What is the Required Pump Horsepower for a Slime Delivery System?

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Discussion Overview

The discussion revolves around determining the required pump horsepower for a slime delivery system designed for an amusement park ride. Participants explore the implications of fluid dynamics, particularly focusing on the behavior of slime as a Bingham plastic, the pressure drop in the pipeline, and the necessary horsepower for effective operation. The conversation includes theoretical considerations, calculations, and assumptions related to the system's design.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether the pressure change is zero since both slime tanks are at 1 atm and the same elevation.
  • Another participant suggests that with a yield stress of τo = 2.76 X 10^-4 psi and a fluid velocity of about 3 inches per second, the slime can be treated as a Newtonian fluid, supporting the initial horsepower estimate of 2.86 hp.
  • A different participant emphasizes the need to specify the pressure drop and considers operational factors, such as whether the system needs to maintain flow continuously or can be started and stopped daily.
  • Concerns are raised about the friction factor calculation, with one participant noting discrepancies in their Reynolds number calculation and suggesting that the pressure drop should be significantly higher than initially calculated.
  • Another participant argues against treating the flow as laminar given the high Reynolds number of 13652, asserting that the flow will exhibit characteristics of plug flow due to the nature of Bingham plastics.
  • One participant proposes that the pressure difference across the pipeline will not be zero and should be calculated based on the yield stress and pipeline dimensions.
  • There is a discussion about the appropriate method for calculating the pressure drop for Bingham plastic flow, with references to established equations for laminar flow in such fluids.

Areas of Agreement / Disagreement

Participants express differing views on whether the pressure change can be considered zero and the appropriate treatment of the fluid flow (laminar vs. Newtonian). There is no consensus on the calculations or assumptions regarding the pressure drop and required horsepower, indicating ongoing debate and uncertainty.

Contextual Notes

Limitations include unresolved assumptions about the flow regime, discrepancies in the Reynolds number calculations, and the need for clarity on the friction factor for Bingham plastics. Participants are working through the complexities of non-Newtonian fluid dynamics without reaching a definitive resolution.

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Homework Statement


You have been employed by Sea World to help design rides for their new amusement park in Palo Alto. Your first task is to design a slime delivery system for a "Alien Adventure" ride. The slime is manufactured at Slimy Corp 225 miles away. The connecting pipeline has a 1.0 ft inside diameter. You need 100 gpm of slime. The slime tanks on both ends of the pipeline are at 1 atm and the same elevation. The pipeline contains no valves or bends. The pump efficiency is 63 percent. The slime behaves as a Bingham plastic at 90 F (the pipeline operating temperature) with τo = 2.76 X 10-4 psi, μ= 27 cP and ρ = 87.3 lbm / ft3. Determine the pressure drop in the line and the required pump horsepower.


Homework Equations





The Attempt at a Solution


For this problem, I am not sure if what I said that the change in pressure is zero, because they are both open to the atmosphere. Also, I got 2.86 hp, which seems small for a pump of this magnitude, pushing slime 225 miles
 

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I think with the yield stress of \tau _0 = 2.76 X 10 ^{-4} psi and a fluid velocity of about 3 inches per second, you can almost treat it as a Newtonian fluid. So 2.86hp could be right.
 
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Don't forget to specify the pressure drop.
You might also want to add some horse power to get the flow started. Do they want this flowing 24 hours per day or do they want to start it up every morning and shut it down at night? If they want the 100gpm within 30 minutes of a start-up, then 2.86hp won't do it.
 
Yes, I said the pressure drop is zero, which is something I am not sure about.
 
For a Reynolds number of 13652, I estimate a friction factor of 0.00732. For this friction factor, I calculate a wall shear stress of 0.056x10-4 psi. This is far less than the yield stress of the Bingham plastic. So, something is very wrong with this calculation. The pressure drop should be much higher than the 26 psi that this calculation implies, and the horsepower of the pump has to be much higher.

Try to do the problem assuming laminar flow for a Bingham plastic. The wall shear stress is going to have to be at least as high as the yield stress.

Chet
 
How can it be treated as laminar with a Reynold's number over 13,000? Also, is my analysis about the pressure difference being zero ill-founded?
 
Maylis said:
How can it be treated as laminar with a Reynold's number over 13,000? Also, is my analysis about the pressure difference being zero ill-founded?

This is going to be very different from what you are envisioning. You are going to have a solid plug of Bingham plastic throughout most of the pipe, in which the shear stress does not exceed the yield stress. Only a small amount of the slime will be shearing, and that will be in a small annular region near the tube wall. So the fluid will not be doing anything like you have with turbulent flow. It will be very close to plug flow, and the shear stress at the wall will be very close to the yield stress (although slightly higher). So the only fluid deformation is taking place by shear near the wall, and the characteristic dimension of this shear flow will be much smaller than the tube diameter. The equivalent viscosity of the flow, even near the wall is going to be much higher than the 37 cp slope of the shear stress-shear rate curve. Please try setting this up and solving it for laminar flow of a Bingham plastic, and you will see what I mean. From the equation of motion, the shear stress is equal to the stress at the wall times the local radius divided by the radius of the tube. If you don't feel like solving the equations yourself, the solution for pipe flow of a Bingham plastic is presented in many books.

The pressure difference from one end of the pipe to the other end is going to be far from zero. It is going to be at least as high as the yield stress times 4L/D.

Chet
 
Where do you come up with your friction factor? The Chart we were given for the friction factor of a Bingham Plastic looks like someone drew it by hand, instead of a computer generated plot.

The reason I say that the change in pressure is zero is because when doing my mechanical energy balance, I am choosing the top of the tanks (which are said in the problem statement to be at the same pressure) as my points that I am doing the energy balance on. Why is the pressure difference not zero, when the problem says both are at 1 atm?

Caught an error: the reynold's number is wrong, I divided by the yield stress, not viscosity! I will keep working on it and post my next solution next time I get home to scan it. That is why you got such a strange calculation no doubt.
 

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Ok here is my attachment. I set my point 1 and 2 and labeled them. There is a pump in between these points, so I do not expect to set the shaft work term to zero. Now my problem is that I have two unknowns and 1 equation. I can solve for the friction head using the chart to find f, since I know L, D, and V.

What I actually want to say is since both point 1 and point 2 are at one atm, then the change in pressure is zero, then the work is just the friction term.

I think I am still shoddy on knowing when to set work terms or pressure change terms to zero, even with your explanation. It still isn't clicking for me.
 

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  • #10
Maylis said:
Ok here is my attachment. I set my point 1 and 2 and labeled them. There is a pump in between these points, so I do not expect to set the shaft work term to zero. Now my problem is that I have two unknowns and 1 equation. I can solve for the friction head using the chart to find f, since I know L, D, and V.

What I actually want to say is since both point 1 and point 2 are at one atm, then the change in pressure is zero, then the work is just the friction term.

I think I am still shoddy on knowing when to set work terms or pressure change terms to zero, even with your explanation. It still isn't clicking for me.
Your approach is correct, because the pressure difference is zero between the two points.

Chet
 
  • #11
Maylis said:
Where do you come up with your friction factor? The Chart we were given for the friction factor of a Bingham Plastic looks like someone drew it by hand, instead of a computer generated plot.

The reason I say that the change in pressure is zero is because when doing my mechanical energy balance, I am choosing the top of the tanks (which are said in the problem statement to be at the same pressure) as my points that I am doing the energy balance on. Why is the pressure difference not zero, when the problem says both are at 1 atm?

Caught an error: the reynold's number is wrong, I divided by the yield stress, not viscosity! I will keep working on it and post my next solution next time I get home to scan it. That is why you got such a strange calculation no doubt.

As I said earlier, this is definitely going to be a laminar flow, if you evaluate the reynolds number in terms of the non-Newtonian viscosity (the shear stress divided by the velocity gradient) rather the by using the Bingham "viscosity parameter" (μ =27Cp), which is not the viscosity of the fluid. The actual non-Newtonian viscosity is going to be greater than 200 Cp at all radial locations.

To solve this, you don't need to determine the friction factor, just as, for a Newtonian fluid, you can get the pressure drop/ flow rate relationship (Hagen Poiseuille) without implementing the intermediate step of determining the friction factor.

I looked up the solution for laminar tube flow of a Bingham plastic fluid, and manipulated the results to the following two equations:

(τ_{wall}/τ_0)-\frac{4}{3}+\frac{1}{3}\frac{1}{(τ_{wall}/τ_0)^3}=\frac{(\frac{8V}{D})μ}{τ_0}
μ_{wall}=\frac{(τ_{wall}/τ_0)}{(τ_{wall}/τ_0)-1}μ
where V is the average flow velocity (0.284 ft/sec), μ is the Bingham plastic viscosity parameter, D is the tube diameter, τ0 is the yield stress, and μ_wall is the non-Newtonian viscosity of the Bingham plastic at the wall.

The first equation is used to solve for the ratio of the shear stress at the wall τwall to the yield stress τ_0. In this equation, for your particular problem, the right hand side of the equation is equal to 0.0322 (dimensionless). Once you solve for the ratio of the shear stresses, you can determine the shear stress at the wall, and then, the pressure drop (which is just 4L/D times the shear stress at the wall).

The second equation is used to determine the non-Newtonian viscosity at the wall, once you know the ratio of the shear stresses. Since a Bingham plastic is shear thinning, the viscosity at the wall will be the lowest of all the locations in the tube. You can then use this viscosity to calculate a Reynolds number for the flow and ascertain that the Reynolds number is indeed less than 2100.

Chet
 
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  • #12
He never mentioned anything close to resembling this in class, which is why I'm hesitant to use it, because I don't think the solution would involve equations never taught nor in the textbook. Something fishy is going on!
 
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  • #13
Maylis said:
He never mentioned anything close to resembling this in class, which is why I'm hesitant to use it, because I don't think the solution would involve equations never taught nor in the textbook. Something fishy is going on!
Very interesting. Did you book cover anything like how to solve for the velocity profile and pressure drop of a non-Newtonian fluid in laminar pipe flow? For example, did your book solve the problem for a power-law fluid. If it did, then you follow exactly the same methodology for a Bingham plastic. That leads you to an equation that's in a similar form to the equation I wrote, with coefficients 1, 4/3, and 1/3. What textbook are you using? If you are interested in seeing the derivation of the equations, google laminar pipe flow of a Bingham plastic.

Incidentally, for your problem, I calculate a pressure drop of about 1500 psi.

Chet
 
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  • #14
This is the textbook we use, the index doesn't even have Bingham plastics listed, so I don't think its in the book. Power law fluids aren't in the index either

https://www.amazon.com/dp/0470128682/?tag=pfamazon01-20

Here are the lecture slides for the Bingham Plastic, which is all I have to go on, and it gives a example problem that doesn't follow any of the concepts you were saying as far as I can tell
 

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  • #15
Maylis said:
This is the textbook we use, the index doesn't even have Bingham plastics listed, so I don't think its in the book. Power law fluids aren't in the index either

https://www.amazon.com/dp/0470128682/?tag=pfamazon01-20

Here are the lecture slides for the Bingham Plastic, which is all I have to go on, and it gives a example problem that doesn't follow any of the concepts you were saying as far as I can tell

OK. Your Bingham plastic slides 3 and 4 have what you need. The first equation on the 3rd slide can be worked into the form of the first equation I presented. Both my equation and this equation are implicit in the friction factor (mine is implicit in the wall shear stress, which is equivalent to being implicit in the friction factor). To make things easier for you, they give you the graph on slide 4, which is the explicit solution for the friction factor. Even though slide 4 is labeled Turbulent Bingham Plastic flow, the graph has a laminar flow region (the portions of the lines to the left of the dashed line labeled "transition") that you can use. You can see the word "laminar" written up top on this figure. This is the figure you need to use to get your answer. There is a worked example in your handout that pretty much shows precisely how to work the problem.

When you solve the problem using this figure, you can check your answer against my estimated value of the pressure drop I gave in my previous post.

Chet
 
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  • #16
How could I use the first equation in the 3rd slide (the nonlinear friction factor calculation) to the equation you wrote on post #11? They look vastly different to me
 
  • #17
Maylis said:
How could I use the first equation in the 3rd slide (the nonlinear friction factor calculation) to the equation you wrote on post #11? They look vastly different to me

They do look different, but they're not. You want to know how to convert the equation on slide 3 to my equation. Here's how.

(1) substitute the equations for the Reynolds number and Hedstrom number into the equation on slide 3
(2) substitute the following for the friction factor: f=\frac{τ_{wall}}{\frac{1}{2}ρV^2}
(3) Manipulate the equation mathematically until it is in the form I presented

You can see in the equation on slide 3 that there is a term with an f3 in the denominator; this will become the term in my equation involving the τwall3 in the denominator. You can also see in the equation on slide 3 that there is a term with an f in the numerator; this will become the term in my equation involving the τwall1 in the numerator.

Notice how much more simple the form of my equation is than the form of the equation on slide 3. This is because they forced themselves to work it into a form involving the friction factor, the Reynolds number, and the Hedstrom number. My equation only involves two parameters: the dimensionless parameter on the right side, and the ratio of the wall stress to the yield stress. However, my equation is customized to laminar flow, and is not easily extendable to turbulent flow.

Chet