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 and Applications
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- A liquid of density 1150 kg/m3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the flow speed is 9.47 m/s and the pipe diameter d1 is 11.7 cm. At Location 2, the pipe diameter d2 is 17.7 cm. At Location 1, the pipe is Ay=8.19 m higher than it is at Location 2. Ignoring viscosity, calculate the difference APbetween the fluid pressure at Location 2 and the fluid pressure at Location 1.arrow_forwardTake the full-blown Couette flow as shown in the figure. While the upper plate is moving and the Lower Plate is constant, flow occurs between two infinitely parallel plates separated by the H distance. The flow is constant, uncompressed, and two-dimensional in the X-Y plane. In fluid viscosity µ, top plate velocity V, distance h, fluid density ρ, and distance y, create a dimensionless relationship for component X of fluid velocity using the method of repeating variables. Show all steps in order.arrow_forwardOil flows between two very long parallel plates, separated from H, with width b. The bottom plate moves with speed U and is isolated. The upper plate is at rest and receives heat from the environment at a rate equal to qs". Consider laminar flow, thermally developed. Due to the high viscosity of the fluid, viscous dissipation is relevant U isolada 1 - Estimate the velocity profile considering the null pressure gradient and determine the average speed. 2. Determine dissipation per volume unit. 3. Set mixing temperature and determine the mixing temperature variation over the plates.arrow_forward
- A fluid with specific weight of 12000 N/m³ and viscosity of 0.002 kg/m.s flows between two parallel plates as shown in the figure. The upper plate moves with the velocity of U and the lower plate is fixed. (a) Find the velocity profile u(y) in terms of h, U, and dp/dx. Assume the flow is steady, 2D, and parallel to the plates. (b) A manometer is used to measure the pressure drop Ap = p2- pi of the flow. Find Ap if the elevation difference in the manometer is H=0.2 cm and ym=16000 N/m³. (c) Assuming dp/dx Ap/L, find the fluid velocity at y=h/2. Consider h=2.5 cm, L=15 cm, and U=0.007 m/s.arrow_forwardConsider a two-dimensional (planar), steady and incompressible flow in the en- trance region of a duct leading to the fully developed region, as shown. Velocity profile at the entrance is U, which is uniform. At length L the flow becomes fully developed. Fluid viscosity is μ and its density of p. Pressure is uniform in the vertical (y) direction. Consider unity width normal to the page. Vo h L เง I x u Find the fully-developed velocity profile u(y) at x ≥ L. Using continuity equa- tion, express the velocity profile in terms of Uo, y and h only. Find the pressure gradient dp/dr and wall shear stress in the fully developed region (x > L) in terms of μ, U₁ and h. Calculate the total friction force acting on the walls in the entrance region (0 < xarrow_forwardP1 A thin layer of water flows down a plate inclined to the horizontal with an angle a = 15° in the shown coordinate system. If the thickness of the water layer is a=0.5 mm, assuming that the flow is laminar and incompressible, (water density p = 1000 kg/m³viscosity µ = 0.001 Pa.s and acceleration of gravity g = 9.81 m/s²) and an air flow shears the layer in a direction opposite to its flow with a shear stress of 1 N /m². Solve the Navier-Stokes equation: air water (a) to find the value of the maximum water velocity in m/s to three decimal points, Answer: (b)and to find the value of water velocity at the layer's surface in m/s to three decimal points, Answer:arrow_forwardHelp me pleasearrow_forwardQA An incompressible fluid (kinematic viscosity, 7.4 x10-7 m2/s, specific gravity, 0.88) is held between two parallel plates. If the top plate is moved with a velocity of 0.5 m/s while the bottom one is held stationary, the fluid attains a linear velocity profile in the gap of 0.5 mm between these plates; the shear stress in Pascals on the surface of top plate is (a) 0.651 x 10-3 (c) 6.51 (b) 0.651 (d) 0.651 x 103arrow_forwardA potential steady and incompressible air flow on x-y plane has velocity in y-direction v= - 6 xy . Determine the velocity in x-direction u=? and Stream Function SF=? ( x2 : square of x ; x3: third power of x ; y2: square of y , y3: third power of y)ANSWER: u= 3 x2 - 3 y2 SF= 3 x2 y - y3arrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
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