Appendix A: Valve Selection and Sizing

Select a linear valve providing close control with a capacity index of 4 and meeting the required pressure and tempera- ture ratings.

NOTE: For steam valves downstream from pressure reducing stations, the steam will be superheated in most cases and must be considered.

EXAMPLE 2:

In Figure 19, a linear valve (V1) is needed for accurate flow control of a steam coil that requires 750 pounds per hour of steam. Upstream pressure in the supply main is 5 psig and pressure in the return is 4 in. Hg vacuum minimum.

 

 

VALVE VI

STEAM

1.96 PSI

SUPPLY

5 PSI

 

 

 

COIL

(VACUUM)

 

 

 

 

 

 

 

 

 

 

 

 

RETURN

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

30% PRESSURE DROP, Cv = 41

 

 

 

 

 

C2336

80% PRESSURE DROP, Cv = 25

 

 

 

 

 

Fig. 19. Linear Valve Steam Application.

Use the steam valve Cv formula to determine capacity index for Valve V1 as follows:

(1 + 0.00075s)Q V Cv = -------------------------------------------------

63.5 h

Where:

Q= Quantity of steam required to pass through the valve is 750 pounds per hour.

h= The pressure drop across a valve in a modulating application is found using:

h= 80% x (Pm – Pr)

and:

Pm

=

Upstream pressure in supply main is 5

 

 

psig.

Pr

=

Pressure in return is 4 in. Hg vacuum.

NOTE:

1 in. Hg = 0.49 psi and 1 psi = 2.04 in. Hg.

Therefore,

4 in. Hg vacuum = –1.96 psig.

h= 0.80 x [5 – (–1.96)]

=0.80 x 6.96

=5.6 psi

The critical pressure drop is found using the following for- mula:

hcritical = 50% x (psig + 14.7 psi)

hcritical= 0.50 x (5 psig upstream + 14.7 psi)

=0.50 x 19.7 psia

=9.9 psi

The pressure drop (h) of 5.6 psi is used in calculating the Cv, since it is less than the critical pressure drop (hcritical) of 9.9 psi.

V= Specific volume (V) of steam, in cubic feet per pound at average pressure in valve (Pavg):

Pavg =

h

Pm --

 

2

=

5.6

5 – ------= 5 – 2.8 = 2.2psig

 

2

The specific volume of steam at 2.2 psig is 23.54 and the square root is 4.85.

63.5= A scaling constant.

s = 0

Substituting the quantity of steam, specific volume of steam, and pressure drop in the Cv formula shows that Valve V1 should have a Cv of 24.17 or the next higher available value (e.g., 25).

Cv =

(---1------+---0.00075--------------------------

0---)---------750---------------4.85---------

 

 

63.5

5.6

 

=

3637.5

= 24.17

 

63.5-----------------2.37---------

NOTE: If Pavg is rounded off to the nearest value in Table 5 (2 psi), the calculated Cv is 24.30.

Select a linear valve providing close control with a capacity index of 25 and meeting the required pressure and tempera- ture ratings.

EXAMPLE 3:

Figure 20 shows the importance of selecting an 80 percent pressure drop for sizing the steam valve in Example 2. This pressure drop (5.6 psi) approximates the linear valve charac- teristic. If only 30 percent of the available pressure drop is used (0.30 x 6.96 psi = 2.10 psi or 2 psi), the valve Cv becomes:

(1 + 0.00075s)Q V Cv = -------------------------------------------------

63.5 h

750 ⋅ 4.85

Cv = ------------------------- = 40.5 63.5 2

APPENDIX

261

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Image 261
Honeywell MS4103 manual = 80% x Pm Pr, In. Hg = 0.49 psi and 1 psi = 2.04 in. Hg, 261