Differential equations Dr. Om Prakash Singh Associate Prof., IIT (BHU) www.omprakashsingh.com

The Differential Continuity Equation Mass conservations To derive the differential continuity equation, the infinitesimal element of Fig. is needed. It is a small control volume into and from which the fluid flows. It is shown in the xy-plane with depth dz. Let us assume that the flow is only in the xy plane so that no fluid flows in the z-direction. Since mass could be changing inside the element, the mass that flows into the element minus that which flows out must equal the change in mass inside the element. This is expressed as

The Differential Continuity Equation Mass conservations where the products u and v are allowed to change across the element. Simplifying the above, recognizing that the elemental control volume is fixed, results in

The Differential Continuity Equation Mass conservations Differentiate the products and include the variation in the z-direction. Then the differential continuity equation can be put in the form The first four terms form the material derivative so above Eq. becomes (4) providing the most general form of the differential continuity equation expressed using rectangular coordinates.

The Differential Continuity Equation Mass conservations The differential continuity equation is often written using the vector operator so that continuity Eq. takes the form (6) where the velocity vector is V = ui + vj + wk. The scalar V is called the divergence of the velocity vector.

The Differential Continuity Equation Mass conservations For an incompressible flow, the density of a fluid particle remains constant as it travels through a flow field, that is, so it is not necessary that the density be constant. If the density is constant, as it often is, then each term in above Eq. is zero. For an incompressible flow, Eqs. (4) and (6) also demand that

Problem Conditions for Incompressible Flow Consider a steady velocity field given by V = (u, v, w) = a(x2y + y2)i + bxy2j + cxk , where a, b, and care constants. Under what conditions is this flow field incompressible? Ans: a = - b

Problem Air flows with a uniform velocity in a pipe with the velocities measured along the centerline at 40-cm increments as shown. If the density at point 2 is 1.2 kg/m3, estimate the density gradient at point 2. Ans: /x = 0.3 kg/m4

The Navier-Stokes Equations Conservation of momentum Rectangular stress components on a fluid element Stresses exist on the faces of an infinitesimal, rectangular fluid element, as shown in Fig for the xy-plane. Similar stress components act in the z-direction. The normal stresses are designated with  and the shear stresses with . There are nine stress components: . If moments are taken about the x-axis, the y-axis, and the z-axis, respectively, they would show that

The Navier-Stokes Equations Conservation of momentum So, there are six stress components that must be related to the pressure and velocity components. Such relationships are called constitutive equations; they are equations that are not derived but are found using observations in the laboratory. Next, apply Newton’s second law to the element of Fig., assuming no shear stresses act in the z-direction (we’ll simply add those in later) and that gravity acts in the z-direction only: These are simplified to

The Navier-Stokes Equations Conservation of momentum If the z-direction components are included, the differential equations become (12) assuming the gravity term ρgdxdydz acts in the negative z- direction. In many flows, the viscous effects that lead to the shear stresses can be neglected and the normal stresses are the negative of the pressure. For such inviscid flows, Above Eq. takes the form

The Navier-Stokes Equations Conservation of momentum In vector form they become the famous Euler’s equation, which is applicable to inviscid flows. For a constant-density, steady flow, above Eq. can be integrated along a streamline to provide Bernoulli’s equation. Constitutive equations relate the stresses to the velocity and pressure fields. For a Newtonian isotropic fluid, they have been observed to be (15)

The Navier-Stokes Equations Conservation of momentum For most gases, Stokes hypothesiscan be used: λ = -2μ/3 . If the above normal stresses are added, there results showing that the pressure is the negative average of the three normal stresses in most gases, including air, and in all liquids in which ∇. V = 0. If Eq. (15) is substituted into Eq. (12) using λ = -2μ/3 there results

The Navier-Stokes Equations Conservation of momentum where gravity acts in the negative z-direction and a homogeneous fluid has been assumed so that, e.g., ∂μ/∂x=0. Finally, if an incompressible flow is assumed so that ∇. V = 0, the Navier-Stokes Equations result: where the z-direction is vertical. If we introduce the scalar operator called the Laplacian, defined by

The Navier-Stokes Equations Conservation of momentum , the Navier-Stokes equations can be written in vector form as The three scalar Navier-Stokes equations and the continuity equation constitute the four equations that can be used to find the four variables u,v,w, and p provided there are appropriate initial and boundary conditions. The equations are nonlinear due to the acceleration terms, such as u∂u/∂x on the left-hand side; consequently, the solution to these equation may not be unique. For example, the flow between two rotating cylinders can be solved using the Navier-Stokes equations to be a relatively simple flow with circular streamlines; it could also be a flow with streamlines that are like a spring wound around the cylinders as a torus; and, there are even more complex flows that are also solutions to the Navier-Stokes equations, all satisfying the identical boundary conditions.

The Navier-Stokes Equations Conservation of momentum The Navier-Stokes equations can be solved with relative ease for some simple geometries. But, the equations cannot be solved for a turbulent flow even for the simplest of examples; a turbulent flow is highly unsteady and three-dimensional and thus requires that the three velocity components be specified at all points in a region of interest at some initial time, say t =0. Such information would be nearly impossible to obtain, even for the simplest geometry. Consequently, the solutions of turbulent flows are left to the experimentalist and are not attempted by solving the equations.

Problem Couette Flow between a Fixed and a Moving Plate Using Navier-Stokes equation, derive the equation of velocity of the moving plate

Solution Consider two-dimensional incompressible plane (/z = 0) viscous flow between parallel plates a distance 2h apart, as shown in Fig.. We assume that the plates are very wide and very long, so that the flow is essentially axial,u  0 but v = w = 0. The present case is Fig. a, where the upper plate moves at velocity V but there is no pressure gradient. Neglect gravity effects. We learn from the continuity equation that

Thus there is a single nonzero axial-velocity component which varies only across the channel. The flow is said to be fully developed (far downstream of the entrance). Substitute u = u(y) into the x-component of the Navier-Stokes momentum equation for two-dimensional (x, y) flow: Most of the terms drop out, and the momentum equation simply reduces to The two constants are found by applying the no-slip condition at the upper and lower plates:

Therefore the solution for this case (a), flow between plates with a moving upper wall, is This is Couette flowdue to a moving wall: a linear velocity profile with no-slip at each wall, as anticipated and sketched in Fig. a.

Problem Flow due to Pressure Gradient between Two Fixed Plates Determine velocity profile Case (b) is sketched in Fig.b. Both plates are fixed (V= 0), but the pressure varies in the x direction. If v= w= 0, the continuity equation leads to the same conclusion as case (a), namely, that u= u(y) only. The x-momentum equation changes only because the pressure is variable: Also, since v= w= 0 and gravity is neglected, the y- and z- momentum equations lead to

Thus the pressure gradient is the total and only gradient: Why did we add the fact that dp/dx is constant? Recall a useful conclusion from the theory of separation of variables: If two quantities are equal and one varies only with y and the other varies only with x, then they must both equal the same constant. Otherwise they would not be independent of each other. Why did we state that the constant is negative? Physically, the pressure must decrease in the flow direction in order to drive the flow against resisting wall shear stress. Thus the velocity profile u(y) must have negative curvature everywhere, as anticipated and sketched in Fig. b.

The solution to above Eq.is accomplished by double integration: The constants are found from the no-slip condition at each wall: Thus the solution to case (b), flow in a channel due to pressure gradient, is The flow forms a Poiseuille parabola of constant negative curvature. The maximum velocity occurs at the centerline y= 0:

Problem Consider a steady, two-dimensional, incompressible flow of a newtonian fluid with the velocity field: u = –2xy, v = y2 – x2, and w = 0. (a) Does this flow satisfy conservation of mass? (b) Find the pressure field p(x, y) if the pressure at point (x = 0, y = 0) is equal to pa.

Problem A belt moves upward at velocity V, dragging a film of viscous liquid of thickness h, as in Figure. Near the belt, the film moves upward due to no-slip. At its outer edge, the film moves downward due to gravity. Assuming that the only nonzero velocity is v(x), with zero shear stress at the outer film edge, derive a formula for: v(x); the average velocity Vavg in the film; and the wall velocity VC for which there is no net flow either up or down. Sketch v(x) for case (c)

Problem Water flows from a reservoir in between two closely aligned parallel plates, as shown. Write the simplified equations needed to find the steady-state velocity and pressure distributions between the two plates. Neglect any z-variation of the distributions and any gravity effects. Do not neglect v(x, y). Solution Continuity eqn. The differential momentum equations, recognizing that

are simplified as follows neglecting pressure variation in the y-direction since the plates are assumed to be a relatively small distance apart. So, the three equations that contain the three variables u, v, and p are To find a solution to these equations for the three variables, it would be necessary to use the no-slip conditions on the two plates and assumed boundary conditions at the entrance, which would include u(0, y) and v(0, y). Even for this rather simple geometry, the solution to this entrance-flow problem appears, and is, quite difficult. A numerical solution could be attempted.

The Differential Energy Equation Most problems in an introductory fluid mechanics course involve isothermal fluid flows in which temperature gradients do not exist. So, the differential energy equation is not of interest. The study of flows in which there are temperature gradients is included in a course on heat transfer. For completeness, the differential energy equation is presented here without derivation. In general, it is where K is the thermal conductivity. For an incompressible ideal gas flow it becomes

The Differential Energy Equation For a liquid flow it takes the form where  is the thermal diffusivity defined by α = K/cp

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