Flow & Viscosity: Estimate Max Temp of Crude Oil for 4Kg/s Flow

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SUMMARY

The discussion focuses on estimating the maximum temperature of crude oil required to maintain laminar flow at a rate of 4 kg/s through a 0.3-meter diameter pipe. The viscosity values provided in the table indicate that at 16°C, the viscosity is approximately 8.49 x 10-3 N·s/m2, which corresponds to a Reynolds number of 1999, thus ensuring laminar flow. The specific gravity of crude oil is noted as 0.885, leading to a density of 855 kg/m3. Participants clarified that the maximum temperature was derived from viscosity values in the table and calculations involving Reynolds number.

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cjm181
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Homework Statement


[/B]
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.

temp 0 20 50 100 200
Viscosity 16 7.5 4 2.5 1.5Crude Oil (sg = 0.885)
Viscosity (x10-3 Nsm-2)https://www.physicsforums.com/threads/mechanical-flow-rate.863262/#post-5418137

Homework Equations


A=PIr^2

V=m/(PA)

Viscosity=(Pvd)/Re

The Attempt at a Solution



Area of pipe = PIr^2=PI*0.15^2=0.07069m^2

V=m/(pA)
V=4/(855*0.07069)
V=0.066m/s

So for the pipe of diameter 0.3m, to achieve 4kg/s we need a fluid velocity of 0.066m/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)(0.066)(0.3) / (1999)

Viscosity=0.00849 or 8.49x10^-3Nsm

so max temp would be around 16deg C. Can anyone confirm?

Kr
Craig
 
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Confirmed. Btw, you don't need to know the specific gravity to solve this.
 
Thanks Ches

Can you show me how to do it without SG?

Kr
Craig
 
$$Re=\frac{\rho vD}{\mu}=\rho v \left(\frac{\pi D^2}{4}\right)\frac{4}{\pi D\mu}=\frac{4\dot{m}}{\pi D\mu}$$
 
hi, I'm a new member, I was just hoping someone could explain why 855 has been used in some equations and why it isn't 0.855. Thanks
 
DC83 said:
hi, I'm a new member, I was just hoping someone could explain why 855 has been used in some equations and why it isn't 0.855. Thanks
HI DC83. Welcome to Physics Forums.

0.855 is the specific gravity, which is the density relative to water. The density of water is 1000 kg/m^3. So the density of the oil is 855 kg/m^3.
 
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Hi
cjm181 said:
so max temp would be around 16deg C
I'm just wondering from the post 1 said
Where do we get that Temperature from ? Is that came from the table or something ? as I have not got a clue.
 
jaff90110 said:
Hi

I'm just wondering from the post 1 said
Where do we get that Temperature from ? Is that came from the table or something ? as I have not got a clue.
Can you please explain your question more precisely?
 
I'm sorry , that was my mistake,
From the post # 1 that said the max temperature is around 16 degree C.
I'm wondering that where did cjm181 get that temperature from?
 
  • #10
jaff90110 said:
I'm sorry , that was my mistake,
From the post # 1 that said the max temperature is around 16 degree C.
I'm wondering that where did cjm181 get that temperature from?
He calculated the Reynolds number at the different temperatures based on the viscosities in the table, and then interpolated to find the temperature at which the Reynolds number exceeds 2100. Use semi-log paper to plot the viscosity as a function of temperature. Please show your graph.
 

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