Crude Oil Flow Rate Calculation: Viscosity & Reynolds Eq.

• cps.13
In summary, the conversation discusses the process of estimating the required maximum temperature of crude oil to flow through a pipe while maintaining a laminar flow. The table provided shows the values for crude oil viscosity and specific gravity. The conversation also touches upon the Reynolds equation for finding laminar flow and the conversion of flow rate to velocity. The conversation ends with a discussion on how to achieve a laminar flow by setting the Reynolds number to 1999 and calculating the required viscosity.
cps.13

Homework Statement

The questions is:
For many liquids the viscosity is strongly dependent on temperature. Use the table below to estimate the required maximum temperature of crude oil to flow at a rate of 4Kg s-1 through a 0.3 metre diameter pipe whilst maintaining a laminar flow.

In the table it has in the rows.

Crude Oil (sg = 0.885)
Viscosity (x10-3 Nsm-2)

Does anybody know what the (sg =0.855) means? There is no explanation of this.

2. Reynolds equation for finding laminar flow

Reynolds # = (density)(velocity)(pipe diameter) / (viscosity)

density = 860kg/3
velocity = ??
pipe diameter = 0.3m
viscosity = 0.016 (others are given in the table but I cannot paste that in here, for now assume 0.016)

The Attempt at a Solution

[/B]
I think I know how to do it, what I am struggling with is converting my 4Kg s-1 to a use able value to get my velocity.

I am trying...

flow rate = area x velocity
therefore
velocity = flow rate/area
velocity = 4/(0.15xπ2) = 56.588

this then goes into my reynolds number equation

(860)(56.588)(0.30) / 0.016 = 912481.5

this answer is far too high as it should be less than 2000!

I'm guessing there is something wrong with my conversion from flow rate to velocity but cannot figure out what I'm doing wrong. Think it is do with the Kg s-1. Needs to be converted first but not sure how!

Thanks

cps.13 said:

Homework Statement

The questions is:
For many liquids the viscosity is strongly dependent on temperature. Use the table below to estimate the required maximum temperature of crude oil to flow at a rate of 4Kg s-1 through a 0.3 metre diameter pipe whilst maintaining a laminar flow.

In the table it has in the rows.

Crude Oil (sg = 0.885)
Viscosity (x10-3 Nsm-2)

Does anybody know what the (sg =0.855) means? There is no explanation of this.

2. Reynolds equation for finding laminar flow

Reynolds # = (density)(velocity)(pipe diameter) / (viscosity)

density = 860kg/3
velocity = ??
pipe diameter = 0.3m
viscosity = 0.016 (others are given in the table but I cannot paste that in here, for now assume 0.016)

The Attempt at a Solution

[/B]
I think I know how to do it, what I am struggling with is converting my 4Kg s-1 to a use able value to get my velocity.

I am trying...

flow rate = area x velocity
therefore
velocity = flow rate/area
velocity = 4/(0.15xπ2) = 56.588

this then goes into my reynolds number equation

(860)(56.588)(0.30) / 0.016 = 912481.5

this answer is far too high as it should be less than 2000!

I'm guessing there is something wrong with my conversion from flow rate to velocity but cannot figure out what I'm doing wrong. Think it is do with the Kg s-1. Needs to be converted first but not sure how!

Thanks

SG = specific gravity of the liquid. The reference fluid for SG is fresh water, which has a SG = 1.000 by definition.

https://en.wikipedia.org/wiki/Specific_gravity

Your velocity calculation cannot produce a valid velocity. Hasn't anyone shown you how to check units in calculations?

If the flow is in kg/s, then dividing this number by the area in m2 will not yield a velocity in m/s. You need to use the density of the oil to convert mass flow (in kg/s) into volumetric flow (in m3/s) before dividing by the area of the pipe. Remember, density = mass / volume and has units of kg / m3.

Hey CPS,

I am actually going to be starting this question on the next few days. I have just completed the first question

kr
Craig

using:
Area of pipe = PIr^2=PI*0.3^2=0.28274334
V=m/(pA)
V=4/(855*0.28274334)
V=0.01654632m/s

So for the pipe of diameter 0.3m, to achieve 4kg/s we need a fluid velocity of 0.017m/s. (sounds really slow?)

Then
Reynolds # = (density)(velocity)(pipe diameter) / (viscosity)

Viscosity=(density)(velocity)(pipe diameter) / (reynolds)

so if the flow is to be laminar, set reynolds no to 1999

Viscosity=(855)(0.017)(0.3) / (1999)

Viscosity=0.00212313 or 2.12x10^-3Nsm

This is as far as i am, but looking at the numbers i thinks the temp will be around 110 to 120?

kr
Craig

cjm181 said:
using:
Area of pipe = PIr^2=PI*0.3^2=0.28274334
V=m/(pA)
V=4/(855*0.28274334)
V=0.01654632m/s

So for the pipe of diameter 0.3m, to achieve 4kg/s we need a fluid velocity of 0.017m/s. (sounds really slow?)

Then
Reynolds # = (density)(velocity)(pipe diameter) / (viscosity)

Viscosity=(density)(velocity)(pipe diameter) / (reynolds)

so if the flow is to be laminar, set reynolds no to 1999

Viscosity=(855)(0.017)(0.3) / (1999)

Viscosity=0.00212313 or 2.12x10^-3Nsm

This is as far as i am, but looking at the numbers i thinks the temp will be around 110 to 120?

kr
Craig
Craig:

I understand you're working on a similar problem as the OP, but it's against the rules at PF to hijack threads, even in this instance.

If you want help on a similar problem, start your own thread. You can always reference another thread at PF created by a different user.

That said, your calculations have a very silly mistake from the get go. You have assumed that the radius of the pipe = diameter of the pipe.

That's one reason your calculations don't make sense.

Hey cjm181,
I'm stuck on that first question. I presume it is the one about ideal straight lines? Any chance of a few pointers. Thanks

1. What is the purpose of calculating crude oil flow rate?

The purpose of calculating crude oil flow rate is to determine the amount of oil that is being transported through a pipeline or well. This is important for monitoring production, predicting future yields, and ensuring that the flow is within safe and efficient operating limits.

2. What is the role of viscosity in crude oil flow rate calculation?

Viscosity is a measure of a fluid's resistance to flow. In the case of crude oil, it refers to how easily the oil can flow through a pipeline or well. A higher viscosity means the oil is thicker and will flow more slowly, while a lower viscosity means the oil is thinner and will flow more quickly. Therefore, viscosity plays a crucial role in determining the flow rate of crude oil.

3. How is the Reynolds equation used in crude oil flow rate calculation?

The Reynolds equation is a mathematical formula used to calculate the flow rate of a fluid through a pipe or channel. It takes into account the fluid's viscosity, density, and velocity, as well as the diameter of the pipe and the pressure drop along its length. In the case of crude oil flow rate calculation, the Reynolds equation is used to determine the flow rate of oil through a pipeline or well, taking into account the oil's viscosity and other relevant factors.

4. What are the units of measurement for crude oil flow rate?

Crude oil flow rate is typically measured in barrels per day (bpd) or cubic meters per day (m3/d). These units represent the amount of oil that is flowing through the pipeline or well in a 24-hour period. In some cases, flow rate may also be measured in gallons per minute (gpm) or liters per second (l/s).

5. How accurate are crude oil flow rate calculations?

The accuracy of crude oil flow rate calculations depends on a variety of factors, including the accuracy of the data used, the assumptions made, and the complexity of the calculation method. In general, these calculations can provide a good estimate of the flow rate, but they may not be exact due to variations in the oil's properties and other external factors. It is important to regularly monitor and update flow rate calculations to ensure accuracy and make any necessary adjustments to maintain efficient and safe operations.

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