V =

(4)

friction head in inches of water column.

length of pipe in feet. internal diameter of pipe in inches.

velocity of flow in feet per second.

or, taking the weight of 1 cu. ft. of water at mean temperature-flow 180° F. and return at 144° F.-equal to 61 lbs.,

or P lbs. per square foot

0.226

0.12 +

+

(0.1

=

1v2

d

0.288

Vd vvd

× 0.893 x

....

(6)

Formulas (4), (5), and (6) hold true for circulation of water in iron pipes commonly used for domestic engineering purposes.

In further developments we shall have to know what must be the velocity of flow of water in pipes in order to transmit a given number of B. T. U. per hour. Taking the weight of 1 cu. ft. of water at mean temperature-flow_being 180° F. and the return at 144° F.— for 61 lbs., we can write

4 x (d/12)2 x 3.14 × v × 60 × 60 × 61 × 36 = B. T. U. per hour...(7) where d

=

V

inches. feet per second.

B. T. U. per hour

or v ft. per second

3,600 × 36 × 61 × (d/12)2 × 3.14 × 4

0.226

1,000. The greatest amount of B. T. U. is placed at 10,000,000. The limits of friction head are taken at from 0.001 lbs. per sq. ft. and 0.5 lbs per square foot, both per running foot. In gravity hot water heating practice, one would hardly have to deal with a friction head less than 1/1,000 lbs. per square foot, or more than 0.5 lbs. per square foot per running foot.

As an illustration showing how the curve was plotted, let us take a piece of straight pipe 1.5 in. in diameter, 1 ft. long. Figuring from the formulas:

Pu

= 0.025 × 0.105 · 0.0026 lbs. per

=

square foot per running foot. 0.31 x 0.1222 = 0.46 lbs. per square foot per running foot. P1 and Pu show us that for friction heads up to 0.0026 lbs. per square foot we shall have to use formula (4). Between 0.0026 lbs. per square foot and 0.46 lbs. per square foot we shall have to use formula (5). Where the friction head is higher than 0.46 lbs. per square foot, formula (6) will have to be used. Then we can write a series of equations as follows:

Vd

0.010

× 0.893 ×

d

0.40

0.08 x 96,966 7,

0.306 × v2; v = 0.114. B. T.

C.

hour per 11,050.

=

0.114 × 96,966

On the chart the B. T. U. per hour are shown on the abscissa axis and friction head, or loss of pressure in pounds. per square foot per running foot, on the ordinate axis. Having solved the above series of equations you get loss of pressure or friction head, on one side, and corresponding B. T. U. per hour transmitted by the pipe on the other. Now it is easy to plot the curve. Only two points are necessary to plot the part of the curve corresponding to the same formula, so that part will be a straight line. Wherever the third point was determined from the same formula, it was to check first two points. In the above example it has seemed unnecessary to use formula (6) as the loss of pressure corresponding to the upper critical velocity is very near the chosen limit of friction head. Should, however, formula (6) be used in this or any other case, the pro

=

Thus the resistance of a 4-in. globe valve can be taken as great as that of R 10 lin. ft. of straight 4-in. pipe; that is, 2 x 10 20 ft., if R = 2. The loss of pressure due to a 1-in. short bend will be as great as that of R x 2 = 0.5 x 2 1 lin. ft. of straight 1-in. pipe, and the loss of pressure due to vena contracta at the flow end of the boiler, if its flow opening and the attached pipe is, say, 6 in., will be as great as that of 2 x 17.5 35 ft. of straight pipe 6 in. in diameter; that is, in the latter case, vena contracta as to resistance is equivalent to 35 ft. of straight pipe. (For other R's see the accompanying chart.) A Typical Example.

=

Now let us see how to use the chart in an example. Assume the flow tem

150°F-

Area

1v2

0.226

d

Vd

FIG. 1.-SKETCH OF EXAMPLE SHOWING USE OF CHART.

perature to be 180° F., and the return 150° F. Imagine two columns of water connected at the bottom as per sketch (Fig. 1). The right-hand column weighs 61.201 lbs., the left-hand column weighs 60.548 lbs., the difference being 0.653 lbs. Apparently, there exists a pull from the right-hand column to the lefthand column, due to that temperature difference, with a force of 0.653 lbs. per square foot. Should the column be 50 ft. high, the pulling force would be as great as 50×0.653-32.65 lbs. per square foot. This pulling force is usually called the "available pressure head."

The

On the sketch of a hot water heating system (Fig. 2) the radiator farthest away and at the lowest point is fIf'. The problem is to figure out the piping so that water will surely circulate. center of this radiator is 5 ft. above the center of the boiler, consequently the pulling force will be equal to 5 × 0.653= 3.265 lbs. per square foot. This pulling

40,000

force must overcome the friction of piping of the whole circuit on which this radiator operates. The total length of this circuit is 285 ft., elbows, valves, etc., not considered.

Experience shows that from 50% to 100% should be allowed to equalize local resistances to straight pipes. Let us take 75%.

PARTS

tributed equally along the total equivalent length of the circuit; that is, the loss of pressure per square foot per running foot can be taken as equal to 3.265 -500-0.0065 lbs.

Referring to the chart (Fig. 3) find on the ordinate axis 0.0065 lbs. and follow the horizontal line, corresponding to 0.0065, up to the ordinate corresponding

PIPING

fg=10'
gig=10' "1 "8
gh =10"-"-
g'h=10'

"/18"

8"/"01=4114

39

ci=10'"/" | "

c'i'= 10' "/" / " ¿Mż-40"/"10"

im=20"-" 3 i'm'=20"-"3

по

n'o'

= 10" 1" 2 "

OVIO' = 10 ́"/" 8 OVILO' 25"/" 10

The total equivalent of the circuit in question will then be 285 x 1.75 = 500 ft. The total friction head, which must be balanced by the available pulling force of 3.265 lbs. per square foot, can be dis

to 20,000 B. T. U. The tip of the finger will lie between 12-in. and 2-in. pipes. You will therefore take the 2-in. pipe for the radiator connection. In a similar manner it will be found that: