=0, y(0) = 6, y'(0) = -4 11) y"-4y+3y=0, y(0)=-1, 13) y'(0) = -5 "+2y+2y=0 15) y"-9y=0 17) y"-4y=0 6) y"-2y+2y=0 8) "+4y+5y=0 10) y"-9y=0, y(0) = 2, y'(0) = 0 12) y"-3y+2y= 0, y(0)=-1, y'(0) = 0 14) 4y+4y+y=0 16) "+6y+12y=0 18) 4y+4y+17y=0arrow_forwardAccess Pearson Mastering Engineering Back to my courses Course Home Course Home Scoresarrow_forward
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