Pressure Loss Through Pipe Size Reduction

You will experience a pressure drop equal to the length of restricted pipe.

There will be numerous opinions on this, I predict.

The sprinklers won't care much, if this is all you've changed.

Without doing calculations, it looks as though the largest component of pressure drop would occur in the 500-meter runs of 50-mm pipe. Another 4 meters (plus losses on reducing and increasing again) ought to be very small by comparison. I doubt you will notice the difference.

If it is only a four meter span just take some of the old pipe that is being replaced and use two sections of that under the new pipe at the creek crossing to make a simple support bridge for the poly pipe.

Agree with Tornado, will make minimal difference.

This site is very useful and the software is free, you may be able to model your system.

Thanks for this link; it looks pretty useful.

Without looking at a flowrate it is not possible to quantify it.

Exactly. He hasn't mentioned the most important thing!

The percentage change in pressure drop from adding 4m to 500m of pipe will be about the same (i.e., insignificant) regardless of the initial flow and pressure drop. So no, we don't really need to know the original Q and ΔP.

It is not necessary to know the flow for a relative estimation of the pressure drop variations: Δp ≈ Q^2*L/(ID)^4 It can be assumed that the small modification will not change the flow in an important manner so that flow conditions will be same. To compare the 2 configurations it is enough to compute the sum: Σ (Qi^2*Li/(IDi)^4 for each one and make the ratio of results. Applying this equation the results is that the replacement of the old pipe (4m of 80mm) with the new ( 4m of 63 mm) leads to an increase of 0.59%. This modification is very small and whichever the pump characteristic is the flow will not be affected very much.

Using your equation, Δp₂/Δp₁ ≈ ID₁⁴/ID₂⁴ (since Q and L are the same) which is about 2.6, so Δp is 2.6 times higher for 63mm than for 80mm. This is in close agreement with 2½" & 3" SCH40 steel pipe, where Δp for 2½" is about 3 times higher than for 3" at the same flow.

However a reduction in size for only 4 m should not cause an appreciable impact to the delivered pressure. Even if the piping losses are the full 25m of head, that is only about 2m loss per 100m of 80mm pipe, or 0.08m loss for 4m of 80mm pipe. Using the above ratio, the loss would increase to 0.21m for 4m of 63mm pipe, which adds only 0.13m to the total system loss.

I would be more concerned with determining why the joints are failing - aged piping, poor workmanship, water hammer, UV degradation (if plastic)?

1- it is not MY formula but a generally used equation in flow -pressure drop analysis.

Δp= k L v² = k L Q²/A² and for a round section A= pi/4*D² thus Δp= k1*L*Q²/ID^4

k depends on turbulence, wall roughness and fluid viscosity.

2- In stead of guesses you may proceed this way perhaps easier to follow:

The length of 500m ID=50 but with Q/2 as flow is equivalent to 819 m of 80 mm tube

Thus total length "equivalent" in 80 mm tube is 260+819=1079m

The 4 m tube ID=63 will have an equivalent length in 80mm of 10.4m

Thus the increase in equivalent 80 mm will be 10.4-4=6.4 m

This represents a total increase of pressure drop (or tube length) of 6.4/1079=0.00593 or 0.593% of initial value.

I do not know why so many engineers prefer to guess in stead of a short and simple estimations which gives numbers. The human brain thinks and extrapolates "linear" and many physical phenomenon we are confronted in engineering are NON-LINEAR. So that "guestimations" can be very wrong if there are out of a narrow domain where a linear approximation is possible. Experience is a false friend due especially to the trend to extrapolate unfortunately many times out of allowed range!

Not that it matters to the end result, but I think you have seriously underestimated the equivalent length of 80 mm pipe for 500 m of 50 mm pipe - should be 1374 m instead of 819 m.

Of course I could make an error but how did you compute the equivalent length?

My computation is

Leq = L1*(Q1/Q2)^2*(ID2/ID1)^4 = 500*1/4*(80/50)^4 = 819.2

and yours ? If you make such a statement please give as well the basis of it if not it is your "estimation" which is not valid. It will be a pity if I cannot correct my error.

Thanks for the attention you gave my comment

nick name

The guest was right. In the basic equation the friction loss is considered for a pipe having same length as ID. This is the reason why the ration L/ID in stead of L appears.

Considering this aspect I did not take care of the equation I used has to be modified the ID ration being at power 5 in stead of 4.

The equivalent length for the 50 mm pipe will be then : 500* (1/2)^2*(80/50)^5= 1310.72 m and not as I first computed 819.2

The 63 mm length will be : 4*(80/63)^5= 13.21m The variation is then 13.21-4= 9.21m.

With respect to a comment considering the losses at the changes from 80 to 63 and from 63 to 80. Both are really very small and can be even more reduced if the 63 mm tube has a chamfer. It is difficult to estimate their "relative value" since for their value flow magnitude has an impact different the one in the tube case.

The original length is 260+1310.7= 1570.7 and the new length 1579.9 which represents an increase of 0.586% not 0.59 as first indicated.

As the comment wrote it does not change the result BUT I hate to use an error equation and wanted to give the correct result for all interested persons.

I am sorry for the first -too fast- response.

Whenever a diameter change occurs there are entry and exit losses to be added to the length. For a given pipe the pressure drop increases as the square of flow rate. My guess is that for normal municipal water flow rates in supply pipes, the drop is only about 0.2 bar/100 m for a 63mm pipe. This is negligible for a 4 m length.