DEV Community: Evgenii Konkin

Demand Charges: Why a 15-Minute kW Spike Costs More Than a Month of kWh

Evgenii Konkin — Fri, 16 Oct 2026 03:23:18 +0000

Your electric bill has two charges. Most people only understand one.

kWh vs kW — two different measurements

Energy charge (kWh):
  How much electricity you used over time.
  Measured in kilowatt-hours.
  Like miles driven.

Demand charge (kW):
  The highest rate at which you used it.
  Measured as peak kilowatts in any 15-min interval.
  Like top speed reached.

You pay for both. Separately.

How demand is measured

The utility records your power draw in 15-minute intervals. The highest 15-minute average in the billing period = your peak demand. You pay a $/kW rate on that single peak.

Bakery example:
  Monthly consumption: 8,400 kWh
  Energy rate: $0.11/kWh
  Energy charge: $924

  Peak demand: 85 kW (all ovens + mixer + HVAC at 4 AM)
  Demand rate: $14.68/kW
  Demand charge: 85 × $14.68 = $1,248

  The 15-minute spike costs MORE than the entire month.

Why it happens

The utility sizes infrastructure — transformers, cables, switchgear — for your peak load. Even if 85 kW only happens once per month, the equipment must handle it. Demand charges pay for that capacity.

Where peaks hide

Simultaneous startup:
  Oven 1: 18 kW
  Oven 2: 18 kW
  Oven 3: 18 kW
  Mixer:  12 kW
  Proofer: 8 kW
  HVAC:   11 kW
  ──────────────
  Total:  85 kW  ← all in one 15-min window

Average load rest of month: 28 kW

The peak is 3× the average. The demand charge reflects the peak, not the average.

The fix: stagger, don't reduce

Before (simultaneous):
  4:00 AM — all equipment starts → 85 kW peak

After (staggered):
  3:45 — Oven 1 preheats (18 kW)
  4:00 — Oven 2 preheats (18 kW), Oven 1 drops to 6 kW
  4:15 — Oven 3 preheats (18 kW), Oven 2 drops to 6 kW
  4:30 — Mixer starts (12 kW), Oven 3 drops to 6 kW
  5:00 — HVAC starts (11 kW)

  Peak 15-min window: 52 kW

Same total kWh. Same production. Different sequence.

The savings

Before: 85 kW × $14.68 = $1,248/mo
After:  52 kW × $14.68 =   $763/mo
Savings:                    $484/mo = $5,813/year

No new equipment. Just a startup schedule.

Demand charge rates by facility type

Facility              Typical Peak    $/kW Rate     Monthly Demand
─────────────────────────────────────────────────────────────────
Small retail           30-50 kW       $8-12/kW      $240-600
Restaurant             50-100 kW      $10-15/kW     $500-1,500
Small manufacturing    100-300 kW     $12-18/kW     $1,200-5,400
Office building        200-500 kW     $12-20/kW     $2,400-10,000
Data center            500-2,000 kW   $15-25/kW     $7,500-50,000

Demand charges are typically 30-50% of the total commercial electric bill.

The check

1. Find "Demand" or "Peak Demand" on your bill (kW, not kWh)
2. Note the $/kW rate
3. Request 15-minute interval data from your utility
4. Find the time window of the peak
5. List every piece of equipment running in that window
6. Stagger startup times across different 15-min intervals

The energy cost per piece of equipment determines when the peak happens. Know each load, schedule the startup, and the demand charge drops without reducing production.

Quick Tip: The 3 Multiplier That Undersizes Three-Phase Generators by 73%

Evgenii Konkin — Thu, 15 Oct 2026 03:11:27 +0000

Single-phase and three-phase kVA formulas differ by one multiplier: √3 (1.732). Use the wrong one and you undersize by 42% — or oversize by 73%.

The two formulas

Single-phase:
  kVA = V × I / 1,000
  kW  = V × I × PF / 1,000

Three-phase:
  kVA = V × I × 1.732 / 1,000
  kW  = V × I × 1.732 × PF / 1,000

The ONLY difference is × 1.732 for three-phase.

The mistake

200A, 480V, three-phase panel:

Wrong (single-phase formula):
  kVA = 480 × 200 / 1,000 = 96 kVA
  kW  = 96 × 0.85 = 81.6 kW

Correct (three-phase formula):
  kVA = 480 × 200 × 1.732 / 1,000 = 166.3 kVA
  kW  = 166.3 × 0.85 = 141.4 kW

Error: 73% undersized

A 100 kW generator ordered. Actual load: 141 kW. Overload shutdown on first test.

How to know which formula

2 poles on breaker → single-phase
3 poles on breaker → three-phase (× 1.732)

Nameplate says "1PH" → single-phase
Nameplate says "3PH" → three-phase

Voltage 208Y/120, 480Y/277, 600Y/347 → three-phase

Quick reference: 480V three-phase at PF 0.85

Amps    kVA      kW
─────────────────────
100A     83 kVA    71 kW
200A    166 kVA   141 kW
400A    333 kVA   283 kW
600A    499 kVA   424 kW

Quick reference: 208V three-phase at PF 0.85

Amps    kVA      kW
─────────────────────
100A     36 kVA    31 kW
200A     72 kVA    61 kW
400A    144 kVA   123 kW

The rule

Before any kVA or kW calculation, check the phase. One multiplier changes the answer by 73%. For generators, that is the difference between a system that runs and one that trips on overload.

Humid Air Is Lighter Than Dry Air — And It Changes Your Fan Selection

Evgenii Konkin — Wed, 14 Oct 2026 03:29:15 +0000

Most engineers assume humid air is heavier than dry air. It feels thick. It feels oppressive. But every physics textbook says the opposite: humid air is lighter. And this matters for fan performance, combustion, and cooling towers.

Why moisture makes air lighter

Dry air is mostly nitrogen (N2, molecular weight 28) and oxygen (O2, molecular weight 32). The effective molecular weight of dry air: 28.97.

Water vapor (H2O) has a molecular weight of 18.02 — 38% lighter than dry air.

When water vapor enters air, it displaces N2 and O2 molecules. Each H2O molecule that replaces a heavier molecule makes the mixture lighter. Total molecular count stays roughly the same (ideal gas law), but average molecular weight decreases.

The density numbers

Air density at 29.92 in. Hg (standard barometric pressure):

70F, 0% RH (dry):   0.0749 lb/ft³
70F, 50% RH:        0.0747 lb/ft³
70F, 100% RH:       0.0745 lb/ft³

95F, 0% RH (dry):   0.0714 lb/ft³
95F, 50% RH:        0.0707 lb/ft³
95F, 100% RH:       0.0698 lb/ft³

At 95F: dry to saturated = -2.2% density

The effect is larger at higher temperatures because warm air holds more moisture.

Impact on fan performance

Fan static pressure and horsepower scale with density:

Actual SP = Rated SP × (ρ_actual / ρ_standard)

Example:
  Fan rated: 2.0 in. w.c. at standard (70F, dry, 0.075 lb/ft³)
  Operating: 95F, 80% RH → ρ = 0.0702 lb/ft³

  Actual SP = 2.0 × (0.0702 / 0.075) = 1.87 in. w.c.

  → 6.5% less static pressure than catalog

Combined effects: altitude + temperature + humidity

Condition                           Density      % of Standard
──────────────────────────────────────────────────────────────
Standard (70F, dry, sea level)      0.0750       100%
Hot dry (95F, dry, sea level)       0.0714        95%
Hot humid (95F, 80% RH, sea level)  0.0702        94%
Denver, hot humid (95F, 80% RH)     0.0617        82%
Flagstaff, hot humid                0.0562        75%

At Denver on a hot humid day, a fan delivers only 82% of its catalog static pressure. All three factors — altitude, temperature, humidity — stack.

Combustion impact

Gas furnaces need O2 to burn fuel.
Humid air has less O2 per cubic foot.

Effect: 1-2% reduction in available oxygen.
Usually compensated by burner excess air.

But at high altitude + high humidity:
  Combined density drop: 5-6%
  Burner may run lean → incomplete combustion

Cooling tower impact

Tower performance depends on air mass flow.
Lower density → less mass through fill at same fan speed.
Less mass → less heat rejection per hour.

Worst case: hot + humid + high altitude
  = exactly when you need max cooling
  = exactly when the tower underperforms

The three variables

Air density depends on:

1. Altitude    → -3-4% per 1,000 ft
2. Temperature → -1.5% per 10F above 70F
3. Humidity    → -0.5 to -2.2% at saturation

All three stack. Ignoring any one gives the wrong density — and every downstream calculation (fan selection, coil capacity, combustion air) uses that density.

Calculate the exact density for your conditions with the air density calculator on CalcEngineer — enter altitude, temperature, and humidity to see how each factor shifts the number.

Quick Tip: Duct Aspect Ratio — Why a 20 6 Duct Has 40% More Friction Than a 12" Round

Evgenii Konkin — Tue, 13 Oct 2026 03:31:14 +0000

A 12-inch round duct doesn't fit in an 8-inch ceiling plenum. The engineer switches to 20×6 rectangular — same cross-sectional area. Same CFM. Should work, right?

The friction penalty

Round ducts have the lowest perimeter-to-area ratio. Less perimeter = less surface contact = less friction.

12" round duct:
  Area: 113 in²
  Perimeter: 37.7 in

20×6" rectangular:
  Area: 120 in²  (slightly larger)
  Perimeter: 52 in  (38% more surface)

That extra perimeter means 40% more friction at the same airflow.

Aspect ratio penalties

Aspect ratio = long side / short side. As it increases, friction climbs:

Aspect Ratio    Friction vs Round
────────────────────────────────────
1:1 (square)    +12%
2:1             +20%
3:1             +35%
4:1             +50%  ← ASHRAE max recommended
6:1             +80%
8:1             +120%

Above 4:1, friction and sealing problems get severe.

The equivalent diameter trap

When converting round to rectangular, use the hydraulic equivalent — not equal area:

De = 1.30 × (a × b)^0.625 / (a + b)^0.250

20×6 rectangular:
  De = 1.30 × 120^0.625 / 26^0.250 = 15.9"

This is NOT equivalent to a 12" round.
It's equivalent to a ~16" round.

To match a 12" round at the same friction, you need something like 14×8 (aspect ratio 1.75:1) — not 20×6 (3.3:1).

When to use each

Round:
  ✓ Lowest friction
  ✓ Strongest structure
  ✓ Easiest to seal
  ✗ Needs clearance height

Rectangular:
  ✓ Fits tight plenums
  ✓ Custom around obstructions
  ✗ Higher friction
  ✗ Keep aspect ratio ≤ 4:1

Flat oval:
  ✓ Lower profile than round
  ✓ Less friction than rectangle
  ✗ More expensive to fabricate

The rule

If space forces a flat duct, increase dimensions to compensate for friction — or accept higher fan power and noise. A 20×6 is not "the same as" a 12" round. The perimeter says otherwise.

ASHRAE 62.1 Ventilation Rates: The Two-Part CFM Formula Most Engineers Simplify Into One

Evgenii Konkin — Fri, 09 Oct 2026 04:25:13 +0000

Ventilation in ASHRAE 62.1 is not a single number per square foot. It has two components, and skipping either one leads to IAQ complaints or energy waste.

The formula

Vbz = Rp × Pz + Ra × Az

Where:
  Vbz = breathing zone outdoor airflow (CFM)
  Rp  = per-person outdoor air rate (CFM/person)
  Pz  = zone population (number of people)
  Ra  = per-area outdoor air rate (CFM/ft²)
  Az  = zone floor area (ft²)

The people component (Rp × Pz) handles CO2, bioeffluents, and body odor. The area component (Ra × Az) handles off-gassing from materials, furniture, and finishes.

Rates by space type

Space Type              Rp (CFM/person)   Ra (CFM/ft²)
──────────────────────────────────────────────────────
Office                  5                 0.06
Conference room         5                 0.06
Classroom (age 9+)      10                0.12
Lecture hall            7.5               0.06
Retail sales floor      7.5               0.12
Restaurant dining       7.5               0.18
Gym / exercise room     20                0.06
Hotel bedroom           5                 0.06
Lobby / waiting         7.5               0.06
Beauty salon            20                0.12

Notice: gym and salon require 20 CFM/person — 4× the office rate.

Example: conference room

Room: 500 ft², seats 20 people
Rp = 5 CFM/person, Ra = 0.06 CFM/ft²

Correct ventilation:
  Vbz = 5 × 20 + 0.06 × 500
      = 100 + 30
      = 130 CFM outdoor air

Sized for "normal" occupancy (5 people):
  Vbz = 5 × 5 + 0.06 × 500
      = 25 + 30
      = 55 CFM outdoor air

55 CFM vs 130 CFM needed. The room gets less than half the required ventilation when full. CO2 climbs to 1,800 ppm by mid-afternoon.

Example: open office vs classroom

Same 1,000 ft² room, 25 occupants:

Open office:
  Vbz = 5 × 25 + 0.06 × 1,000 = 125 + 60 = 185 CFM

Classroom:
  Vbz = 10 × 25 + 0.12 × 1,000 = 250 + 120 = 370 CFM

Same room, same people count. The classroom needs double the ventilation because both Rp and Ra are higher.

The energy cost of getting it wrong

Excess outdoor air at 95F cooled to 75F:

Per 100 CFM excess:
  Q = 1.08 × 100 × 20 = 2,160 BTU/hr
  Cost at $0.12/kWh ≈ $0.08/hour
  Cooling season (2,000 hrs) ≈ $160

Per 100 CFM deficit:
  CO2 rises → complaints → productivity loss
  Estimated cost: $50-100/person/year in lost focus

The check

1. Identify space type → look up Rp and Ra
2. Determine design occupancy (not average — peak)
3. Calculate: Vbz = Rp × people + Ra × area
4. Verify the outdoor air damper delivers this CFM
5. For variable occupancy: consider DCV (CO2 sensor)

Ventilation is not just about comfort. It is a code requirement. ASHRAE 62.1 is referenced by IBC, UMC, and most local codes.

Quick Tip: Delta-T Diagnostic — Three Zones That Tell You What's Wrong Before You Open a Panel

Evgenii Konkin — Thu, 08 Oct 2026 03:40:34 +0000

Return air temperature minus supply air temperature = delta-T. This single number points to one of three problem categories before you touch gauges or open panels.

The three zones (cooling)

Delta-T         Status        What to check
──────────────────────────────────────────────────────
Below 14F       LOW           Refrigerant, coil, compressor
16 to 22F       NORMAL        Coil is working properly
Above 24F       HIGH          Airflow is restricted

Low delta-T (below 14F)

The supply air is barely cooler than return air. The coil is not absorbing enough heat.

Possible causes:
  → Low refrigerant charge (suction pressure low)
  → Dirty evaporator coil (insulated by dirt layer)
  → Failed compressor valve (low mass flow)
  → Excessive airflow (air moves too fast through coil)

High delta-T (above 24F)

The supply air is very cold. Sounds good — it is not. The coil is overcooling a reduced volume of air.

Possible causes:
  → Dirty filter (most common)
  → Collapsed or undersized duct
  → Closed damper or blocked return grille
  → Low fan speed (wrong tap, failing capacitor)

Danger: coil surface drops below 32F → ice forms
        → further restricts airflow → more ice
        → complete coil freeze in 30-60 minutes

The capacity check

Delta-T alone is not capacity. You need airflow too:

Q = 1.08 × CFM × Delta-T

Normal:     1.08 × 1,200 × 20 = 25,920 BTU/hr (2.16 tons)
Restricted: 1.08 ×   800 × 28 = 24,192 BTU/hr (2.02 tons)

High delta-T partially compensates for low CFM in the formula — but the system is stressed, the coil risks freezing, and humidity removal fails.

Heating reference

Gas furnace:
  Below 40F → low output (check gas pressure, ignition)
  40-70F    → normal
  Above 70F → DANGER: heat exchanger overheat, CO risk

Heat pump:
  Below 20F → check charge, defrost cycle
  25-35F    → normal

Two thermometers. One subtraction. The answer points you to the right system before you open anything.

Calculate delta-T and system capacity with the delta T calculator on CalcEngineer.

Mixed Air Conditions: The Calculation Step That Undersizes Cooling Coils by 20%

Evgenii Konkin — Wed, 07 Oct 2026 03:30:36 +0000

When an AHU brings in outdoor air for ventilation, the air entering the cooling coil is not return air. It is a mixture of return air and outdoor air. The properties of this mix determine the actual coil load.

The mixing formulas

Mixed air properties are mass-weighted averages of the two streams:

T_mix = (CFM_return × T_return + CFM_outdoor × T_outdoor) / CFM_total

W_mix = (CFM_return × W_return + CFM_outdoor × W_outdoor) / CFM_total

h_mix = (CFM_return × h_return + CFM_outdoor × h_outdoor) / CFM_total

Where T = dry-bulb temperature, W = humidity ratio (grains/lb), h = enthalpy (BTU/lb).

Example: 30% outdoor air in Atlanta

Return air (from conditioned space):
  T = 75F, RH = 50%
  W = 65 grains/lb
  h = 28.1 BTU/lb

Outdoor air (ASHRAE design day):
  T = 95F, WB = 78F
  W = 118 grains/lb
  h = 38.5 BTU/lb

Mix ratio: 70% return / 30% outdoor
Total: 5,000 CFM (3,500 return + 1,500 outdoor)

The mixed air result

T_mix = 0.70 × 75 + 0.30 × 95 = 81.0F
W_mix = 0.70 × 65 + 0.30 × 118 = 80.9 grains/lb
h_mix = 0.70 × 28.1 + 0.30 × 38.5 = 31.3 BTU/lb

The coil sees 81F air at 81 grains/lb — not 75F at 65 grains/lb.

The capacity impact

Coil load if sized for return air only:
  Q = 4.5 × 5,000 × (28.1 - 22.7) = 121,500 BTU/hr = 10.1 tons

Coil load at actual mixed air condition:
  Q = 4.5 × 5,000 × (31.3 - 22.7) = 193,500 BTU/hr = 16.1 tons

Difference: 72,000 BTU/hr = 6 tons undersized

The 22.7 BTU/lb is the enthalpy of 55F supply air at 95% RH (coil leaving condition).

Where outdoor air fraction matters most

OA Fraction    T_mix    W_mix      h_mix     Extra Load vs Return
─────────────────────────────────────────────────────────────────
0% (recirculate) 75.0F   65 gr/lb   28.1      baseline
10%              77.0F   70 gr/lb   29.1      +3%
20%              79.0F   76 gr/lb   30.2      +8%
30%              81.0F   81 gr/lb   31.3      +11%
50%              85.0F   92 gr/lb   33.3      +19%
100% (all OA)    95.0F  118 gr/lb   38.5      +37%

Hospitals and labs running 100% outdoor air have coil loads 37% higher than the same space with recirculated air.

The psychrometric shortcut

On a psychrometric chart, the mixed air point lies on a straight line between the two state points. The position along the line corresponds to the mixing ratio.

30% outdoor air → point is 30% of the way from return to outdoor
50% outdoor air → point is exactly halfway

This visual method lets you read off all mixed properties — temperature, humidity ratio, enthalpy, wet-bulb — without calculating each one separately.

The rule

Never size a cooling coil for return air conditions when the system includes outdoor air. Calculate the mixed air state point first, then size the coil from mixed air to supply air.

The difference is small in dry climates with low outdoor air fractions. It is enormous in humid climates with high ventilation rates.

Quick Tip: Why an Oversized AC Makes Your Building More Humid, Not Less

Evgenii Konkin — Tue, 06 Oct 2026 04:17:17 +0000

An oversized air conditioner cools the temperature faster — but removes less moisture. The result is a space that reads 72F on the thermostat but feels clammy.

How it works

An AC does two things at the evaporator coil:

1. Sensible cooling — lowers air temperature
2. Latent cooling — condenses moisture from the air

Moisture removal requires sustained coil contact time. The first minutes of a cycle mostly lower temperature. Dehumidification ramps up after the coil is fully cold and wet.

Short cycling kills dehumidification

Properly sized system:
  Cycle length: 15-20 minutes
  SHR: 0.75 (25% of cooling removes moisture)
  Indoor humidity: 45-50%

Oversized system (2× design load):
  Cycle length: 5-8 minutes
  SHR: 0.90 (only 10% removes moisture)
  Indoor humidity: 60-70%

SHR = Sensible Heat Ratio. Lower SHR = more moisture removal.

The compressor damage

Properly sized:  4-6 starts per hour
Oversized:      10-15 starts per hour

Each start draws 5-7× running current.
More starts = more wear = shorter compressor life.

The sizing rule

Design cooling load (Manual J or ASHRAE): X tons
Correct equipment size: 100-110% of X

At 150%+ → short cycling begins
At 200%+ → humidity control fails completely

Real example

4,000 sq ft office:
  Design load:     4.8 tons
  Installed:      10.0 tons (208% of design)
  Cycle time:      5-8 minutes
  Indoor humidity: 65%

Replaced with 5-ton variable speed:
  Cycle time:     20-30 minutes
  Indoor humidity: 48%

Bigger is not better. Match the tonnage to the load.

How Altitude Reduces Cooling Tower Capacity and Chiller Efficiency at the Same Time

Evgenii Konkin — Mon, 28 Sep 2026 04:59:27 +0000

A cooling tower rejects heat by evaporating water into the air. At altitude, the air is thinner. Less mass flow = less evaporation = less cooling.

The capacity loss

Altitude    Air Density     Tower Capacity (vs sea level)
──────────────────────────────────────────────────────────
Sea level   1.225 kg/m³     500 tons (100%)
2,000 ft    1.155 kg/m³     475 tons (95%)
5,000 ft    1.005 kg/m³     425–450 tons (85–90%)
7,500 ft    0.900 kg/m³     390–410 tons (78–82%)

A 500-ton tower at sea level delivers 425–450 tons at 5,000 ft. That is 50–75 tons of missing heat rejection.

Why: the density equation

ρ = P / (R × T)

Sea level:  P = 101,325 Pa → ρ = 1.225 kg/m³
5,000 ft:   P = 84,556 Pa  → ρ = 1.005 kg/m³

Ratio: 1.005 / 1.225 = 0.82 (18% less mass per ft³)

The tower fan moves the same CFM but each cubic foot carries 18% less air mass. Less mass = less evaporative cooling.

The chiller penalty

The chiller rejects heat to condenser water. The tower cools that water. If the tower cannot reach design temperature:

Design condenser water:  85°F leaving tower
Actual at altitude:      88°F leaving tower
Delta:                   +3°F

Every extra degree of condenser water:

+1°F → +1–2% compressor energy
+3°F → +3–6% compressor energy

Over a cooling season (3,000 hours):

500-ton chiller at 0.55 kW/ton = 275 kW
+5% penalty = 13.75 kW additional
13.75 kW × 3,000 hrs = 41,250 kWh
41,250 × $0.12 = $4,950/year extra energy

The double hit

At altitude, both sides of the HVAC plant lose capacity simultaneously:

Air side:
  Fan delivers same CFM but 18% less mass flow
  Sensible heat formula: Q = 1.08 × CFM × ΔT
  The 1.08 drops to ~0.89 at 5,000 ft
  Result: 18% less cooling from air handling units

Condenser side:
  Cooling tower rejects 10–15% less heat
  Condenser water returns 2–4°F warmer
  Chiller efficiency drops 3–6%
  Result: higher energy cost per ton of cooling

Both hit at once. The building cannot hold setpoint AND the energy bill is higher than projected.

When to correct

Below 2,000 ft:  correction usually unnecessary
2,000–4,000 ft:  check tower selection data for altitude
Above 4,000 ft:  always correct towers, fans, and gas equipment
Above 6,000 ft:  significant — 20%+ air-side, 10%+ tower

The correction method

1. Calculate actual air density at altitude and design temp
2. Apply density ratio to tower selection (derate published capacity)
3. Apply density ratio to air-side calculations (adjust 1.08 constant)
4. Check chiller selection at corrected condenser water temp
5. Verify total plant capacity covers building load at altitude

Enthalpy vs Temperature: Why Sensible Load Alone Undersizes Cooling Coils

Evgenii Konkin — Fri, 25 Sep 2026 02:59:52 +0000

Temperature tells you half the cooling story. Enthalpy tells you the whole thing.

What enthalpy captures that temperature misses

Sensible heat = heat from air temperature (what a thermometer reads)
Latent heat   = heat from moisture content (invisible, but real energy)
Enthalpy      = sensible + latent combined (BTU/lb of dry air)

The formula:

h = 0.24 × T + W × (1061 + 0.444 × T)

Where:

h = enthalpy, BTU/lb
T = dry-bulb temperature, °F
W = humidity ratio, lb moisture / lb dry air

Same temperature, different enthalpy

Two air conditions at 80°F:

Condition A: 80°F, 40% RH
  Humidity ratio: 62 gr/lb
  Enthalpy: 28.5 BTU/lb

Condition B: 80°F, 70% RH
  Humidity ratio: 110 gr/lb
  Enthalpy: 37.8 BTU/lb

Same temperature. 33% more total heat in Condition B. A coil sized for A cannot handle B.

The coil load comparison

5,000 CFM outdoor air cooled from 95°F/75°F-wb to 55°F/54°F-wb:

Sensible-only calculation:

Q_sensible = 1.08 × CFM × ΔT
Q_sensible = 1.08 × 5,000 × (95 − 55)
Q_sensible = 216,000 BTU/hr = 18.0 tons

Enthalpy-based calculation (total load):

Q_total = 4.5 × CFM × (h_in − h_out)
Q_total = 4.5 × 5,000 × (38.5 − 22.7)
Q_total = 355,500 BTU/hr = 29.6 tons

The gap:

Sensible:  18.0 tons (61%)
Latent:    11.6 tons (39%)
Total:     29.6 tons (100%)

Sizing by temperature alone: 18 tons. Actual need: 29.6 tons. The coil is 39% undersized.

When latent load dominates

Climate              Latent %    Enthalpy matters?
──────────────────────────────────────────────────
Phoenix (hot-dry)    5–15%       minimal
Denver (moderate)    10–20%      moderate
Atlanta (humid)      25–35%      yes
Houston (very humid) 35–50%      critical
Miami (tropical)     40–50%      critical

In Phoenix, sensible-only sizing is close enough. In Houston, it undersizes by 35–50%.

The practical shortcut

If you know the entering and leaving wet-bulb temperatures, you can look up enthalpy from psychrometric tables or a calculator:

Entering: 95°F db / 75°F wb → h = 38.5 BTU/lb
Leaving:  55°F db / 54°F wb → h = 22.7 BTU/lb

Total load = 4.5 × CFM × (38.5 − 22.7)

No need to separate sensible and latent. Enthalpy captures both in one subtraction.

For quick psychrometric property lookups including enthalpy from any two known air properties, use the psychrometric calculator on CalcEngineer.

Quick Tip: Heat Pump Tonnage — Why Cooling Capacity Does Not Equal Heating Capacity

Evgenii Konkin — Thu, 24 Sep 2026 03:30:05 +0000

A heat pump is an AC that reverses direction. Same compressor, same refrigerant. But the tonnage it delivers in heating mode is not the same as in cooling mode.

Why capacity drops in heating

In cooling mode, the system rejects heat to outdoor air at 95°F. The temperature lift is moderate.

In heating mode, the system extracts heat from outdoor air at 30°F. The temperature lift is larger and there is less heat available to extract.

Mode       Outdoor    Indoor    Lift     Capacity
──────────────────────────────────────────────────
Cooling    95°F       75°F      20°F     3.0 tons (rated)
Heating    47°F       70°F      23°F     2.8 tons
Heating    30°F       70°F      40°F     2.0 tons
Heating    10°F       70°F      60°F     1.5 tons

The same 3-ton unit delivers only 2 tons at 30°F and 1.5 tons at 10°F.

The sizing dilemma

Mild climate (Houston, Atlanta):
  Cooling load: 3.5 tons
  Heating load: 2.0 tons
  → size for cooling (3.5T), heating is covered

Cold climate (Minneapolis, Chicago):
  Cooling load: 2.5 tons
  Heating load: 4.0 tons
  → size for heating, but then oversized for cooling

Moderate climate (Nashville, Charlotte):
  Cooling load: 3.0 tons
  Heating load: 3.0 tons
  → depends on specific house and insulation

Four options for cold climates

Option                           Pros                    Cons
────────────────────────────────────────────────────────────────
1. Oversize for heating          covers cold nights       short cycles in summer
                                                         poor humidity control

2. Auxiliary heat strips         simple, no extra unit    expensive to operate
                                                         $0.15-0.25/kWh electric

3. Dual-fuel (HP + gas furnace)  efficient above 30°F    two systems to maintain
                                 gas below 30°F          higher install cost

4. Cold-climate heat pump        rated to -15°F          higher equipment cost
                                 no backup needed        newer technology

The balance point

The outdoor temperature where heat pump capacity equals building heat loss:

Above balance point: heat pump covers the load alone
Below balance point: supplemental heat required

Typical balance points:
  Standard heat pump:        30–35°F
  Cold-climate heat pump:    5–15°F

If your winter design temperature is below the balance point, the heat pump alone cannot maintain setpoint on the coldest nights.

The quick check

1. Calculate cooling tonnage (area × climate factor / 12,000)
2. Calculate heating load (BTU/hr from Manual J or simplified method)
3. Check heat pump capacity at winter design temperature
4. If heating load > HP capacity at design temp → add supplemental heat

The cooling calculation is the starting point, but heating is what determines whether the system needs backup.

Heating Season Delta-T: The Ranges That Change When You Switch From Cooling to Heating

Evgenii Konkin — Wed, 23 Sep 2026 03:26:34 +0000

Delta-T in heating mode uses the same formula but different expected ranges.

ΔT = T_supply − T_return

Note: in heating mode, supply is warmer than return (reversed from cooling).

Heating delta-T ranges by equipment type

Equipment              ΔT (°F)     ΔT (°C)
───────────────────────────────────────────
Gas furnace            40–70       22–39
Electric furnace       40–60       22–33
Heat pump (heating)    25–35       14–19
Hot water coil         20–40       11–22

Compare to cooling:

Residential DX coil    14–22       8–12
Chilled water          10–12       5.5–6.7

Heating delta-T is 2–3x higher than cooling for the same equipment because combustion and electric resistance generate much higher temperatures than refrigeration.

Gas furnace: high delta-T is dangerous

In cooling, high delta-T means restricted airflow — uncomfortable but safe.

In heating on a gas furnace, high delta-T means the heat exchanger is overheating:

ΔT < 40°F  →  low output (gas pressure, ignition, flame issue)
ΔT 40–70°F →  normal range
ΔT > 70°F  →  DANGER: heat exchanger overheat
                → limit switch trips
                → if limit bypassed: exchanger cracks
                → cracked exchanger = CO in supply air

Most common cause of high delta-T in heating:

1. Dirty filter (restricts airflow)
2. Collapsed return duct
3. Undersized return grille
4. Blower at wrong speed setting
5. Failed blower capacitor

A dirty filter from last winter is the #1 pre-season find.

Heat pump: low delta-T is normal

Gas furnace supply air:   130–150°F
Heat pump supply air:      95–105°F
Body temperature:           98.6°F

Heat pump air feels "cool" compared to a furnace because the delta-T is only 25–35°F vs 40–70°F. The system compensates by running longer:

Gas furnace:  6–8 hrs/day in cold weather
Heat pump:    12–16 hrs/day in cold weather

Total daily heat delivered is similar. Delivery rate per hour is lower.

If heat pump ΔT drops below 20°F:

→ low refrigerant charge
→ defrost cycle running too frequently
→ auxiliary heat strip not energizing

Heating capacity from delta-T

Same formula as cooling:

Q (BTU/hr) = 1.08 × CFM × ΔT

Gas furnace at 1,200 CFM, 55°F delta-T:

Q = 1.08 × 1,200 × 55
Q = 71,280 BTU/hr

If nameplate says 80,000 BTU output:

71,280 / 80,000 = 89% of rated capacity → normal (duct losses)

At 35°F delta-T (something wrong):

Q = 1.08 × 1,200 × 35
Q = 45,360 BTU/hr = 57% of rated → investigate

Pre-season check

1. Replace air filter
2. Run furnace for 15+ minutes (steady state)
3. Measure supply temperature 6 ft from plenum
4. Measure return temperature before filter
5. Subtract: ΔT = supply − return
6. Compare to range for your equipment type
7. If outside range → schedule service before first cold night