A hydrostatic bearing is to support a load of 1000 lbI/ft of length perpendicular to the diagram

A hydrostatic bearing is to support a load of 1000 lbI/ft of...

Key Concepts

Flow Rate in Lubrication

Flow rate in lubrication refers to the volume of fluid passing through a section per unit time. It is directly linked to the dimensions of the bearing, the viscosity of the lubricant, and the applied pressure. By controlling the flow rate, engineers can ensure a consistent and adequate supply of lubricant to maintain the appropriate film thickness and pressure necessary for load support.

Pressure Gradient

The pressure gradient is the rate of change of fluid pressure along the direction of flow. In the context of lubrication and hydrostatic bearings, it is fundamental for driving the viscous flow that sustains the lubricating film. Determining the pressure gradient is essential for designing systems that can generate sufficient pressure to support the applied loads.

Fully Developed Flow

Fully developed flow in a confined space occurs when the velocity profile of the fluid no longer changes along the flow direction. This simplifies the analysis since the balance between viscous shear and pressure gradient becomes constant along the length of the bearing. Recognizing this condition allows for the use of simplified models to accurately predict pressure distributions and flow characteristics.

Hydrostatic Bearings

Hydrostatic bearings support loads using an externally pressurized fluid film between surfaces. This concept involves designing a bearing such that the fluid pressure, controlled by supply pressure and geometry, counteracts the mechanical loads. It is essential in ensuring stability and precise control in applications where low friction and high load capacity are required.

Viscous Fluid Flow in Narrow Gaps

Viscous fluid flow in narrow gaps pertains to the behavior of a lubricant confined between two closely spaced surfaces. In these conditions, the flow is dominated by viscous forces, and the thin film of fluid must sustain the applied loads. This concept is a cornerstone of lubrication theory and is critical in analyzing and predicting performance in bearings and similar mechanisms.

Transcript

00:01 We have problem number 15 in which we have a bedding hydrostatic bearing which can support a load of thousand pounds per feet okay and of length perpendicular to the diagram which is being given and a bearing is being supplied with from here it is the oil is being supplied which is of the configuration being given and 33 psi that is the initial pressure that is inlet pressure pi will be equal to 35 psi so we have to calculate the required bit of the bearing pad the resulting pressure gradient dp by dx and the gap height of the flow okay and gap height and the flow rate is q equal to 202 .5 gallon per hour per feet.

01:04 Okay, so we have to first write what all things being given.

01:11 Okay, so things being given are the force, which is equal to 1000 lbf and it is applied on a length of one feet.

01:30 Okay now inlet pressure b i is equal to 35 p s i inlet pressure is being given 31 okay and mass flow rate q sorry volume flow rate is 2 .5 gallons per r per feet okay and one more thing is being given is that at the given configuration that is given condition that 10w 30 oil at 220 bfurnhe.

02:11 So the viscosity that is kinematic viscosity at this will be equal to 0 .0 from the table this is from the table according to specification of the oil which is being given 0 .01 newton second per meter is and if we convert it into because everything is in foot pound second so we have to convert it in fps okay so to convert this in fps unit we have to multiply this with 0 .02089 so it will become lbf second divide by pitt square okay so this comes out to be 5 .4 .2 .089 into 10 to power minus 4 pound second per feet square.

03:21 Now we'll be using all these things, all these things to calculate the given parameters.

03:29 Okay, so part a we have to find first the width of the bearing pad.

03:37 So to get the width of the bearing pad we have to first write basic equation q by l equal to minus h q by 12 mu and pressure gradient dp by d x okay dp by d x now if we just assume the flow to be laminar for laminal flow we will be having for laminar flow we will be having for laminar flow we know pressure gradient is constant pressure gradient is constant okay so we can write px that is pressure as a pressure as a function of x equal to pi 1 minus 2 x by w where pi is 35 p s i that is inlet pressure is being given and x equal to 0 to w by 2 okay 0 to w by 2 because this will be half so this is w by 2 and this is w by 2 so 0 to w by 2 x equal to 0 to w by 2 okay now total force in x direction f will be equal to so total force in x or in y direction that is vertically a quad direction could be found out by f equal to l this is px and dx now l into px value of px is p i minus 2x by w where w is our pad length okay and dx okay one more thing to be noted that the limit will be 0 to w by 2 so since these are in two directions that is these are in halves there are two halves so let us multiply this with 2 okay now this is 2 l 0 to w by 2 so p i is 35 so let us just concentrate with all which all things are constant p i is constant that is 35 psi yes so we can take it outside so 0 2 w by 2 1 minus 2 x by w d x now we have to integrate it so 2 into l into pi 02 w by 2 and x minus 2 x squared by 2 into w into w so these 2 will get cancelled out so this is 2 l p i 0 2 w by 2 x minus x square by w now let us plug in the upper limit first and then the lower limit so 2 l p i upper limit that is w by 2 minus w square by 4 w minus 0 that is 2 l 2 that is 2 l2 p i will be cancelled out so w by 2 minus w by 4 so 2 l p i into w by 4 2 times okay so we have force equal to net force equal to 1 by 2 p i into l into w this is the net force from here w could be found out that is the width of the pad okay yes the weight of the bearing pad would be found out just by writing 2 into f 2 into f by l into bai now f by l is already being given which is 35 p s i now f by l is 1 000 l b a okay 1 ,000 l per feet so this is 2 into 1000 lbf per feet divided by pi that is 35 p s i k so then this is per square inch so let us convert it into feet so this is 2 000 lbf divide by 35 okay okay so psi it is 35 lbf by inch square and here it is feet so 2000 lbf divide by 35 lbf by inch square means 1 by 12 feet whole square into feet so these two will get cancelled out and we will be having 2 ,000 into 144 k 2 ,000 into 144 divide by no 144 will be in the denominator itself 2 ,000 divide by 35 into 144 feet that is 0 .39682 approximately 0 .397 feet so this is the pad length okay now part b part b is we have to find we have to just find the value of raised everything that is the pressure gradient value of pressure gradient we have to find out dp by dx okay so dp by dx will be equal to minus delta p that is change in pressure and divide by d x we should write w by 2.

11:48 Okay, so minus 2 delta p by w.

11:57 So let us write the let us plug in the values minus 2 into 35 lbf by this is w by 2.

12:16 Inch square because this is psi into here it will be w w is 0 .397 feet 0 .397 feet so minus 70 l bf divided by 1 by 12 feet square into 0 .397 feet feet so minus 70 lbf divide by 1 by 12 feet square into 0 .397 feet feet so minus 70 lbf into 144 divide by feet square into 0 .397 feet okay into 0 .397 feet okay anything i'm missing on okay no problem we'll be just calculating it so 70 to 144 divided by 0 .397 397.

13:56 So this becomes 25 ,390.

14:12 4 .2 lbf by p2.

14:17 Now this could have been done as, this could have been written as negative with negative sign minus 17.

14:32 26 psi by feet if we convert this feet equal to 20 feet equal to 12 inches so it will be 144 so we have to multiply we have to divide this like we will be doing like this right okay yes if we want that the pressure should be in psi is itself so we could write it at 35 25 000 390 point 4 2 l bf by lbf by this is feet square so if feet square so 12 inches whole square and here it will be feet so 12 inches whole square so 12 inches whole square so minus 25390 42 by 144 lbf by square feet feet is in denominator if we divide with 144 we'll be getting 176 with negative sign point 3 2 this comes out to be psi per p so this is dp by t s that is the pressure gradient part c part c is we have to find what was part three we need to find the gap height and flow rate is being given so we will be using the basic equation which i have written earlier that q by l equal to minus h q by 12 mu d p by d x okay so from here we have to find this h so we can write we can write as by cross multiplication 12 mu by l equal to minus h cube d p by d x okay so this is 12 mu into q by l so minus hq will be equal to 12 mu q by l divide by d p by d x x cube will be equal to minus 12 mu q by l divide by d p by d x h cube will be equal to minus 12 mu q by l d p by d x h will be equal to minus 12 mu q by l d p by d x raised to the power 1 by 3 so let us plug in all the values minus 12 into mu that is viscosity is being given where is viscosity viscosity viscosity viscosity viscosity is 2 .089 into 10 to power minus 4 2 .089 2 .089 into 1080 minus 4 and the unit will be l bf s by feet square k lbf s by feet square q by l value of q by l we need to get so it is q is given as 2 .5 gallon per hour per feet 2 .5 gallon per hour per feet and this l is one feet.

19:04 So divide by one feet.

19:06 We should write feet here.

19:10 Is this cool? let us cross check gallon per hour per feet.

19:14 Okay.

19:15 Yes, this is good...

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