Energy Recovery

4% of exam

Sensible and total (enthalpy) recovery effectiveness, enthalpy wheels, heat pipes, and run-around loops, with recovered-energy and economizer savings.

3 concepts

Exam areas

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

C_{min}

, the smaller of the two air-capacity rates

\dot m c_p

. The driving potential is the full inlet difference between the two entering streams,

t_3 - t_1

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

75\%

effective in January is about

75\%

effective in July.

\varepsilon_s = \frac{\dot q_s}{\dot q_{s,\max}} = \frac{\dot m_s c_{ps}\,(t_2 - t_1)}{C_{min}\,(t_3 - t_1)} = \frac{\dot m_e c_{pe}\,(t_3 - t_4)}{C_{min}\,(t_3 - t_1)}

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

\varepsilon_s

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

\varepsilon_L

built on humidity ratio

w

, and a total (enthalpy) effectiveness

\varepsilon_t

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

\varepsilon_s

, a humidity swing needs

\varepsilon_L

, and a load expressed in Btu/h of mixed air needs

\varepsilon_t

\varepsilon_L = \frac{\dot m_s\,(w_2 - w_1)}{\dot m_{min}\,(w_3 - w_1)}, \qquad \varepsilon_t = \frac{\dot m_s\,(h_2 - h_1)}{\dot m_{min}\,(h_3 - h_1)}

From effectiveness to leaving conditions

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

C_{min}/(\dot m_s c_{ps})

. 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

t_2 = t_1 + \varepsilon_s (t_3 - t_1)

. 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.

t_2 = t_1 + \varepsilon_s\,\frac{C_{min}}{\dot m_s c_{ps}}\,(t_3 - t_1) \;\xrightarrow{\;\dot m_s = \dot m_{min}\;}\; t_1 + \varepsilon_s\,(t_3 - t_1)

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 (

\rho \approx 0.075\ \text{lb/ft}^3

, moist-air

c_p = 0.24 + 0.45w \approx 0.244\ \text{Btu/lb·°F}

) the handbook's standard-air coefficients apply (PE Mechanical Reference Handbook v2.1 §9.1.7, p. 454):

1.10

for sensible Btu/h per cfm·°F,

4{,}840

for latent Btu/h per cfm per

\text{lb}_w/\text{lb}_{da}

(built on

h_{fg}\approx1{,}076\ \text{Btu/lb}

; with

\Delta w

in grains of moisture per lb of dry air use

4{,}840/7{,}000 = 0.691

), and

4.5

for total Btu/h per cfm·(Btu/lb). These bake in standard density — at altitude use the handbook's factors (

C_{x,alt} = C_{x,0}\,P/P_0

, §9.1.8:

0.92

4{,}027

, and

3.74

at 5,000 ft), and in hot supply ducts rebuild them from

60\,\dot m\,\Delta

with the actual density. The maximum, against which effectiveness is defined, uses the smaller airflow

Q_{min}

and the full inlet difference.

\dot q_s = 1.10\,Q_s\,(t_2-t_1), \quad \dot q_L = 4{,}840\,Q_s\,(w_2-w_1), \quad \dot q_t = 4.5\,Q_s\,(h_2-h_1)

The supply/exhaust airflow ratio

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

C_{min}

and

\dot 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

\dot q_{max}

on it, and apply the capacity-ratio form of the leaving-state equation.

\dot q_{s,\max} = 60\,\dot m_{min} c_p (t_3 - t_1) = 1.10\,Q_{min}\,(t_3 - t_1), \qquad \dot m_{min} = \min(\dot m_s,\, \dot m_e)

Frost control basics

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

t_4

; the more effective the device and the colder the outdoor air, the lower

t_4

and the greater the frost risk. Designers hold a frost-threshold outdoor temperature (often near

5

-10\,^\circ\text{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

t_4

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.

t_4 = t_3 - \varepsilon_s\,\frac{C_{min}}{\dot m_e c_{pe}}\,(t_3 - t_1) \;\Rightarrow\; \text{frost risk when } t_4 < 32\,^\circ\text{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

\Rightarrow \varepsilon_s

, moisture

\Rightarrow \varepsilon_L

, mixed/enthalpy load

\Rightarrow \varepsilon_t

. Find the supply leaving state, then multiply the supply airflow by the change using

1.10

4{,}840

(or

0.691

per grain), or

4.5

— but only at sea-level standard air; at altitude use the handbook's

C_{x,0}\,P/P_0

factors, otherwise rebuild from

60\,\dot m\,\Delta

. If flows are unequal, the smaller one owns both

\dot q_{max}

and the capacity ratio. A fast sanity check: leaving supply temperature must land between the two inlets (

t_1 \le t_2 \le t_3

), and recovered energy can never exceed

\dot q_{max}

Key equations

Sensible effectiveness

\varepsilon_s = \frac{\dot m_s c_{ps}(t_2 - t_1)}{C_{min}(t_3 - t_1)} = \frac{\dot m_e c_{pe}(t_3 - t_4)}{C_{min}(t_3 - t_1)}

Ratio of actual to maximum sensible recovery. $t_1$ outdoor in, $t_2$ supply out, $t_3$ exhaust in, $t_4$ exhaust out (°F); $C_{min}=\min(\dot m_s c_{ps},\dot m_e c_{pe})$ is the smaller air-capacity rate. Because numerator and denominator share the same $\dot m$ basis, the ratio is dimensionless regardless of whether $\dot m$ is taken in lb/min or lb/h.

Latent (moisture) effectiveness

\varepsilon_L = \frac{\dot m_s(w_2 - w_1)}{\dot m_{min}(w_3 - w_1)}

Built on humidity ratio $w$ ( $\text{lb}_w/\text{lb}_{da}$ ). $\dot m_{min}$ = smaller dry-air mass flow (lb/min). Zero for an HRV; nonzero only for vapor-permeable ERV media.

Total (enthalpy) effectiveness

\varepsilon_t = \frac{\dot m_s(h_2 - h_1)}{\dot m_{min}(h_3 - h_1)}

Built on moist-air enthalpy $h$ ( $\text{Btu/lb}_{da}$ ). Total recovery equals the sum of sensible and latent recovery.

Supply leaving temperature

t_2 = t_1 + \varepsilon_s\,\frac{C_{min}}{\dot m_s c_{ps}}\,(t_3 - t_1)

Conditioned outdoor air handed to the downstream coil. Reduces to $t_2 = t_1 + \varepsilon_s(t_3-t_1)$ when the supply is the smaller (or balanced) stream.

Exhaust leaving temperature

t_4 = t_3 - \varepsilon_s\,\frac{C_{min}}{\dot m_e c_{pe}}\,(t_3 - t_1)

Used for the frost check: $t_4$ below 32 °F at the exhaust dew point freezes condensate inside the device.

Recovered sensible heat

\dot q_s = 1.10\,Q_s\,(t_2 - t_1) = 60\,\dot m_s c_{ps}(t_2 - t_1)

Btu/h. The $1.10$ coefficient assumes sea-level standard air ( $\rho=0.075$ lb/ft³, $c_p = 0.24 + 0.45w \approx 0.244$ ; handbook §9.1.7). $Q_s$ = supply cfm.

Recovered latent heat

\dot q_L = 4{,}840\,Q_s\,(w_2 - w_1) = 0.691\,Q_s\,(\text{gr}_2 - \text{gr}_1)

Btu/h. Use $4{,}840 = 60(0.075)(1{,}076)$ with $\Delta w$ in $\text{lb}_w/\text{lb}_{da}$ (handbook §9.1.7), or $0.691 = 4{,}840/7{,}000$ with $\Delta w$ in grains per $\text{lb}_{da}$ — the two embed the same $h_{fg}\approx1{,}076\ \text{Btu/lb}$ . Standard air.

Recovered total heat

\dot q_t = 4.5\,Q_s\,(h_2 - h_1) = \dot q_s + \dot q_L

Btu/h. The $4.5$ coefficient ( $60\times0.075$ ) assumes standard air; total = sensible + latent.

Maximum sensible recovery

\dot q_{s,\max} = 1.10\,Q_{min}\,(t_3 - t_1)

Denominator of $\varepsilon_s$ . Built on the smaller airflow $Q_{min}$ and the full inlet temperature difference.

Capacity rate

C = 60\,\dot m c_p = 60\,Q\,\rho\,c_p, \qquad C_{min} = \min(C_s, C_e)

Air-side heat capacity rate (Btu/h·°F) with $\dot m$ in lb/min and $Q$ in cfm; the $60$ converts min $\to$ h. The smaller stream caps recovery. In the dimensionless effectiveness ratios the $60$ cancels, so $C_{min}/(\dot m_s c_{ps})$ may be evaluated on either basis.

Effectiveness decomposition

\varepsilon_t\,(h_3 - h_1) = \varepsilon_s\,c_p(t_3 - t_1) + \varepsilon_L\,h_{fg}(w_3 - w_1)

Total recovery is sensible plus latent; with $h_{fg}\approx1{,}076\ \text{Btu/lb}$ (the value behind the handbook's $4{,}840$ latent coefficient) this ties $\varepsilon_t$ back to $\varepsilon_s$ and $\varepsilon_L$ at balanced flow.

Worked examples

HRV leaving temperature and recovered heating

Problem. A sealed plate HRV serves a building with balanced airflow of $4{,}000\ \text{cfm}$ each way. On a winter design day the outdoor air enters at $t_1 = 10\,^\circ\text{F}$ and the building exhaust enters at $t_3 = 72\,^\circ\text{F}$ . The HRV has a sensible effectiveness $\varepsilon_s = 0.75$ . Find the temperature of the preheated supply air and the rate of heat recovered.

Solution. Flows are balanced, so

C_{min}/(\dot m_s c_{ps}) = 1

and the supply leaves at

t_2 = t_1 + \varepsilon_s(t_3 - t_1) = 10 + 0.75\,(72-10) = 10 + 46.5 = 56.5\,^\circ\text{F}

. Recovered sensible heat at standard air:

\dot q_s = 1.10\,Q_s\,(t_2 - t_1) = 1.10\,(4000)(56.5 - 10) = 1.10\,(4000)(46.5) = 204{,}600\ \text{Btu/h}

\dot q_s \approx 2.05\times10^{5}\ \text{Btu/h}

(about

17

tons-equivalent of preheat). Sanity check:

t_2 = 56.5\,^\circ\text{F}

lies between the two inlets (

10

and

72

), and the recovery equals

0.75

of the maximum

1.10(4000)(62) = 272{,}800\ \text{Btu/h}

, confirming

\varepsilon_s = 0.75

t_2 = 10 + 0.75\,(72-10) = 56.5\,^\circ\text{F}

Unbalanced flow — supply and exhaust leaving temperatures

Problem. A summer-mode device has supply (outdoor) airflow $Q_s = 5{,}000\ \text{cfm}$ and exhaust airflow $Q_e = 4{,}000\ \text{cfm}$ . Outdoor air enters at $t_1 = 95\,^\circ\text{F}$ , room exhaust at $t_3 = 75\,^\circ\text{F}$ , and the sensible effectiveness referenced to $C_{min}$ is $\varepsilon_s = 0.70$ . Treating $c_p$ as equal on both sides, find the supply and exhaust leaving temperatures.

Solution. The smaller airflow is the exhaust, so

\dot m_{min} = \dot m_e

and

C_{min}/(\dot m_s c_{ps}) = \dot m_e/\dot m_s = Q_e/Q_s = 4000/5000 = 0.80

. Supply leaving:

t_2 = t_1 + \varepsilon_s\,(0.80)(t_3 - t_1) = 95 + 0.70\,(0.80)(75 - 95) = 95 + 0.70(0.80)(-20) = 95 - 11.2 = 83.8\,^\circ\text{F}

. Exhaust leaving (exhaust is

C_{min}

, ratio = 1):

t_4 = t_3 - \varepsilon_s\,(t_3 - t_1) = 75 - 0.70(75 - 95) = 75 + 14 = 89.0\,^\circ\text{F}

. Energy balance check: heat gained by supply

= 1.10(5000)(95-83.8) = 61{,}600\ \text{Btu/h}

; heat lost by exhaust

= 1.10(4000)(89.0-75) = 61{,}600\ \text{Btu/h}

— equal, as required. Supply precooled to

\mathbf{83.8\,^\circ\text{F}}

, exhaust warmed to

\mathbf{89.0\,^\circ\text{F}}

t_2 = 95 + 0.70\,(0.80)(75-95) = 83.8\,^\circ\text{F}

ERV total recovery from sensible and latent parts

Problem. An enthalpy wheel runs balanced at $4{,}000\ \text{cfm}$ . Outdoor air (state 1) is $95\,^\circ\text{F}$ db / $76\,^\circ\text{F}$ wb ( $h_1 = 39.3\ \text{Btu/lb}$ , $104.6$ grains/lb), and return air (state 3) is $75\,^\circ\text{F}$ / $50\%$ RH ( $h_3 = 28.1\ \text{Btu/lb}$ , $64.6$ grains/lb). The wheel has $\varepsilon_s = 0.75$ and $\varepsilon_L = 0.65$ . Find the recovered sensible, latent, and total cooling, and the implied total effectiveness.

Solution. Recovered sensible:

\dot q_s = 1.10\,(4000)(0.75)(95 - 75) = 1.10(4000)(0.75)(20) = 66{,}000\ \text{Btu/h}

. Recovered latent: grain difference

= 104.6 - 64.6 = 40.0

grains/lb, i.e.

\Delta w = 40.0/7000\ \text{lb}_w/\text{lb}_{da}

, so

\dot q_L = 4{,}840\,(4000)(0.65)(40.0/7000) = 71{,}910\ \text{Btu/h}

. Recovered total:

\dot q_t = \dot q_s + \dot q_L = 66{,}000 + 71{,}910 = 137{,}910\ \text{Btu/h} \approx 1.38\times10^{5}\ \text{Btu/h}

(

11.5

tons). Implied total effectiveness:

\dot q_{t,\max} = 4.5\,(4000)(39.3 - 28.1) = 4.5(4000)(11.2) = 201{,}600\ \text{Btu/h}

, so

\varepsilon_t = 137{,}910/201{,}600 = 0.684

. Sanity check:

\varepsilon_t = 0.68

falls between

\varepsilon_s = 0.75

and

\varepsilon_L = 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.

\varepsilon_t = \frac{66{,}000 + 71{,}910}{4.5(4000)(11.2)} = 0.684

Common pitfalls

•Confusing sensible, latent, and total. A temperature swing needs $\varepsilon_s$ , a humidity swing needs $\varepsilon_L$ , and a mixed-air Btu load needs $\varepsilon_t$ — they are different numbers for the same device. Fix: identify the property that changes (t, w, or h) before picking the effectiveness.
•Building $\dot q_{max}$ on the supply airflow when the exhaust is smaller (or vice versa). Effectiveness is referenced to $C_{min}$ . Fix: always compare $\dot m_s$ and $\dot m_e$ and use the smaller for both $\dot q_{max}$ and the capacity ratio.
•Using the sea-level $1.10/4{,}840/4.5$ coefficients at altitude or on hot supply air. They embed sea-level standard density (0.075 lb/ft³). Fix: at altitude use the handbook's $C_{x,alt} = C_{x,0}\,P/P_0$ (§9.1.8; $0.92/4{,}027/3.74$ at 5,000 ft); at elevated temperature rebuild from $60\,\dot m\,\Delta$ with the actual air density.
•Treating effectiveness as a function of the weather. $\varepsilon$ is set by the device and the flow ratio, not by $t_1$ or $t_3$ . Fix: hold $\varepsilon$ fixed across seasons; only the inlet difference (the driving potential) changes.
•Putting $\Delta w$ in $\text{lb}_w/\text{lb}_{da}$ into the $0.691$ per-grain latent coefficient (which expects grains), or grains into $4{,}840$ . Fix: use $4{,}840$ with $\text{lb}_w/\text{lb}_{da}$ , or $0.691$ with grains — never mix the two.
•Ignoring frost. On the cold design day a high-effectiveness device drives the exhaust leaving temperature below freezing and ices up. Fix: check $t_4$ against 32 °F at the exhaust dew point and credit the frost-control derate (preheat, bypass, or defrost).
•Setting the capacity ratio to 1 in both leaving-state equations when the flows are unequal (only truly balanced mass flows make both ratios 1). The ratio $C_{min}/(\dot m_e c_{pe})$ equals 1 only for whichever stream is $C_{min}$ . Fix: apply the correct ratio to each leaving-temperature equation.

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.

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