# Pressure drop at power station pumping power law fluid

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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Mentor

## 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

Gold Member
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:
Mentor
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

Gold Member
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:
Mentor
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