# Transport Processes problem regarding viscosity/fluid flow and energy.

1. Aug 15, 2011

### nefizseal

Hey guys, there is this superhard question (atleast for me). Ive been trying at it for days but I seem to get nowhere.

The Trans-Alaska Pipeline System (TAPS) carries around 100,000m3 of oil per day from the Northern Alaskan oil fields to the nearest ice-free port of Valdez, around 1300km away. The pipe has an outer diameter of 1.22m and a wall thickness of 12mm. Eleven pumping stations are used along the total length of the pipeline to transport the oil.

Note: Assume that the pumping stations are equally spaced along the pipeline, that the pipe is roughly straight and horizontal, and that the flow with the pipe is laminar, Newtonian and steady-state. Also assume that the pumps are 100% efficient so that all energy consumed by the pumps is dissipated by the fluid. The density and kinematic viscosity of the oil are (rho)=890 kg/m3 and (nu)= 7.17 x 10-4 m2 /s respectively.

(a) Starting from first principles, estimate the pressure increase that must be generated by each of the eleven pumping stations to maintain the flow.

(b) The rate that energy is dissipated D(W) by a fluid when it flows through a horizontal pipe under the influence of a pressure difference is given by

D = (delta)P x Q

where (delta)P is the difference in pressure between the inlet and outlet to the pipe (Pa), and Q is the volumetric flowrate through the pipe (m3/s). How much power (rate of energy use) is required to maintain the flowrate of oil through the entire pipeline?

If oil is burned to power the pumps, and 3.6 x 104 MJ of energy can be harnessed from burning 1 m3 of oil, what percentage of the total flowrate needs to be burnt to maintain the flow?

2. Aug 15, 2011

### nefizseal

Well here is what I have done so far (which isnt much to be honest)

I know that the density is 890 kg/m3 and I also know that the kinametic viscosity is 1.17 x 10-4 m2/s

I know from basic principles that kinametic visc. = viscosity/ density

So I can get, absolute viscosity = density x kinametic visc.

So my absolute viscocity comes to 0.638 Pa s = (mu)

I dont exactly know what to do from here. I mean, they said how much pressure do I need to maintain flow. But then there are 11 pumps, do I divide something by 11 1st and then get it and what do I do? Will I need the measurements of the pipe for it?

I dont really want the answer as much as I would want someone to tell me what to do?

And im completely clueless about the 2nd part.....I mean I can solve the whole differential equation thing to get Vx, Vmax, Q etc. for a general flow in a cylindrical pipe, but I dont know what exactly I should do....

Thanks

3. Aug 16, 2011

### nefizseal

Actually, I got the answer to the first part. It was pretty easy. Just had to solve the equation for the Horizontal pipe. Can someone help me with part B? COME ON MAN!!!

4. Aug 22, 2011

### abc007

Given :
Q = 100,000 m3/day * 1 day/24 h * 1 h/60 s = 69.444 m3/s
OD = 1.22 m , ID = 1.196 m , Lt = 1300 000 m , Laminar flow
L = 1300 000 m/11 = 118182 m
µ = 0.638 Pa.s , ρ =890 Kg/m3

Starting from first principles, estimate the pressure increase that must be generated by each of the eleven pumping stations to maintain the flow.

from Hagen-Poiseuille Law can be rephrased as :
Q = π*ID4*(-ΔP)/(8*µ*L)
-ΔP = 8*Q*µ*L/ (π*ID4)
 ΔP = 8*69.444*0.638*118182/(3.14*1.196^4) = 6519921.438 Pa = 65.2 bar

The rate that energy is dissipated D(W) by a fluid when it flows through a horizontal pipe under the influence of a pressure difference is given by
D = (delta)P x Q
D= 6519921.438 * 69.444 = 452.769424 MW

How much power (rate of energy use) is required to maintain the flowrate of oil through the entire pipeline?

power for pump = Q*ΔP
since we have 11 pumps then
Total Power = 11* 6519921.438 * 69.444 = 4980.4637 MW

If oil is burned to power the pumps, and 3.6 x 104 MJ of energy can be harnessed from burning 1 m3 of oil, what percentage of the total flowrate needs to be burnt to maintain the flow?

(4980.4637 M J/s needed)/(3.6 x 10^4 MJ/m^3 provided) ==0.1383 m^3/s

so 0.1383 m3/s of oil have to be burn to provide enough energy for pumps.

% of flow needed to be burned = 0.1383/69.444 *100 = 0.199 %

Please let me know If you find that my answer is wrong .

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook