To equal the flow area of two six inch, multiply 6 by square root of 2.

Ted

You really need to use ID, but if you assume 6" is the ID then the basic rule to calculate the change of one thing without changing any other is (D1/d2)^5.

Assuming what you mean is you have the same pressure drop and overall you want the same flow through the system, then the flow rate of one 6" pipe needs to be 0.5 of your bigger pipe.

(0.5)^0.2 = 0.87.

Hence your bigger pipe is 6/ 0.87 = 6.9 ID.

To do it right you need the ID and actual pressure drops/unit length.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

My inside diameter is 6 inch. The i.d. of 7 is 7 inch. The i.d. of 8 is 8 inch. Here is a pressure drop chart. I would like to pump water 2 miles at the flow of 1250 gpm. I was thinking of running 2-6 inch lines the 2 miles, but if those 2 lines equal 1-7 inch line then I will probably just run the one 7 inch line. On the other hand if the 2 lines will produce more than the 7 inch hose, I would like to run the 2 6's. I have some 6 inch hose but not enough. Should I buy more 6 inch hose and run 2 lines or upgrade to the 7 inch hose and run one line? Hope this is the info you are needing, LittleInch. Thanks!

• http://files.engineering.com/getfile.aspx?folder=08715bb4-5a30-41e8-b1a0-82

Based on the chart, you're better off with two 6 inch hoses

If you take 1500GPM, then your 7 inch hose needs 15 psi per 1/8 mile.
the 6 inch hoses need less pressure drop for the same total flow (750 x 2) at 8.6 psi or for the same pressure drop can flow 1000 gpm each.

The d/d ratio can sometimes get a bit tight at small sizes and the IDs don't exactly match the data provided, but it's pretty close.

If this pressure drop chart is for your actual hoses then go for two 6 inch

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

Thank You, LittleInch!!! This is the chart for my hose. I have seen multiple charts for the same hose with data varying a bit, but it's close like you said. I appreciate your time. :)

I don't think you are reading the chart correctly. Extrapolating (with Excel, using a power function to approximate the trendline) at 625 gpm (the flow through one 6" hose) I get a pressure drop of 13.9 ft of head, or 6.1 PSI, every 1/8 mile (at ~7.1 fps). This adds up to ~223 feet of head (~98 PSI) over your 2 mile run. For 7", the pressure drop is read directly from the table, and comes out to 384 feet of head (168 PSI) for 2 miles (at 10 fps). 8" gives 208 feet (88 PSI) (8 fps). Assuming you want to use the same pump, if you want to run 1 line you have to jump up to 8" to keep the pressure drop constant (or nearly so) compared to a twin 6" line setup.

Matt

Quality, quantity, cost. Pick two.

You have to go to one 8" or stay with two 6". Unless you pump can handle the increased head of 7".

Ted

One 7-Inch hose has a headloss of almost twice the headloss for 2 6-Inch hoses.

At 1250 gpm:

Headloss in 2 - 6-Inch hose = 6 * 10,560/660 = 96 psi

Headloss in 1 - 7-Inch hose = 10.5 * 10,560/660 = 168 psi

Headloss in 1 - 8-Inch hose = 5.5 * 10,560/660 = 88 psi

It is a good idea to oversize piping as the extra cost to up one pipe size is minor compared to the overall installed cost.

It is pretty obvious that 1 x 8" is near enough the same head loss as 2 x 6" so it becomes an exercise in costing, purchase 4 miles of 6" and installation compared to the purchase of 2 miles of 8" and installation.
You should look at 9" if available or 10" and run the calculations on material / installation costs and the difference in pump, and pumping costs compared to the 6/8 configurations. Reducing head to the minimum will reduce the pump purchase cost, cheaper electrics (smaller motor, starter and cabling)and something like a 1/3 of the power costs.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

I often use the "diameter ratio to the power of 5" rule proposed by LittleInch in his first post above. Provided that the flow is turbulent, this rule has always served me well and I was surprised that it was so wrong here. On closer inspection, LittleInch has left out a step and if it is included the results stack up well with the readings from the chart - so my faith in the rule is restored.

If we use subscript 1 for the existing line and subscript 2 for the proposed line then we can express the rule relating the differential pressure (dP) to the diameter (ID) as

dP₂ / dP₁ = (ID₁ / ID₂)⁵

But this assumes that the flows are the same through the two lines.

In turbulent flow the pressure drop is proportional to approximately the flow squared. We only know the pressure drop through the 6" line with half the total flow. To use the "power of 5" rule above we need to compare the lines with equal flow. If we double the flow in the 6" line the pressure drop (dP) increases by a factor of 4 so we need to rewrite the equation as

(dP₂ / (4 x dP₁)) = (ID₁ / ID₂)⁵

We want the pressure drops to be equal, so dP2 = dP1 and this simplifies to
0.25^0.2 = ID₁ / ID₂

or ID₂ = 6" / 0.758 = 7.92"

which agrees closely with the values taken from the chart.

Katmar Software - AioFlo Pipe Hydraulics
http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

katmar - thanks - I was wondering where it had gone wrong. I think I hadn't thought it through as well as you had and assumed if the PD was the same, the flow rate was half, then the diameter ratio was 0.5^0.2.

If this works better for identical flows I'll bear that in mind the next time I use it.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.