ef6e9839-de38-41a8-9c36-6183830a403a 60 Javier Ochoa 4 5 false false 0 0 decimal

&[DATE] &[TIME] - &[FILENAME]

Flow parameters and constant definitions with units, for water flowing in a pipe.

W 700 lb s / * : 317.5147 kg s / *

μ 8.67 104 -^* lb ft s * / * : 0.0013 s * Pa *

d 1.5 ft * : 0.4572 m *

ρ 62.5 lb ft 3^/ * : 1001.154 kg m 3^/ *

ΔP 3.6 psi * : 24821.1263 Pa *

N.Re 4 W π d * μ * / * : 6.8533 105^*

β a.0 π β 2^* d 2^4 / * :

g.c 32.174 lb ft * (lbf s 2^* / * :

Y 1 :

1 lbm * 0.4536 kg *

factor ρ ΔP * 2 * sqrt : 7049.7899 s Pa * m / *

β K 0.593 0.4 β 4^* + 1.5 β sqrt * 12 β 4^* + (N.Re sqrt / + :

Solution 1: Solve the orifice meter diameter based on an Iterative method with the following equations:

β.0 1.5228 W d 2^Y * ρ ΔP * sqrt * / * sqrt : 0.6812

β f β.0 2^0.6745 β 6^* - β 2^2.5295 β sqrt * 20.236 β 4^* + N.Re sqrt / (* - sqrt - β + :

res β f β 0.25 0.72 solve :

res a.0 0.0679 m 2^*

W1 res K res a.0 * Y * factor * : 318.4701 kg s / *

res 0.6431

res K 0.6653

Solution 2: Non iterative method after averaging γ β0 and β equations.

β β.0 2 / 11 0.67455 β.0 4^* - 2.5295 β.0 sqrt * 20.236 β.0 6^* + (N.Re sqrt / - (12 /^+ (* :

1 atm * 1.0133 105^* Pa *

1 psi * 6894.7573 Pa *

β a.0 0.0703 m 2^*

W2 β K β a.0 * Y * factor * : 332.5252 kg s / *

β 0.6546

β K 0.6706

The last equation is valid for the following parameters:

N.Re 106^≤

N.Re 5104^* ≥

ΔP 50 psi * ≤

ΔP 1 psi * ≥

β 0.72 ≤

β 0.25 ≥

Both methods gave the same result, which are not different from the values for β = 0.6525 and β0 = 0.6793, found by the author in the article. Source. Chemical engineering , January 6, 1984. Denis J. Zirang and D. Sylvester.The equations were converted in SI units in order to solve the same problem, which used different constant coefficients for the equations presented in the article.

0.6416 K 0.6647

0.6416 a.0 0.0676 m 2^*

W3 0.6416 K 0.6416 a.0 * Y * factor * : 316.6843 kg s / *

M Equation β Mass flux (kg/s) Solution 1 0.6431 318.5 Solution 2 0.6546 332.5 Article Sol. 1 0.6525 330 Article Sol. 2 0.6416 316.7 53 mat :

M W3 W2 β W1 res f K factor g.c a.0 N.Re ΔP ρ d μ W M W3 W2 β W1 res f K factor g.c a.0 N.Re ΔP ρ d μ W W3 W2 β W1 res f K factor g.c a.0 N.Re ΔP ρ d μ W

M
Equation	β	Mass flux (kg/s)
Solution 1	0.64	318.5
Solution 2	0.65	332.5
Article Sol. 1	0.65	330
Article Sol. 2	0.64	316.7
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The estimated values of β are within 1% in both equations, but the final mass flux values found are almost 10% different, when checking from the initial value 317.51 Kg / s.