# Pressure drop at power station pumping power law fluid

• gfd43tg
In summary: P3 - P2)/ρ = hfP3 = P2 + hfρ... which would make the pressure at the end of the 68.3 mile section (P3) greater than the pressure at the start of the 68.3 mile section (P2)?In summary, the Black Mesa pipeline transports a 50 wt % slurry of solid coal and water at a rate of 660 tons/hr for 273 miles through northern Arizona. The pipeline has an 18 inch ID and is divided into four sections with four pumping stations. The slurry behaves as a power law fluid with n=0.2 and K=0.58 kg/m s^(2-n). Using Bernoulli's equation
gfd43tg
Gold Member

## Homework Statement

The Black Mesa pipeline transports 660 tons / hr of solid coal ground to 8 mesh (2.4 mm) as a 50 wt % slurry in water with an estimated sg of 1.26 for 273 miles across northern Arizona. The pipeline ID is 18 inches and the slurry flow rate (coal and water) is 4200 gal/min. There are four pumping stations which divide the 273 miles into four sections. The slurry behaves as a power law fluid with n = 0.2 and K (kg / m s (2-n) )= 0.58.

Determine:
a) The pressure drop in each 68.3 mile section.
b) The total pumping power required for all four stations assuming the pump efficiency is 70 percent.

## The Attempt at a Solution

For this problem, I calculated the reynold's number and found that it is turbulent flow. Then I used the attached chart to find the fanning friction factor, which I knew I needed for the head friction for Bernoulli's equation. From there, I can't find the pressure drop using any experimentally derived formulas since those deal with laminar flow. At this point I wasn't sure how to calculate the pressure drop, so I took a peek at the solution.

What they do is use bernoulli's equation (I was doing the same), but the peculiar thing that was done was setting the work per unit mass to zero. I was under the impression that there is a pump, therefore W should not be zero.

I posted the solution, which corresponds to question 3 on that set
http://www.cchem.berkeley.edu/cbe150a/Old HW/HW2-Solutions.pdf

#### Attachments

• 2.3 attempt 1.pdf
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• turbulent friction factor power law fluid.docx
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Maylis said:

## Homework Statement

The Black Mesa pipeline transports 660 tons / hr of solid coal ground to 8 mesh (2.4 mm) as a 50 wt % slurry in water with an estimated sg of 1.26 for 273 miles across northern Arizona. The pipeline ID is 18 inches and the slurry flow rate (coal and water) is 4200 gal/min. There are four pumping stations which divide the 273 miles into four sections. The slurry behaves as a power law fluid with n = 0.2 and K (kg / m s (2-n) )= 0.58.

Determine:
a) The pressure drop in each 68.3 mile section.
b) The total pumping power required for all four stations assuming the pump efficiency is 70 percent.

## The Attempt at a Solution

For this problem, I calculated the reynold's number and found that it is turbulent flow. Then I used the attached chart to find the fanning friction factor, which I knew I needed for the head friction for Bernoulli's equation. From there, I can't find the pressure drop using any experimentally derived formulas since those deal with laminar flow. At this point I wasn't sure how to calculate the pressure drop, so I took a peek at the solution.

What they do is use bernoulli's equation (I was doing the same), but the peculiar thing that was done was setting the work per unit mass to zero. I was under the impression that there is a pump, therefore W should not be zero.

I posted the solution, which corresponds to question 3 on that set
http://www.cchem.berkeley.edu/cbe150a/Old HW/HW2-Solutions.pdf
I looked over what you did, and it looks OK. As far as their solution is concerned, I don't see where they set the work per unit mass to zero. It looks like they did just about the same thing that you did. What they calculated was the shaft work that would need to be done.

I'm interested in the formula that you used for the Re. This looks similar to the Metzner Reed Re, but not quite. I used a similar approach to get the Re (based on the viscosity evaluated at the laminar flow shear rate at the wall), and it came out to be about 22000. I see where most of the factors in your Re come from, but not the first two.

chet

Yes, the crux of my question is the following: How can the shaft work be zero if a pump is present? Isn't a pump doing shaft work?? Do you see the mechanical energy balance, where I set W to zero and equate the change in pressure over density to the friction head? That is what the solution says, but I don't understand how W is zero if there is a pump present.

Below is the equation for the Reynold's number of a power law fluid given to us, since you are interested.

#### Attachments

• Reynold's number Power Law Fluid.png
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Last edited:
Maylis said:
Yes, the crux of my question is the following: How can the shaft work be zero if a pump is present? Isn't a pump doing shaft work?? Do you see the mechanical energy balance, where I set W to zero and equate the change in pressure over density to the friction head? That is what the solution says, but I don't understand how W is zero if there is a pump present.

Below is the equation for the Reynold's number of a power law fluid given to us, since you are interested.

Oh. Now I understand. It all depends on which two points you choose for applying the Bernoulli equation. If you choose the two ends of the pipeline between the exit of the upstream pump and the entrance to the next downstream pump, there is no pump in-between, so the shaft work between them is zero. You get the equation that both you and they got.

If you choose the inlet to the upstream pump and the inlet to the downstream pump, then there is the upstream pump in-between, but the Δp/ρ between these two points is zero. This then says that the shaft work per unit mass is equal to the frictional loss per unit mass. If you choose the inlet to the upstream pump and the exit of the upstream pump, then the frictional loss is zero, and the shaft work per unit mass is equal to Δp/ρ across the pump (which the same as the Δp/ρ down the pipe).

Chet

If I'm saying that the shaft work is zero in calculating part a, then would it not be the case that the answer to part b is zero as well?? The way I interpret it, we are saying the work is zero, and then we're saying its not zero. It doesn't make sense to me.

Also, since no diagram is present, I don't really know where I should choose my two points to do my mechanical energy balance. Apparently where no pump is present.

Last edited:
Maylis said:
If I'm saying that the shaft work is zero in calculating part a, then would it not be the case that the answer to part b is zero as well?? The way I interpret it, we are saying the work is zero, and then we're saying its not zero. It doesn't make sense to me.

Also, since no diagram is present, I don't really know where I should choose my two points to do my mechanical energy balance. Apparently where no pump is present.

What you do is, you first choose the two points between two of the pumps, and get the pressure drop in the section of pipe. Next you choose the two points right before and right after one of the pumps. The pressure change across each pump is equal to the pressure drop in the section of pipe between pumps. From the first law of thermo, the change in enthalpy per unit mass of fluid passing through the pump is equal to the shaft work per unit mass (since the pump is assumed adiabatic). Also, since the temperature rise across the pump is going to be close to zero (assuming negligible viscous heating in the pump), the change in enthalpy per unit mass is just equal to the Δ(PV)=VΔP. So the shaft work per unit mass is just VΔP.

Chet

## What is pressure drop?

Pressure drop is the decrease in pressure that occurs as fluid flows through a system, such as a power station. This can be caused by friction, changes in elevation, or obstructions in the system.

## How is pressure drop calculated?

The pressure drop at a power station for a power law fluid can be calculated using the Darcy-Weisbach equation, which takes into account the fluid's viscosity, velocity, and the characteristics of the system, such as pipe diameter and length.

## What is a power law fluid?

A power law fluid is a non-Newtonian fluid, meaning its viscosity is not constant and can change with shear rate. This type of fluid follows a power law relationship between shear stress and shear rate, rather than the linear relationship seen in Newtonian fluids.

## Why is pressure drop important in a power station?

Pressure drop is an important consideration in a power station because it can affect the efficiency of the system. A high pressure drop can result in energy loss and increased pumping costs, while a low pressure drop can lead to inadequate flow and potential equipment damage.

## How can pressure drop be reduced?

Pressure drop can be reduced by increasing the diameter of pipes, decreasing the length of the system, and using smoother materials for the walls of the pipes. Proper maintenance and cleaning of the system can also help to minimize pressure drop.

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