Losses through long annulus. 1

[dbecker]

Hello,

If I have an annulus that has a large ratio of annulus radius to annulus gap (meaning the gap is small compared to the radius) what is the method of calculating K?

I am assuming that K = f (L/D) where D is hydraulic diameter.

The problem I see though is that the flow area is not well represented with hydraulic diameter and this will be problematic if i use equation 3-20 of cranes because it requires a value for "D^4". And D is hydraulic diameter which may be many times smaller than what is represented by the actual flow area. Do you see what I mean?

Thank you!!

 

[BigInch]

That's exactly why hydraulic diameter is used. With a very very small annular space, you can see immediately that you have doubled the frictional surface area of the annular flow.

Start with a 10" ID pipe and another pipe inside with an interior radius of 0, then begin increasing the interior radius to almost 10". Annular flow contact surface (friction) increases as a function of the square of the interior radius, while at the same time flow area decreases by the square of the interior radius. Thus, trying to maintain the same flow Q, velocity Q/A is being squared and friction is being squared again, so looks like D^4 pretty well covers the expected effect on head loss in an annular space.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[dbecker]

Thank you for the explanation.

So is it safe for me to use Hydraulic diameter in equation 3-20 as Dh^4? Even if my Dh = 1" and my inner and outer radii can be on the order of 100+ inches (very large flow area but small hydraulic diameter Dh=1")?

 

[BigInch]

How safe is safe?
You could consider this method,

http://www.eng-tips.com/faqs.cfm?fid=1142

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[cvg]

sounds a lot like this thread...

thread378-235573

 

[dbecker]

Thank you for the useful material.

But I am still confused about something,

I made a quick spreadsheet on annulus geometry and hydraulic diameter.

I start with a D=100" ID outer pipe and work my way up from D=0" inner pipe back up to 100".

My results are the following


Dout Din Aactual Dequiv Dh A(Dh)
100 0 7853.98 100.00 100.00 7853.98
100 5 7834.35 99.87 95.00 7088.22
100 10 7775.44 99.50 90.00 6361.73
100 15 7677.27 98.87 85.00 5674.50
100 20 7539.82 97.98 80.00 5026.55
100 25 7363.11 96.82 75.00 4417.86
100 30 7147.12 95.39 70.00 3848.45
100 35 6891.87 93.67 65.00 3318.31
100 40 6597.34 91.65 60.00 2827.43
100 45 6263.55 89.30 55.00 2375.83
100 50 5890.49 86.60 50.00 1963.50
100 55 5478.15 83.52 45.00 1590.43
100 60 5026.55 80.00 40.00 1256.64
100 65 4535.67 75.99 35.00 962.11
100 70 4005.53 71.41 30.00 706.86
100 75 3436.12 66.14 25.00 490.87
100 80 2827.43 60.00 20.00 314.16
100 85 2179.48 52.68 15.00 176.71
100 90 1492.26 43.59 10.00 78.54
100 95 765.76 31.22 5.00 19.63
100 100 0.00 0.00 0.00 0.00


What we see here is that the actual area (Aactual) of the annulus and the area due to hydraulic diameter (ADh) diverge GREATLY.

On a purely flow area basis, the flow rate CANNOT be correct if hydraulic diameter is used for a very short annulus (low friction). Since friction for a short annulus is negligible and the flow rate is almost entirely dependant on the flow area, hydraulic diameter used in Eq 3-20 for a short annulus would give me an innaccurate solution. Please review this and get back to me as I am curious about this topic.

Thank you

 

[dbecker]

All,

I have figured out a way to use Eq. 3-20. I use Equivalent Diameter instead of Dh.

Equivalent Diameter is the actual area times 4 over pi. This is going to give me a pseudo pipe diameter to use for Eq. 3-20.

I verified the legitimacy of this method with Flowmaster V6.5.2. Here are the results.

K Y w EquivD w Dh Flowmaster
2 0.69 1230.72 31.56 1097.47
10 0.81 641.42 16.45 626.495
20 0.83 465.76 11.94 463.189
40 0.84 332.77 8.53 335.657
100 0.84 212.38 5.45 215.594

w EquivD and Flowmaster are 12% different with low K values, this is expected because flowmaster probably uses some entrance and exit effects on the component whereas I am using purely K = 2 and Eq 3-20.

When K increases, both flowmaster and the EquivD method converge and become more accurate, because now friction is dominant and it dilutes the innaccuracy of the entrance effect.

I think this seems valid, no?

Thank you,

 

[vpl]

You mention equation 3-20 of cranes. My version of Crane Technical Paper 410 "Flow of Fluids Through Valves, Fittings, and Pipe" (or the Metric version, which is Technical Paper 410M) has on the bottom right-hand side of page 3-5 a note which says "See page 1-4 for limitations." When I go to page 1-4 and the discussion on hydraulic radius it says "This applies to any ordinary conduit (circular conduit not flowing full, oval, square, or rectangular) but not to extremely narrow shapes such as annular or elongated openings, where width is small relative to length. In such cases, the hydraulic radius is approximately equal to one-half the width of the passage."

If you were looking at page 3-4 (where equation 3-20 resides), it's a shame you couldn't have read one page over...


Patricia Lougheed

Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of the Eng-Tips Forums.

 

[dbecker]

I understand the limitations. Which is why I did the calculations, if you read those you will see that hydraulic diameter nor hydraulic radius cannot be used in Eq 3-20 and I am resorting to augmenting D with an effective area or an equivalent diameter. I just want to know the validity of this, if anyone has done this?

That is what my inquiry is about, calculating flow rate.

Yes, page 1-4 says hydraulic radius is equal to one half the width of the passage. This still doesnt answer my question as to what my value of D should be in Eq 3-20 (should not be hydraulic diameter). Actually, I sort of answered my own question.


Yes it is a shame I didn't read one page over, but I am not an imbecile.

Thank you,

 

[hydtools]

dbecker, wouldn't your equivalent diameter be the square root of (area x 4/pi) ?

Ted

 

[katmar]

I agree that for flow in an annulus the hydraulic diameter will give wrong answers if a pipe of a diameter equal to the hydraulic diameter is substituted for the annulus. The "secret" of how to apply the hydraulic diameter is hinted at in the last paragraph on page 1-4 in Crane 410. The diameter used for calculating the equivalent K value for the pipe (i.e. K = f(L/D) ) and the diameter used for calculating the fluid velocity are NOT the same.

An important factor in determining the pressure drop along a pipe is the ratio of the cross sectional area to the wall area because it is the wall that provides the friction. The hydraulic diameter concept provides a value for a regular round pipe that would have the same ratio of cross sectional area to wall area as the particular annulus. The hydraulic diameter is therefore the right thing to use to calculate the K value for the annulus.

However, this does not mean that we should regard all the fluid flowing down the annulus as flowing through a pipe of the hydraulic diameter. The average velocity in the annulus is easily calculated, and it is this velocity which should be applied to the K value calculated for the pipe of equivalent hydraulic diameter. If you want to use a standard pipe pressure drop spreadsheet or program to calculate the pressure drop then simply divide the cross sectional area of the annulus by the cross sectional area of the pipe with the equivalent hydraulic diameter to get the fraction of the flow that should be used in the calculation. This is difficult to put into words but will be obvious in the example below. Another way of looking at the annulus is as a bunch of pipes in parallel, where each pipe has the hydraulic diameter of the annulus.

I believe that calculating an equivalent pipe diameter on the basis as having the same cross sectional area as the annulus (as per your example earlier) will significantly under estimate the pressure drop because the wall area of such a pipe would be much less than that in the annulus.

As an example I will continue with the example you have already started. I took the outer pipe as 100" ID and the inner pipe as 99" OD. To avoid the complication of the Y factor I have assumed the fluid to be water flowing at 6 ft/sec in the annulus. The annulus has a cross sectional area of 156.3 inch² and a wetted perimeter of 625.2 inch. At 6 ft/sec this gives a flow rate of 2920 USgpm. In all the calculations below I have assumed the absolute roughness of the pipe is 0.002 inch.

Alternative 1 - Pipe of same cross sectional area as annulus
A standard round pipe of 14.1" ID would have the same cross sectional area as the annulus. Water at 6 ft/sec gives Re = 655000, f_M = 0.01446, Pressure drop = 0.2985 psi/100ft.

Alternative 2 - Pipe of Hydraulic Diameter
The hydraulic diameter of the annulus is 1", and a pipe with this diameter will have a cross sectional area of 0.7854 inch². It would require 156.3 / 0.7854 = 199 pipes of this diameter to provide the same cross sectional area as the annulus. We can regards the flow in the pipe as being 2920 / 199 = 14.67 USgpm. This gives Re = 46400, f_M = 0.02663, Pressure drop = 7.73 psi/100ft.

Alternative 3 - Pipe of Equivalent Diameter according to "Petroleum Engineering" method
This method is described by ZDAS04 in faq378-1142. D_e = (199² x 1³)^0.2 = 8.3 inch. Using the full flow of 2920 USgpm this gives Re = 1113000, f_M = 0.015, Pressure drop = 4.4 psi/100ft, Velocity = 17.3 ft/sec.


Comparing these three alternatives I would regard the first one as totally unrealistic. But I find it amazing that the third method is able to predict an equivalent diameter that gives a value that is usable for calculating a friction factor and a velocity that gives an answer so close to alternative 2.

Some other references (eg Transport Phenomena by Bird, Stewart and Lightfoot) give another way of calculating the equivalent diameter. This is D_e = R x (1 - k) where R = Outside diameter of inner pipe, and k = Outside diameter of inner pipe / Inside diameter of outer pipe. If this diameter is used in the same way as Alternative 2 above, very similar results are obtained as with the hydraulic diameter.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[zdas04]

Annular gas flow is a ubiquitous phenomena in Oil & Gas. I've measured actual dP in a lot of gas flows and the "Petroleum Method" that I describe in the FAQ always gets calculated values acceptably close to measured values.

Let's move into a world where you can purchase pipe (never seen 99" diameter pipe). Let's say you have a 12" pipe (OD 12.75") inside a 16" Sched 30 pipe (ID 15.250"). Let's assume that the pressure at the head of 1 mile of pipe is 100 psia and we're flowing 1 MMCF/d.
MMCF/d.

The wetted perimeter is 7.33 ft. This is the key factor, but it isn't simple. The flow at the surface of the pipe is zero. More surface area means more of the flow is subject to surface friction. It also means that for a given mass flow rate, the velocity will be less than for a round pipe with the same cross sectional area. Lower velocity means less friction. So the goal needs to be to find a pipe diameter that can be used in the standard flow equations that balance these competing factors.

Using simple difference in cross sectional area, you would get a 8.367 inch diameter and the wetted perimeter would be 2.19 ft. The downstream pressure (using the AGA equation and a pipe roughness of 0.0015) would be 99.9 psia

Using the Hydraulic Diameter, the effective diameter would be 2.5 in, and the wetted perimeter would be 0.654 ft. Downstream pressure would be 54.3 psia.

Using the Petroleum Method, the effective diameter is 6.571 in, and the wetted perimeter is 1.72 ft. Downstream pressure would be 99.754 psia.

None of the three calculations is anywhere close to the actual surface area, but the simple difference in area will understate dP, Hydraulic diameter will overstate it, and the Petroleum Method will be the closest for every dP I've ever measured.

Nobody has ever built this line or anything like it, but I would be very confident designing it using the Petroleum Method, in this case using the simple difference method would probably be ok, using the hydraulic diameter would be dumb.

I would expect liquid-full pipes to act like a gas (i.e., the Petroleum Method should match measured data closest). Partially full pipes will always behave different from gas.

David Simpson, PE
MuleShoe Engineering

http://www.muleshoe-eng.com

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"Everyone is entitled to their own opinion, but not their own facts" Patrick Moynihan

 

[dbecker]

Great info guys, give me some time to review these methods.

Thanks for the support!!!!

 

[dbecker]

In response to hydtools, yes you are right it is square rooted!! Typo on my part on my previous post.


Thanks!

 

[hydtools]

Hydraulic diameter is defined as the flow area divided by the wetted perimeter.
In the case of an annulus: hydraulic diameter = (D^2-d^2)/4*(D + d) or (D-d)/4.

Area = (pi/4)*(D^2-d^2)
Wetted perimeter = pi*D + pi*d
Also D-d = diametral clearance
Hydraulic diameter = 1/4 * diametral clearance = 1/2 * radial clearance

Ted

 

[zdas04]

Not according to my reference books (see Lindberg, page 36-8 for example). The equation you referenced is the one that Patricia pointed out that on Page 1-4 of Crane 410 that it is not applicable to annular flow. Lindberg referenced the Nusalt derivation to come up with:

HydDiam = 4{area)/(wetted Perimeter)

Area = {pi/4)(D^2-d^2) = (pi/4)(D-d)(D+d)

Wetted Perimeter = pi(D+d)

HydDiam = 4(pi/4)(D-d)(D+d) / [pi(D+d) = D-d

If you use the Area/Perimeter formulation then the area (which the Lindberg formulation results in a diameter that is far too small) is 1/4 as large.

Since the Hydraulic Diameter method is ubiquitous, I would guess that in partially full ducts it might have real value. In totally full annular spaces it is worse than worthless.

David

 

[25362]

The definition by hydtools corresponds to the hydraulic radius, not the hydraulic diameter.

 

[katmar]

zdas04 - thanks for the input and your more feasible example. I haven't seen any 99" pipe either!

The point I was trying to make is that when using the Hydraulic Diameter method it is wrong to assume that the full flow is flowing through a standard round pipe of diameter equal to hydraulic diameter. This gives nonsense results, as you have shown. It is necessary to use the actual velocity in the annulus together with the K (=fL/D) value from the hydraulic diameter.

Applying this to your example with 1 MMSCF/d - the cross sectional area of the annulus is 54.98 inch² and the cross sectional area of the 2.5" pipe (i.e. hydraulic diameter) is 4.909 inch². This ratio is 54.98/4.909 = 11.2. The flow through the pipe should therefore be taken as 89285 SCF/d (=10^6 / 11.2) to give the same velocity as in the annulus. On this basis the downstream pressure is 99.63 psia which is much more reasonable.

The flow of 1 MMSCF/d gives a velocity in the annulus of only 4.4 ft/sec so the pressure drops are small and differences between the methods are not so easily seen. If we increase the flow to 5 MMSCF/d the pressure drop using the Petroleum Engineering method would be 6.1 psi and the pressure drop using the Hydraulic Diameter method (assuming flow is 446400 SCF/d (=5x10^6/11.2) ) is 7.8 psi. This is a small difference and I believe both are realistic.

From a theoretical standpoint I prefer the Hydraulic Diameter method because is has the same cross sectional area to wall area ratio as the annulus, and the fluid velocity is the same whereas the Petroleum Engineering method is completely empirical (as far as I know). If you have any real life examples where you could test this it would be very interesting.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[katmar]

hydtools and 25362 - my references also use this definition for hydraulic radius, and then define hydraulic diameter as 4 times the hydraulic radius.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[BigInch]

This looks interesting,

http://www.roughneckcity.com/uploads/Bit_Hydraulics.xls

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/