Crude Oil Flow Rate Calculation: Viscosity & Reynolds Eq.

[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

 

[SteamKing]

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.

 

[cjm181]

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

 

[cjm181]

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

 

[SteamKing]

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.

 

[Ianlamb]

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