BYPASS FLOW AND ORIFICE SIZE CALCULATIONS

MGT INC., BOULDER CO 90303-3607

BYPASS FLOW CALCULATION BASED ON ALLOWABLE PRESSURE DROP Results in blue

SUBSTITUTE YOUR DATA IN THE EXAMPLE FOR YOUR SPECIFIC HEATER

TO CALCULATE YOUR REQUIRED BYPASS FLOW AND ORIFICE SIZE EXAMPLE WITH 20% OF TUBES PLUGGED

INPUT DATA CALCULATIONS PARTS OF FORMULA THE RESULTS REPRESENT

W = Total Feedwater Flow, #/hr 4,301,500

1,040

208

832

1.2500

15.0 1.25 raised to the 1.8 power 1.4943

17.0 Inverse of 1.4943 0.6692

1.1333

1.1333 x 0.6692 = 0.7584

0.7584 to the 0.556 power 0.8575

1 - 0.8575 = 0.1425

The linear velocity through the tubes is proportional to fhe feedwater flow

divided by the number of active tubes:

Required bypass #/hr flow is:

0.1425 x 4301500 = 612,964

ORIFICE SIZING CALCULATION BASED ON A SHARP EDGED ORIFICE FOR

CALCULATED BYPASS FLOW

62.40

4.074

CALCULATION RESULTS

WB = Bypass Flow, #/hr

Nnew

= Number of Tube Ends in Inlet Pass

Nplugged

= Number of Plugged Tubes in Inlet Pass

Nnew - Nplugged = Remaining Active Tubes

(Nnew/(Nnew - Nplugged) = 1/(1 - Fraction plugged)

DPnew

= Pressure Drop in New Condition with W Flow, Psi [Nnew

/(Nnew

- Nplugged

)]1.8

DPmaximum

= Maximum Allowable Pressure Drop, Psi 1/[Nnew

/(Nnew

- Nplugged

)]1.8

DPB = Pressure Drop Through N

new tubes with W - W

B Flow,

Psi

DPBplugged

= Pressure Drop Through Nnew

- Nplugged

tubes with W - WB Flow,

Psi DP

Bmaximum/DP

new = (DP

Bplugged/DP

new)

V = linear velocity through Nnew

tubes with W flow, ft/sec (DPBplugged

/DPnew

)*1/[Nnew

/(Nnew

- Nplugged

)]1.8

VB = linear velocity through N

new tubes with W - W

B flow, ft/sec {(DP

Bplugged/DP

new)*1/[N

new/(N

new - N

plugged)]1.8}0.556

VBplugged

= linear velocity through Nnew

- Nplugged

tubes with W - WB flow, ft/sec 1 - {(DP

Bplugged/DP

new)*1/[N

new/(N

new - N

plugged)]1.8}0.556

V is proportional to W/Nnew, VB is proportional to (W - WB)/Nnew,

VBplugged is proportional to (W - WB)/(Nnew - Nplugged)

WB = W(1 - {(DP

Bplugged/DP

new)*1/[N

new/(N

new - N

plugged)]1.8}0.556)

WB = W(1 - {(DP

Bplugged/DP

new)*1/[N

new/(N

new - N

plugged)]1.8}0.556)

Assumed feedwater density, #/ft3 =

D = Hole diameter, inch, r = Feedwater density, #/ft3

D = 0.0297*SqRt(WB/[rDPplugged]0.5), inch

Required diameter of orifice in pass partition plate >>

INPUT DATA

From the other sheet 208 tubesTube diameter 3/4 " ODTube wall 0.05 " wall

Flow area one tube 0.332

Total flow area 69.0

Diameter of hole that area 9.37 inches

THIS IS ANOTHER METHOD OF DETERMINING THE ORIFICE SIZE. THE AREA OF THE ORIFICE IS THE SAME AS THE FLOW AREA OF THE TUBES PLUGGED. THIS METHOD IS NOT RECOMMENDED; IT IS PROVIDED FOR COMPARISON PURPOSES ONLY.

CALCULATION RESULTS

in2

in2