Please solve part (a) and (b) **D. Explanation of Lubrication System Flow Dynamics****Overview:**Figure D illustrates a section of a lubrication system comprising two circular disks with a radial flow of lubricant. The flow is driven by a modified pressure difference \( p_1 - p_2 \) between the inner and outer radii \( r_1 \) and \( r_2 \), respectively. This system experiences laminar, steady-state flow of a Newtonian fluid, characterized by a constant viscosity \( \mu \) and density \( \rho \), without the influence of gravity.**Tasks:****(a) Identifying Zero Velocity Components:**In a cylindrical coordinate system (radial, azimuthal, axial), determine which velocity components vanish for this flow configuration.**(b) Continuity Equation Formulation:**- Develop the continuity equation for this radial flow.- Calculate and analyze the result, providing insight into the flow dynamics.- Justify why the radial component of velocity depends on both the radial and axial directions.**(c) Momentum Equation in Axial Direction:**- Formulate the momentum equation for the axial direction.- Identify and explain the cancellation of specific terms.- Deliver the outcome of this analysis and interpret its implications for the flow.**(d) Momentum Equation in Azimuthal Direction:**- Establish the momentum equation for the azimuthal direction.- Highlight which terms cancel out.- Summarize the conclusions drawn from this analysis and its significance in understanding the flow behavior. **Title: Analysis of Fluid Flow in a Circular Geometry**The illustration demonstrates the flow of fluid in a circular disk geometry, both in 3D and 2D representations. The diagram on the left represents a 3D view, while the right shows a 2D cross-section.**3D Diagram Description:**- A stream of fluid enters from above and impacts a circular disk.- The central vertical line indicates the axis of symmetry (z-axis), with the radial direction labeled \( r \).- Fluid spreads radially outward from the point of impact on the disk.- The disk has two significant planes: \( z = +d \) (above the disk plane) and \( z = -d \) (below the disk plane).- Several arrows illustrate the radial flow and rotational symmetry around the z-axis.**2D Diagram Description:**- The entrance of fluid is marked differently, showing the impact on a linear plane.- Arrows indicate the flow direction, spreading symmetrically from the point of impact.- The axial position is labeled along the z-axis, with two planes: \( z = +d \) and \( z = -d \).- Two radii \( r_1 \) and \( r_2 \) represent different flow regions along the radial direction.- The 2D view highlights "Flow of fluid," showing the progression along lines perpendicular to the primary flow axis.This educational diagram clarifies fluid dynamics in circular geometries and different dimensional perspectives, useful for understanding symmetrical flow patterns in radial turbines, pumps, or similar engineering applications.

Part-a - Continuity equation in cylindrical co-ordinates - ∂p∂t + 1r∂∂tρrvr + 1r∂∂θρvθ + ∂∂zρvz = 0 Flow is steady, laminar, incompressible and Newtonian. Hence, ∂p∂t = 0andvz =0, vθ = 0 So, 1r∂∂tρrvr = 0⇒ρr∂∂trvr = 0Part-b - r component of the momentum equation - ρvr∂vr∂r= -∂p∂r +μ ∂∂r1r∂∂rrvr + ∂2vr∂z2 Hereϕ = rvr .........1 which is function of z only⇒∂∂r1r∂∂rrvr = 0Hence,ρvr∂vr∂r= -∂p∂r +μ ∂2vr∂z2 from equaion 1,ρϕr∂∂rϕr= -∂p∂r +μ ∂2∂z2ϕr ⇒ρϕr-ϕr2= -∂p∂r +μr ∂2ϕ∂z2 ⇒-ρ(ϕ2r3)= -∂p∂r +μr [∂2ϕ∂z2 ] ϕ is the function of z only and it is not affected by r.Part-c - -ρϕ2r3= -∂p∂r +μr ∂2ϕ∂z2 ϕ2 is neglected for strokes's flow⇒-∂p∂r +μr ∂2ϕ∂z2 = 0⇒∂p∂r = μr ∂2ϕ∂z2 Integrating the equation w.r.t. r,⇒∫P1P2∂p = μ ∂2ϕ∂z2 ∫r1r2drr⇒P2 - P1 = μ ∂2ϕ∂z2 ×lnrr1r2⇒P2 - P1 = μ ∂2ϕ∂z2 ×lnr2r1⇒∆P + μ [∂2ϕ∂z2 ]×ln(r2r1) = 0

Engineering

Civil Engineering

D. Figure D shows a part of a lubrication system that consists of two circular disks between which a lubricant flows radially. The flow takes place because of a modified pressure difference p1 – P2, between the inner and the outer radii r, and r, respectively. This is a laminar, steady state flow of a Newtonian liquid of constant viscosity µ and constant density p without gravity effects. (a) Identify which velocity components of a cylindrical coordinate system are zero for this flow. (b) Formulate the continuity equation for this flow, compute the result and comment on the result of your analysis. Argue why the radial component of the velocity vector is a function of both the radial and the axial direction. (c) Formulate the momentum equation for this flow in the axial direction, explain which terms cancel, give the result of your analysis and express what you learn from this result.

D. Figure D shows a part of a lubrication system that consists of two circular disks between which a lubricant flows radially. The flow takes place because of a modified pressure difference p1 – P2, between the inner and the outer radii r, and r, respectively. This is a laminar, steady state flow of a Newtonian liquid of constant viscosity µ and constant density p without gravity effects. (a) Identify which velocity components of a cylindrical coordinate system are zero for this flow. (b) Formulate the continuity equation for this flow, compute the result and comment on the result of your analysis. Argue why the radial component of the velocity vector is a function of both the radial and the axial direction. (c) Formulate the momentum equation for this flow in the axial direction, explain which terms cancel, give the result of your analysis and express what you learn from this result.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

Two springs of different properties, A and B, are connected in series that support a force P as shown. Select all equations that generally hold true. No partial points. Note: P is force, ō is change in length. O PA + PB = P O PA=PB = P O PA > PB O PA PB O A = OB O OA # OB A T B
All Elements have a fixed value of E and I. Do load modelling for 1 meter strip width in the water tank section of wwwwww www.www the system.Draw the M N T diagram of the entire system. n=47 6 m. A ΔΒ B 7 water 8 m D 2 m. 47ton 3 m. E 1/10
Read the question carefully and give me complete right solution according to the question. Don't miss anything.
find the general solution
2- For the beam given in the figure, a Find the equation of elastic curve by integration method. b- Draw (T), (M.) diagrams of the beam. c- Find the support reactions. Explanation: Mumbers in the last dinit of har A q=a+b+c 2q Xh B
! Required information NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. A frame and loading are shown. B P 2 120 mm 120 mm D of c 120 mm 90 mm ↓ Determine the force acting on member ABC at C. Take P= 320 N. The force acting on member ABC at C is N.
Use 4 decimal places during computation and 2 decimal places for the final answer. Please write you’re your solution on a paper and draw the FBD.
NOTE: 1. The following are grounds for not considering your solution and final answers: a)Figures not properly labeled. b)Equations that are not consistent with the firgure. C)Equilibrium equations do not have sign conventions. d) unsequenced solution. e) answers without units 2. Computations accurate to the nearest tenths. Problem. Show that the following structures are statically determinate and determine the support reactions. b) B is pin connected, C is fixed, hinge supports A and D. 8 kN 8 kN 2 m -2 m 3 kN/m C 4 m D -3 m
A=12 B=12 C=7 D=7
Given: The uniaxial bar shown below is homogeneous, prismatic and has a distributed axial load of p,(x)= px, where p=constant (note: P, is not a constant!) applied along its length as shown. E=constant Left end is fixed (u(x=0)%3D0) → → - → → A=constant Required: 1) Using uniaxial bar theory, derive an expression for each of the following: a) P= P(x, p;) b) o, 3D о, (х, р,, А, L, E) c) E = E (x, p, 4,L, E) d) и %3D и(х, р%, 4, L, E) 2) Plot the results of a)-d) on four different graphs: P=P(x), Om=0xx(x), Ex=Ex(x), and u=u(x) (for a given value of the input loads, geometry, and material properties) 3) Determine the reactions at each end, P(x=0) and P(x=L) 4) Determine the maximum axial deflection umax and determine its coordinate location
Solve the given DE via Variable Separable
Figure below shows a uniformly tapering bar of rectangular cross-section of length L and thickness t. The width of the bar at one end is b₁ and the width at the other end is b2, where (b₂> b₁). The bar is subjected to an axial force P. The axial deformation due to load P is given by _mm. b₂ PL atE b₁ = 20 mm b₂-40 mm -In () then the value of (a + c) is