The largest area, 24–36 items: choosing and checking the machines of an HVAC plant, from cooling towers and boilers to fans, coils and energy recovery. The October 2025 outline adds thermal storage, and cavitation, curves and NPSH for pumps.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamPE Mechanical HVAC & Refrigeration
questions on the exam24–36
worked problems here22
areas, October 2025 outline3 of 4
Checked
What the area covers
Seven topics in the specifications, in our words:
Cooling towers and fluid coolers.
Boilers and furnaces: efficiencies, fuels and combustion.
Heat exchangers: shell-and-tube and plate-and-frame.
Condensers and evaporators: chillers, variable refrigerant flow, heat pumps and thermal storage.
Pumps, compressors and fans: the laws, efficiency, selection, curves, cavitation and NPSH.
Cooling and heating coils.
Energy recovery: enthalpy wheels, heat pipes and run-around loops.
22 equipment and components problems
0 / 22 answered · 0 correct
Question 1 of 22 · HVAC Equipment and Components · 24–36 items
Airflow (cfm)
Fan Static Pressure (in. wg)
8,000
1.60
8,500
1.45
9,000
1.31
9,500
1.16
10,000
1.00
An existing air distribution system operates at an original equilibrium point of 10,000 cfm at 1.00 in. wg static pressure. A proposed indoor air quality upgrade adds a high-efficiency filter bank that contributes exactly 0.50 in. wg of static pressure drop independent of the existing system flow dynamics. The fan operates at a fixed rotational speed, and the new system curve assumes the added filter pressure drop is strictly additive to the existing parabolic dynamic losses. Using the tabulated fan performance data, determine the new operating airflow rate.
Given
Original operating point = 10,000 cfm at 1.00 in. wg
Added filter drop = 0.50 in. wg
Fan curve: 8,000 cfm 1.60; 8,500 cfm 1.45; 9,000 cfm 1.31; 9,500 cfm 1.16; 10,000 cfm 1.00 in. wg
Find
New operating airflow (cfm)
Solution
k = 1.00/10,0002 = 1.0 × 10−8
SPsys = 1.0 × 10−8·Q2 + 0.50
Q = 9,000: SPsys = 0.81 + 0.50 = 1.31 in. wg = fan SP → operating point
10,500 cfm would need more flow than before; adding resistance at a fixed fan speed moves the operating point left of 10,000 cfm.
At 9,500 cfm the system needs 1.40 in. wg but the fan makes only 1.16, so the fan cannot hold this flow.
9,000 cfm: the new system curve 1.0 × 10−8 Q² + 0.50 gives 1.31 in. wg there, exactly what the fan develops at 9,000 cfm.
At 8,500 cfm the system needs only 1.22 in. wg while the fan makes 1.45, so the flow keeps rising past this point.
Question 2 of 22 · HVAC Equipment and Components · 24–36 items
Property
Natural Gas
#2 Fuel Oil
HHV to LHV Ratio
1.11
1.06
NOx Guarantee (LHV basis)
0.105 lb/MMBtu
0.108 lb/MMBtu
An industrial facility must select between firing Natural Gas or #2 Fuel Oil in a new boiler to meet a state site-wide NOx permit limit of 0.100 lb/MMBtu (evaluated on a Higher Heating Value basis). The burner manufacturer provides emission performance guarantees exclusively on a Lower Heating Value (LHV) basis. Based on the provided fuel properties, which option complies with the regulatory limit?
Given
NOx permit limit = 0.100 lb/MMBtu (HHV basis)
Natural gas HHV/LHV ratio = 1.11
#2 fuel oil HHV/LHV ratio = 1.06
Natural gas NOx guarantee = 0.105 lb/MMBtu (LHV basis)
This compares the LHV guarantees directly with an HHV limit; converted to HHV, natural gas passes.
Oil has the smaller HHV/LHV ratio, so it gains less from the conversion and stays above the limit.
The table gives both HHV/LHV ratios, which is all the conversion needs.
Oil still exceeds the limit after conversion, 0.1019 against 0.100 lb/MMBtu.
Natural gas: 0.105/1.11 = 0.0946 lb/MMBtu on an HHV basis, under the 0.100 limit; oil converts to 0.108/1.06 = 0.1019.
Question 3 of 22 · HVAC Equipment and Components · 24–36 items
An engineer must select a supply fan for an industrial facility located at an elevation of 5,000 ft, where the actual air density is 0.060 lb/ft3. The system requires an actual airflow of 15,000 CFM to overcome an actual total static pressure of 2.0 in. w.g. The manufacturer provides the standard performance data table below based on standard sea-level air density (0.075 lb/ft3). Based on the required performance and the provided standard catalog data, what will be the selected fan's operating RPM and actual brake horsepower (BHP) at the high-elevation site?
RPM
CFM
Static Pressure (in. w.g.)
Standard BHP
700
15,000
1.5
4.5
800
15,000
2.0
6.0
900
15,000
2.5
7.8
1000
15,000
3.0
9.5
Given
Site air density = 0.060 lb/ft³
Required airflow = 15,000 acfm
Actual total static pressure = 2.0 in. w.g.
Catalog density = 0.075 lb/ft³
Catalog at 15,000 cfm: 700 rpm 1.5 in. 4.5 bhp; 800 rpm 2.0 in. 6.0 bhp; 900 rpm 2.5 in. 7.8 bhp; 1000 rpm 3.0 in. 9.5 bhp
The speed is right, but 7.80 BHP is the standard-air power; the lighter air needs 0.8 times as much.
1000 RPM applies the density ratio twice (2.0 × 1.25² ≈ 3.1 in. w.g.) and then skips the power correction.
900 RPM and 6.24 BHP: 2.0 in. w.g. at 0.060 lb/ft³ equals 2.5 in. w.g. at standard density, which the catalog meets at 900 RPM; its 7.8 BHP then scales down by 0.060/0.075.
800 RPM and 6.00 BHP reads the catalog at the actual 2.0 in. w.g. without the density correction, so the fan falls short at altitude.
4.80 BHP corrects the power for density but keeps the uncorrected 800 RPM selection.
Question 4 of 22 · HVAC Equipment and Components · 24–36 items
A ventilation system is served by two identical fans operating in parallel. The single fan performance curve is defined by SP = 5.0 − 2.5 × 10−8·Q2. The system's static pressure requirement is strictly proportional to the square of the volumetric flow rate, requiring 1.00 in. w.g. at 10,000 cfm. Determine the reduction in total volumetric airflow if one of the two parallel fans fails, assuming the system curve remains constant.
5,589 cfm: two fans meet the system at 17,541 cfm, one fan alone at 11,952 cfm.
8,771 cfm assumes flow halves when one fan stops; the remaining fan moves up its curve to a higher flow than half.
9,112 cfm treats the system curve as linear in flow (1.00 in. w.g. per 10,000 cfm): two fans then meet it at 21,394 cfm and one fan at 12,283 cfm; the stem says pressure varies with the square of flow.
11,952 cfm is the flow left with one fan running, not the reduction.
17,541 cfm is the original flow of both fans together, not the loss when one fails.
Question 5 of 22 · HVAC Equipment and Components · 24–36 items
A commercial water-cooled chiller serves a 400-ton building load. At design conditions, the compressor motor draws 285 kW of electrical power. The condenser water system is designed for a 12°F cooling tower range. Assuming 100% of the electrical energy input to the compressor is rejected as heat to the condenser water along with the evaporator load, calculate the required design condenser water circulation flow rate.
Given
Building load = 400 tons
Compressor power = 285 kW
Cooling tower range = 12°F
Find
Condenser water flow (gpm)
Solution
q = 400 × 12,000 + 285 × 3,412 = 5,772,000 Btu/h
gpm = q/(500·ΔT) = 5,772,000/(500 × 12) = 962 gpm
162 gpm covers only the 285 kW of compressor heat and leaves out the evaporator load.
800 gpm sizes the condenser water for the 400-ton evaporator load alone; the compressor heat also goes to the tower.
962 gpm: the tower rejects 4,800,000 Btu/h of building load plus 972,000 Btu/h of compressor heat at a 12°F range.
1,154 gpm uses a 10°F range instead of the stated 12°F.
1,200 gpm is the 3 gpm/ton rule of thumb, which assumes about a 10°F range and a typical heat of compression.
Question 6 of 22 · HVAC Equipment and Components · 24–36 items
Water Source
Chloride Concentration (ppm)
Make-up Water
35
Basin Water
140
A cooling tower system circulates 2,000 gpm with a calculated evaporation rate of 24.0 gpm. Damaged internal eliminators result in a severe drift loss equivalent to 0.15% of the total circulation rate. To determine the actual cycles of concentration (COC), a field water analysis was performed, yielding the chloride values in the table. Assuming chloride does not evaporate, what controlled blowdown rate (gpm) is required to maintain this exact COC?
Given
Circulating flow = 2,000 gpm
Evaporation = 24.0 gpm
Drift = 0.15 % of circulation
Make-up chloride = 35 ppm
Basin chloride = 140 ppm
Find
Controlled blowdown rate (gpm)
Solution
COC = 140/35 = 4.0, D = 0.0015 × 2,000 = 3.0 gpm
B + D = E/(COC − 1) = 24.0/3 = 8.0 gpm → B = 5.0 gpm
32.0 gpm is the total makeup (24.0 evaporation + 8.0 blowdown and drift), not the controlled blowdown the question asks for.
11.0 gpm adds the drift on top of the 8.0 gpm total instead of subtracting it.
8.0 gpm is the total of blowdown and drift; the drift already removes 3.0 gpm of concentrated water.
6.0 gpm divides evaporation by the cycles (24/4) instead of by cycles minus one.
5.0 gpm: at 4 cycles, blowdown plus drift must equal 24/3 = 8.0 gpm, and the damaged eliminators already lose 3.0 gpm as drift.
Question 7 of 22 · HVAC Equipment and Components · 24–36 items
An air handling unit delivers 12,000 cfm of supply air against an initial total static pressure of 3.0 in. w.g. The fan operates with a constant static efficiency of 65%. Over time, severe particulate fouling on the cooling coil increases the total system static pressure drop by 25%. If the fan speed is adjusted to maintain the original 12,000 cfm design flow rate, what is the approximate additional brake horsepower (bhp) required to overcome the fouled coil?
2.18 hp: the fouling adds 0.75 in. w.g. at the same 12,000 cfm, and 12,000 × 0.75/(6,356 × 0.65) = 2.18.
3.46 hp scales power with pressure to the 1.5 power, a fan-law relation for a speed change on a fixed system; here flow is held and the system changed.
4.90 hp scales the 8.71 hp original power with the square of the pressure ratio, 8.71 × (1.25² − 1); at constant cfm power rises in direct proportion to pressure.
8.71 hp is the original fan power at 3.0 in. w.g., not the increase.
10.89 hp is the total power at 3.75 in. w.g.; the question asks only for the added power.
Question 8 of 22 · HVAC Equipment and Components · 24–36 items
Parameter
Value
Expansion tank gauge pressure
12.0 psig
Atmospheric pressure
14.7 psia
Water temperature
200°F
Water vapor pressure at 200°F
11.5 psia
Water specific weight at 200°F
60.1 lbf/ft³
Elevation of tank water level above pump centerline
8.5 ft
Total friction loss from tank to pump
2.3 ft
A closed-loop hydronic heating system has a pressurized expansion tank connected directly upstream of the circulating pump suction. The system operates with the following parameters. Calculate the available Net Positive Suction Head (NPSHa) at the pump inlet. Assume the velocity head at the pump suction is negligible.
Given
Expansion tank pressure = 12.0 psig
Atmospheric pressure = 14.7 psia
Water temperature = 200°F
Vapor pressure = 11.5 psia
Specific weight γ = 60.1 lbf/ft3
Tank level above pump = 8.5 ft
Friction loss = 2.3 ft
Find
NPSH available (ft)
Solution
hp = (12.0 + 14.7) × 144/60.1 = 63.97 ft, hv = 11.5 × 144/60.1 = 27.55 ft
NPSHa = hp + z − hf − hv = 63.97 + 8.5 − 2.3 − 27.55 = 42.62 ft
7.40 ft uses the 12.0 psig gauge pressure; NPSH needs absolute pressure because vapor pressure is absolute.
25.62 ft subtracts the 8.5 ft elevation; the tank level is above the pump, so it adds head.
34.12 ft leaves out the 8.5 ft of positive static head; the tank level is above the pump: 63.97 − 2.3 − 27.55.
41.28 ft converts psi to feet with 62.4 lbf/ft³; at 200°F water weighs 60.1 lbf/ft³.
42.62 ft: absolute tank pressure 26.7 psia is 63.97 ft, plus 8.5 ft of elevation, minus 2.3 ft of friction and 27.55 ft of vapor pressure.
Question 9 of 22 · HVAC Equipment and Components · 24–36 items
Duct System Component
Equivalent Length (ft)
Existing supply and return ductwork
300
Existing diffusers and fittings
100
New balancing dampers (retrofit)
50
New high-efficiency filters (retrofit)
150
An air distribution system originally delivers 5,000 CFM. The fan operates at 1,200 RPM with a performance curve defined by Pfan = 4.0 - 0.00000008 Q2, where Pfan is in inches of water gauge (in. wg) and Q is in CFM. The system undergoes a retrofit that adds new components, increasing the total equivalent length of the ductwork. Assuming system static pressure is strictly proportional to total equivalent length at a given flow, calculate the new fan speed required to maintain the original 5,000 CFM.
1,200 RPM assumes the speed stays the same because the airflow is the same; the higher system resistance needs more speed.
1,342 RPM: the system now needs 3.0 in. wg at 5,000 cfm, and the speed-scaled curve 4.0(N/1,200)² − 2.0 reaches it when (N/1,200)² = 1.25.
1,470 RPM scales speed with the square root of the pressure ratio, 1,200 × √1.5, as if the whole fan curve shifted with the system.
1,800 RPM scales speed with the 1.5 pressure ratio directly, a linear rule the fan laws do not support.
2,700 RPM multiplies by the square of the pressure ratio, which is the wrong direction for the pressure–speed relation.
Question 10 of 22 · HVAC Equipment and Components · 24–36 items
A water-cooled condenser rejects 1,500,000 Btu/h to a cooling tower. The condenser water circulates at 300 gpm and enters the tower at 95°F. Using the heat-balance relation for cooling towers in USCS units, compute the required tower Range (°F). (Use the approximation that 1 gpm = 8.33 lb/min and Cp for water ≈ 1.0 Btu/lb·°F.)
Given
Heat rejection Q = 1,500,000 Btu/h
Condenser water flow = 300 gpm
Entering tower water = 95°F
1 gpm = 8.33 lb/min
cp = 1.0 Btu/lb·°F
Find
Cooling tower range (°F)
Solution
Range = q/(500·gpm) = 1,500,000/(500 × 300) = 10.0 °F
600.2 leaves out the 60 min/h conversion: 1,500,000/(8.33 × 300).
105.0°F adds the range to 95°F as if 95°F were the cold-water temperature; water enters the tower hot at 95°F.
85.0°F is the tower's leaving water temperature (95 − 10), not the range the question asks for.
10.0°F: 1,500,000 Btu/h over 500 × 300 gpm, where 500 is 8.33 lb/gal × 60 min/h × 1.0 Btu/lb·°F.
5.0°F uses a constant of 1,000 instead of 500, doubling the assumed heat capacity of the flow.
Question 11 of 22 · HVAC Equipment and Components · 24–36 items
What characteristic most distinguishes modular boilers from conventional single large boilers?
Solution
Modular plant: N small boilers staged by load → better seasonal efficiency + redundancy.
Operating temperature is set by the boiler design and code class, not by splitting capacity into modules.
More units mean more burners and controls to service; maintenance is not reduced.
Multiple modules add headers, controls and venting; first cost is often similar or higher.
Several small modules stage on as load rises, so each runs near its best efficiency instead of one large boiler cycling at low fire.
Question 12 of 22 · HVAC Equipment and Components · 24–36 items
A seawater cooling application requires a compact heat exchanger with high thermal performance. The water carries moderate biofouling, and the design must allow periodic cleaning with little downtime. Which type is more appropriate, a brazed-plate or a gasketed-plate unit, and why?
Solution
Fouling service → serviceable exchanger (gasketed plates, titanium for seawater). Brazed packs cannot be opened, and copper brazing corrodes in chlorides.
A seawater loop runs at modest pressure; the deciding issue is fouling, where a brazed unit cannot be opened.
Strainers remove debris, not biofilm, so fouling still has to be cleaned off the plates.
Pressure drop is similar for similar plate patterns; the reason to choose gasketed plates is that they can be opened.
Biofouling and silt in seawater often need mechanical cleaning, which a sealed brazed pack cannot get.
A gasketed plate-and-frame unit can be opened and its plates scrubbed, and titanium plates resist seawater; the trade is gasket replacement.
Question 13 of 22 · HVAC Equipment and Components · 24–36 items
Which type of chiller would be most appropriate for a high-capacity application (over 1000 tons) with limited machine room space?
Solution
Above ~1,000 tons: centrifugal (single or dual compressor) gives the smallest footprint per ton.
Reciprocating chillers top out at a few hundred tons; reaching 1,000 tons would take several large machines.
Scroll compressors are small; scroll chillers are modular units far below 1,000 tons each.
Absorption chillers reach large sizes but are bulky and heavy per ton, a poor fit for a tight machine room.
Centrifugal machines deliver over 1,000 tons per unit with the highest capacity per floor area.
Question 14 of 22 · HVAC Equipment and Components · 24–36 items
In a heat pump system operating in heating mode, what component serves as the condenser?
The compressor raises refrigerant pressure; it does not condense refrigerant.
The expansion device drops the liquid pressure ahead of the evaporator; it is not a heat exchanger.
In heating mode the reversing valve sends hot discharge gas to the indoor coil, which condenses it and heats the room air.
The outdoor coil is the condenser in cooling mode; in heating mode it is the evaporator, absorbing heat from outside air.
Question 15 of 22 · HVAC Equipment and Components · 24–36 items
What is the primary advantage of using a direct expansion (DX) evaporator compared to a flooded evaporator in refrigeration systems?
Solution
DX: small charge, simple oil return. Flooded: higher h, closer approach, large charge.
DX coils can suffer poor distribution and superheat hunting at low load; low-load behavior is not their advantage.
A DX evaporator feeds only the refrigerant that boils in the tubes, so it holds far less charge than a shell flooded with liquid.
Flooded evaporators have the higher coefficient because every tube is wetted; the DX superheat region transfers heat poorly.
DX circuits push refrigerant through long tube passes, usually with more refrigerant-side pressure drop than a flooded shell.
Question 16 of 22 · HVAC Equipment and Components · 24–36 items
Which design feature makes a tubular heat exchanger particularly suitable for handling fluids with high fouling tendencies?
Solution
Fouling service → put the dirty fluid in straight tubes (removable bundle or tube-side access) for mechanical cleaning.
Thin walls slightly improve conduction but do nothing against fouling and reduce the allowance for cleaning and corrosion.
A high coefficient does not help once deposits form; cleanability is what matters for fouling service.
Compact, high-area designs have narrow passages that foul quickly and are harder to clean.
Straight tubes can be rodded, brushed or hydro-blasted end to end, which is what heavy-fouling fluids need.
Question 17 of 22 · HVAC Equipment and Components · 24–36 items
What is the primary advantage of a variable geometry diffuser in a centrifugal compressor?
Solution
Surge limit at low flow → variable geometry (diffuser/inlet vanes) or speed control widens the stable range.
Narrowing the diffuser passage at low load keeps gas velocity up and delays surge, so the compressor turns down further.
At full load the diffuser is fully open, so it brings no efficiency gain there.
A movable diffuser adds an actuator and control sequence; it makes control more complex, not simpler.
Maximum capacity is set by impeller size and speed; the variable diffuser matters at low load.
Question 18 of 22 · HVAC Equipment and Components · 24–36 items
Which type of fan is most appropriate for kitchen exhaust applications with grease-laden air?
Solution
Grease exhaust: centrifugal, backward-inclined (upblast or utility set), motor out of the airstream.
A vane-axial fan usually puts the motor in the airstream, exposing it to hot grease vapor.
A backward-inclined centrifugal fan can have its motor and belt outside the grease-laden airstream, and its blades shed grease better than forward-curved wheels.
Forward-curved wheels have many small cupped blades that load up with grease and lose balance.
Propeller fans cannot develop the static pressure of a hood-and-duct system, and their motor sits in the airstream.
Question 19 of 22 · HVAC Equipment and Components · 24–36 items
A shell‑and‑tube heat exchanger in operation shows a measured shell‑side pressure drop significantly lower than the design prediction, and the overall heat transfer coefficient U is 20% below the design value. The design specified baffle spacing of 24 inches, but field inspection reveals baffles installed at 36‑inch spacing. Which of the following is the most likely cause of the reduced U, and what corrective action should be recommended?
Solution
Shell-side h ∝ (crossflow velocity)≈0.6; velocity ∝ 1/baffle spacing → 36 in vs 24 in lowers both ho and ΔPshell.
Going to 12 in. overshoots the design, sharply raising pressure drop and vibration risk; the fix is the specified 24 in.
Baffle cut was not found to deviate; changing it would leave the spacing error in place.
Nothing in the inspection points to sealing strips, while the 50% wider spacing fully explains the symptoms.
Fouling would cut U, but it cannot explain the low shell-side pressure drop, and the inspection found the real deviation.
With baffles 36 in. apart the shell fluid crosses the bundle more slowly, and the lower velocity cuts both the film coefficient and the pressure drop, matching both symptoms.
Question 20 of 22 · HVAC Equipment and Components · 24–36 items
Parameter
Coil A
Coil B
Airflow (cfm)
10,000
10,000
Face velocity (fpm)
400
800
An engineer evaluates two finned-tube cooling coils for an air handling unit processing identical airflows. Coil A has an airside pressure drop of 0.50 in. wg. Based on the operating differences in the table, what is the expected approximate pressure drop for Coil B, and by what factor does the fan power needed to overcome the coil increase compared with Coil A?
Given
Airflow = 10,000 cfm (both coils)
Coil A face velocity = 400 fpm
Coil B face velocity = 800 fpm
Coil A pressure drop = 0.50 in. wg
Find
Coil B pressure drop (in. wg) and fan power factor
Solution
ΔPB = ΔPA(VB/VA)2 = 0.50 × (800/400)2 = 2.00 in. wg
P ∝ Q·ΔP → at equal cfm, ×4
4.00 in. wg cubes the velocity ratio; the cube law is for fan speed changes, not coil resistance.
2.00 in. wg: doubling face velocity quadruples the drop, and at the same cfm the power to push air through the coil rises by the same factor of 4.
Same cfm does not mean same drop; the smaller face area of coil B doubles the velocity through it.
The drop is right, but ×8 applies the fan-law cube; at constant cfm power follows the pressure drop directly.
This scales pressure drop linearly with velocity; coil drop grows roughly with the velocity squared.
Question 21 of 22 · HVAC Equipment and Components · 24–36 items
A chilled-water coil processes 8,000 cfm of standard air entering at 80°F DB and a humidity ratio of 0.0120 lbw/lbda. The leaving air is at 55°F DB and a humidity ratio of 0.0085 lbw/lbda. Using qL = 4,840 × cfm × ΔW for the latent load and 8.34 lb/gal for water, calculate the latent load removed (Btu/h) and the condensate rate (gal/h).
135,520 Btu/h and 15.1 gal/h: 36,000 lb/h of dry air drops 0.0035 lb/lb, condensing 126 lb/h of water.
Both values use the entering humidity ratio 0.0120 instead of the 0.0035 drop across the coil.
The latent load is right, but 126.0 is the condensate in lb/h; it still needs dividing by 8.34 lb/gal.
Both values are per minute: the 60 min/h in the dry-air mass flow was left out.
216,000 Btu/h is the sensible load, 1.08 × 8,000 × 25°F, not the latent load.
Question 22 of 22 · HVAC Equipment and Components · 24–36 items
Option
Configuration
Primary Material
Joint/Seal Material
Option 1
Brazed Plate
316 Stainless Steel
Copper
Option 2
Plate & Frame
304 Stainless Steel
FKM (Viton)
Option 3
Shell & Tube
Carbon Steel
Welded (None)
Option 4
Brazed Plate
316 Stainless Steel
Brass
An engineer is selecting a heat exchanger to serve as an evaporator for a large industrial ammonia (NH3, R-717) refrigeration system operating at a design pressure of 250 psig. Based on the off-the-shelf manufacturer specifications provided in the table, which option is the most appropriate selection for this application?
Solution
Ammonia (R-717): no copper or copper alloys; steel or stainless with welded joints or ammonia-compatible seals.
An all-welded carbon-steel shell-and-tube unit has no copper, brass or elastomer seal for ammonia to attack, the standard choice for R-717.
FKM (Viton) gaskets swell and break down in ammonia; plate units for R-717 use compatible elastomers or welded plate pairs.
Brass is a copper alloy, so it has the same incompatibility with ammonia as copper brazing.
Both use copper-based joining, the one family of materials that must be kept out of ammonia systems.
Copper brazing is attacked by ammonia once any moisture is present, so the joints fail.
Answer every question to see your score
Ratings you convert between
Equipment data comes in several currencies. One ton of refrigeration is 12,000 Btu/h, and one watt is 3.412 Btu/h, so
EER = 3.412 × COP, kW/ton = 12/EER
For a heat pump on the same cycle, the heat delivered is the heat absorbed plus the work, so the heating COP is the cooling COP plus one. A boiler's AFUE is a seasonal figure that includes off-cycle losses, while combustion efficiency is a steady-state flue-gas measure; the two are not interchangeable in an annual fuel estimate.
Common traps
Five rating and selection mistakes
Approach is to the wet bulb. A cooling tower's approach is leaving water minus entering-air wet bulb, not dry bulb, and no tower cools water below that wet bulb (ASHRAE Handbook—HVAC Systems and Equipment, Cooling Towers).
Pump head in feet of the pumped fluid. Converting to psi needs the specific gravity:
Δp (psi) = H (ft) × SG/2.31
Effectiveness on the smaller stream. Energy-recovery and heat-exchanger effectiveness divide the actual transfer by the maximum possible, which is set by the stream with the smaller capacity rate Cmin.
LMTD for the real arrangement. Counterflow LMTD is the reference; multipass shell-and-tube units need a correction factor F below one.
A heat pump's condenser moves. In heating mode the coils swap jobs; check which coil rejects heat before applying a capacity.