01 POWER ISLAND / 01 CCPP / V. Ganapathy-Industrial Boilers and HRSG-Design (2003)

Source: Ref. 5.

A:

Assume that k is nearly the same for both air and ammonia. Hence for the valve, CKAPa is a constant. For air, use C ¼ 356, K ¼ 0.98, A ¼ 0.503, MW ¼ 29, and T ¼ 560.

¼ 4990 lb=h

(An acc value of 0.10 was used above.) From Eq. (6), substituting MW ¼ 17 and T ¼ 510 for ammonia, we have

Hence

4990

Wamm ¼ 1:246 ¼ 4006 lb=h

7.10a

Q:

How are the relieving capacities for liquids determined?

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A:

An expression for relieving capacity at 25% accumulation [5] is

where

P1 ¼ set pressure, psig

Pb ¼ backpressure, psig

p = ;

Ks ¼ 1 s s being the specific gravity

A ¼ orifice area, in.2

q ¼ capacity, gpm

7.10b

Q:

Determine the relieving capacity of a relief valve on an economizer if the set pressure is 300 psig, backpressure is 15 psig, and s ¼ 1. The valve has a G orifice (A ¼ 0.503 in.2).

A:

Using Eq. (7), we have

¼231 gpm

¼231 500 ¼ 115;000 lb=h

At 10% accumulation, q would be 0.6 231 ¼ 140 gpm and the flow W ¼ 70,000 lb=h (500 is the conversion factor from gpm to lb=h when s ¼ 1.)

7.11

Q:

A safety valve bears a rating of 20,017 lb=h at a set pressure of 450 psig for saturated steam. If the same valve is to be used for air at the same set pressure and at 100 F, what is its relieving capacity?

A:

For a given valve, CKAPa is a constant if the set pressure is the same. (See Q7.09a for definition of these terms.)

For steam,

20;017 ¼ 50 KAPa

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Hence

Converting to acfm, we have

15

q ¼ 32;430 560 0:081 492 465 60 ¼ 244 acfm

(The density of air was estimated at 465 psia and 100 F.)

7.12

Q:

How is the size of control valves for steam service determined?

A:

Control valves are specified by Cv or valve coefficients. The manufacturers of control valves provide these values (see Table 7.4). The Cv provided must exceed the Cv required. Also, Cv at several points of possible operation of the valve must be found, and the best Cv characteristics that meet the load requirements must be used, because controllability depends on this. For example. a quick-opening characteristic (see Fig. 7.2) is desired for on–off service. A linear characteristic is desired for general flow control and liquid-level control systems, whereas equal percentage trim is desired for pressure control or in systems where pressure varies. The control valve supplier must be contacted for the selection and for proper actuator sizing.

For the noncritical flow of steam (P1 < 2P2) [7],

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FIGURE 7.2 Typical control valve characteristics.

where

t; ts ¼ steam temperature and saturation temperature (for saturated steam,

t ¼ ts)

W ¼ steam flow, lb=h

Pt ¼ total pressure (P1 þ P2), psia

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7.13a

Q:

Estimate the Cv required when 60,000 lb=h of superheated steam at 900 F, 1500 psia flows in a pipe. The allowable pressure drop is 30 psi.

A:

Since this is a case of noncritical flow, from Eq. (8), substituting t ¼ 800 and ts ¼ 596, we have

60;000 ½1 þ 0:00065 ð900 596Þ&

Cv ¼ p

2:11 30 ð1500 þ 1470Þ

¼ 114

If the steam is saturated, t ¼ ts and Cv ¼ 95. We have to choose from the valve supplier’s catalog a valve that gives this Cv or more at 90–95% of the opening of the trim. This ensures that the valve is operating at about 90% of the trim opening and provides room for control.

7.13b

Q:

In a pressure-reducing station, 20,000 lb=h of steam at 200 psia, 500 F is to be reduced to 90 psia. Determine Cv.

A:

Use Eq. (9) for critical flow conditions:

(382 is the saturation temperature at 200 psia.)

7.14

Q:

Determine the valve coefficient for liquids. A liquid with density 45 lb=cu ft flows at the rate of 100,000 lb=h. If the allowable pressure drop is 50 psi, determine Cv.

A:

The valve coefficient for liquid, Cv, is given by [8]

r

s

Cv ¼ q D ð10Þ

P

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where

q ¼ flow, gpm

DP ¼ pressure drop, psi s ¼ specific gravity

Q:

How is cavitation caused? How is the valve sizing done to consider this aspect?

A:

Flashing and cavitation can limit the flow in a control valve for liquid. The pressure distribution through a valve explains the phenomenon. The pressure at the vena contracta is the lowest, and as the fluid flows it gains pressure but never reaches the upstream pressure. If the pressure at the port or vena contracta should drop below the vapor pressure corresponding to upstream conditions, bubbles will form. If the pressure at the exit remains below the vapor pressure, bubbles remain in the stream and flashing occurs.

A valve has a certain recovery factor associated with it. If the recovery of pressure is high enough to raise the outlet pressure above the vapor pressure of the liquid, the bubbles will collapse or implode, producing cavitation. Highrecovery valves tend to be more subject to cavitation [9]. The formation of bubbles tends to limit the flow through the valve. Hence the pressure drop used in sizing the valve should allow for this reduced capacity. Allowable pressure drop

where

Km ¼ valve recovery coefficient (depends on valve make) P1 ¼ upstream pressure, psia

rc ¼ critical pressure ratio (see Fig. 7.3)

Pv ¼ vapor pressure at inlet liquid temperature, psia

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FIGURE 7.3 Critical pressure ratios for water.

Full cavitation will occur if the actual DP is greater than DPall and if the outlet pressure is higher than the fluid vapor pressure. If the actual DP is less than DPall, the actual DP should be used for valve sizing. To avoid cavitation, select a valve with a low recovery factor (a high Km factor).

7.16

Q:

How are valves selected for laminar flow and viscous liquids?

A:

Calculate the turbulent flow Cv from Eq. (10) and the laminar Cv from [10]

Use the larger Cv in the valve selection. (m is the liquid viscosity in centipoise.)

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7.17a

Q:

Determine the pressure loss in a 3 in. schedule 80 line carrying water at 100 F and 2000 psia if the total equivalent length is 1000 ft. Flow is 38,000 lb=h.

A:

The expression for turbulent flow pressure drop of fluids (Reynolds number > 2100) is [11]

where

DP ¼ pressure loss, psi

f ¼ Darcy friction factor W ¼ flow, lb=h

Le ¼ equivalent length, ft (Q7.26 shows how the equivalent length can be computed)

v ¼ specific volume of fluid, cu ft=lb di ¼ tube inner diameter, in.

For water at 100 F and 2000 psia, from Table A4 (Appendix 3), v ¼ 0.016. In industrial heat transfer equipment such as boilers, superheaters, economizers, and air heaters, the fluid flow is generally turbulent, and hence we need not check for Reynolds number. (Q7.24 shows how Re can be found.) However, let us quickly check Re here:

Referring to Table 7.5, water viscosity, m, at 100 F is 1.645 lb=ft h

38,000

Re ¼ 15:2 2:9 1:645 ¼ 121;070

The inner diameter of 2.9 for the pipe was obtained from Table 5.6. For turbulent flow and carbon or alloy steels of commercial grade, f may be obtained from Table 7.6. Here f for a tube inner diameter of 2.9 in. is 0.0175. Substituting into Eq. (13) yields

DP ¼ 3:36 10 6 0:0175 ð38;000Þ2 100 02:016:95 ¼ 6:6 psi

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TABLE 7.5 Viscosity of Steam and Water (lbm=h ft)

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