Today we're going to touch on layout and hydraulic calculations.

Prior to 1985 most hydraulic calculations were done by hand but today they are done by special computer programs that can do in a few minutes what used to take hours to do.

Below is a sketch of our 70'x100' building. Sprinkler heads are the red circles, pipe is red, walls are black.

Since this is an Ordinary Hazard Group II Occupancy we will use a design density of .20 gpm over the hydraulically most remote 1,500 sq. ft..

What a designer does is lay out a system he *thinks* will work based on past experience. You do this enough and in a few years you'll be able to guess and get it pretty close.

I don't think the layout I sketched will work, I think that second piece of 1" pipe from the end kills our chances, but we will try it anyway.

Click image to enlarge.

With sprinklers 12'-6" apart on lines and lines 10'-0" apart each sprinkler covers 125 sq. ft..

The number of sprinklers needing to be in the calculated area is 1,500/125=12. If heads were spaced 124 sq. ft. we would need 13 heads in the calculated area. We always round up to the nearest whole sprinkler.

Our hydraulic remote area has to be 1,500 sq. ft. and must be rectangular in shape with the long direction being at least 1.2 * sqrt of the area of application.

Rectangular length: 1.2*sqrt 1,500 or 46.47 feet. The long side of the rectangle must be parallel to the lines and you add sprinklers until the length is equal to or exceeds 46.47 feet.

The calculated area is located in the lower left corner of the building and is bounded by two walls and two green lines. The green numbers are "Nodes" or identifiers we will use in our hydraulic calculations.

The 12 heads we will calculate as open will be #1 through #4, #9 through #12 and #17 through #20. Sprinklers used will be Viking VK100 1/2" orifice brass upright sprinkler having a K-Factor equal to 5.6.

A K-Factor is a coefficient of discharge used to find the discharge of a sprinkler head at varying pressures. The formula is:

Q=K*P^.5 or Q=K*sqrtP

Where Q=Water in Gallons Per Minute, K=K-Factor and P is the Pressure at the sprinkler head in pounds per square inch or psi.

Sprinkler #1 covers an area of 125 sq. ft. and in order to provide a density of .20 per square foot over the area of sprinkler coverage it must discharge 25.0 gpm. 125*.20=25.0

Using Q=K*sqrt P we can determine the pressure required to discharge 25.0 gpm by:

P=(Q/K)^2

In order to discharge 25.0 gpm a sprinkler with a K-Factor=5.6 must be supplied with 19.9 psi.

When hand calculating we will round our numbers to the nearest tenth.

When calculating sprinkler systems we proceed from the end sprinkler inwards. We will start with sprinkler head #1.

Ever connect two or three identical yard sprinklers together in series fed by a garden hose only to notice the sprinkler farthest away discharged the least water? When water travels through pipe it encounters pressure or friction loss.

In accordance with NFPA #13 Section 14.4.2.1 this head loss loss is calculated using the Hazen-Williams formula shown to the right.

For wet pipe systems using black steel or galvanized pipe the friction loss coefficient or C-Value we use is 120.

For our purposes we will use schedule 40 black steel pipe that has the following inside pipe diameters:

1"=1.049"

1 1/4"=1.380"

1 1/2"=1.610"

2"=2.067"

2 1/2"=2.469"

3"=3.068"

4"=4.026"

6"=6.065"

For calculating friction loss the Engineering Toolbox has a little online calculator or you can do what I do and that's use a small scientific calculator similar to the Casio fx-115MS which I am using right now and can be purchased at K-Mart for around $20.00.

There is another way to calculate pressure loss in pipe besides using the Engineering Toolbox and that is using Pipe Table B shown here.

On most wet pipe sprinkler systems C-120.

To calculate friction loss through 1 1/4" pipe Sch. 40 pipe while flowing 32.5 gpm use a scientific calculator for 32.5^1.85*1.34x10^-4

The answer is 0.084

Using the Engineering Toolbox how much head or friction loss will we encounter through the 1" pipe flowing from Node #1 to Node #2 while flowing 25.0 gpm?

I got a head loss of 0.197 psi per linear foot. For 12'-6" our total head loss between Node #1 and Node #2 will be 2.46 psi or we will call it 2.5 psi after rounding to nearest tenth.

What we are seeing is we require 19.9+2.5=22.4 psi pressure at Node #2 for us to be able to discharge 25.0 gpm at Node #1.

Now we have to add the flow of sprinkler Node #2 but we just can't add 25.0 gpm because we've already determined we must have 22.4 psi pressure at sprinkler Node #2.

How much water will be discharged from sprinkler Node #2 when supplied with 25.7 psi pressure?

q=K*sqrtP

q=5.6*sqrt22.4

q= 26.504 or 26.5 gpm after rounding.

Our minimum required flow from sprinkler Node #2 to sprinkler Node #3 is 25.0+26.5=51.5 gpm.

Remember, everything we do is backwards. With calculations we are moving from the farthest point on the system to the center which is the same way a system is drawn up or designed.

What is the head or friction loss with 51.5 gpm flowing through 1" pipe from sprinkler Node #2 to sprinkler Node #3?

I got 0.749 psi per linear foot or 12.5*0.749=9.4 psi for the 12'-6" length of 1" pipe.

At sprinkler Node #3 our system so far requires 35.7 psi (25.7+9.4=35.1 psi) at sprinkler Node #3 to adequately supply water to both sprinkler Node #1 and sprinkler Node #2 insuring each node discharges at least 25.0 gpm.

With 35.7 psi required at sprinkler Node #3 how much water will sprinkler Node #3 discharge?

q=K*sqrtP

q=5.6*sqrt35.1

q= 33.18 or 33.2 gpm after rounding.

Adding the 33.2 gpm to the required combined flow of 51.5 gpm the total water demand for sprinkler Node #1, #2 and #3 is 84.7 gpm.

Earlier was was talking about "theoretical minimum" water supply requirements mentioning it was always more in actual practice. For example a 1,500 square foot area with a minimum density of .20 gpm per sq. ft. with sprinklers spaced 125 sq. ft. would require 300.0 gpm but you can already see why it would be more. In what we've done with three heads, each requiring a minimum discharge of 25.0 gpm, we should only need 75.0 gpm but we require 84.7 gpm.

By this time you should be pretty well lost. It isn't something you master in a day, a week or even a year but once you've done it a while it does come easy.

It is more than most people realize and it isn't your normal everyday cad job.

We've only calculated three sprinkler heads and it's getting unwieldy. We do have a regular form we can use so let's go to that.

## Sunday, August 23, 2009

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I couldn't help myself to just leave! So Thank you for this. It helped me a lot for my HW

ReplyDeleteThis comment has been removed by the author.

ReplyDelete

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How came Node #2 with 25.7 psi pressure?

ReplyDeleteYour calculation is

q=5.6*sqrt22.4

It should be 22.4, shouldn't it?

NODE #2: 22.4

DeleteNODE #3: 25.7

Those are some pretty intense calculations. I didn't know how much went into where and how to place the fire sprinklers. I'm glad there are people that can do this sort of thing and make buildings a little more safe.

ReplyDeleteGerald Vonberger | http://www.inlandfireprotectionco.com/fire-sprinklers--hydrants---pump-systems.html

Thank you for your sharing. Really appreciate the effort of the writer.

ReplyDeleteJust want to pinpoint some calculation mistake-

At sprinkler Node #3 our system so far requires 35.7 psi (25.7+9.4=35.1 psi).

And instead of (25.7+9.4=35.1 psi), it should be 22.4 +9.4 =31.8 psi.