FLAME SAFEGUARn CONTRULS phần 9 pdf

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FIG. 1- PRESSURE DROP VS. CAPACITY CHART SHOWING SIZING PROCEDURE OF A DOUBLE SSOV

PIPED IN SERIES.

c. It point X 1a/(s only a very short disTance to the right

of the diaoonal line, increasing the pressure drop

might allow use of the next smaller valve size.

Remember that we estimated only 45 percent of the

available SSOV pressure drop lor each valve. A margin of

safety or 5 percent per valve remains to allow Sllghl over·

sizing. Therefore, we can increase the pressure drop 5

percent 01 12.0. or 0.6 inch we Without exceeding the esti·

mated lotal pressure drop over bolh SSOVs.

5% of 12.0 = O.S in. we

To determine lhe extra pressure drop, move upward

110m X to the next dia~llineZ. Then move horizontally

to the Pressure Drop Scale fo find the pressure drop

needed for a 2 inch valve. In this case, the actual pressure

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drop is 6,4 inches we, whiCh is 1 inch wc greater than the

esllmated individual pressure drop at 5,4 inches we.

6.4 - 5.4 = 1.0 in. we

Since our excass pressure drop is greater than the 0.6

inch wc allowed, we should use the 2,1/2 inch valve.

Using a 2 inch varve may still be p::>ssitJle. The actual

lolal SSOV pressure drop would be 2 x 6.4 or 12.8 inches

wc. The diHerence between the eslimatad and actual

pressure drop, 0.8 must be taken from another part 01 the

manifold to keep the required burner inlet pressure. We

cannOI reOOce lhe Firing Rate Control pressure drop pasr

our minimum estimated fi\}Jre of 30 percent of the mani￾fold pressure drop. Therefore, we musl increase the pres·

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FIG. 8- PRESSURE DROP VS. CAPACI'N FOR A 'NPICAl BUlTERFlY VALVE. MAXIMUM OPENING

ANGLES SHOWN 'ARE USEO AS TRIAL SETTINGS WHEN AOJUSTING VALVES FOR HtGH FIRE.

sure regul!'llor oullel pre,ssure by O~ in. \Of<: to allow for the

additional pressure drcp across the 2 inch SSQVs.

This requires estimating a new pressure dfcp for the

pressure regulator. SSOVs and Firing Rate Control, plus

repealing the sizing procedure.

SIZING CHART FOR FIRING RATE

CONTROL VALVE

Butterfly Valves are otlen used for Firing Rale Control

Valves. SinCe a Butterfly Valve ctleS no! provide light clo￾sure, a safely shutoff can!rol valve (SSOV) must be used

l4)slream.

Inaddilion 101M buUerlly control valve size, we neeO 10

know lhe mallimum QQenino angle (in degrees) U'OOd as a

trial selting when adjusting a BUllerfly Valve to hi~ fire.

In our ellample. the eslimaleO pressure drq:J across the

liring rate control is 6.5 inches wC wilh a capaCity of 10,500

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cfh at standard conditions.

a. If SSOV estimaled <md actual pressure drop are lha

~. Mark the intersection oflhe estimated Butter￾fly Valve pressure drq:J and capacity, X on Fig. 8.

Use Ihe valve size and opening angle indicaled IYy

lhe nearest slanted line below X. In this case, ar;ply·

ing the estimated pressure drq:J results 'n a 2 inch

valve wHh a mallimum opening angle at 45 degrees.

b. If SSOV aclual pressure droo is less than lbe esli·

malad pressure droo.

Add lhe pressure drq:J difference (0.8 inch we) to the

pressure Orq:J available for the Bullerlly Valve.

0.8 + 6.5 '" 7.3 in. we

Mark the intersection 017.3 in. wc and lhe capacity

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Use the valve size and opening angle indicated by lhe

nearest slanteO line below. In lhis case, we have a 2 inch

valve with a 40 degree opening.

V5055 VALVE SIZING NOMOGRAPH

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l4UOO V5055 VALVE

100.000 SIZING CHART

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CIlA"'TY COIlAtel'ON

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The correct size V505 5 Industrial Gas Valve can be

quiCkly selected usinog this nomograph. The nomograph is

available in pads at 25 under form number 70-86.27.

Tha 1oll0winog 8)(amples show how to use the VS055

Valve Sizing Char1.

EXAMPLE 1

These are the specifications from our original example:

Type of Gas Milmd

SpeCifiC Gravity 0.72 sp gr

Gas Flow Capacity 7.200,000 Btuh

Heat Content of Gas 800 Btu per Cu. fl.

Estimated Pressure Drcp across V5055

Single Valve 12.2 in. we

2 Valves Piped in Series 5.5 in. we

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For this chart, we need to know the inlel and outlet pres￾sure for the V5055.

1. Determine inlet pressure for V5055. When 2 valves are

used, lind the inlel pressure of the first valY'6. From Fig.

2 we see thai the V5055 inlel pressure equalS the pres￾sure regulator oullet pressure less lhe piping pressure

loss from the pressure regulator outlet to the V5Q55 il'l￾Ie!. Find the piping lOSS flQJre tram Fig. 3. It we haY'6 5

leet at 2 incll pipe, tM pressure drcp will be 1 inch wc.

Regulator Outlet Pressure - Piping Pressure Drcp

= V5055 Inlet Pressure

27.6 in. we - 1 In. we = 26.6 in. WC

2. Determine tile oullel pressure for V5055. The outlet

pressure equals lhe inlel pressure minus the pressure

drcp across the valve.

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