Welcome to the last lecture of video course

on Tribology. So, final lecture, we are going to complete bearing design or hydrodynamic

bearing design based on short bearing approximation assuming bearing length is a much shorter

than length diameter of bearing much short term is may be say 0.2 5 times of the diameter.

And most often in the present situation, we prefer short bearings, even then previous

lecture it was pointed out increasing length is going to increase load carrying capacity

and sensitivity of the length is much larger or so stronger; but increasing length has

a some problem like a misalignment more heat decipitation or more heat generation increase

on a temperature wise increase in a coefficient of friction.

So, we need to account this, when we are talking about the design is not only the load carrying

capacity, we need to think from temperature point of view, coefficient of friction point

of view. And again we are always aiming for the miniaturization shorter is the better

that means, smaller size is the better, is from that angle length need to be reduced. So we are continuing with first slide from

the previous lecture you say attitude angle which was shown or which was mentioned in

previous lecture can be given as a 10 inverse of 1 minus epsilon square square root of that

divided by epsilon. By using this relation we can find out attitude angle. And once we

know the extensity and attitude angle we say that, shaft center is fixed is been located

and it shows that, when you try to plot this e verses attitude angle or we say that, epsilon

verses attitude angle what we are going to get, we can change extensity from 0.0 1 to

say 0.9 9 and try to find out what will be attitude angle this is turning out to be one

extensity ratio is very low attitude angle is very large, it is almost 90 degree.

When extensity is very high, attitude angle is much smaller in real, in practical situation

we do not want much smaller attitude angle, that is why from designer point of view keep

extensity ratio up to 0.8 8 or say 0.80 more than that will not be recommended because,

we know there will be some sought of fluctuation in a load, that may increase extensity ratio.

So, from design point of view I prefer 0.8 and minimum value as a 0.5 or 0.4 that means,

this should be the operating zone for the bearing under operation. Reason being lesser

the attitude angle more will be in a stability of operation, that is why whenever there is

attitude angle as much larger extensity is much smaller, we change the bearing, we reduce

the bearing length, we make a necessary grooves in the bearing, so that extensity ratio is

greater than 0.5. And bearing is going to be little more stable

and this is often the case with bearing operating at the high speed applications, because we

know that, as speed is increasing load carrying capacity of the bearing is going to increase;

and if the load carrying capacity of bearing is going to increase, because of the increase

in the speed extensity will decrease, extensity will decrease for the same equilibrium load

and that decrease is in a extensity is going to increase attitude angle which is unfavorable,

that is why many times for high speed operation, we do not go higher with a cylindrical bearings,

we will go higher with a half set bearings, we go higher with elliptical bearings, four

load bearings, we try to disturb the bearing clearance, we try to reduce the bearing clearance

and try to increase stability, increase extensity in those ratio in those zones.

But that is a totally a separate subject for us, we are going to discuss about the bearing

design based on the short bearing approximation and this is a gives an indication try to keep

extensity ratio, we say 0.4 more than 0.4, but lesser than 0.8. If any time bearing extensity

is increasing beyond the 0.8 try to modify the design change the parameter, if it is

a turning out to be lesser than 0.4 try to change the parameter, increase this extensity,

this should be more operating zone for the bearing. Now, we discussed about the load carrying

capacity, we can use a short bearing approximation to the find out that and the addition mention

the bearing length should not be extended, because its that is going to increase the

friction force. Question comes how? That is a what we are

trying to find out how to drive the friction force we know, we have discussed Petroff equation,

but for the bearing we are saying that is inaccurate equation I will demonstrate that,

using the after deriving this friction force formula. Friction force will be depending

on whether there is a metal to metal contact and we know for hydro dynamic bearing there

will not be any metal contact that means there is a only sharing of liquid lubricant, that

sharing can be sharing the stress can be given by using this relation is eta higher the viscosity

higher will be sharing, higher the velocity higher will be the sharing, lesser the film

thickness higher will be the sharing, then comes the pressure gradient and a h by 2.

To find out friction force, naturally we need to integrate this sharing is fix over its

area and that area is R d theta because coordinate into design and extra length and we have only

to demonstrated, that bearing effective bearing they say only half of the bearing, that is

why the integration for the theta will be theta equal to 0 to theta is equal to point.

We again mentioned about the z that at the mid plain we are assuming z is equal to 0

that means extreme will be minus L but, 2 2 plus L by 2, L is a bearing length.

Once we integrate this what we are going to get this expression it is interesting to note

that, F friction force itself depends on the load, this is a W sign 5, whatever the load

applied load that is going to affect the friction force, may be to lesser extent but, it is

going to affect friction force. And we try to see what is this friction force

or relation that is a W sign 5, that can be given in terms of other parameters like eta,

U, L, cube, c, square and we are able to see they are smaller terms are common in this

two, so to simplify it what we can do we can take some common terms and compare this term,

first term with a second term in the second term is negligible for our simple calculation

we can neglect this term, but if it is not negligible, then we need to account slightly

more complex problem. And we talk about the Petroff equation, the Petroff equation can

be derived from this relation itself, if I use epsilon is equal to 0 we are going to

get the Petroff equation same which we have derived in our lecture or earlier lectures.

Now, gives when am using the epsilon equal to 0, that is giving the Petroff equation

but, is the situation there is no extensity or we say that the shaft in that bearing surface

is a center are conceding there is no load carrying capacity of that bearing, how friction

force will be generated? Friction force cannot be generated that means Petroff equation which

is the predicting the friction force, because of that its inaccurate no force, no normal

force, but still there is a friction force that is a that is a that is not a good option

that is why, we say that Petroff equation cannot be used for the bearing design. But

with the modification when we say using an integration of shear stress can give provide

a good results, off course this is based on the short bearing approximation, so it cannot

be 100 percent reliable but, it gives a reliable results.

Now, as I mentioned there number of terms common in this two relations, this expression

we can take common term this common term turn out to be 2 pie eta U L R divided by c and

in bracket also in this square root 1 minus epsilon square. So this will be 1 plus epsilon

square into L e square divided by 16 L R square. So, if I assume 4 R square is equal to d square

that means L by d square term is somewhere here and we are talking about the short bearing

approximation so length maybe equal to 0.25 times of the diameter, so what will turn out

to be this L by d square will turn out to be 0.25 square.

We know extensity, may be 0.8 maximum value we are choosing, so 0.8 square will turn out

to be 0.64, so 0.64 into 0.25 into 0.25 divided by 4 that means 0.25 into 0.25 into 0.25 into

0.64 will turn out to be much lesser than 1, it can be neglected for our simple calculations

for detail analysis this can be incorporated for simple class room calculation it can be

neglected. That is why in our calculation we are going to treat this F or we are going

to write F or estimate F as 2 pie into eta into L into e R U into R divided by c in square

root of 1 minus epsilon square, so this is a simple friction expression we can be utilized

yeah we can utilize this. Now, even though we were discussing of the

friction force from temperature point of view, we want to estimate temperature so that, we

can modify viscosity we can properly estimate the load carrying capacity of bearing as viscosity

is going to affect load and temperature is going to affect viscosity and friction is

going to affect temperature. And this is that common change, so naturally

we required a good approximation do a complete iteration procedure to find out what will

be the final result, prefect combination of temperature viscosity, friction force and

the load that is why we require a flow rate also. If there is a possibility the temperature

or heat generation is there and it is getting convicted and conducted simpler one we say

assume the conduction is 0, that is going to give slightly conservative design to us

but, that is fine for us. And to proceed in that direction we require

what is a flow rate, we discussed earlier if there is a hole arrangement or slot arrangement

we can estimate what will be the flow rate, that is a Q P in addition to that flow generally

occurs because of the velocity, that velocity the circumferential flow can be given by this

relation that is a Q theta flow rate passing from any location, theta location can be given

in this term again when this situation also this term is a almost negligible compared

to this dominating feature or dominating term. So, for our calculation we are going to neglect

this second term we are going to account only for first term and this overall leakage we

say that, when we are finding the flow is coming in may be theta is equal to 0 and theta

pie film thickness is minimum the most of the liquid is a leaked out not most of most

of the whatever the liquid is a some portion of that liquid is a leaked out and remaining

portion is getting circulated. So, we require this leakage rate, that is

going to give us cooling effect because this liquid will be again cool and come back or

whether we can be return back using the pump arrangement and that Q leakage as it is happening,

because of the hydro dynamic action can be given in an other form also we say that instead

of writing Q leakage I can write Q H due to hydro dynamic action and what we say that

here the feed pressure flow due to the feed pressure will be Q P.

Overall may be a combination of Q H plus Q P there will be some sought of a disturbance

when the pressure comes there will be slightly decrease in a Q H or we say overall flow rate

for time being we are neglecting that we are saying the Q H can be calculated by integrating

both the length for two situation what is a the exit condition that Q theta is equal

to pie and what is the entrance condition that is a theta is equal to 0, so theta is

equal to 0 theta is equal to pie that difference is going to give us what will be the leakage

from the bearing Q H coming out of the bearing which is going to give cooling effect.

And in fact, it has been observed 80 to 90 percent heat is been carried away by liquid

lubricant which are going to get leaked from the surface, so this is going to give us a

reliable results reliable in the sense 80 to 90 percent results when we integrate it

we know the U is not depending on z h is not depending on z, so it will be turn out moved

out and the d z will turn out to be complete length that means L by 2 minus minus L by

2 that minus minus will turn out to positive, that is a L by 2 plus L by 2 is equal to 1

that comes out here this will be maximum film thickness or we say that, Q as another that

will be h as a at theta is equal to 0. So this will be c into 1 plus epsilon this

point h will be minimum so that will be c 1 minus epsilon and when we arrange it is

turn out to be Q H is equal to velocity, clearance extensity ratio into length it depends on

all and this is a volume flow rate we can find out there relation here this is the meter

per second meter meter so that will turn out to be meter cube per second.

But if you are interested in mass flow rate, what we are going to do we are going do multiply

with the density, that is generally k g divided by meter cube and meter meter cube will be

cancelled out they will turn out to be k g per unit second, the mass flow rate this is

a important for calculating the temperature raise. So, finding the temperature raise we know

we need to find out what will be the heat generation, that the friction force into velocity

is going to give us what will be the generated heat and that is going to be carried away

by liquid lubricant. Assumption we are saying that there is a no much conduction and most

of the heat is been affected from using the liquid lubricants. So, we required a thermal

equilibrium we say rate of heat generated is equal to rate of heat converted by oil,

heat rate of heat generation can be given it as a F into U, F is given like this 2 eta

U L R into pie divided by c square root of minus epsilon square that is been given over

here and there is a velocity that is can be given as a 2 pie R into N or say that pie

d N is equal to as a m C p as a mass flow rate specific heat into temperature raise

for heat connection and that is going to give us the total temperature raise is equal to

this relation. Now, this temperature raise is going to depend

on viscosity and viscosity is going to depend on the temperature, that is why we are keeping

close loop of temperature and viscosity, then it depends on radius very sensitive temperature

raise is very sensitive towards the temperature radius larger the radius more and more temperature

raise and it depends on the clearance, larger the clearance lesser will be the temperature

raise, this is a overall relation for temperature raise we can calculate using this relations.

And off course, we derived m as a mass flow rate, in previous slide that can be incorporated

over here mass flow rate which was derived in previous slide it was given as row U c

that clearance epsilon into length, substitute rearrange after substitution we are rearranging

this what we are going to get this is a ratio R by c square, that means and this is a related

to the liquid specific heat and density this is a speed and viscosity.

So, larger viscosity larger temperature raise, larger in this ratio R by c larger will be

the temperature raise, larger density which is generally not happen by large this row

and C p will remain constant it is not going to much affect this whatever the temperature

is that if I change the liquid lubricant it is not going to change, because of this product,

but because of viscosity it will change. Now, we can think what we have done we have

derived the relation for load relation for flow, relation for friction, relation for

temperature and flow incorporate the relation of sorry temperature involves the relation

of the flow that is why, we say how to do design first guess extensity ratio extensity

ratio been guess may be 0.5 will be an initial approximation.

We know minimum value 0, maximum value 1 I will take 50 percent of that, 50 percent is

a epsilon ratio, epsilon is equal to 0.5 or extensity ratio as a 0.5 that is a initial

guess we need to go ahead with that. Once we know extensity ratio we can find out the

load we can find out the friction force flow has been already been incorporated in temperature

raise. So we do not have to calculate separate flow rate if there is no flow due to supply

pressure or we say supply pressure is 0 in those situation, otherwise when we are supplying

with a some feed pressure naturally need flow rate need to be counted and separately accounted

for the temperature raise. So, for present case we are calculating the

load for given or estimate extensity ratio, estimate the friction force and temperature

raise. Once we know the temperature raise we need to use lubricant viscosity relation

or we say temperature viscosity relation, to modify to modify the viscosity. Once we

modify again we need to do calculation, we can think about extensity calculation, load

calculation, friction calculation, temperature raise, so it will be continuously iterated

that is why we say repeat steps 1 to 3, so that average viscosity and load are going

to convert after that, even you repeat results are not going to change that is a convergence

it may be 3 step may be 5 step may be 7 step may be 10 step depend on 15 steps depends

on your initial approximation of this. But most of the complicated situation we use

this procedure to provide initial good approximation, however if I use a finite difference method

and start with a some orbiter extensity ratio is going to take long time to convert, but

if I use this kind of short bearing approximation, estimate the results and use those results

as input to the finite difference method, number of calculation will reduce significantly

and overall there will be better we say the overall there will be lot of on that.

So. what we say is we assume some extensity ratio substitute this find out the load carrying

capacity, that is one find out the friction force, then find out the temperature raise.

Once we know the temperature raise delta t modify the viscosity that is a what we have

mentioned there are 3 steps extensity is a guessing, so that is does not require any

calculation as such it is a guess than load calculation friction force calculation temperature

calculation then finally, viscosity calculation. And this viscosity is going to be input to

the load, naturally it requires iterations it requires overall iteration to convert to

one final solution and demonstrate to demonstrate this procedure which is been discussed over

here, let us take one example. What this example means see in this case we

are trying to find out number of bearing parameters, see its determined estimate or calculate minimum

film thickness maximum pressure coefficient of friction we are not done coefficient of

friction, but we know if we are able to estimate friction force and we know the normal load

take that ratio F by W that is going to give me the coefficient of friction.

And this is been, what we say that it is been required to estimate all this, when bearing

is supporting 600 Newton load or the rotational speed of 2000 rpm, rpm rotational speed is

defined, applied load is been defined, shaft diameter is a 40 mm there is a similarly,

bearing dimension also will turn out to be almost 40 mm. Assuming the bearing length

is a 10 mm that the L by D ratio is 0.25 all viscosity is a room temperature is been defined

as a 15 mille Pascal second, beta that is a constant for calculating the viscosity at

any temperature raise is been given as a 0.029. And radial clearance c is been defined as

a 20 micron, it shows clearly the radius of the shaft is 20 micron was 20 mm and clearance

will be 0.1 percent of that that is a 20 micron so we are following that same scheme radial

clearance is a 0.1 percent of radius. Now, some important calculation we say that

U will be utilized again and again, so instead of directly calculating in a formula you separately

calculate what is the Q and that is a calculate a pie d n, we try to find out that is a 4

0.19 meter per second or say for 0.2 meter per second, factor U i Q pie 0.25 divided

by clearance square can be should be calculated separately, because we are going to do iterations

we do not know exactly what will be the extensity ratio, we are going to assume and calculate

W based on that once we calculate W we are going to find out the friction force temperature

raise and viscosity based on that again that means there will be iterations.

That is why we use this calculation separately U L Q pie 0.25 divided by c square this is

not going to be calculated again and again, we can directly use this factor that 8 2 2

7 meter per second directly. Same way for the friction force we use a separate

factor that is a 2 pie R L into R into pie divided by c square again this will not be

calculated again and again, but friction force will be calculated again and again, because

we require iterative scheme we require perfect iterations to evaluate what will be the friction

force. To demonstrate it as I say the first is approximation for me will be extensity

as a 0.5, maximum value is 1, minimum value is 0 ill take 1 plus 0 divided by 2 that it

will turn out to be 0.5, I will be using that as first approximation.

So, when epsilon is 0.5 what we are going to get W as a 1 1 8 lower, I do not have any

other idea we can use some sought of a interpolation, but for that purpose what we required again

the mean value for interpolation we require two values at least first load and then subsequent

second load, so what do we do we know maximum value of extensity 1 this epsilon mid value

0.5 and we are not getting the desirable load load carrying capacity which is 600 Newton

and what we are getting is a 1 1 8 Newton what I will do I will again do 0.5 plus 1

divided by 2, take a mean value that is turning out to be 0.75.

So, extensity 0.75 what we are going to get W as a 562 meter, now I can use some sought

of interpolation to evaluate or I can find out again averaging value 0.75 plus 1 divided

by 2 I can go ahead with that, but that rough approximation says that, generally we do not

recover more than 0.8, so am just taking the value as a 0.8, epsilon value as a 0.8, am

want to find out whether the load capacity is a really exceeding 600 Newton and not.

So extensity 0.8 what we are going to get W as a 900 Newton which is a 50 percent higher

than this naturally we will be we can decrease this take intermediate value of these two

but, we know viscosity which is been used in this calculation is a 15 mille Pascal second

and that is happening at the room temperature, we are not calculated that operating temperature

and viscosity is going to be lesser than this at a operating temperature that is going to

reduce a load carrying capacity. So, for times being we are saying this was

assuming extensity is a 0.8 calculate W calculates friction force, find out the temperature raise

and that is turning out to be 8. 65 degree centigrade.

Now, what is the next step to find out what will be the viscosity for this. Off course,

we have use this relation we say factor one how it is been utilized factor two how it

is been utilized we are not doing this calculation again and again we are simply substituting

the factor, we know epsilon is going to change viscosity is going to change, that is why

except these two we have assuming one constant factor. Similarly, for friction force viscosity

is going to change epsilon is going to change so keep apart from these two factor thing

all other will be constant so, that is what the factor 2, this is required for simple

calculation to reduce efforts. And temperature raise again can be given in this case again

we can find out viscosity sorry speed is a constant 4, 5 is a constant R by c is constant

row C p is constant we can take this as the factor 3.

And here nothing is been mentioned, but we are using the word density of the liquid as

a 860 kilo gram meter Q and c P as a 1 c 1 6 0 joule per unit kilo gram per centigrade,

this is been utilized and most often for liquids these are the results for the liquid lubricant

these are the relations or we say that these are the parameters for liquid lubricant. So,

we can substitute we can find out another factor 3 in this case which will be 4 pie

into n into R by c square divided by row c P, that will cannot be remain constant and

after that we can we can keep changing eta and epsilon to find out final temperature

raise. Now, once we know the temperature raise we

can use this to find out what will be the viscosity, what will be the operating viscosity,

that is viscosity is given as this is defined eta in is already known, beta is defined,

delta t we calculated that is going to give viscosity as 11.7 mille Pascal second, in

our calculation earlier we assume this is a 15 mille Pascal second, so almost 20 to

30 percent change in a viscosity. Now, we need to substitute this value this

viscosity in the load say for epsilon is equal to 0.8 with a modified viscosity what we are

going to get W as a 7 0 3 Newton, earlier we got 900 applied load is a 630 by 50 percent

variation, but after incorporating this viscosity change what we are getting W as a 7 0 3 Newton,

which is still higher than applied load, naturally we need to change extensity we need to decrease

extensity ratio for here for this kind of extensity and viscosity what we are getting

F as a 5.13 Newton calculate temperature raise. So it has reduced from 8.65 percent to 6.75

percent am sorry there are not percent is a 68.65 degree centigrade to 6.75 degree centigrade

it is reduced. Now, at this temperature we will again try

to find out what will be the operating viscosity and that is slightly more, that is turning

out to be 12.3 mille Pascal second, we did calculation for 11.71 Pascal second naturally

this viscosity W will be on a higher side. That is why now it is a time has come to take

an average 0.75 and 0.8 take an average of that turn out to be 0.75 for time being we

are taking extensity ratio as a 0.78. So 0.78 when you calculate this temperature

raise will turn out to be 7 degree centigrade, calculate load at this is turning out to be

fortunately it is turning out to be 599.1 Newton very close, we know we have done approximation

if you want we can keep slightly more load we can increase this 0.78 from 0.7 8 to 0.79

and keep a higher load or depends what we want. In this case, friction force is almost

a same temperature raise is also almost coming to the 0.7 degree centigrade, viscosity operating

viscosity is turning out to be 12.2 mille Pascal second this is a fine design for me.

Or we say if you want to go for the finite difference method this design is going to

give reliable results or we assume that load carrying capacity estimated by short bearing

will be slightly higher, then we can think about 0.7 also 0.79 also which will give roughly

may be say around 640 Newton load capacity applied load is on a 600 we say that is fine

for us we can go ahead, so this is a word we say the how we do calculation.

Next comes how to find out attitude angle we are following the short bearing approximation

we have calculated epsilon as a 0.78 substitute this value find out attitude angle and attitude

angle is turning out to be 302.2 degree that is fine say extensity 0.78 attitude angle

32.2 degree. So this is all about, but problem what we have asked in question find out minimum

film thickness? Find out maximum pressure? Find the coefficient of friction? And we are

not discussed those things, till now this is a first step locate the shaft, find out

extensity and attitude angle and after that do remaining calculations. So, this is we

have completed we have converged to the final results extensity 0.7 8 attitude angle 32.2

degree now time has come to find out the minimum film thickness. We know clearance film thickness can be determined

based on epsilon as a c minus in bracket 1 minus epsilon, substitute this value 20 micron

is a radial clearance 0.78 is a epsilon value is turning out to be 4 .4 microns. Question

comes whether we have done everything right? We are talking about hydro dynamic lubrication

and we know hydro dynamic lubrication we need to will be valid for the specific film thickness

by enlarge more than 5. If surface softness of shaft and bearing surfaces are given to

us or provided to us, we can find out this is specific film thickness using this relation,

that will be ratio of minimum film thickness to the composite self surface softness.

Now, if I assume roughness of the shaft is a 0.4 micron may be on higher side and bearing

may be even 0.6 microns, take a composite one and find out what will be the composite

surface softness if this ratio is turning out to be more than five is say hydro dynamic

lubrication is fine bearing design is as per the hydro dynamic lubrication there is no

problem. If that is not the case then we can think

about adding hydro static lubrication, we can supply pressure that is going to reduce

temperature and that is going to reduce film thickness or sorry reduce extensity when epsilon

is decreased naturally h main is going to increase. So, that way feed pressure is going

to increase film thickness is going to bring is going to bring you say mixed lubrication

domain to hydro dynamic domain so we can do calculation when we do calculation we find

out whether everything is fine and not if it is not then we should use some sought of

pressure to supply liquid to cooler liquid to cool the liquid lubricant can reduce the

temperature make into a factor viscosity more than what is been estimated and.

So this is a film thickness then next comes a maximum pressure, how to find our maximum

pressure this is a short bearing approximation, we know maximum pressure will occur when z

is equal to 0 that is condition which we are used to derive this relation the mid plain

d P by d z will be equal to 0. Now, we have used z equal to 0, but we need

to find out at which angular position maximum pressure is going to be there or maximum pressure

will be generated and pressure profile what will be location of maximum pressure, that

mean theta o max theta o, because of short bearing or we say theta 0 as a we are using

a short bearing approximation. So start with the pressure relation first you say that this

is a pressure relation we say that maximum pressure will occur at z equal to 0, substitute

this, rearrange this and after that in h differentiate with the respective theta. After differentiating

with respective theta equate to 0 that is going to give us that is going to give us

theta for maximum pressure once we know substitute those value over here and find out maximum

pressure. In our case, this been done and we find this

location or maximum pressure theta 0 max depends only on extensity ratio that is a what is

shown over here, it is a 1 minus in square root of 1 plus 24 epsilon square divided by

4 pie. Now we know this will be always greater than 1, that means theta is going to be greater

than 90 degree this is going to be negative. Now epsilon can be 0.1, 0.2, 0.3, 0.4, 0.5,

0.6, 0.7, 0.8 whatever you take this will be negative, if this is a negative our value

will turn out to be negative sorry this will be more than 1 and if it is more than 1 minus

more than 1 naturally will be negative that is why the theta will be always greater than

90 degree. In present case, this is turning out to be

105.43 degree and this is a different than theta is equal to pie, this is difference

that theta is equal to pie that means location of minimum film thickness is different than

location of maximum pressure. Otherwise in many books there is been confuse have been

treated the maximum pressure location and minimum film thickness location are the same

it is not, there locations are different. Now, based on then we substitute this value

theta o max in this expression what we are going to get maximum pressure as 7.24 bar

not very high value is well within permissible limit mega Pascal it is only 0.724 mega Pascal

its not very big, this is a big quantity or larger any material can be used any material

can be used in this situation. Now finally, comes the coefficient of friction

we needed to find out coefficient of friction, we know what is the friction force, we know

what is the applied normal load, take a ratio I have divided the W that is what is going

to get, we are going to get that is a 0.0086, this is very low coefficient of friction 0.0086

almost negligible pressure that is why the hydro dynamic bearings are most popular, whenever

we require good good damping local self friction 0 wear hydro dynamic bearings are on the top

place. Now, what we can do to find out if length

is doubled what is going to happen with this parameters, just now of a present case we

took length is a 0.25 times of the diameter, but if we are going to increase this length,

we say doubled the length instead of 10 mm we can think about 20 mm what is going to

happen. That is shown over here, the same data which

we defined the same viscosity, same load, same speed, same everything right. Now only

the variation is bearing length instead of 10 mm we are considering as a 20 mm, if we

do that, again we need to assume as a first step what is the extensity ratio we need to

guess as I mentioned earlier for me simple case is summation of 1 and 0 divided by 2

that is an average value at the first step and that is epsilon giving as me as a 0.5

this is turning out to be 943 Newton. Now, I can take a again average 0.5 plus 0

divided by 2 that is a epsilon as a 0.25, but we feel epsilon lesser than 0.4 should

not be recommended for in this case, we have recommended a 0.45 minimum value, it should

not be lesser than that. However, we need to change the design. So epsilon if I assume

the 0.45, we can find out the load as a 700 40 Newton, what is a more than 600 we can

reduce it further, but we what we want to do that calculation we know this 741 is at

that 15 mille Pascal second viscosity and if we account the temperature raise this is

will be this will be lesser than that any b if viscosity will be 12 mille Pascal second

or 30 mille Pascal second so this will be reduced.

So, for time being we can assume the extensity of initial approximation as a 0.45 calculate

the friction force and friction force calculation gives the result as a 8.84 Newton, based on

that we can find out the temperature raise, that is turning out to be 10.33 degree centigrade.

Based on this we will modify the viscosity this is a 15 mille Pascal second temperature

raise is 10.33 degree centigrade and beta is given to us in example or we say in a question

so based on that, what we can find out the viscosity that is turning out to be 11.1 mille

Pascal second, we substitute this value to find out what will be the load carrying capacity.

Now, with this viscosity extensity 0.45 load turn out to be 548 Newton which is a lower

than applied load, I can change viscosity or I can change extensity immediately here.

Other one is that calculate again the friction force recalculate temperature raise re calculate

viscosity and again see whether that is going to increase this load carrying capacity reaching

to the final value which we require, if it a case you say that we do the calculation,

now find the friction force evaluate temperature raise based on that find out the viscosity

that is a instead of 11.1. Now is turning out to be 11.9 mille Pascal second substitute

W is turning out to be 588 Newton, again we can do calculation, but we know this extensity

is not going to increase to 600. I can think about this extensity may be say

0.46 and find 4 6 also can give me some result which is slightly more than that, now it is

my choice whether I go for slightly higher value or convergent you say the 588 is a value

within 5 percent of the load I can converge it or we can think about slightly more than

that so when we are talking about the epsilon as a 0.46 and the load carrying capacity is

a 606 Newton, F is a 6.94 Newton. So, I can find out, I can close this example

with extensity as a 0.46 or extensity of 0.45 both the options are open to me this is giving

me high load carrying capacity, that is giving slightly lower load carrying capacity. So

from conservative point of view I can choose this or for the continuity we say no we do

not require too many iterations then iteration is itself is giving a reasonable good results

choose those results. So, depends on the whether we have freezing

0.46 or 0.45 we can find out attitude angle and attitude angle for 0.45 is turning out

to be 57.3 degree, which is a higher value we can increase this we can change the bearing

length, we can reduce the bearing length, we can change the parameters so that extensity

is coming out to be 0.7, 0.6, 0.7, 0.8 that will be the more desirable level, that is

a giving indication bearing length increasing from 10 mm to 20 mm is not full filling the

purpose extensity ratio is going down its more like we have a capabilities and but,

we are not doing our best we are not giving our best performance.

So, bearing have a lot of capabilities but, because of the larger length bearing is not

able to give the best, which is not desirable first thing is a increasing a going for a

larger bearing length naturally cost of the manufacturing is going to be increased, cost

of the material is going to increase, in addition there will be some sought of the some sought

of misalignment in addition there will be friction force I use the word there will be

additional friction force, but we need to check it we say that for a this if I assume

0.45 as a freezing point calculate what will be the maximum pressure.

For that purpose we require theta o max or theta 0 max and that is turning out to be

110 degree 110 degree in this case, that is going to give me value of something like a

P max or the 13 bar or 1 point here mega Pascal, if you remember the length was 10 mm this

pressure maximum pressure was only 0.7 mega Pascal.

So, what we are doing increasing the bearing length maximum pressure is been increased

to 1.3 mega Pascal, one way it is a negative side another way is that bearing material

mostly they are able to sustain more than 5 mega Pascal, so why to worry about 1.3 mega

Pascal let it be like that, if bearing length is giving all other advantages we should go

ahead with a larger length, but temperature from coefficient of friction point of view,

say that coefficient of friction in this situation, because friction force is a 7 Newton in this

case divided by 600 is going to give us a result as 0.0117 coefficient of friction is

a 0.0117 while in earlier example this coefficient of friction was a lesser than this 0.0086.

So, increasing length is not full filling lot of purposes, first extensity is going

down lesser than, we say that 0.5 may be recommended fine, but that is increasing maximum pressure

larger length is giving lesser pressure compared to sorry smaller length is giving larger pressure,

smaller pressure compared to larger length which is a quite reverse. While coefficient

of friction for the smaller length is lesser than larger length, naturally I will prefer

lesser length smaller length for the bearing it rest it is says a space, say the cost gives

optimum performance. And that is what we gain from a studying lubrication

mechanism understanding tribology. Otherwise, if somebody says the load carrying capacity

is go ahead with that maximum length that is a wrong, here lesser length, smaller length

is giving more benefits lesser, maximum pressure, lesser coefficient of friction, lesser space,

lesser cost, so that is essential for us or we say that that is going to give us overall

economics. So, with this am trying to close the course

I hope you understood the course and in future you will be able to say lot of cost incorporating

tribological principles, tribological guidelines thank you, thank you for your attention.

nice

Very nice course! Thanks Prof. Hirani, thanks NPTEL and thanks IIT…

wonderful thanks indeed

What is Capital R? is it R1 or R2?

thanks a lot sir for describing the bearings so well

Why can't we increase radial clearance about 30 microns, by keeping the axial length of bearing constant and we have the film thickness of 5microns as well?

good sir,for u hand some thing happened