Equipment performance monitoring boundaries

A 300 ton chiller that draws 180 kW has a COP of 5.86 on its own supply. Add its 20 kW of chilled and condenser water pumps and the COP is 5.28. Nothing in the plant changed, but the figure fell by 10 %. Performance is output divided by input, and the boundary decides the answer. Write the boundary down before you calculate anything, or two people will quote both numbers for the same week.

Download the boundary record (.txt). Print it, or fill it in on site.

Four questions for every method

1. What equipment is inside the boundary? List what is included and what is not: drives, pumps, fans, dryers, cooling towers, controls and auxiliary heaters. Include a standby unit that shares the supply.
2. What is the useful output? Hydraulic power for a pump, delivered air for a compressor, heat or cooling for thermal plant. Measure it on the same physical system as the input. A flow meter downstream of a bypass or a mixing valve does not see the pump flow.
3. Which operating state is being compared? Speed or stage, load, and the pressure or temperature the machine works against explain most of the spread. The same pump at 60 % and at 90 % speed draws power in a ratio of about 1 to 3.4 (0.6³ to 0.9³). Compare like with like.
4. Do all values cover the same time? A ratio of rates needs both values averaged over the same interval on one clock. A daily, batch or seasonal ratio needs energy totals with the same missing-data rule on both sides.

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Use the energy monitoring project worksheet to record the boundaries, the data destination and the acceptance owners for a first site.

Pumps

Pump performance is wire-to-water efficiency: hydraulic power divided by the electrical input at the drive supply. Hydraulic power is ρ·g·Q·H. Measure the head as differential pressure across the pump; 1 bar is 10.2 m of water at 20 °C. A pump that moves 100 m³/h against 30 m of head delivers 1,000 × 9.81 × (100 ÷ 3,600) × 30 = 8.2 kW of hydraulic power. If the drive supply reads 12 kW, the wire-to-water efficiency is 68 %. This figure includes the drive, motor and pump losses, so it is lower than the efficiency on the pump curve.

The affinity laws scale flow with speed, head with the square of speed and power with the cube of speed. At 80 % speed, power falls to 0.8³ = 51 %, so 30 kW becomes about 15.4 kW:

The cube law holds only when the system curve passes through zero head at zero flow. That is a closed loop with friction losses only. An open system with static head, such as a lift to a tank or an open cooling tower circuit, needs a minimum head before any water moves. Slowing that pump saves much less than the cube law predicts. Below a certain speed the pump delivers no flow and still draws power. Check the estimate against measured flow and kW before you report a saving.

Compressed air

Specific power is electrical input divided by delivered air. A compressor that draws 100 kW for 16.7 m³/min has a specific power of 6.0 kW per m³/min:

Three things change that number without any change to the compressor. The first is the flow unit. Compressor datasheets state free air delivery (FAD) to ISO 1217, which is the flow at the stated inlet conditions, often 1 bar(a) and 20 °C. Many thermal mass flow meters report normal cubic metres (Nm³), commonly at 0 °C and 1.01325 bar. One Nm³ occupies about 1.09 m³ at 20 °C and 1 bar. If the 16.7 above is read in Nm³/min, the flow is about 18.2 m³/min FAD and the specific power is 5.5 kW per m³/min. Record the unit the meter reports before you compare with a datasheet.

The second is discharge pressure. In a system that runs near 7 bar(g) (100 psig), each extra 2 psi (0.14 bar) adds about 1 % to compressor power, or roughly 7 % per bar. A figure taken at 7.5 bar is not comparable with one taken at 6.5 bar.

The third is the boundary. A package kW read at the compressor terminals leaves out the dryer, the condensate drains and the ventilation fans. A heatless desiccant dryer also purges part of the compressed air to regenerate, so the air delivered after the dryer is less than the compressor FAD. State whether the flow meter is before or after the dryer.

Boilers and thermal plant

Boiler efficiency is useful heat divided by fuel input, and the fuel basis alone moves the result. The gross calorific value (GCV) of natural gas is about 10.7 % higher than the net calorific value (NCV), because GCV includes the latent heat of the water vapour in the flue gas. Condensing boilers are often quoted above 100 % on a net basis. EU ecodesign rules for boilers up to 400 kW (Regulation 813/2013) state seasonal efficiency on a gross basis, and gas bills in Ireland and the UK convert volume to kWh with the gross value. A boiler that delivers 870 kWh of heat from 1,000 kWh of gas (gross) is 87 % efficient on a gross basis. The same input is 903 kWh net (1,000 ÷ 1.107), so the boiler is 96 % efficient on a net basis. Write the basis next to every figure.

A monitoring system uses the direct method: measured input and measured output. The indirect method, which subtracts measured flue, casing and blowdown losses from 100 %, belongs to a combustion test. Take the useful heat from a heat meter, or calculate it from flow and temperature difference as Q = ρ·cp·V·ΔT. At 10 m³/h and 70/50 °C that is 228 kW:

Most of the error is in the temperature difference. At 20 K, a 0.1 K mismatch between the flow and return sensors is 0.5 % of the result. At 5 K, the same mismatch is 2 %. That is why heat meters to EN 1434 use a matched pair of sensors. Two installation faults are common. Swapped flow and return sensors give a negative or near-zero heat flow, which some meters do not register at all. A flow meter on the secondary side of a low-loss header measures the circuit flow, not the boiler flow, so the ratio describes a different boundary. The heat meter data guide covers register and status checks.

Compare boilers at similar firing rates. At low fire and in summer, standing losses from the casing and pipework are a larger share of a small output, so a July efficiency is not comparable with a January one.

Chillers

Chiller COP is cooling output divided by electrical input. The US metric kW per ton is the inverse, equal to 3.517 ÷ COP, so a lower value is better. The 300 ton (1,055 kW) chiller drawing 180 kW has a COP of 5.86, or 0.60 kW/ton. With 20 kW of chilled and condenser water pumps inside the boundary, the COP is 5.28, or 0.67 kW/ton:

Declare the cooling tower fans too. A figure that includes the pumps and towers describes the chilled water plant, not the chiller.

A measured COP is one operating point. The integrated part load value (IPLV) in AHRI 550/590 is a weighted blend of four points: IPLV = 0.01A + 0.42B + 0.45C + 0.12D, where A to D are the efficiencies at 100, 75, 50 and 25 % load. Each point has a standard condenser water or air temperature that falls with load. Full load carries 1 % of the weight. NPLV uses the same weights at the project design conditions. A site reading at 60 % load with 27 °C condenser water cannot be compared with either value until both load and condenser temperature are matched.

Heat pumps

Heat pump COP is delivered heat divided by electrical input at one time. Over a period the same ratio of energy totals is the seasonal performance factor (SPF). The SEPEMO-Build project defined four boundaries for it, and they are the usual way to state which SPF you mean:

EU guidance on counting renewable heat from heat pumps (Commission Decision 2013/114/EU, section 3.4) uses the H2 boundary and SCOPnet from EN 14825. With 300 kW of delivered heat, 90 kW to the unit and its source fan and 10 kW to the backup heater, the COP is 3.33 at H2 and 3.00 at H3:

Keep defrost in the totals. An air-source unit that reverses its cycle to defrost takes heat back from the heating circuit for a few minutes. If the heat meter or the calculation ignores negative heat flow, the SPF is too high. Store the operating mode (heating, hot water, defrost, standby) and the outdoor temperature with each interval value. Then an annual SPF can be split by mode and by outdoor temperature band.

Time alignment and stale values

Divide energy by energy where you can. The change in an electricity register and a heat meter register over the same 15 minutes gives a ratio that tolerates small timing differences. A ratio of instantaneous rates needs both values averaged over the same interval on one clock.

The common failure is a stale input. On an EpiSensor system the electrical input usually comes from a ZEM on the plant supply. A heat meter or flow meter reaches the ZGW-20 Gateway over M-Bus, Modbus or pulses, often at a slower polling rate. If the heat meter stops answering, its last value can sit beside fresh power readings and give a COP that looks valid and is wrong. Set a maximum age for each input, for example two of its own polling intervals. Withhold the ratio when any input is older.

Missing data is not zero. Apply one rule to both sides: if an hour of heat is missing, remove the same hour of electricity from the period total. Keep the formula, the date it last changed, the unit conversions and the equipment list with every result. A change of boundary then appears in the record as its own dated entry.

Boundary record

Keep one record for each asset. The downloadable template covers:

Build a system from the asset, its useful output and the auxiliary loads inside the boundary.