D6 · HVAC and Refrigeration Distribution and Systems

Foundation lesson · 18 minute study session

Make the control action match the disturbance.

Connect sensors, actuators, valve pressure drop and reset schedules to stable system behavior.

By the end of this lesson

Explain feedback, proportional action and integral action.
Calculate liquid-valve pressure drop from a supplied flow coefficient.
Interpret an explicit outdoor-temperature reset schedule.

Follow the feedback loop.

A sensor measures the controlled variable, a controller compares it with a setpoint, and an actuator changes a valve, damper or speed. State whether increasing the output opens or closes the device. Fail-open and fail-closed choices depend on the stated safety objective; they are not universal.

Two meanings of reset.

Integral action accumulates control error and can remove a steady offset. An outdoor reset schedule instead changes a temperature setpoint as outdoor conditions change. They solve different problems. Identify the process, setpoint and disturbance before choosing a controller response.

A valve needs pressure drop to control flow.

For a liquid valve, $C_v$ relates flow to valve pressure drop and specific gravity. At the stated design flow, valve authority compares the valve drop with the valve-plus-branch drop. Low authority can distort installed behavior; pump control and the rest of the network still matter.

A decision map for this lesson

Choose a control action that matches the disturbance

Keep valve sizing, authority, reset and failure response as separate decisions.

Choose the governing relationship and catch the wrong route
Condition or question	Governing decision	Error to avoid
Liquid valve flow or pressure drop	Use the stated $C_v$ , specific gravity and pressure-drop relation.	Ignoring the liquid property basis or mixing head with psi.
Valve authority at design flow	Use $a_v=\Delta p_v/(\Delta p_v+\Delta p_{other})$ on one branch basis.	Using valve drop alone as authority, or comparing drops at different flows.
A linear outdoor-reset schedule	Interpolate between the supplied outdoor and supply-temperature pairs.	Interpolating only the supply values without locating the outdoor state.
Offset, saturation or loss of power	Match integral action, anti-windup or the specified failure position to the stated problem.	Assuming one universal valve fail position serves every safety sequence.

On a narrow screen, scroll across the table to compare the decision and the error.

Relationships and their limits

Liquid valve coefficient

\dot V=C_v\sqrt{\Delta p_v/SG};\qquad \Delta p_v=SG(\dot V/C_v)^2

gpm and psi with the conventional U.S. $C_v$ definition. A liquid relation; use a separate method for gas or steam.

Stated branch authority

a_v=\frac{\Delta p_v}{\Delta p_v+\Delta p_{branch}}

Both pressure drops at the same stated design flow; branch term excludes the valve.

Linear reset between supplied endpoints

T_s=T_{s,1}+\frac{T_o-T_{o,1}}{T_{o,2}-T_{o,1}}(T_{s,2}-T_{s,1})

Interpolate only within the specified operating interval unless extrapolation is authorized.

Symbols and unit basis
Symbol	Meaning	Units or basis
$C_v$	Conventional liquid valve flow coefficient	gpm at 1 psi drop for SG = 1
$\Delta p_v$	Valve pressure drop	psi
$SG$	Liquid specific gravity	dimensionless
$a_v$	Valve authority	dimensionless
$T_s$	Supply-temperature setpoint	°F
$T_o$	Outdoor temperature	°F
$\dot V$	Liquid volumetric flow	gpm
$\Delta p_{branch}$	Other-branch pressure drop excluding valve	psi

One decision, fully worked

Apply the method.

Water flows at 20 gpm through a valve with $C_v$ = 10. Other branch components lose 8 psi at that flow. The reset schedule specifies 180°F supply at 0°F outdoors and 120°F supply at 60°F outdoors. Determine valve authority and the setpoint at 30°F outdoors.

Use SG = 1 for the stated water calculation.
Add the valve and other branch drops in the authority denominator.
Interpolate the supplied reset schedule at its midpoint.

\Delta p_v=(20/10)^2=4\ \mathrm{psi};\qquad a_v=\frac4{4+8}=0.333

T_s=180+\frac{30-0}{60-0}(120-180)=150^{\circ}\mathrm F

Decision

Valve authority is 0.333 and the scheduled supply setpoint is 150°F. These values do not establish controller tuning or prove the full system is stable.

Independent check

At half the outdoor interval, the supply setpoint lies halfway between 180°F and 120°F. The valve contributes one-third of the 12 psi branch drop.

Specialist extension

Check valve authority on the actual design-flow boundary

Valve authority compares valve loss with total controlled-branch loss at the defined design condition. It is not a universal recommendation for a single fixed value. Greater authority can improve the controllability of a stated valve characteristic while increasing pumping burden. Separately, a saturated actuator can allow a PI integral term to accumulate; an explicitly specified anti-windup strategy addresses that condition rather than changing a sensor calibration.

A new applied case

At design flow, a water valve loses 2 psi and the other branch components lose 6 psi. A replacement valve would lose 6 psi at that same target flow. Use the supplied authority definition.

a_{old}=2/(2+6)=0.25

a_{new}=6/(6+6)=0.50

\Delta p_{branch,old}=8;\quad\Delta p_{branch,new}=12\ \mathrm{psi}

Decision. Authority rises from 0.25 to 0.50, while required design-flow branch pressure drop rises from 8 to 12 psi.

Cross-check. Changing the valve does not prove the existing pump will preserve the target flow. Check the revised system operating point.

Unit basis. Pressure drops are psi at the same design flow; authority is dimensionless and uses the controlled-branch boundary defined by the problem.

NCEES PE Mechanical Reference Handbook v2.1; see the module reference path.

Open targeted practice in the study studio →

Transfer the method

HVAC and Refrigeration Distribution and Systems · HV-LC-D6

Try a new decision.

Select one

Water with SG = 1 flows at 30 gpm through a liquid valve with $C_v$ = 15. Other branch components lose 6 psi at this flow. Define valve authority as valve drop divided by valve-plus-other-branch drop. The valve authority is most nearly:

A0.600B0.250C0.667D0.400

A lesson check supports practice, not a readiness score. Your response stays in this page session and clears on reload.

Catch the common errors

Applying the liquid $C_v$ equation to steam without the required flow model.
Including the valve twice in total branch pressure drop.
Confusing integral reset with outdoor setpoint reset.

Handbook navigation drill

Find it, then explain it.

Search: Valve Flow Coefficient

Locate the liquid valve relation and identify its units. Then distinguish a proportional-plus-integral controller from an outdoor reset schedule.

Check your navigation

Section 11.3.8 gives valve-flow guidance. A supplied reset schedule defines its own endpoints; do not invent them from the valve equation.

Reference path: §11.1–11.3, pp. 555–561; §11.3.8, p. 560, Valve Flow Coefficient. These are printed handbook page numbers; section names and search terms remain the primary navigation method. See technical review status.

Deeper application · Decision Lab

A safety latch changes what the controller may do

Read the stated sequence as a set of permissions. A normal temperature error does not authorize bypassing an active protective state.

Choose the method, predict the changed-condition result and inspect the complete cross-check. Sign in before starting to keep your first response and later reflection in the separate reasoning notebook.

Open the original workshop →

Theory sources

NCEES PE Mechanical Reference Handbook v2.1: §9.3.1, p. 486; §11.3.8, Valve Flow Coefficient
Mechanical and Electrical Equipment for Buildings, 12th ed.: mechanical systems and energy principles. Supporting theory; not an examination reference.

Worked examples and lesson prose are original ReasonSpan^™ material. Obtain authorized publications separately; this preview does not reproduce the supplied books or handbook.