Energy Recovery · Study · PE Mechanical: HVAC and Refrigeration · FE → PE Prep
Energy Recovery
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Sensible and total (enthalpy) recovery effectiveness, enthalpy wheels, heat pipes, and run-around loops, with recovered-energy and economizer savings.
3 concepts
G. Energy Recovery
Sensible and Total Energy-Recovery Effectiveness
How sensible, latent, and total effectiveness define an HRV/ERV, how to find leaving conditions and recovered energy, and how the supply/exhaust airflow ratio and frost control bound it.
Ventilation air is a relentless load: every cfm of outdoor air you bring in must be dragged from outdoor conditions to room conditions, and in a tight, code-ventilated building that load can rival the envelope. Air-to-air energy recovery answers it by letting the air you are throwing away precondition the air you are bringing in — the exhaust stream gives up heat (and, in an enthalpy device, moisture) to the incoming outdoor stream before either leaves the building. The single number that rates how well it does this is effectiveness, and the PE exam tests whether you can move fluently between effectiveness, leaving conditions, and recovered energy. The NCEES PE Mechanical Reference Handbook builds the framework in §9.2.7 (Heat-Recovery Ventilator) and §9.2.8 (Energy-Recovery Ventilator), and that is the notation you will have on exam day.
What effectiveness actually measures
Effectiveness is the ratio of the energy you actually recovered to the most you could ever recover if the device were infinitely large. The ceiling is set by the smaller of the two airstreams — you cannot transfer more heat into the supply than the exhaust can give up, or vice versa — so the maximum always carries Cmin
, the smaller of the two air-capacity rates
m˙cp
. The driving potential is the full inlet difference between the two entering streams,
t3−t1
for sensible heat (state 1 = outdoor air entering, state 3 = exhaust air entering, following the handbook's airstream numbering). Effectiveness is therefore dimensionless, between 0 and 1, and — crucially — it is a property of the device and the flow rates, not of the weather: the same wheel that is
Sensible, latent, and total — three effectivenesses, one device
A bare heat-recovery ventilator (HRV) — a sealed plate exchanger or heat-pipe — moves only sensible heat, so it has a single sensible effectiveness εs built on temperature. An energy-recovery ventilator (ERV) also moves moisture across a vapor-permeable medium, so it earns two more numbers: a latent effectiveness εL built on humidity ratio w, and a total (enthalpy) effectiveness εt built on enthalpy h. They are not independent — total recovery is the sum of the sensible and latent parts — but the three effectivenesses generally differ because a given medium transfers heat and mass at different rates. Always read which one a problem gives you: a temperature swing needs εs, a humidity swing needs εL, and a load expressed in Btu/h of mixed air needs εt.
Most exam questions hinge on the supply leaving state — the air the device hands to the cooling or heating coil downstream. Rearranging the definition gives it directly: the supply leaves at its inlet condition plus effectiveness times the available swing, scaled by the capacity ratio Cmin/(m˙scps). When the supply stream is itself the smaller one (or flows are balanced), that ratio is 1 and the result collapses to the clean form t2=t1+εs(t3−t1). The same algebra in w and h gives the leaving humidity and enthalpy. Compute the supply leaving state first; everything else — coil load, recovered energy, frost check — follows from it.
Recovered energy with the standard-air coefficients
Once you have the supply leaving state, the recovered duty is just the enthalpy (or temperature, or moisture) change carried by the supply airflow. At sea-level standard air (ρ≈0.075lb/ft3, cp≈0.24Btu/lb⋅°F) the familiar HVAC coefficients apply: 1.08 for sensible Btu/h per cfm·°F, 0.68 for latent Btu/h per cfm·grain (with Δw in grains of moisture per lb of dry air, or equivalently 4760Δw=0.68×7000Δw with Δw in lbw/lbda — both embed hfg≈1058Btu/lb), and 4.5 for total Btu/h per cfm·(Btu/lb). These bake in standard density — at altitude or in hot supply ducts they drift, and you must rebuild them from 60m˙Δ with the actual density. The maximum, against which effectiveness is defined, uses the smaller airflow Qmin and the full inlet difference.
Recovery devices rarely see exactly equal flows. Buildings run a positive pressure (supply > exhaust) to keep infiltration out, and some exhaust is taken locally (restrooms, hoods) and never reaches the device. The imbalance matters two ways. First, it fixes Cmin and m˙min — the smaller stream caps the recovery, so a device starved of exhaust air recovers less even at high rated effectiveness. Second, manufacturer-rated effectiveness is quoted at balanced flow; pushing the supply/exhaust ratio away from 1.0 changes the actual effectiveness, and AHRI 1060 ratings come with correction factors (OACF for outdoor-air correction, EATR for exhaust-air transfer ratio). On the exam, when flows are unequal, identify the smaller airflow, build q˙max on it, and apply the capacity-ratio form of the leaving-state equation.
In cold climates the warm, moist exhaust stream can be chilled below its dew point and then below freezing inside the device, so condensate freezes on the exhaust-side surfaces — blocking flow on a plate or heat-pipe core, or icing the cold sector of a wheel. The threshold depends on exhaust humidity and on how cold the outdoor air drives the exhaust leaving temperature t4; the more effective the device and the colder the outdoor air, the lower t4 and the greater the frost risk. Designers hold a frost-threshold outdoor temperature (often near 5 to −10∘F depending on indoor RH) and below it invoke a control: preheat the entering outdoor air, modulate the wheel speed or bypass part of the supply to keep t4 above freezing, or run a periodic defrost. Each costs recovered energy or fan/heater energy, so frost control is a real derate on cold-design-day performance, not an afterthought.
t4=t3−εsm˙ecpeCmin(t3−t1)⇒frost risk when t4<32∘F at the exhaust dew point
Exam strategy
Lock the airstream numbering first — 1 outdoor in, 2 supply out, 3 exhaust in, 4 exhaust out — then the equations write themselves. Decide which effectiveness the problem invokes: temperature only ⇒εs, moisture ⇒εL, mixed/enthalpy load ⇒εt. Find the supply leaving state, then multiply the supply airflow by the change using 1.08, 0.68, or 4.5 — but only at sea-level standard air; otherwise rebuild from 60m˙Δ. If flows are unequal, the smaller one owns both q˙max and the capacity ratio. A fast sanity check: leaving supply temperature must land between the two inlets (t1≤t2≤t3), and recovered energy can never exceed q˙max.
Problem. A sealed plate HRV serves a building with balanced airflow of 4,000cfm each way. On a winter design day the outdoor air enters at t1=10∘F and the building exhaust enters at t3=72∘F. The HRV has a sensible effectiveness εs=0.75. Find the temperature of the preheated supply air and the rate of heat recovered.
Solution. Flows are balanced, so Cmin/(m˙scps)=1
t2=10+0.75(72−10)=56.5∘F
Unbalanced flow — supply and exhaust leaving temperatures
Problem. A summer-mode device has supply (outdoor) airflow Qs=5,000cfm and exhaust airflow Qe=4,000cfm
ERV total recovery from sensible and latent parts
Problem. An enthalpy wheel runs balanced at 4,000cfm. Outdoor air (state 1) is 95∘F db / 76∘F wb (h1=39.3Btu/lb
Common pitfalls
•Confusing sensible, latent, and total. A temperature swing needs εs, a humidity swing needs εL, and a mixed-air Btu load needs εt — they are different numbers for the same device. Fix: identify the property that changes (t, w, or h) before picking the effectiveness.
•Building q˙max on the supply airflow when the exhaust is smaller (or vice versa). Effectiveness is referenced to Cmin. Fix: always compare m˙s
•Using the 1.08/0.68/4.5 coefficients at altitude or on hot supply air. They embed sea-level standard density (0.075 lb/ft³). Fix: at altitude or elevated temperature rebuild from 60m˙Δ with the actual air density.
•Treating effectiveness as a function of the weather. ε is set by the device and the flow ratio, not by t1 or t3. Fix: hold ε fixed across seasons; only the inlet difference (the driving potential) changes.
•Putting Δw in lbw/lbda into the 0.68 latent coefficient (which expects grains). Fix: use 0.68
•Ignoring frost. On the cold design day a high-effectiveness device drives the exhaust leaving temperature below freezing and ices up. Fix: check t4 against 32 °F at the exhaust dew point and credit the frost-control derate (preheat, bypass, or defrost).
•Assuming balanced flow makes the capacity ratio drop out for the exhaust side too. The ratio Cmin/(m˙ecpe)
References
NCEES PE Mechanical Reference Handbook — §9.2.7 Heat-Recovery Ventilator and §9.2.8 Energy-Recovery Ventilator
ASHRAE Handbook—HVAC Systems and Equipment, Ch. 26 Air-to-Air Energy Recovery Equipment — Source of the airstream numbering and the sensible/latent/total effectiveness framework.
AHRI Standard 1060 — Performance Rating of Air-to-Air Exchangers for Energy Recovery Ventilation — Defines rated effectiveness, OACF, and EATR at balanced flow.
ASHRAE Handbook—Fundamentals, Ch. 1 Psychrometrics — Moist-air enthalpy and humidity-ratio properties used in latent/total calculations.
Enthalpy Wheels, Heat Pipes and Run-Around Loops
How rotary wheels, fixed-plate cores, heat pipes, and pumped run-around loops differ in what they recover, how they leak, and the fan power they cost.
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Economizer and Recovered-Energy Savings
Air-side and water-side economizers, dry-bulb versus enthalpy changeover, free-cooling hours, and how to turn recovered or free cooling into an annual energy-and-dollar saving.
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exhaust out (°F);
Cmin=min(m˙scps,m˙ecpe)
is the smaller air-capacity rate. Because numerator and denominator share the same
m˙
basis, the ratio is dimensionless regardless of whether
m˙
is taken in lb/min or lb/h.
Built on humidity ratio w (lbw/lbda). m˙min = smaller dry-air mass flow (lb/min). Zero for an HRV; nonzero only for vapor-permeable ERV media.
Built on moist-air enthalpy h (Btu/lbda). Total recovery equals the sum of sensible and latent recovery.
Conditioned outdoor air handed to the downstream coil. Reduces to t2=t1+εs(t3−t1) when the supply is the smaller (or balanced) stream.
Used for the frost check: t4 below 32 °F at the exhaust dew point freezes condensate inside the device.
Btu/h. The 1.08 coefficient assumes sea-level standard air (ρ=0.075 lb/ft³, cp=0.24). Qs = supply cfm.
Btu/h. Use 0.68 with Δw in grains per lbda, or 4760=0.68×7000 with Δw in lbw/lbda — the two embed the same hfg≈1058Btu/lb. Standard air.
Btu/h. The 4.5 coefficient (60×0.075) assumes standard air; total = sensible + latent.
Denominator of εs. Built on the smaller airflow Qmin and the full inlet temperature difference.
Air-side heat capacity rate (Btu/h·°F) with m˙ in lb/min and Q in cfm; the 60 converts min→h. The smaller stream caps recovery. In the dimensionless effectiveness ratios the 60 cancels, so Cmin/(m˙scps) may be evaluated on either basis.
Total recovery is sensible plus latent; with hfg≈1058Btu/lb (the same value behind the 0.68/4760 latent coefficients) this ties εt back to εs and εL at balanced flow.
as equal on both sides, find the supply and exhaust leaving temperatures.
Solution. The smaller airflow is the exhaust, so m˙min=m˙e and Cmin/(m˙scps)=m˙e/m˙s=Qe/Qs=4000/5000=0.80.
Supply leaving: t2=t1+εs(0.80)(t3−t1)=95+0.70(0.80)(75−95)=95+0.70(0.80)(−20)=95−11.2=83.8∘F.
Exhaust leaving (exhaust is Cmin, ratio = 1): t4=t3−εs(t3−t1)=75−0.70(75−95)=75+14=89.0∘F.
Energy balance check: heat gained by supply =1.08(5000)(95−83.8)=60,480Btu/h; heat lost by exhaust =1.08(4000)(89.0−75)=60,480Btu/h — equal, as required.
Supply precooled to 83.8∘F, exhaust warmed to 89.0∘F.
t2=95+0.70(0.80)(75−95)=83.8∘F
,
104.6
grains/lb), and return air (state 3) is
75∘F
/
50%
RH (
h3=28.1Btu/lb
,
64.6
grains/lb). The wheel has
εs=0.75
and
εL=0.65
. Find the recovered sensible, latent, and total cooling, and the implied total effectiveness.
Solution. Recovered sensible: q˙s=1.08(4000)(0.75)(95−75)=1.08(4000)(0.75)(20)=64,800Btu/h.
Recovered latent: grain difference =104.6−64.6=40.0 grains/lb, so q˙L=0.68(4000)(0.65)(40.0)=70,720Btu/h.
Recovered total: q˙t=q˙s+q˙L=64,800+70,720=135,520Btu/h≈1.36×105Btu/h (11.3 tons).
Implied total effectiveness: q˙t,max=4.5(4000)(39.3−28.1)=4.5(4000)(11.2)=201,600Btu/h, so εt=135,520/201,600=0.672.
Sanity check: εt=0.67 falls between εs=0.75 and εL=0.65, as the enthalpy-weighted blend must — and the latent part is the larger share, exactly why an ERV (not a bare HRV) is specified in a humid climate.
εt=4.5(4000)(11.2)64,800+70,720=0.672
and
m˙e
and use the smaller for both
q˙max
and the capacity ratio.
with grains, or
4760
with
lbw/lbda
— never mix the two.
equals 1 only for whichever stream is
Cmin
. Fix: apply the correct ratio to each leaving-temperature equation.