10.3 Valve Sizing, Pressure Drop, and Split-Range

What the exam will actually ask you to size

Topic 3.C is valve sizing, pressure drop, and split-range. The 2027 PE Control Systems exam still does not hand you IEC 60534-2-1 or ISA-75.01.01 as supplied standards. You are expected to treat the qualitative equations and factors as professional practice: liquid $C_v$, choked versus unchoked gas, $F_L$, $x_T$, piping geometry, rangeability, and the idea that two valves can share one analog signal.

ISA-5.1 (2024) may show a split-range note on a P&ID. IEC 61511-1 (2018) may constrain whether a throttling valve is even allowed as a safety final element. Neither document computes $C_v$.

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Incompressible liquid: the $C_v$ you must be able to use

For turbulent, non-choked, incompressible liquid in US customary units, the working relation is:

$$C_v = Q \sqrt{\frac{SG}{\Delta P}}$$

where $Q$ is U.S. gallons per minute, $SG$ is specific gravity relative to water at 60 °F, and $\Delta P$ is psi across the valve at that flow. $C_v$ is the flow of 60 °F water in gpm that the valve passes at 1 psi drop when fully open (manufacturer’s definition). Metric $K_v$ is the SI cousin; do not mix units inside the square root.

The full IEC/ISA equation multiplies in a numerical constant $N_1$ for other unit systems, a piping factor $F_p$, and Reynolds-number corrections for viscous flow. On the exam, the qualitative message is: capacity scales with $Q$ and with $1/\sqrt{\Delta P}$. Double the required flow at the same ΔP and you need roughly double $C_v$. Quarter the ΔP at the same flow and you need twice the $C_v$ because $\sqrt{1/4} = 1/2$ in the denominator’s inverse.

Worked numeric example

A hydrocarbon liquid, $SG = 0.81$, must pass 80 gpm with 9 psi across the valve at that flow (non-choked, turbulent, no reducers).

$$C_v = 80 \sqrt{0.81 / 9} = 80 \sqrt{0.09} = 80 \times 0.30 = 24$$

Select a body whose rated $C_v$ is larger than 24 so design flow occurs at a sensible travel (often about 70–80% open for a globe, not 100%, and not 20%). If the only catalog globe nearby is $C_v = 40$, design travel is roughly in the middle of the characteristic — usable. If you only have $C_v = 25$, you are sizing on the stop and you have no margin for fouling, gravity tolerance, or a pump that is a few psi low.

Second check with water: $Q = 200$ gpm, $SG = 1.0$, $\Delta P = 25$ psi $\Rightarrow C_v = 200 \sqrt{1/25} = 200 \times 0.20 = 40$. Same algebra, cleaner numbers.

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When liquid flow chokes: $F_L$

$F_L$ (liquid pressure-recovery factor) measures how much pressure recovers from the vena contracta to the valve outlet. A globe might have $F_L \approx 0.85$–$0.9$ (low recovery). A butterfly or ball might have $F_L \approx 0.55$–$0.7$ (high recovery).

Choked (or fully cavitating) liquid flow occurs when $\Delta P$ reaches a limit of the form $F_L^2 (P_1 - r_c P_v)$. After that, lowering $P_2$ does not increase flow. High-recovery valves hit that limit at a smaller $\Delta P$. That is another reason a butterfly is a poor flashing/cavitating letdown valve: it recovers, collapses cavities, and erodes, and its effective $C_v$ saturates.

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Gas and vapor: subcritical versus critical (qualitative)

For compressible flow the driving “drop” is not simply $P_1 - P_2$ once the jet goes sonic at the vena contracta.

Define the pressure-drop ratio $x = \Delta P / P_1$. The expansion factor $Y$ declines as $x$ increases. Choked (critical) flow occurs when $x$ reaches $F_k x_T$ (specific-heat ratio factor times the valve’s pressure-drop ratio factor $x_T$). Beyond choke, mass flow no longer rises as $P_2$ falls.

$x_T$ is small for high-recovery rotaries (choke early) and larger for globes (more of the drop is usable before choke). If a stem gives $P_1 = 100$ psia and $P_2 = 20$ psia, $x = 0.80$. A globe with $x_T \approx 0.7$ is choked; a butterfly with $x_T \approx 0.3$ was already choked at a much higher $P_2$. You cannot “create more steam flow” by opening a downstream vent once the valve is choked — you can only raise $P_1$ or install a larger $C_v$ / different trim.

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Piping geometric factors

Valves are often smaller than the line: 6 in pipe, 4 in globe, reducers on both ends. $F_p$ (piping geometry factor) is less than 1.0; the assembly passes less flow than the same valve on a matched test stand. There is a related $F_{LP}$ when reducers and recovery interact near choke. Exam implication: if you compute $C_v = 24$ from line ΔP and then drop in reducers, you may need the next body size or a higher-capacity trim. Do not ignore $F_p$ because “the tag says 4 inch.”

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Rangeability and split-range travel versus $C_v$

Inherent rangeability is the ratio of maximum to minimum controllable flow at constant ΔP (often 30:1 to 50:1 for a good globe). Installed rangeability is smaller because ΔP collapse at high flow and seat/plug geometry at low flow both bite. When the process needs 100:1, one valve is usually the wrong answer.

Split-range maps one 4–20 mA (or 0.2–1.0 bar) signal onto two travels. Example:

• Valve A (small), equal-percentage, rated $C_v = 15$: 0–100% travel over 4–12 mA.
• Valve B (large), equal-percentage, rated $C_v = 120$: 0–100% travel over 12–20 mA.

At 12 mA, A is fully open ($C_v = 15$ available) and B is still closed. Combined wide-open capacity is about 135, not 120. If A can still control down to about $C_v \approx 0.5$, the pair’s rangeability is on the order of $135 / 0.5 \approx 270$, which is why split-range exists.

Travel versus $C_v$ is not linear with milliamps unless you characterize it. Two equal-percentage valves butted at 12 mA without overlap produce a gain bump at the handoff. Typical practice overlaps (for example A 4–13 mA, B 11–20 mA) so combined $C_v$ is smooth. Split-range can also be two valves doing opposite jobs (steam 4–12 mA opening, coolant 12–20 mA opening) — that is a control strategy split, still the same travel-versus-signal idea.

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Pressure-drop allocation: do not put ALL the drop in the valve

A pump, a static lift, piping, and a valve share one energy balance. If you put all frictional drop in the valve at design flow, you paid for a larger pump, you may cavitate or make noise, and the valve is working as an orifice, not a controller. If you put almost none of the drop in the valve, $C_v$ becomes huge, travel sits near the seat, and installed gain and rangeability collapse (the classic oversized valve).

Professional practice is a substantial but not total share at the design flow: often on the order of one-third of the dynamic (frictional) losses, or a minimum of about 10 psi when the arithmetic would otherwise give a few psi on a large liquid valve — rules of thumb, not IEC 61511 clauses. The pump must still deliver valve drop + piping + static at $Q_{\max}$.

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Exam trap: $Q_{\max}$ at $\Delta P_{\max}$ when they do not coexist

Size using coincident conditions: the ΔP that actually exists at that flow.

Worked trap: a pump curve gives 80 psig at the high-flow operating point. Static lift is 20 psi. Piping at $Q_{\max} = 150$ gpm drops 35 psi. Valve ΔP at 150 gpm is $80 - 20 - 35 = 25$ psi. Water, $SG = 1$:

$$C_v = 150 \sqrt{1/25} = 30$$

At $Q_{\min} = 30$ gpm, piping drop is about $35 \times (30/150)^2 = 1.4$ psi, pump head is higher, and valve ΔP might be 50–60 psi. If you wrongly size 150 gpm at 60 psi:

$$C_v = 150 \sqrt{1/60} \approx 19.4$$

That valve cannot pass 150 gpm when only 25 psi is left at high flow. The trap stem sounds conservative (“use the maximums”). The coincident physics is the opposite: maximum flow occurs with minimum valve ΔP on a pump-plus-pipe system.

Size the valve at $Q_{\max}$ with $\Delta P$ at $Q_{\max}$, then check low-flow cavitation, actuator thrust at shutoff ΔP, and whether split-range is needed for turndown.