Area 2 is 20–30 items: how air, water, steam, fuel and refrigerant are moved, cleaned and controlled. Since October 2025 it also holds ventilation and air quality, control concepts and whole refrigeration systems, which the old outline spread across other areas.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamPE Mechanical HVAC & Refrigeration
questions on the exam20–30
worked problems here20
areas, October 2025 outline2 of 4
Checked
What the area covers
Five topics in the specifications, in our words:
Air distribution and ductwork: duct sizing and losses, terminal units, diffusers and room air patterns.
Air quality and ventilation: outdoor air rates, filtration, dilution and mixing.
Fluid distribution and piping: hydronic loops, fuel oil and fuel gas, steam and condensate.
Control concepts: valves, dampers, sensors and actuators, reset schedules and control loops.
Refrigeration systems: cycles and system types, food processing and storage, refrigerant properties and machinery-room ventilation.
20 distribution and systems problems
0 / 20 answered · 0 correct
Question 1 of 20 · HVAC and Refrigeration Distribution and Systems · 20–30 items
An interior space with a design occupancy of 30 people relies on demand-controlled ventilation. The average CO2 generation rate per person is 0.0106 cfm. If the ambient outdoor CO2 concentration is 400 ppm, determine the minimum outdoor airflow required to maintain a steady-state zone CO2 setpoint of 1000 ppm.
17.7 cfm is the airflow for one person; the generation rate was not multiplied by 30 occupants.
53.0 cfm reads 600 ppm as 0.006 instead of 0.000600, a decimal slip that makes the allowed rise ten times too large: 0.318/0.006.
227.1 cfm divides by the sum of the concentrations, 1,400 ppm, instead of their difference.
318.0 cfm divides by the full 1,000 ppm setpoint, ignoring that outdoor air already carries 400 ppm.
530.0 cfm: 30 people generate 0.318 cfm of CO2, diluted by the 600 ppm allowed rise above outdoor air.
Question 2 of 20 · HVAC and Refrigeration Distribution and Systems · 20–30 items
A constant-volume air handling system delivers 12,000 cfm. The facility upgrades the existing filter bank, which had an operating pressure drop of 0.4 in. w.g., to a high-efficiency array with an operating pressure drop of 1.1 in. w.g. The combined efficiency of the fan and motor is 65%. Assuming airflow remains constant, calculate the resulting increase in electrical power required by the motor.
Given
Airflow = 12,000 cfm
Old filter pressure drop = 0.4 in. w.g.
New filter pressure drop = 1.1 in. w.g.
Fan and motor efficiency = 65 %
Find
Increase in motor electrical power (kW)
Solution
ΔP = 1.1 − 0.4 = 0.7 in. w.g.
P = Q·ΔP/(6,356·η) = 12,000 × 0.7/(6,356 × 0.65) = 2.033 hp
P = 2.033 hp × 0.746 kW/hp = 1.517 kW
0.641 kW multiplies by the 0.65 efficiency instead of dividing, so input power comes out lower than air power.
0.986 kW is the air power for the added 0.7 in. w.g.; it leaves out the 65% fan and motor efficiency.
1.517 kW: the extra 0.7 in. w.g. at 12,000 cfm is 2.033 hp of input at 65% combined efficiency, times 0.746 kW/hp.
2.033 is the increase in horsepower; it was not converted with 0.746 kW/hp.
2.385 kW uses the whole new 1.1 in. w.g. drop; the question asks for the increase over the old 0.4 in. w.g.
Question 3 of 20 · HVAC and Refrigeration Distribution and Systems · 20–30 items
Filter Model
Stage Type
E2 Efficiency
Pressure Drop (in. wg)
Model A
Prefilter
20%
0.15
Model B
Prefilter
40%
0.25
Model C
Final Filter
75%
0.35
Model D
Final Filter
85%
0.45
A rooftop air handling unit supplies conditioned air to a laboratory. The design requires a two-stage filter configuration that specifically maximizes the removal of particles in the E2 range (1.0 to 3.0 µm) while ensuring the combined initial static pressure drop strictly does not exceed 0.65 in. wg. Based on the manufacturer performance data provided in the table, select the optimal configuration.
B + C is within the limit at 0.60 in. wg, but its combined efficiency, 1 − 0.60 × 0.25 = 85%, is below A + D.
B + D would reach 91%, but its 0.70 in. wg drop breaks the 0.65 in. wg limit.
Three of the four pairs stay within 0.65 in. wg, so a compliant choice exists.
A + C has the lowest drop, 0.50 in. wg, but only 80% combined efficiency.
A + D gives 0.60 in. wg and 1 − 0.80 × 0.15 = 88% E2 efficiency, the best of the pairs that stay at or under 0.65 in. wg.
Question 4 of 20 · HVAC and Refrigeration Distribution and Systems · 20–30 items
A main pipe anchor must be designed for a carbon steel piping system incorporating a single untied expansion joint. The system operates at an internal pressure of 150 psig. The expansion joint has a bellows effective area of 50 in2 and a spring rate of 1,200 lb/in.
Under peak operating conditions, the joint will undergo an axial compression of 1.5 inches. The frictional force from the pipe guides is calculated to be 500 lbf.
Calculate the total axial force the main anchor must resist, accounting for both pressure thrust and thermal forces.
2,300 lbf leaves out the 7,500 lbf pressure thrust, which an untied bellows passes straight to the main anchor.
6,200 lbf subtracts the spring force from the thrust; compressing the bellows adds to the anchor load.
8,000 lbf adds thrust and friction but leaves out the 1,800 lbf needed to compress the bellows.
9,300 lbf adds thrust and spring force but leaves out the 500 lbf of guide friction.
9,800 lbf: pressure thrust 150 × 50 = 7,500 lbf, bellows spring force 1,200 × 1.5 = 1,800 lbf, and 500 lbf of guide friction all act on the anchor.
Question 5 of 20 · HVAC and Refrigeration Distribution and Systems · 20–30 items
A contractor submits a design for a 400-ft high-temperature hot water (HTHW) line. The piping will be installed at 50°F and operate at 350°F. The submittal states that because the pipe is heavily insulated, the effective pipe wall temperature will only reach 150°F, and it calculates the thermal expansion from this reduced temperature difference using the copper coefficient from the table below.
Identify the errors in the submittal and determine the correct estimated total axial thermal expansion for this carbon steel piping segment.
Material
α (in/in·°F)
Carbon steel
6.5 × 10−6
Copper
9.4 × 10−6
Given
Pipe length = 400 ft carbon steel
Installation temperature = 50 °F
Operating temperature = 350 °F
Carbon steel α = 6.5 × 10−6 in/(in·°F)
Find
Total axial thermal expansion (in)
Solution
ΔL = α · L · ΔT
ΔL = 6.5 × 10−6 in/(in·°F) × (400 × 12) in × (350 − 50) °F = 9.36 in
9.36 inches: 4,800 in of carbon steel at 6.5 × 10−6 in/in·°F over the full 300°F rise; insulation does not keep the wall below the water temperature.
4.68 inches halves the growth as if each end of the run moved half; the total axial expansion of the 400-ft segment is the full 4,800 × 6.5 × 10−6 × 300 = 9.36 in.
4.51 inches repeats both submittal errors: the copper coefficient and the 100°F rise to an assumed 150°F wall.
3.12 inches has the right coefficient but the submittal's 100°F rise; an insulated pipe wall runs at the fluid temperature.
0.78 is the expansion in feet; the 400-ft length was not converted to 4,800 inches.
Question 6 of 20 · HVAC and Refrigeration Distribution and Systems · 20–30 items
A chilled water pump delivers 450 gpm through a closed-loop piping network. The system comprises a chiller evaporator with a pressure drop of 15.0 ft wg and a cooling coil with a pressure drop of 25.0 ft wg. The straight piping and fittings are detailed in the table below. The piping has an inside diameter of 0.3355 ft and the water velocity is 11.3 ft/s. Over time, scaling on the water side increases the original Darcy-Weisbach friction factor of 0.018 by 25%. Calculate the new total dynamic head required by the pump.
Component
Quantity
Equivalent Length (ft/each)
Straight 4-inch pipe
1
300.0
Standard 90° elbow
10
10.0
Gate valve (fully open)
4
2.5
Given
Flow = 450 gpm
Chiller evaporator pressure drop = 15.0 ft wg
Cooling coil pressure drop = 25.0 ft wg
Straight 4-inch pipe equivalent length = 1 × 300.0 ft
10.9 ft wg is only the increase in head caused by scaling, not the new total head.
54.5 ft wg is the piping friction alone; the 15.0 ft chiller and 25.0 ft coil drops are in series and must be added.
79.9 ft wg counts only the 300 ft of straight pipe and leaves out 110 ft of fitting equivalent length.
83.6 ft wg uses the clean-pipe f = 0.018; the 25% scaling increase is not applied.
94.5 ft wg: 40.0 ft of fixed drops plus 54.5 ft of friction over 410 ft equivalent length at the scaled f = 0.0225.
Question 7 of 20 · HVAC and Refrigeration Distribution and Systems · 20–30 items
An industrial freezer utilizes an electric resistance defrost system that initiates a 20-minute timed defrost cycle every 4 hours, regardless of actual frost accumulation. The electric defrost heater draws 8.0 kW. During the defrost cycle, the compressor is turned off, but after the cycle, the compressor must run to remove the heat introduced by the heater. The refrigeration system operates with a coefficient of performance (COP) of 2.0. Assuming all heater energy directly enters the refrigerated space, calculate the total additional electrical energy consumed (heater plus compressor) over a 30-day month solely due to these timed defrosts.
Given
Defrost cycle = 20 min every 4 h
Heater power = 8.0 kW
Refrigeration COP = 2.0
Period = 30 days
Find
Additional monthly electrical energy (kWh)
Solution
N = (24/4) × 30 = 180 defrosts
Eh = 180 × 8.0 kW × (20/60) h = 480 kWh
Ec = Eh/COP = 480/2.0 = 240 kWh → E = 720 kWh
240 kWh counts only the compressor energy to remove the defrost heat, not the heater's own consumption.
480 kWh counts only the heater; the same heat then has to be removed by the refrigeration system.
720 kWh: 180 defrosts × 8.0 kW × 1/3 h = 480 kWh of heater energy, plus 240 kWh of compressor energy to pump that heat out at COP 2.0.
960 kWh doubles the heater energy, treating the compressor penalty as equal to it (COP of 1).
1,440 kWh multiplies the heater energy by the COP; compressor energy is the heat removed divided by COP.
Question 8 of 20 · HVAC and Refrigeration Distribution and Systems · 20–30 items
A VAV terminal serves a 20 ft × 24 ft conference room (9 ft ceiling, return grille on a 20 ft wall) with a design flow of 1,200 cfm and a minimum flow of 360 cfm (30%). Two ceiling diffusers are evaluated in the table below. The target is NC ≤ 35, and the throw at minimum flow must be at least 0.75 × (room length/2) to prevent short-circuiting. Which diffuser best meets the criteria?
Diffuser
Design flow (cfm)
Throw (ft)
NC
Min flow (cfm)
Throw (ft)
NC
X
1,200
15
30
360
6
25
Y
1,200
18
32
360
9
28
Given
Design flow = 1,200 cfm
Minimum flow = 360 cfm
Diffuser X: 15 ft throw, NC 30 at 1,200 cfm; 6 ft, NC 25 at 360 cfm
Diffuser Y: 18 ft throw, NC 32 at 1,200 cfm; 9 ft, NC 28 at 360 cfm
Room = 20 ft × 24 ft, 9 ft ceiling
Target NC ≤ 35
Minimum throw at low flow ≥ 0.75 × room length/2
Find
Best diffuser
Solution
throwmin = 0.75 × (24 ft/2) = 9 ft
X: 6 ft < 9 ft (fails); Y: 9 ft ≥ 9 ft, NC 32/28 ≤ 35 (passes)
A larger diffuser would lower outlet velocity and shorten throw at 360 cfm, making the low-flow problem worse.
Diffuser Y: its 9 ft throw at 360 cfm meets the 0.75 × 12 = 9 ft minimum, and NC 32 and 28 stay under 35.
NC 30 and NC 32 at design flow are both under the NC 35 target.
Both pass the NC limit, but only Y meets the low-flow throw requirement.
Diffuser X is quieter, but its 6 ft throw at minimum flow falls short of 9 ft, so supply air short-circuits to the return.
Question 9 of 20 · HVAC and Refrigeration Distribution and Systems · 20–30 items
Time
SP Sensor (in. w.g.)
VFD Speed (%)
Avg. Zone Damper (% open)
10:00
1.5
60
65
11:00
1.8
50
75
12:00
2.1
40
85
13:00
2.5
30
100
A facility engineer is reviewing a Building Management System (BMS) trend log for a variable air volume (VAV) air handling unit. The unit is programmed to maintain a duct static pressure setpoint of 1.5 in. w.g. by modulating the supply fan VFD. Over a 4-hour period, the logged data in the table is observed. Based on the system's response, which fault best explains the observed behavior?
Solution
Rising measured SP + falling VFD speed + dampers opening to 100 % = the controller is chasing a false high signal.
A sensor reading high makes the loop slow the fan (60% to 30%) while real duct pressure falls, so starved zones drive their dampers fully open.
Space pressurization does not raise the supply-duct static reading while fan speed falls.
One thermostat could open one damper; it cannot raise the duct static reading or slow the fan.
A stuck drive would not trend smoothly from 60% down to 30% over four hours.
A blowout would drop the measured static pressure and make the fan speed up, the opposite of the trend.
Question 10 of 20 · HVAC and Refrigeration Distribution and Systems · 20–30 items
In a chilled beam system, what safeguard is most important to prevent condensation on the chilled beam surfaces?
Solution
Condensation-free operation: chilled-water supply ≥ space dew point (typically about 57–60 °F) with automatic reset or shutoff.
Manual flow changes cannot follow humidity swings in real time, and lowering flow does not guarantee the coil stays above dew point.
Chilled beams normally have no drain pan to overflow, and a sensor that trips after water appears is a backup, not prevention.
Surface temperature alone says nothing without the space dew point, and IR monitoring is not a control strategy for the water loop.
Keeping the chilled water above the measured space dew point stops condensation before it forms; this is the primary safeguard for beams without drain pans.
Question 11 of 20 · HVAC and Refrigeration Distribution and Systems · 20–30 items
A 20-ft horizontal run of 4-inch Schedule 40 steel piping (OD = 4.5 inches, I = 7.23 in4) terminates at a perpendicular chiller nozzle. The run experiences an unconstrained axial thermal growth of 0.15 inches, which acts as a point load deflecting the perpendicular 60-inch connecting pipe segment. No anchor was installed to prevent this movement.
Assuming the connecting segment acts as a cantilever beam with the nozzle as the fixed support, calculate the maximum bending stress transferred to the nozzle connection.
Use the formula for cantilever bending stress: σ= (3 E c Δ)/L2, where E = 29 × 106 psi.
Given
Pipe = 4-inch Schedule 40, OD = 4.5 in., I = 7.23 in4
680 psi converts the 0.15 in. growth to feet while keeping L in inches.
2,719 psi drops the factor of 3 from the cantilever stiffness relation.
8,156 psi: σ = 3 × 29 × 106 × 2.25 × 0.15/60², with c as half the 4.5 in. OD and L in inches.
16,312 psi uses the full 4.5 in. OD for c; the extreme fiber is at the radius.
1,174,500 psi uses L = 5 ft against E in psi and Δ in inches, so the length term is 144 times too small.
Question 12 of 20 · HVAC and Refrigeration Distribution and Systems · 20–30 items
A 2-inch hydronic branch runs 25 feet of straight pipe with the fittings listed below. Using the equivalent length table provided, calculate the total effective length and compare it to a straight-pipe-only assumption. What is the percentage increase in friction head loss due to fittings, and how does this affect pump selection?
Fitting Type
Quantity
Equivalent Length (ft) Each
90° standard elbow
3
5
Gate valve (full open)
1
1
Tee (flow through branch)
2
15
Check valve
1
12
Given
2-in branch straight length = 25 ft
3 × 90° elbow, 5 ft each
1 × gate valve, 1 ft
2 × tee through branch, 15 ft each
1 × check valve, 12 ft
Find
Total effective length (ft) and % increase in head loss
Solution
Lfit = 3 × 5 + 1 + 2 × 15 + 12 = 58 ft
Leq = 25 + 58 = 83 ft
Δh/hstraight = 58/25 = 2.32 → +232 %
108 ft adds the 25 ft run twice, and fittings raise friction loss, not velocity head or the pressure rating.
The 83 ft length is right, but 58 ft is the added length, not a percentage; relative to 25 ft it is a 232% increase.
40 ft adds only the three elbows (15 ft) and leaves out the tees and valves; 60% is that partial 15 ft over 25 ft.
25 ft ignores the fittings, which in a short run add more resistance than the pipe itself.
83 ft: the fittings add 15 + 1 + 30 + 12 = 58 ft to 25 ft of pipe, and friction scales with length, so 58/25 = 232% more head loss.
Question 13 of 20 · HVAC and Refrigeration Distribution and Systems · 20–30 items
A two-way valve modulates chilled water flow through an air-side cooling coil that must hold a precise leaving-air temperature. Which control valve characteristic is most appropriate?
Solution
Coil capacity vs flow is concave (≈ 50 % flow → ≈ 80–90 % capacity); an equal-percentage valve is the inverse curve, giving a near-linear system response.
Quick-opening valves pass most of their flow in the first part of the stroke, suited to on/off service, not precise modulation.
A linear valve paired with a coil gives large capacity jumps near closed, because the first fraction of flow delivers most of the heat transfer.
Equal percentage offsets the coil's curve: a coil gives most of its capacity at low flow, so the valve opens that flow gently and gives a near-linear capacity response.
On-off control cycles between full and no flow and cannot hold a precise leaving-air temperature.
Question 14 of 20 · HVAC and Refrigeration Distribution and Systems · 20–30 items
A rooftop unit (RTU) control sequence specifies the following startup order: (A) verify smoke detector status and interlocks, (B) open the outdoor air (OA) damper to minimum position, (C) energize the supply fan, (D) stage compressors based on cooling demand. During commissioning, the building automation system (BAS) event log shows: 08:00:00 – smoke detector verified (A); 08:00:15 – compressor 1 energized (D); 08:00:45 – OA damper command issued (B); 08:01:10 – supply fan started (C). Inspection finds no mechanical or electrical faults, and every point was in automatic mode. What is the most likely root cause of this misordered sequence?
Solution
Required A → B → C → D; logged A → D → B → C with sound hardware in auto → program logic (missing proof/delay between steps).
Electrical faults were ruled out, and the BAS logged the compressor as a commanded event.
With hardware healthy and all points in auto, the BAS itself let step D run before B and C were proven — a missing interlock or delay in the program.
All points were in automatic mode, so no hand override could have bypassed the sequence.
The detector was verified first and the log shows no reset; a false alarm would stop the unit, not energize a compressor.
The damper command was issued and the stem rules out mechanical faults; a stuck actuator would not start the compressor early.
Question 15 of 20 · HVAC and Refrigeration Distribution and Systems · 20–30 items
A large blast-chill cold storage room requires a defrost strategy that minimizes product temperature rise while maintaining energy efficiency and operational reliability. The facility operates continuously with product temperatures maintained at -10°F. Evaluate the four defrost methods in the table below and recommend the most appropriate strategy with supporting justification including control requirements.
Defrost Method
Energy Use (kWh/cycle)
Defrost Time (min)
Product Temp Rise (°F)
Control Complexity
Off-cycle (natural)
0
120-180
8-12
Low
Electric resistance
28
25
6-8
Medium
Hot-gas
12
18
4-6
High
Reverse-cycle
15
22
5-7
High
Solution
Rank by the stated goals: product temperature rise (hot gas 4–6 °F lowest) → energy (hot gas 12 kWh vs electric 28) → time (hot gas 18 min).
Hot gas wins on the two stated priorities — 18 min and a 4–6°F rise — and uses 12 kWh per cycle, far below electric.
Off-cycle defrost relies on room air above freezing; in a −10°F room it barely works, and the table shows the largest product temperature rise.
Electric defrost uses 28 kWh per cycle, more than twice hot gas, and lets the product warm 6–8°F.
The table rates reverse-cycle control complexity as high, the same as hot gas, and it is slower with a larger temperature rise.
Control complexity is a cost, not a stated priority; hot gas is still better on product temperature rise and energy.
Question 16 of 20 · HVAC and Refrigeration Distribution and Systems · 20–30 items
What is the primary cause of pressure surges in HVAC liquid piping systems during pump startup?
Solution
Δp ≈ ρ·a·Δv (Joukowsky)
A sudden change of velocity Δv in a liquid column produces a pressure wave; soft starts and slow valves limit it.
Thermal expansion develops over minutes to hours and moves the pipe; it does not produce a surge at pump start.
Starting a pump accelerates a long column of water; its inertia turns the sudden velocity change into a pressure surge.
Entrained air causes noise and poor heat transfer; it is not what creates the startup pressure surge.
Cavitation is vapor collapse from low suction pressure; it damages the impeller but is not the startup surge mechanism.
Question 17 of 20 · HVAC and Refrigeration Distribution and Systems · 20–30 items
What is the primary purpose of an electronic expansion valve (EEV) in a modern HVAC refrigeration system?
Solution
Expansion device = pressure drop + metering; EEV holds superheat by electronic control of the opening.
An EEV meters refrigerant into the evaporator with a stepper motor to hold the superheat setpoint as load and head pressure change.
Noise reduction is not the function of an expansion valve; it is a metering device.
An EEV still needs clean, dry liquid; the filter drier stays in the liquid line.
Condensing pressure is held by fan cycling or a head-pressure control valve, not by the expansion device.
Question 18 of 20 · HVAC and Refrigeration Distribution and Systems · 20–30 items
A space CO2 sensor used for demand-controlled ventilation outputs 0–10 V for a nominal 0–2,000 ppm, but calibration shows it is nonlinear: 0 V = 0 ppm, 5 V = 900 ppm and 10 V = 2,000 ppm, linear between these points. The BAS still converts the signal with the nominal linear scaling of 200 ppm/V. When the space is actually at 1,500 ppm, what concentration does the BAS display?
Given
Nominal scaling = 0–10 V for 0–2,000 ppm (200 ppm/V)
Calibration: 0 V = 0 ppm, 5 V = 900 ppm, 10 V = 2,000 ppm
1,364 ppm scales 1,500 by 200/220, which converts slopes instead of finding the actual output voltage first.
1,455 ppm has the 45 ppm error with the wrong sign; the sensor sends more volts than nominal at 1,500 ppm, so the reading is high.
1,500 ppm assumes the nominal scaling is right; the calibration shows the sensor departs from it above 5 V.
1,545 ppm: on the 5–10 V segment (220 ppm/V) the sensor sends 5 + 600/220 = 7.73 V, which the BAS reads as 7.73 × 200.
1,667 ppm uses the 0–5 V slope of 180 ppm/V for the whole range; 1,500 ppm lies on the upper segment.
Question 19 of 20 · HVAC and Refrigeration Distribution and Systems · 20–30 items
A machinery room houses a chiller utilizing R-454B. The room measures 30 ft by 40 ft with a 12 ft ceiling. A catastrophic leak of 150 lb of refrigerant occurs, assuming instantaneous and perfect mixing. The table below provides standard parameters for common refrigerants. Determine the minimum mechanical ventilation rate required to ensure the room concentration drops below 25% of the lower flammability limit (LFL) exactly 15 minutes after the leak. Use the standard dilution equation C = C0 e−Qt/V.
114.1 cfm targets 50% of the LFL instead of the required 25%.
338.5 cfm uses log base 10 instead of the natural log that the exponential decay equation requires.
533.8 cfm treats the decay as linear, removing the excess concentration at a constant rate instead of exponentially.
779.4 cfm: the room must dilute from 10.42 to 4.63 lb/1,000 ft³, a ratio of 2.25, so Q = (14,400/15) × ln 2.25.
960.0 cfm is one room volume every 15 minutes; the required rate depends on the concentration ratio, not a single air change.
Question 20 of 20 · HVAC and Refrigeration Distribution and Systems · 20–30 items
A refrigeration system utilizing a fixed capillary tube restrictor is originally sized for a design ambient temperature of 95°F. The relationship between mass flow rate and pressure drop across a fixed restrictor is generally approximated by:
ṁ = CA √(2 ρΔP)
If the ambient temperature unexpectedly increases to 110°F, how does this specifically affect the system's overall cooling capacity?
Solution
qe = ṁ·(h1 − h4), h4 = h3 rises with condensing temperature
Higher condensing temperature sends warmer, higher-enthalpy liquid to the evaporator, so each pound absorbs less heat; capacity falls.
A higher compression ratio lowers volumetric efficiency, so the compressor moves less refrigerant, not more.
Nothing balances the two effects exactly; the loss of refrigeration effect dominates and capacity drops.
A fixed restrictor does not hold flow constant; flow changes with the pressure difference across it.
The capillary does pass a little more flow at the higher pressure difference, but the compressor's volumetric efficiency falls and the refrigeration effect shrinks.
Answer every question to see your score
Pressure bookkeeping in ducts and pipes
Most distribution problems are an energy balance written as pressures. In a duct the total pressure is the sum of static and velocity pressure,
TP = SP + VP, VP = (V/4,005)2 in. w.g. for standard air, V in fpm
and fans are rated on one or the other, so check which one a curve uses. Equal-friction sizing holds the loss per 100 ft constant; static-regain sizing slows the air after each branch so that the recovered velocity pressure offsets the next run's friction (ASHRAE Handbook—Fundamentals, Duct Design).
In water piping, head in feet of the pumped fluid converts to pressure through 2.31 ft of water per psi at standard density. Steam work adds one more step: steam tables are in absolute pressure, so convert a gauge reading first (gauge versus absolute is explained on FE Environmental fluid mechanics).
Common traps
Five pressure and piping mistakes
4,005 off standard air. The velocity-pressure constant assumes 0.075 lb/ft3; for other densities use VP = ρ(V/1,097)2.
Two-way and three-way valves. Two-way valves make the loop variable-flow, so the pump is usually controlled to a differential pressure (or a bypass is added); three-way valves keep the flow through the pump nearly constant.
Valve sizing in the wrong units. The flow coefficient relates gallons per minute to a pressure drop in psi,
Q = Cv · √(ΔP/SG)
Superheat and subcooling. Superheat is measured at the evaporator outlet against the saturation temperature at suction pressure, subcooling at the condenser outlet against saturation at liquid pressure; both need the same refrigerant's pressure–temperature data.
Flash at the trap. Condensate dropping to a lower pressure flashes part of itself to steam, which is why return lines are sized for two-phase flow.
No. Since October 2025 refrigeration systems are part of distribution and systems, and refrigeration equipment such as condensers and evaporators is in equipment and components.
How many distribution and systems questions are on the exam?
20–30 items of the 80, per the October 2025 specifications.
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