Lec 17: Herschel Bulkley Fluids Flow through Pipes

Welcome to the MOOCs Course Transport Phenomena of Non-Newtonian Fluids. The title of this lecture is Herschel-Bulkley Fluids Flow through Pipes. In the last class we have started discussing how to obtain the velocity profile and then volumetric flow rate equations etcetera for the case when viscoplastic fluids flowing through pipes.

In the previous class we have derived such equations for the case of Bingham plastic fluids. Now, what we try to do? We try to do this similar case for the case of a Herschel-Bulkley fluids ok.

So; however, before going to the ah today's class details of today's class we will be having a recapitulation of what we have seen in the last class. So, flow of Bingham plastic fluids through circular tubes due to pressure difference. Then if you follow the ah process that we have been ah doing in order to get the velocity profile etcetera.

We get ah velocity profile for the annular region that is between r is equals to R p to capital R we get this expression. And this is the region where the deformation of the fluid is taking place and then region where the deformation is not taking place. There the material is flowing like a solid plug and then that region is between r is equals to 0 to R p and then corresponding plug velocity is given by this one this is a constant velocity.

Plug velocity is constant one single value for a given pressure drop ok whereas, v z the above expression that changes with respect to the r values ok. Corresponding equation for the volumetric flow rate we obtained this one this is for the entire flow region ok. It is not individually for annular or plug region, but the entire flow region.

So, this is what we have derived in the last class and in addition to this one we also derive the average velocity by dividing the volumetric flow rate by cross section area of the circular tube whichever we have taken. So, then we got this expression then using this expression we further calculated the friction factor as well. The friction factor that we have got this one which we have written in terms of the properties of a material that is density, viscosity and then ah yield stress etcetera ok.

Yield stress is also ah property of the material characteristic property of the material like density etcetera that we have for other ah properties like. So, what we have seen it is a implicit equation. So, we cannot solve directly we have to go for a ah trial and error approach that we have seen right.

The same equation we have written in terms of some dimensionless ah numbers like Reynolds number Hedstrom number and Bingham number and then corresponding equations are given here. This is when you write in terms of Hedstrom number and then Reynolds number. If the same equation if you write in terms of ah Bingham number and Reynolds number then we have this equation all three equations or these thre equations are same whichever equation you we you use to get the friction factor you will get the same same value.

Here for a Bingham plastic fluids Reynolds number is rho v average D by mu B plastic viscosity ok. Similarly, we had Bingham number for Bingham plastic fluids is given as tau naught B D divided by mu B v average which can be treated as a kind of a dimensionless characteristic yield stress of the material ok. So, Hedstrom number is nothing but the product of these two dimensionless numbers that is Reynolds number and Bingham number and then you get rho D square tau naught B divided by mu B square.

Ah Because of the linear nature of tau versus ah gamma dot for Bingham plastic fluids after crossing ah tau naught B value. ah We could develop the volumetric flow rate velocity profile etcetera easily in the previous class. But in the case of Herschel-Bulkley fluids we have ah the tau versus gamma dot ah expression non-linear like a power law ah fluid right after crossing this tau ah tau naught H value right.

When the applied stress is more than tau naught H value then tau versus gamma dot expression is non-linear like a power law fluid. So, then derivation may become slightly complicated or maybe lengthier rather complicated. So now, that is what we are going to see now Herschel-Bulkley fluids flow through pipes due to pressure difference right.

So, now let us say this is the pipe that we have and then this is the centre of the pipe or circular tube whichever you are taking. So, coordinate system if you are taking like ah this at the centre. So, then this is z this is r and then this is theta direction.

So, now so we are taking the region where it is fully developed flow and then at z is equals to 0 ah we are taking pressure P naught and z is equals to L we are taking pressure P L right. So, under these conditions we are taking 1D laminar flow. So, where v z is ah dominating velocity component because flow is taking place in the z direction and then this v z is function of r.

Because at centre it is maximum velocity and then at r is equals to capital R; it is 0 velocity because of the no slip condition. So, when we move r is equals to 0 to r is equals to capital R velocity decreases. So, that is what we are taking.

And then compared to v z other component of velocity v r and then v theta are very small. So, then we can neglect them right we do not need to consider them ok. So, that is what we have ah you know these are the assumptions.

So, now, as long as the geometry is the pipe geometry and then if you take 1D laminar motion. So, then what we have seen in last ah few classes we have seen that the ah shear stress whatever the shear stress tau r z is there that is nothing but minus delta p by L r by 2 and this is irrespective of the nature of the fluid; however, we are going to derive in this class again once again. So, in order to have a kind of a continuity right, so what we understand the shear stress is ah linearly increasing with r.

So, let us say if um ah this is from here to here as you are moving. So, at this point at r is equals to 0 you are having tau is equals to tau ah tau r z is equals to 0. And then at this location at r is equals to capital R you are having ah you are having tau r z is equals to maximum value of tau w and then in between these two limits linearly it is increasing.

So, now in this case of Herschel-Bulkley fluids what we have we have the characteristic ah yield stress tau naught H it is having some value in between 0 to tau w value that is sure ok. So, if it is not ah or the applied value is not more than tau naught H so then flow will not take place ok. Now let us say a different location if you try to find out the tau value at this value some value at this location some tau value at this location, some tau value like that gradually it is increasing right that is what we know we understand now from this profile.

So, now let us say this location is the location where this tau r z is becoming tau naught H for example, ok. So, we do not know let us say that is the location. So, then these location whatever is there, these location we can take it as R p r is equals to R p right.

So, now, let us say this is the location ah at which tau r z is having a value equal to tau naught H which is nothing but the characteristic yield stress of this Herschel-Bulkley fluid right. What does it mean? The material ah for any visco plastic fluid the material does not flow as long as tau r z are applied ah ah stress is less than tau naught characteristic yield stress.

So, then gamma dot is 0 there is no deformation ok. So, but when ah; that means, if there is no deformation the material should flow like a ah solid plug like this, it will be flowing like a solid plug without any kind of deformation fine. So, but when ah applied stress is ah greater than tau naught H.

What will happen? Deformation will take place and then depending on the behaviour or tau was depending on the tau versus gamma dot behaviour of that material after crossing this tau naught H depending on the rheological behaviour of the material. Then ah the volumetric flow rate ah the velocity profile will change ok will change with respect to r.

So, if you draw it here ah let us say this location if you draw the ah you know velocity profile for this fluid. So, then it will be having solid plug like ah a motion up to r is equals to R p and then after that the velocity gradually decreases and then reaches a ah you know ah value of ah 0 value at r is equals to capital R. And then now actually if it is not having any viscoplastic behaviour the material would be having the maximum velocity at r is equals to 0 only at r is equals to 0 only you may be having the ah maximum velocity right.

So, but now here it will be having the maximum velocity for this entire range of r is equals to ah 0 to r is equals to R p that is we call v z p which is nothing but v z max right and then after r is equals to R p it not necessarily be the profile as shown here it depends on the value of n because of the Herschel-Bulkley nature. So, accordingly we will have the profile. So, that is what we are going to derive now.

So, what we understand ah for a Herschel-Bulkley fluid or for that matter any visco plastic material the velocity profile would be having two regions. One region where the solid plug like region is there and another region where now you are having ah deforming ah fluid kind of region. So, this this region is deforming ah fluid region and then this is solid plug plug like region.

I have drawn it only for a half of the domain the same is true for the bottom half of this you know tube cross section as well ok. So, now, we are going to derive the required equation for this case. So, so whatever I have described the same thing is ah pictorially shown here ok.

Now before going into the details of the derivation we have to list out the constraints of the problem. So, that these because these constraints are going to be useful in simplifying the equations of motion ok. From where we are going to get a relation which will be helpful for us to get the velocity profile.

What are the assumptions that we have? We have that cylinder is infinitely long that is L by D is very large. Then flow is laminar and incompressible, gravity we are not considering isothermal conditions we are considering and then further steady state.

So, dou by dou t of anything is 0. And then symmetric in theta direction. So, dou by dou theta of anything is 0.

Especially this is for the flow variable not for the ah scalars like temperature and pressure and then fully developed flow flow is taking place in the z direction. So, in the z direction dou by dou z of any flow variable is 0. So, this is also for the flow variables not for the scalars or you know ah like temperature pressure.

Further we have a velocity dominating in the v z direction which is changing ah as we change the r value from 0 to capital R whereas, the other component of velocity v r v v theta r ah very small compared to v z or negligible we can take them as 0. And then only shear stress component existing is tau r z other components of shear stress are not existing right. Then first we have to see whether the continuity is being satisfied for this problem or not.

So, because of the steady state first term is 0 then v r is not existing dou by dou theta of any flow variable is 0. Because of the symmetry v theta is not existing and then because of the fully developed flow dou by dou z of any flow variable is ah 0. So, then um what we understand from here from by simplifying this continuity equation the continuity is satisfied ok.

Next equation of motion r component we have written here. So, steady state this term is 0, v r is not existing v theta is not existing or they are very small and dou by dou theta of anything is 0. Because of the symmetry v theta is not existing and then v z is existing, but ah because of the fully developed flow dou by dou z of anything is 0.

So, that is cancelled out last term. So, left hand side all the terms have been cancelled out. Pressure we do not have any generalized conclusions or boundary conditions in general.

So, then we cannot cancel out this term and then only tau r z is existing. So, it is 0 and then because of symmetry this term is 0, and then only tau r z is existing. So, this is 0 and then because of the fully flow dou by dou z of any flow variable is 0, so this is also 0 gravity we are not considering here.

So, then what we get dou p by dou r is equals to 0; that means, pressure is not function of r. Similarly, if ah theta component of ah equation of motion if we simplify what we get we will see. Because of the steady state this term is 0 v r is not existing v theta is not existing because of symmetry this term is 0 v r v v r v theta both are 0 v z is existing, but v theta is not existing and then dou by dou z of ah any flow variable is 0 because of the fully developed flow.

And then pressure we do not know any ah ah you know generalized conclusions we cannot arrive without ah doing the simplification. So, let is ah let us keep it as it is then only tau r z is exiting. So, this is 0 ah because of symmetry this term is 0, because of the fully developed flow this term is 0 and then for a simple laminar flow these two terms are ah same equal to each other.

So, then this difference is 0 gravity we are not considering. So, here also we get dou p by dou theta is equals to 0; that means, pressure is not function of theta as well. So; that means, pressure is not function of both r and theta; that means, pressure has to be function of z.

Because the flow is taking place because of the pressure difference so, but what is that function we do not know we will get it. So, z component of momentum equation if you simplify what we get we will see. So, steady state this term is 0 v r is not existing v theta is not existing and then because of symmetry this term is 0 v z is existing, but v z is not function of ah z it is function of r only as well as because of the fully developed flow dou by dou z of anything is 0.

So, this is 0 we cannot take dou p by dou z is equals to 0 because of the fully developed flow. Because fully developed flow that is for the flow variables not for the scalars like temperature and pressure. So, then we cannot cancelled out cancel it out ok.

And then tau r z is existing and then it is function of r. So, then we cannot cancel out this term also and then because of symmetry this term is 0. Because of the fully developed flow this term is 0 we are not taking gravity into the consideration.

So, what we have? We have only these two terms remaining. So, that is dou p by dou z is equals to 1 by r dou by dou r of r tau r z.

So, this equation will give you expression for v z as function of r when you substitute what is tau r z for a given fluid. So, till now the rheology of the fluid has not been brought into the picture. So, it is generalized one ok only thing that the flow has to be laminar and symmetric, laminar symmetric fully developed incompressible flow should be there.

So, this equation what we understand we already realized that p is not function of r and theta. So, whatever the ah value information whatever the ah expression that is there in the right hand side that is not going to affect pressure. So; that means, dou p by dou z can be treated as a constant similarly tau r z is function of r only it is not function of z ah because of the fully developed flow.

So, then whatever the left hand side term is there that can be taken as a constant when you integrate the ah right hand side term. That means both these terms individually can be consider and then ah get the ah and and do the integration to get the required ah velocity profile. So, first what we do?

We take this dou p by dou z then integrate. So, then we get P is equals to c 1 z plus c naught right. Now, apply the boundary conditions z is equals to 0 p is equals to P naught.

So, c naught is P naught at z is equals to L p is equals to P L so; that means, P L ah is equals to c 1 L plus c naught; that means, c 1 is P L minus P naught by l. So, p you get this expression P is equals to minus delta p by L z plus P naught ok. So P what we get?

We get a linear profile what we understand we know that the p is function of z, but which kind of function it is we do not know in the previous slide. Now we realize that it is a linear ah profile ok. Now, we can integrate this one.

So, where we are taking dou p by dou z as a constant because the other side we wanted to integrate. Because the other side ah is independent of z ok. So, now, 1 by r dou by dou r what we had this 1 by r dou by dou r of r tau r z we were having and then that was equals to dou p by dou z.

So, now, this r we take to the other side, so r dou p by dou z is equals to dou by dou r of r tau r z right. When you integrate r tau r z is equals to r square by 2; d p by d z plus c 2 that is tau r z is equals to r by 2 d p by d z plus c 2 by r right. So, now, what happens here it ah for any value of r tau r z cannot be infinite as per the continuum hypothesis and then we are doing all these problems under the assumption of continuum hypothesis right.

So, c 2 has to be 0 because if you substitute r is equals to 0 ah then it will become infinite which is not acceptable. So; that means, we get tau r z this expression minus delta p by L r by 2 ok. So, now, this equation is valid.

Now this equation is valid for entire range of r that is r is equals to 0 to capital R. So, in this equation if you substitute r is equals to capital R then you get tau r z is equals to tau w and then when you ah substitute r is equals to R P. What should you get?

Tau r z should be what because R p is the location where applied stress is becoming equal to the yield stress characteristic yield stress of the material. So, at r is equals to R p tau r z should be equals to tau naught H. So, that is what we are they ah writing for plug region that is ah at r is equals to R p tau naught H is equals to minus delta p by L R p by 2 and then for wall tau w is equals to minus delta p by L capital R by 2.

And then if their ratio this tau naught H by tau w we can write it as R p by R we can also call it phi for our calculation purpose right. So, now till now we have not included any information about the rheology of the material. So, now we bring that ah rheological information of the material from this point onwards ok.

For Herschel-Bulkley fluids ah between 0 to R p the material flows like a solid plug with constant velocity that v z p that we have to do one region. And then remaining region between R p to capital R velocity ah profile would be there that will be gradually decreasing from a constant plug velocity at r is equals to R p to 0 at r is equals to capital R that we already realized ah when we were discussing the ah physics of the problem at the beginning of the class. So, we have to get the velocity profile for these two regions independently.

When the applied stress is more than the characteristic yield stress then that region we call it deforming region for that region we know that for Herschel-Bulkley fluid tau r H is equals to tau naught H plus m minus d v z by d r whole power n right. So, this is only for this region between r is equals to 0 to R p gamma dot is equals to 0 that is the equation that we already know we have seen in ah basics of non-Newtonian fluids. So, now this equation I am rewriting like this I am just keeping minus d v z by d r one side and rest all other things other side ok.

Now in place of a tau r z in the previous equation I am going to write minus delta p by L r by 2 which we have derived in previous slide. So, in place of tau r z I have substituted this thing and then rest everything I am keeping same. Now I am writing minus d v z is equals to whatever the these term d r and then now we can integrate this equation.

Before integrating this equation for simplicity what I am writing minus delta p by 2 L m as alpha and then whatever whatever the tau naught H by m as beta. So, that I do not need to write. So, many terms repeatedly ok that is alpha r minus beta power 1 by n d r is equals to minus d z.

When you integrate what you get alpha r minus beta whole power n plus 1 by n divided by n plus 1 by n 1 by alpha is equals to minus v z plus C 3 integration constant. Now, at r is equals to capital R the v z should be 0 because of the no slip boundary condition and then you get C 3 is equals to this value right. So, now, this value we are going to substitute here in the next step.

So, when you do you get this one v z is equals to this value C 3 and then this value whatever is there right. So now what we do? We ah take these n by alpha n plus 1 as a common term.

So, then we have this term. Next in the next step what we are taking alpha capital R minus beta power n plus 1 by n common. So, then we have 1 minus this term.

So, next step we are going to write the corresponding expression for alpha and beta alpha is nothing but minus delta p by 2 L m and now I am multiplying by R and then dividing by R this expression. So, that I have n R by n plus 1 multiplied by 1 by minus delta p r by 2 L m that is nothing but this part only this fraction ok. And then remaining part alpha capital R minus beta power n plus 1 by n is this one.

So, this is your alpha and this is your beta right same we can do here also, but I wanted to do step by step a step by step. So, that ah you know calculations are easier and then easy to follow. So, then now in the next step what I am trying to do wherever a minus delta p by L capital R by 2 is there I am writing tau w.

So, here in this place I can write tau w by m. So, tau w by m I am writing and then in this place I can write here tau w by m and then this is again tau w by m is as it is there. So, then tau w minus tau naught H whole power n plus 1 by n and then whatever divided by m power n plus 1 by n is there.

So, that I have written here right. So, this term I am keeping here as it is right. So, now, this this m if I join these terms together what I get ah m power 1 by n only I will be getting.

So, that is m power 1 by n and then tau w as it is and then. So, next what we are trying to do whatever the parentheses terms what we were having we were having tau w power n plus 1 by n minus tau naught H power n plus 1 by n this was we are having. So, then tau w power n plus 1 by n I am taking common.

So, that I can write this expression here like this. Remember this ah H is not a power it is a superscript indicating Herschel-Bulkley fluid only ok and in last parentheses term I am keeping as it is. So, now, these term these 2 tau w terms if I combine what I can get I can get tau w power 1 by n.

So, that tau w power 1 by n and then divided by m power 1 by n I can combine and then I can write tau w by m whole power 1 by n and then this is tau naught H by tau w is nothing but R p by R right. Tau naught H we have seen it is nothing but minus delta p by L capital R p by 2 and then tau w is nothing but minus delta p by L just capital R by 2. So, when you divide these two you get R p by R that R we write it as see in the next step.

Now here again what we are doing, so since this term is simplified. So, we are concentrating in the next term in the next term again now we are substituting for values of alpha and beta. So, then we have here these terms right.

So, here in place of R p by R we have written phi now what we are trying to do like previously we have done wherever minus delta p by L capital R by 2 is there. So, we are writing tau w in this here. So, then we have tau w by m r by R minus tau naught H by m whole power n plus 1 by n and then here tau w by m minus tau naught H power m whole power n plus 1 by n right.

So, next step what we do? We take tau w by m whole power n plus 1 by n common from these two terms. So, that what we can have here tau naught H divided by tau w we can had we can have in in this in the second term right.

So, so here when you take these tau w by m whole power n plus 1 by n ah taking common you get this expression right. So, this tau w by m whole power n plus 1 here and then here we can cancel out. So, then here tau naught H by tau w here also tau naught H by ah tau w I write it as phi.

So, I have r by R minus phi divided by 1 minus phi whole power n plus 1 by n right. So, this already we know. So, R p by R is equals to phi we are writing.

So, in the next step what we do? We do LCM and then we cancel out this 1 minus phi power n plus 1 by n and then whatever the denominator 1 minus phi power n plus 1 by n right. So, v z we are getting this expression; v z is equals to n R by n plus 1 tau w by m whole power 1 by n multiplied by 1 minus phi power n plus 1 by n minus r by r minus phi power n plus 1 by n ok.

So, now, this is the velocity profile which is valid for r is equals to R p to R only. In the deforming region what is the distribution of ah velocity how it is being distributed as function of r that we got it ok. So now what we have to get?

We have to get what is that constant value between r is equals to 0 to R p like a plug solid plug. So, in since this equation is also valid for r is equals to R p. If you substitute r is equals to R p here then you get this equation in place of R we have substituted R; R p.

So, R p by R is nothing but phi, so then this term is 0. So, v z p is nothing but these term, n R by n plus 1 tau w by m whole power 1 by 1 by n 1 minus phi whole power n plus 1 by n. This is what we get this is plug region this is valid for only plug region.

So, now we got both ah velocity profile in the deforming region and then constant velocity expression in the non deforming plug like region ok. So, next step we move to the volumetric flow rate calculations or obtaining equation for the volumetric flow rate. So, volumetric flow rate Q is given by this expression actually integral 2 pi r v z d r it should be right.

But integration now has to be done in two parts one part between 0 to R p another part between R p to R. Because in these two parts the corresponding velocity expressions are different. So, v z p is this constant value v z p is a constant value and then v z is function of r.

So, those things we are substituting here this is v z p and then this is for ah v z. Only thing that whatever 2 pi n r by n plus 1 tau w by m whole power 1 by n is there that we are taking common from these two terms right. So, now, this whatever the 2 pi n r by n plus 1 tau w by m power 1 by n is there that I am calling it as A ok.

So, that I do not need to ah repeatedly write this expression again and again ok. So, now, this is a constant and then when you multiply this one with r and integrate. So, you get r square by 2 limits 0 to R p.

So, here again um this term these two terms we are going to integrate. So, what I am trying to do? I am integrating only the this part first ok.

So, then I get ah r square by 2 actually r square by 2 and then integration limits, but here R p to R whereas, here integration limits 0 to R p. Whereas, the third term I am keeping as it is we can do similar way, but we will be doing after simplifying these two terms ok. So, now, you ah take you know here in this term what last term what we have done we have taken r by R as zeta and d r is capital R d zeta.

So, then we have this equation. So, this integration will be doing slightly after some time. So, ah first two terms when you substitute the limits you get these expressions and then when you combine these two terms you have these terms because here you are taking ah R square common.

So, that wherever R p square is there you will be getting divided by R square. So, R p by R we can write phi. So, we have in place the R p by R phi, so phi square by 2 here.

Similarly, here 1 minus ah R p square by R square that is 1 minus phi square, we are getting remaining everything same without ah any further simplification right. So, now next step what we are doing? These two terms we are joining.

So, then we get A R square divided by 2 because the two is also we have taken common A R square divided by 2 multiplied by 1 minus phi whole power n plus 1 by n and then phi square plus 1 minus phi square these two cancelled out. And then since two we have taken common. So, this last term would be now multiplied by 2 and then integration limits for the last terms are R p by R to R by R.

So, that is nothing but phi to 1. So, that is phi to 1 right, so this is what we have now. So, now, these term we can integrate.

So, that you know we can finalize we can so that we can get the final expression. We know this expression integral f x g x d x and then ah following this equation what you take f you take as zeta and g you take as zeta minus phi whole power n plus 1 by n. So, here minus 2 and then integration of zeta minus phi power n plus 1 by n is nothing but zeta minus phi power 2 n plus 1 by n divided by 2 n plus 1 by n and then zeta g zeta that is f is as it is limits phi to 1 minus integral integration of ah this part is again the same.

Whereas, a differentiation of ah zeta is nothing but 1 now and d zeta right. So, in the next step when we integrate this this term as well then we get zeta minus phi whole power 3 n plus 1 by n divided by 3 n plus 1 by n and then limits phi to 1. So, you substitute the limits in the next step.

Then we have this expression whereas, the other terms are you know as it is we are not doing anything with them right. So now what we do? Next step we ah take 1 minus phi power n plus 1 by n common from the remaining this term as well.

Then we get we get 1 minus 2 n divided by 2 n plus 1; 1 minus phi plus 2 n square divided by 2 n plus 1 by 3 n plus 1; 1 minus phi whole square this is what we get right. So, next step what we do? This term we divide into two terms then we get 2 n by 2 n plus 1 minus 2 n by 3 n plus 1 in place of this term and then whatever 1 minus phi whole square ah being multiplied as it is right.

So, now, in the next step what we do? We combine these two terms. So, that 2 n by 2 n plus 1 if you take common 1 minus 1 minus phi that is what you get.

Then we get next step 2 n 1 minus phi divided by 2 n plus 1 and then from here one phi is there. So, that is this term whereas, the remaining terms are as it is now right. So, now, whatever the A expression is there that is 2 pi n R by n plus 1 tau w by m R square term is there.

So, that we are substituting per A right. So, this A ah this 2; this 2 is cancelled out. And then next step what we do this n plus 1 we bring inside the parenthesis here to do further simplification.

So, what we have here n pi r cube divided by n plus 1 tau w by m whole power 1 by n; 1 minus phi whole power 2 n plus 1 by n is there. So, then whatever n plus 1 we are bringing into the parenthesis then we have this term right. So, in the next step what we are trying trying to do?

This term we are dividing into two terms. So, that we have this one. Similarly, this term we are dividing into two parts.

So, then we have this term 1 by n plus 1 minus 1 by 3 n plus 1 is nothing but 2 n by 3 n plus 1 n plus 1 ok right. So, in the next step what we are going to do? We are multiplying this with these two steps um two terms.

Similarly we are multiplying this 1 minus phi whole square with the two terms within this parenthesis. So, then we have these 5 terms now 4 plus 1; five terms right. So, next step what we try to do?

We try to join these three terms where we are dividing by n plus 1 right then we get phi square by n plus 1 whereas, the remaining two terms are as it is. So, this is the final expression for the volumetric flow rate right exactly similar as we have done for the case of Bingham plastic fluids, but the calculations as slightly lengthier. Average velocity you can get divide by dividing the volumetric flow rate with pi r square.

So, then when we do you get this expression for the average velocity right. So, the same thing you can write in a simplified manner like this where ah the parenthesis whatever the terms within this parenthesis is there that we you can call it as F function of phi comma n ok. So, that function is nothing but this one.

Now, once we have the v average we can get the friction factor friction factor ah we can get by using this phi definition phi is equals to tau naught H by tau w tau naught H is characteristic of the material that we can keep as it is. So, now next step wherever in this equation phi is there in place of phi you can substitute tau naught H by ah 1 by 2 f rho v average square or 2 naught H by f rho v average square like this. So, here also what you see this friction factor is not explicit it is implicit.

So, then you may be needing to do lot of trial and error approach right. Now before concluding today's lecture we take an example problem on this ah derivation that we have done for the case of Herschel-Bulkley fluid. So, one material density is having 1500 kg per meter cube is ah flowing through a pipe under laminar flow conditions pipe dia is 40 mm length is 500 meters ok.

The material can be represented by Herschel-Bulkley ah fluid model where tau naught H is 17 pascal, m is equals to 0. 83 and n is equals to 0. 5 right.

So, now, the question is estimate the pressure drop when this slurry is flowing under laminar condition with a mean velocity v average is given; v average is given as 0. 5 meter per second. So, when this is the average velocity.

What is minus delta p? That is what we have to find out. The second part of the question what is the plug velocity?

What is v z p when minus delta p is whatever the value obtained in the first case right? And then finally, calculate the size of the plug. So, what is the value of R p?

That we have to find out. So, v average we have this expression we are going to use this expression because v average is given here ok. Now in the right hand side we see what we know we know n it is given R is given 20 mm tau naught H is given m is also given right ah phi is not known right.

So, only thing unknown here in this equation is phi if you know the phi; phi is nothing but tau naught H by tau w and then both these ah tau w is related to minus delta p right. So, then you can find out minus delta p. So, what we do?

First we simplify this equation to get phi and then from that phi value we try to find out tau w and then from that tau w value we try to find out minus delta p ok. So, when we substitute all these values and then simplify and then you follow the trial and error approach you get phi is equals to 0. 58 ok.

So, that is R p by R is 0. 58 that is almost half of the cross section of the pipe is flowing like a plug half of the cross section is flowing like a solid plug that is what we understand ok. That means, applied stress is not sufficiently large enough to make the material to deform for a longer ah for a higher ah cross section region for a higher portion of the pipe cross section.

So, phi is this one. So, which is phi definition is nothing but tau naught H by tau w right. So, from here tau w is nothing but tau naught H by phi.

So, tau naught H is given as 17 phi you got is 0. 58 so; that means, 29. 31 pascals is tau w, but tau w is minus delta p by L D by 4 or R by 2.

So, from here minus delta p is nothing but 4 tau w L by D 4 tau w L by D then you get minus delta p as 1. 46 into 10 power 6 pascals approximately right. So, the first part of the problem is solved second part of the problem.

If this is the delta p what is the corresponding plug velocity? So, v z p is this one. So, now, here in this equation everything known including tau w tau w is 29.

31. So, you substitute all these values simply and then calculate you get it 0. 62 meter per second.

So, second part of the problem is also done. Last part of the problem we have to find out R p. So, phi is equals to R p by R which we got it as 0.

58. So; that means, R p is equals to 0. 58 multiplied by 20 multiplied by 10 power minus 3 which is nothing but R value actually.

So, you get 11. 6 mm see; R is 20 mm whereas, R p is 11. 6 mm that is more than half of the ah cross section of the pipe the material is flowing like a solid plug ok.

So, deformation is taking place only in a small fraction of the pipe cross section. References: This lecture is prepared from this lecture where the final equations are given we have done the derivation and then we have solve the problems. Other useful references are given here.

Thank you.