Viscous liquid, at volume flow rate Q, is pumped through the central opening into the narrow gap

Viscous liquid, at volume flow rate Q, is pumped through the...

Key Concepts

Creeping (Stokes) Flow

This concept refers to fluid motion at very low Reynolds numbers where viscous forces completely dominate inertial forces. In creeping flow conditions, the convective acceleration terms in the Navier–Stokes equations are negligible compared to the viscous terms, allowing linearization of the equations. This simplification is crucial for obtaining analytical solutions in problems involving very slow or highly viscous flows.

Fully Developed Flow Between Parallel Plates

In the context of laminar flow between two closely spaced stationary plates, the flow becomes fully developed when the velocity profile does not change along the flow direction. The resulting profile is typically parabolic, a fundamental result in viscous flow theory. This profile is used as a model to analyze the velocity distribution in narrow gaps, making it essential for solving problems involving lubrication and thin-film flows.

Relationship Between Flow Rate and Average Velocity

The volume flow rate is directly related to the average velocity across a cross-section of the flow, especially in channels or narrow gaps. Knowing the velocity profile allows one to integrate over the cross-sectional area to obtain the flow rate, which in turn is used to determine the spatial variation of the average velocity. This relationship is fundamental when connecting experimental measurements or design specifications of a flow to theoretical models.

Net Force Balance on a Boundary

In viscous fluid problems, determining the net force required to hold a moving or stationary boundary in place involves integrating the pressure distribution over the surface area in contact with the fluid. This force balance takes into account the contribution from both the pressure forces and the viscous shear stresses, and it is crucial for the design and analysis of systems involving fluid-structure interaction in the context of lubrication and bearing theory.

Pressure Gradient in Viscous Flows

The pressure gradient is the driving force in viscous flow problems, particularly under creeping flow conditions, where the movement of the fluid is primarily resisted by viscous forces. Evaluating the spatial variation of pressure is achieved by integrating the momentum balance equations, and it directly relates to the viscous shear stresses in the fluid. This concept is vital to determining the flow rate, velocity distribution, and forces on boundaries in viscous flows.

Transcript

00:01 We have problem number 18, sorry 19 in which there are two parallel disks.

00:10 These are two parallel disks and the oil is pumped from here, oil is bumped through here.

00:17 And this is r not, these things are being given.

00:20 This gap between two plates is h and this is r.

00:25 So everything is being given and we need to find, okay, so we need to find.

00:32 What all things we need to find? optin general for the variation of average velocity in the gap.

00:39 So we have to find average velocity variation, okay, that is, we have to find average velocity variation and in the gap, okay, okay, and we have to find the pressure gradient d p by d r as a function of radius, okay.

01:03 Obtain xp .r and pr pressure as a function of the radius so we have to find these things okay so let us start with let us start with okay now we know that mean velocity we can say mean velocity sorry we have to find mean velocity so this is simply mean velocity mean velocity is always the mass flow rate and divided by that is quantity of oil it is flowing divided by the curve surface area of this pipe and if it has radius r so this will be 2 pi r h so in this case vr so v will be function of r that is q by 2 pi r h so this is the expression for v okay now we know that we know that q by l we should say this is the pressure chain with radius okay q by l equal to one by minus one by 12 mu where mu is the dynamic viscosity curly p by curly x into we have formula for this okay now if we have l that is length of the pipe this to be the curve surface area of the pipe so we can say l equal to 2 pi r okay okay okay one more thing to be noted here here l will be equal to 2 pi r where r is the radius okay so in this case we would be writing q by 2 pi r equal to 1 by 12 mu curly p by curly x sorry curly okay no problem we should be writing curly p by curly r we can convert it into h cube but we have to find d p by d r so in this case curly p by curly r will be equal to 12 mu q divided by with negative sign of course 2 pi r hq this is 6 the pressure gradient you should write d p by d r equal to minus 6 mu q divided by pi r h q so this is equation number 2 this is the equation number 2 this is the pressure gradient okay third we need to find we are pressure as a function of radius okay we know that pressure okay here we can see that this is the pressure gradient so from here we can find pressure so for that we have to first rewrite this as dp equal to minus 6 mu q by pi r hq d r so let us integrate let us integrate so this will be pressure and this will be atmospheric pressure okay so this p is the pressure through which by which the oil is being pumped and this is at most pressure that is oil is exiting in atmosphere that's why atmospheric pressure small r to capital okay so this will be p from lower limit p to upper limit p atmosphere and six mu and q pi and h cube are constants so can be taken out so minus six mu q by by h q and it will be one by it will be integration of from r to capital r smaller to capital r one by h cube h minus 3 t r okay so here we will be writing p atmospheric pressure minus p equal to minus 6 mu cube by pi hq this is h minus 3 so from basic formula of integration minus 3 plus 1 small r to capital r okay so minus 6 mu q by pi h cube into h raised to the power minus 2 by minus 2 from small r to capital r now minus 2 this can be written as minus 6 mu cube divide by pi h cube into minus 1 by 2 1 h square small r to capital r this minus minus will get cancelled out and it will be three so and one more thing to be noted here that this is uh what we are doing is for half of this r so to get complete r we have to multiply here with two into two into two into two so it will be into 2 so this will be 6 mu q by pi h cube 1 by r square minus 1 by r square okay okay i'm extremely sorry because we had to multiply with d r and i multiply we have to we have to integrate with with respect to d r but we are i am integrating with respect to r so this whole thing we need to repeat sorry for this okay okay, sorry for the delay.

08:28 We have to find pr, that is b pressure as the function of r.

08:32 So let us rewrite this equation as dp equal to minus 6 mu q by pi hq into r, dr.

08:42 Since this is dr, this is operator of integration...

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