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1Fundamental Principles of Convective Heat Transfer - Governing equationsDr. Om Prakash SinghAsst. Prof., IIT Mandiwww.omprakashsingh.com

Course PolicyCourse type: Self study mode/Teach othersContact Hours: 3 hours/weekCredits: 3Assessment: Presentations: 20% Assignment/Project/Wikipedia article: 20%End Exam- 3 hrs: 30%Quiz 1, and 2: 20%Attendance: 10%

Class Hours: 2.00 5 PM Monday

Modes of heat transferHeat transfers in three ways:ConductionConvectionRadiation

Temperature difference must for heat transfer to occur

(understanding heat transfer from philosophical point)Life is all about difference

Difference in thinking creates excitement, happiness sadness etc.Dress to reduce temperature difference

Difference in force bends a structureEinstein relativity theory based on velocity differenceLife is all about difference

Brilliant = Your talent talent of average person

MathematicsDifferenceCorruption = Black money white money (tendency to create abrupt large difference in money leads to corruption)

Modes of Mass transferConcentration difference leads to mass transfer (kg/s) by

Diffusion (solids, liquids, gas)Melecular phenomenon like heat conduction/diffusionConvection (Liquid, Gas)Transfer of mass by bulk motion and diffusion

There is no Radiation-LIKE counterpart in Mass transferDouble-diffusive convection exhibits both diffusion and convection phenomenonEffect of Rayleigh numbers on the evolution of doublediffusive salt fingers, 2014, O. P. Singh, J. Srinivasan, Phys. Fluids, 26(6), pp. 1-18

Why study heat transfer?Design for failure free products (safety)

High temperature in products leads: - to catastrophic failure - reduced product life - human discomfort

Designs are becoming more compact (economy) - heat dissipating devices closely pact - high heat concentration - challenge to remove heat

Fundamental principlesMass balance (continuity equation)

Force balances (momentum equations)

Energy balance (laws of thermodynamics)Any system should follow the following conservation principles

The Differential Continuity EquationMass 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 asThe Differential Continuity EquationMass 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

Differentiate the products and include the variation in the z-direction. Then the differential continuity equation can be put in the form

The Differential Continuity EquationMass conservationsThe first four terms form the material derivative so above Eq. becomes

providing the most general form of the differential continuity equation expressed using rectangular coordinates.(4)The differential continuity equation is often written using the vector operator

so that continuity Eq. takes the form

The Differential Continuity EquationMass conservationswhere the velocity vector is V = ui + vj + wk. The scalar V is called the divergence of the velocity vector. (6)The Differential Continuity EquationMass conservationsFor 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

ProblemAir 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

ProblemConditions for Incompressible FlowConsider 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

The Navier-Stokes EquationsConservation of momentumStresses 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

Rectangular stress components on a fluid elementThe Navier-Stokes EquationsConservation of momentumSo, 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 Newtons second law to the element of Fig., assuming no shear stresses act in the z-direction (well simply add those in later) and that gravity acts in the z-direction only:

These are simplified to

If the z-direction components are included, the differential equations becomeThe Navier-Stokes EquationsConservation of momentum

assuming the gravity term gdxdydz 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

(12)The Navier-Stokes EquationsConservation of momentumIn vector form they become the famous Eulers equation,

which is applicable to inviscid flows. For a constant-density, steady flow, above Eq. can be integrated along a streamline to provide Bernoullis 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 EquationsConservation of momentumFor most gases, Stokes hypothesiscan be used: = -2/3 . 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 = -2/3 there results

where gravity acts in the negative z-direction and a homogeneous fluid has been assumed so that, e.g., /x=0.The Navier-Stokes EquationsConservation of momentumFinally, 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

Navier-Stokes equations can be written in vector form asThe Navier-Stokes EquationsConservation of momentum

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 uu/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 EquationsConservation of momentumThe 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.ProblemCouette Flow between a Fixed and a Moving Plate

Using Navier-Stokes equation, derive the equation of velocity of the moving plateSolutionConsider 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.ProblemFlow due to Pressure Gradient between Two Fixed PlatesDetermine 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:

ProblemWater 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 thatare 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.Material Derivative (or Total derivative)The Material Derivative, also called the Total Derivative or Substantial Derivative is useful as a bridge between Lagrangian and Eulerian descriptions. Definition of the material derivative - The material derivative of some quantity is simply defined as the rate of change of that quantity following a fluid particle. It is derived for some arbitrary fluid property Q as follows:

In this derivation, dt/dt = 1 by definition, and since a fluid particle is being followed, dx/dt = u, i.e. the x-component of the velocity of the fluid particle. Similarly, dy/dt = v, and dz/dt = w following a fluid particle.Note that Q can be any fluid property, scalar or vector. For example, Q can be a scalar like the pressure, in which case one gets the material derivative or substantial derivative of the pressure. In other words, dp/dt is the rate of change of pressure following a fluid particle. Or, using the same equations above, Q can be the velocity vector, in which case one gets the material derivative of the velocity, which is defined as the material acceleration, i.e. the rate of change of velocity following a fluid particle. Material Derivative (or Total derivative)Note also the notation, DQ/DT, which is used by some authors to emphasize that this is a material or total derivative, as opposed to some partial derivative. DQ/DT is identical to dQ/dt.

The material derivative is a field quantity, i.e. it is expressed in the Eulerian frame of reference as a function of space and time (x,y,z,t). Thus, at some given spatial location (x,y,z) and at some given time (t), DQ/Dt = dQ/dt = the material derivative of Q, and is defined as the total rate of change of Q with respect to time as one follows whatever fluid particle happens to be at that location at that instant of time.

Q changes for two reasons: First, if the flow is unsteady, Q changes directly with respect to time. This is called the local or unsteady rate of change of Q.

Second, Q changes as the fluid particle migrates or convects to a new location in the flow field. This is called the convective or advective rate of change of Q. The first term on the right hand side is called the local acceleration or the unsteady acceleration. It is only non-zero in an unsteady flow. The last three terms make up the convective acceleration, which is defined as the acceleration due to convection or movement of the fluid particle to a different part of the flow field. The convective acceleration can be non-zero even in a steady flow! In other words, even when the velocity field is not a function of time (i.e. a steady flow), a fluid particle is still accelerated from one location to another. Example - the material acceleration, following a fluid particle - The material acceleration can be derived as follows:Material Derivative (or Total derivative)

N-S Equation in various flow situationsThe Navier-Stokesequations forincompressible flowinvolve four basicquantities: Local (unsteady)acceleration. Convectiveacceleration. Pressure gradients. Viscous forces.The ease with whichsolutions can be obtainedand the complexity of theresulting flows oftendepend on whichquantities are importantfor a given flow.

(steady laminar flow)

(impulsively started)

(boundary layer)

(inviscid)(inviscid, impulsively started)

(steady viscous flow)

(unsteady flow)Steady laminar flow Steady viscous laminar flow in ahorizontal pipe involves abalance between the pressureforces along the pipe andviscous forces. The local acceleration is zerobecause the flow is steady. The convective acceleration iszero because the velocityprofiles are identical at anysection along the pipe.

Pressure gradient and Viscous forces

39

Flow past an impulsively started flat plate Flow past an impulsively startedflat plate of infinite lengthinvolves a balance between thelocal (unsteady) accelerationeffects and viscous forces. Here,the development of the velocityprofile is shown. The pressure is constantthroughout the flow. The convective acceleration iszero because the velocity doesnot change in the direction of theflow, although it does changewith time.

Local acceleration and Viscous forces

15Impulsively started flow of an inviscid fluidImpulsively started flow of an inviscid fluid in a pipe involves a balance between local (unsteady) acceleration effects and pressure differences.The absence of viscous forces allows the fluid to slip along the pipe wall, producing a uniform velocity profile.The convective acceleration is zero because the velocity does not vary in the direction of the flow.The local (unsteady) acceleration is not zero since the fluid velocity at any point is a function of time.

Local acceleration and Pressure gradient Boundary layer flow along a flat plate Boundary layer flow along a finiteflat plate involves a balancebetween viscous forces in theregion near the plate andconvective acceleration effects.

The boundary layer thicknessgrows in the downstreamdirection. The local acceleration is zerobecause the flow is steady.

Convective acceleration and Viscous forcesInviscid flow past an airfoil Inviscid flow past an airfoilinvolves a balance betweenpressure gradients andconvective acceleration. Since the flow is steady, the local(unsteady) acceleration is zero. Since the fluid is inviscid (=0)there are no viscous forces.

Convective acceleration and Pressure gradient 16Steady viscous flow past a cylinder Steady viscous flow past acircular cylinder involves abalance among convectiveacceleration, pressure gradients,and viscous forces. For the parameters of this flow(density, viscosity, size, andspeed), the steady boundaryconditions (i.e. the cylinder isstationary) give steady flowthroughout. For other values of theseparameters the flow may beunsteady.

Convective acceleration, Pressure gradient and Viscous forcesUnsteady flow past an airfoil Unsteady flow past an airfoil at alarge angle of attack (stalled) isgoverned by a balance amonglocal acceleration, convectiveacceleration, pressure gradientsand viscous forces. A wide variety of fluid mechanicsphenomena often occurs insituations such as these whereall of the factors in the Navier-Stokes equations are relevant.

Local acceleration, Convective acceleration, Pressure gradient and Viscous forcesThe Differential Energy EquationMost 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 EquationFor a liquid flow it takes the form

where is the thermal diffusivity defined by = K/cp

When dealing with extremely viscous flows of the type encountered in lubrication problems or the piping of crude oil, the model above is improved by taking into account the internal heating due to viscous dissipation,The Differential Energy Equation

In three dimensions, the viscous dissipation function is expressed as follows:

End