Plant Energy Balance
Introduction
After analyzing electricity and fuel billing data, the next step is to understand where energy is
used within a facility. In some cases, detailed sub-metered data may be available that can
precisely define how much energy is used by various pieces of equipment or subsystems.
However, a useful approximation can be derived from a list of the major pieces of energy-using
equipment and estimates of loading and operation time.
This chapter shows how to calibrate these approximations of energy use by equipment against
total energy input using the energy balance concept that the sum of the energy used by the
various pieces of equipment must equal the total energy input to the plant as defined by the
utility bills. The result is an approximation of energy use by equipment. This information is
helpful in focusing efforts on large energy users and in calibrating engineering models of
equipment energy use used for calculating savings.
Rated Output Power, Efficiency and Input Power
In most energy using equipment, useful output power, Puseful,output, is less than total input
power, Pinput; some power is lost, Plost, during energy conversion. In many cases, lost power is
rejected as heat to the environment. Thus, a steady state energy balance gives:
Pinput – Plost – Puseful,output = 0
Energy efficiency is defined as the ratio of useful power output to total power input.
Efficiency = Puseful,output / Pinput
Many pieces of energy using equipment are rated by energy or power output. Thus, energy
input is:
Einput = Euseful,output / Efficiency
For example, the horsepower rating of a motor refers to shaft output power. In addition, most
motors are about 90% efficient. Further, most motors are slightly oversized so that they don’t
burn up if the load is slightly greater than anticipated. The load on most motors is about 75% of
the design output. Putting this together, the input power for a 75% loaded, 100-hp motor that
is 90% efficient is about:
Pin = Poutput / Efficiency
Pin = (100 hp x 75% loaded x 0.746 kW/hp) / 90% efficiency = 62.17 kW
The rate of heat loss from the motor is:
Ploss = Pinput (1 – Efficiency) = 62.17 kW x (1 – 0.90) x 3,413 Btu/kWh = 21,218 Btu/hr
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Similarly, chillers are typically rated by heat output. Thus, a 100-ton chiller removes 100 tons of
heat (100 tons x 12,000 Btu/hr = 1,200,000 Btu/hr) of heat when operating at full load. Input
power to chillers varies according to chiller type and age, but typically ranges from 0.5 kW/ton
to 1.2 kW/ton. Thus, the input power for a 60% loaded, 100-ton chiller requiring 0.6 kW/ton is
about:
Pin = 100 tons x 60% loaded x 0.7 kW/ton = 42 kW
The heat output rate of boilers is typically listed on the boiler nameplate. Thus, the rate of fuel
input for a 60% loaded, 80% efficient boiler with rated heat output of 5,000,000 Btu/hr is about:
Pin = 5,000,000 Btu/hr x 60% loaded / 80% efficient x ( 1 mmBtu / 1,000,000 Btu)
Pin = 3.75 mmBtu/hr
Electricity Use by Equipment
Most facilities can compile lists of the major energy using equipment, along with rated power
output and operating hours. Combining this information with an estimate of the fraction of full
load power that each piece of equipment operates at can yield a useful approximation of energy
use by equipment. The sum of the total energy use by equipment can be compared to total
annual energy use derived from utility bills, to adjust estimates of operating hours and fraction
loaded. The result is an approximation of energy use by equipment. This information is helpful
in focusing efforts on large energy users and in calibrating engineering models of equipment
energy use used for calculating savings.
Consider for example, the following table of major electricity-using equipment, rated output
power, fraction loaded, and operating hours. In this list, the lights draw .465 kW each; this this
is actually input power not output power. This type of list can be compiled by most facilities,
even in advance of an energy assessment, and is an important component of the baseline
analysis.
Number
10
4
200
1
1
2
Type
40-hp stamping machines
50-hp chilled water pumps
400-W MH Lights
100-ton chiller
20-hp cooling tower fan
100-hp air compressors
Rated
Output Power
(hp, kW or tons)
40
50
0.465
100
20
100
Fraction
Loaded
0.50
0.75
1.00
0.50
0.75
0.90
Operating
Hours
(hours/year)
4,000
6,000
7,000
4,000
3,500
6,000
Next the efficiency of each category of equipment can be estimated and the annual electricity
use calculated using the proper unit conversions. In this example, chiller power input is
estimated to be 0.8 kW/ton. The sum of the annual electricity use by equipment can then be
compared to the total electricity use of the plant, and if necessary, estimates of fraction loaded
and operating hours can be adjusted. The difference between the sum of the listed electricity
use and the total plant electricity use from the utility bills is attributed to “Other Equipment”. In
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the example shown here, the following table represents a calibrated estimate of electricity use
by equipment type.
Number
10
4
200
1
1
2
Type
40-hp stamping machines
50-hp chilled water pumps
400-W MH Lights
100-ton chiller
20-hp cooling tower fan
100-hp air compressors
Rated
Output Power
(hp, kW or tons)
40
50
0.465
100
20
100
Total Listed Equipment
Other Equipment
Measured Total
Fraction
Loaded
0.50
0.75
1.00
0.50
0.75
0.90
Operating Efficiency
Hours
(hours/year)
4,000
0.90
6,000
0.90
7,000
1.00
4,000
1.00
3,500
0.90
6,000
0.90
Conversion
0.746
0.746
1.000
0.750
0.746
0.746
Annual
Energy Use
(kWh/year)
663,111
746,000
651,000
150,000
43,517
895,200
Fraction
Total
Energy
25.5%
28.7%
25.0%
5.8%
1.7%
34.4%
2,485,717
114,283
2,600,000
4.4%
This list can be sorted from large to small and graphed to give a concise and visual estimate of
electricity use by equipment.
Fuel Use by Equipment
The same methodology can be applied to estimate fuel use by equipment. Consider for
example, the following table of major fuel-using equipment, rated output power, fraction
loaded, and operating hours. In this list, the lights draw .465 kW or 465 W each. As before, this
type of list can be compiled by most facilities, even in advance of an energy assessment, and is
an important component of the baseline analysis.
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Number
2
15
2
1
Type
Boilers
Hanging unit heaters
Make-up air units
Dry-off ovens
Rated
Output
(Btu/hour)
3,000,000
100,000
2,000,000
1,600,000
Fraction
Loaded
0.80
1.00
0.70
0.60
Operating
Hours
(hours/year)
6,000
3,000
3,000
6,000
As before, the efficiency of each category of equipment can be estimated and the annual fuel
use calculated using the proper unit conversions. The sum of the annual fuel use by equipment
can then be compared to the total fuel use of the plant, and if necessary, estimates of fraction
loaded and operating hours can be adjusted. The difference between the sum of the fuel use of
the listed equipment and the total plant fuel use from the utility bills is attributed to “Other
Equipment”. Using this methodology, the following table represents a calibrated estimate of
fuel use by equipment type.
Number
2
15
2
1
Type
Boilers
Hanging unit heaters
Make-up air units
Dry-off ovens
Total Listed Equipment
Other Equipment
Measured Total
Rated
Output
(Btu/hour)
3,000,000
100,000
2,000,000
1,600,000
Fraction
Loaded
0.80
1.00
0.70
0.60
Operating
Hours
(hours/year)
6,000
3,000
3,000
6,000
Efficiency
0.80
0.80
1.00
0.85
Annual
Energy Use
(mmBtu/year)
36,000
5,625
8,400
6,776
Fraction
Total
Energy
60.0%
9.4%
14.0%
11.3%
56,801
3,199
60,000
5.3%
This list can be sorted from large to small and graphed to give a concise and visual estimate of
fuel use by equipment.
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