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Calcualation for right sizing a check valve per Crane Technical Paper No. 410

CFM ft 3^min / :

FPM ft min / :

in.wc 248.6 Pa * :

DN 0.5 0.75 1 1.25 1.5 2 2.5 3 3.5 45681012141618202224121 mat (1 * in * :

f.T 0.027 0.025 0.023 0.022 0.021 0.019 0.018 0.018 0.017 0.017 0.016 0.015 0.014 0.014 0.013 0.013 0.013 0.012 0.012 0.012 0.012 121 mat :

Calculation assume the use of water at standard conditions

ρ 62.3 lb ft 3^/ * :

Density of water at STP

Q v.el A.rea * ≡

Continuity equation

Q ṁ ρ / ≡

Mass flow rate

A.rea π D 2^4 / * :

Δ°F 59 / K * :

Δ°C 1 K * :

Cross sectional area of a circle/pipe

With algebraic manipulation

V.el Q v.el A.rea * ≡ v.el solve maple : 4 Q * π D 2^* /

Parallel flow swing check valve (pg. A-28)

Vel.min 601 ρ / lb ft 3^/ * sqrt * ft s / * :

Minimum Velocity (Note that the constant 60 should be changed for a UL listed valve)

v.el Q A.rea / ≡

Where Q (mass flow rate) can be redefined as

K 50 f.t * :

Representative resistance coeffiecient for valves and fittings

Vel.min ft s / 7.6016

Minimum velocity of a parrallel flow swing check valve. Notice that physical size is not a factor

h.Lvalve K 8 Q 2^* π 2^D 4^* g.e * / * ≡

Head loss through a valve

Sudden enlargements and contractions (pg. A-27)

Resistance to flow due to sudden enlargement/contraction (θ=180 degrees). The value of the resistance coefficient is determined by dividing the equations below by the ratio of the small over big diameter to the 4th power. This is a worst case scenario

K.contraction .5 1 D 2^D max (2^/ - (* D D max / (4^/ ≡

Sudden contraction'

K.enlargement 1 D 2^D max (2^/ - (2^D D max / (4^/ ≡

Sudden enlargment

Given the following parameters:

Q 5700 gal min / * :

Flow of pump previously selected for the project

Minimum diameter calculated

Dmin 4 Q * π Vel.min * / sqrt :

Vel.min ft s / 7.6016

Dmin in 17.5016

i 1 DN length range Dmin DN i el ≤ jj i 1 - : break 21 line continue if for #

Find the position of the minor nominal diameter close to the minimum diameter calculated

jj 17

test jj 1 - :

z 0 :

i 1 DN length range v.el i el 4 Q * π DN jj el (2^* / 1 * : jj jj 1 + : z z 1 + : 31 line for #

Calculate the speed from the smaller diameter selected, to the greatest

z 5

i 1 z range Vel.min v.el i el > z1 i 1 - : break 21 line continue if 11 line for #

Select the best diameter

z1 1

D.s DN z1 test + el :

Diameter selected

D.s in 16

From here, you can calculate the friction losses , according to the main pipe diameter .

Lets look at the velocity for probable nominal pipe sizes.

D 1416182041 mat in * :

to which

f.t .013 .012 .012 .012 41 mat :

from page A-27

Vectors - Column matrix are multiplied getting their scalar product (sum of products of their elements) in this program. To get the elemental product we need to use "for" loops

i 1 D length range v.el i el 4 Q * π D i el (2^* / : for

v.el ft s / 11.8798 9.0955 7.1865 5.8211 4.0424 51 mat

I'd like to program this. In this case, 20 in is too big and fails min vel as calculated above.

i 1 D length range h.Lvalve i el K i el 8 Q 2^* π 2^D i el (4^* g.e * / * : for

h.Lvalve ft 1.4256 0.7714 0.4816 0.316 41 mat

i 1 D length range K.enlargement i el 1 D i el (2^D max (2^/ - (2^D i el D max / (4^/ : for

K.enlargement 1.0833 0.3164 0.055 041 mat

i 1 D length range K.contraction i el .5 1 D i el (2^D max (2^/ - (* D i el D max / (4^/ : for

K.contraction 1.0621 0.4395 0.1448 041 mat

Therefore, the total headloss, with in terms of the largest diameter

i 1 D length range h.Lassembly i el K K.contraction + K.enlargement + (i el 8 Q 2^* π 2^D i el (4^* g.e * / * : for

h.Lassembly ft 6.1308 1.7431 0.6419 0.316 41 mat