Energy-Efficient Heating, Ventilating and Air Conditioning
Introduction
Many manufacturing facilities require large quantities of space-conditioning energy use due to:
· large envelope areas
· large openings in the building envelope required for moving material into and out of the facility
· large ventilation rates
· the difficulty of effectively delivering heating and cooling where it is needed
Thus, careful consideration of heating, ventilating and air conditioning (HVAC) systems can lead to substantial energy savings.
This chapter begins by using an energy balance of heating fuel use and the inside-out approach as guides to space conditioning energy saving opportunities. It then develops methods of modeling space conditioning energy use which take into account the large internal loads characteristic of manufacturing facilities. One version of these methods can be derived directly from the statistical modeling of existing energy use data to improve modeling accuracy. Finally, the chapter presents many examples of energy saving opportunities with examples of how to quantify expected savings.
Principles of Energy-Efficient HVAC
Energy Balance Approach
Consider an energy balance on space-heating energy use. The space-heating fuel use, Qf, must equal the net heat loss out of the building divided by the efficiency of the heating system. The net heat loss out of a building is the sum of the conduction heat loss through the building envelope and the heat required to warm ventilation air to room temperature reduced by the internal heat gain from equipment and people.
Qf = [ Qenv + Qair – Qint ] / Eff,h
Qf = [ A/R (Tia – Toa) + V pcp (Tia – Toa) – Qint ] / Eff,h
Qf = [ (A/R + V pcp) (Tia – Toa) – Qint ] / Eff,h
where A is the area of the envelope, R is the thermal resistance of the envelope, V is the air flow rate into/from a facility, pcp is the product of air density and specific heat, Tia is the inside air temperature, Toa is the outside air temperature, Qint is internal heat gain and Eff,h is the efficiency of the heating equipment. Following this energy balance, the opportunities for reducing heating energy use in existing facilities are:
· Increasing R
· Decreasing V
· Decreasing Tia
· Increasing utilization of Qint
· Increasing Eff,h
The first option, increasing the thermal resistance by adding insulation is attractive because insulation is simple, passive and long lasting. However, the incremental gain of adding insulation diminishes as more insulation is added; thus adding insulation to a well-insulated envelope has less effect than adding insulation to an under-insulated envelope. Managing ventilation and infiltration air is critical to energy efficient space conditioning; careful analysis of this option is likely to result in significant saving opportunities. Inside air temperature can be reduced during non-occupied periods using programmable thermostats. In addition, effectively delivering heating to needed areas can reduce overall inside air temperature without reducing comfort. Some ventilation patterns improve utilization of internal heat gain more than others. The efficiency of space conditioning equipment varies, and utilizing the most efficient equipment for the task reduces energy use. This chapter provides many specific opportunities for acting on these variables and reducing heating and cooling energy use. The principles of energy-efficient space cooling are similar.
Opportunities for Improving The Energy-Efficiency of HVAC Systems
These principles can be organized using the inside-out approach, which sequentially reduces end-use energy, distribution energy, and primary conversion energy. Combining the energy balance and inside-out approaches, common opportunities to improve the energy efficiency of HVAC systems include:
· Reduce end use loads
o Add insulation to under-insulated walls
o Add double-polycarbonate sheets over windows
o Replace dark roofs with white roofs in air-conditioned facilities
o Seal exhaust fan openings
o Close dock doors using motion sensors or garage door openers
o Turn off dust collectors when not in use
o Turn off unnecessary exhaust air fans during winter
o Introduce make-up air near exhaust air locations
o Reduce temperature set points during unoccupied periods
· Improve efficiency of distribution system
o Reduce excess temperature stratification using high-volume fans
o Install radiant heaters in high ventilation areas
o Reverse direction of exhaust air fans to utilize internal heat
· Improve efficiency of energy conversion
o Pre-heat makeup air with solar panels
o Use 100% efficient makeup air units to heat make up air
o Use unit heaters rather than MAUs when outdoor air not required
o Control MAUs with differential pressure control
o Use economizers to reduce cooling loads
o Use high efficiency cooling systems
o Pre-heat or pre-cool outdoor air with exhaust air
Calculating Hourly Heating Energy Use
Energy Balance on Heating Energy Use
Heating loads in manufacturing facilities are primarily driven by heat loss through the building envelope and the requirement to heat ventilation or infiltration air. In most manufacturing facilities, heat loss through the ground and solar loads are negligible compared to heat loss through the envelope and to air. Thus, heat loss through the ground and solar loads are not considered in this analysis. Further, energy storage effects are neglected except when considering temperature set back.
The steady-state rate of heat loss through the building envelope, Qenv, is:
Qenv = A / R (Tia – Toa) (1)
where A is the area of the envelope, R is the average thermal resistance, Tia is the inside air temperature and Toa is the outside air temperature.
The steady-state rate of heat lost to ventilation or infiltration air, Qair, is:
Qair = V pcp (Tia – Toa) (2)
where V is the air flow rate, pcp is the product of air density and specific heat, Tia is the inside air temperature and Toa is the outside air temperature. For air at standard conditions, the product of air density and specific heat is about:
pcp = 0.018 Btu/F-ft3
The heating load is reduced by heat discharged by electrical equipment, process heating equipment and occupants, Qi. In manufacturing facilities, these sources of internal heat gain can be significant. The net heating load, Qh, is:
Qh = [ Qenv + Qair – Qi ]+ (3)
The superscript + indicates that the quantity evaluates to zero if the value within the parentheses is negative.
The total rate of heat loss can characterized in terms of the heating coefficient, HC, as:
HC = A / R + V pcp (5)
Thus, the net heating load, Qh, is:
Qh = [ HC (Tia – Toa) – Qi ]+ (6)
The fuel consumed by the heating system, Qf, must take into account the efficiency of the heating system, Effh.
Qf = Qh / Effh = [ HC (Tia – Toa) – Qi ]+ / Effh (7)
Equation 7 calculates fuel use by explicitly considering indoor air temperature and internal heat gain. An alternative but equivalent method of calculating heating energy use employs the concept of balance temperature. The balance-temperature method is easier to interpret graphically and can be used in conjunction with statistical data analysis.
The balance temperature is the outside air temperature at which no heating is required. Based on this definition, balance temperature, Tb, can be calculated as:
Qh = [ HC (Tia – Toa) – Qi ]+
0 = [ HC (Tia – Tb) – Qi ]+
Tb = Tia – Qi / HC (8)
Thus, the net heating load, Qh, can be calculated as:
Qh = [ HC (Tia – Toa) – Qi ]+ = HC (Tb – Toa)+ (9)
And, heating fuel use, Qf, can be calculated as:
Qf = HC / Effh (Tb – Toa)+ (10)
Example
Calculate hourly fuel use for a facility with the following characteristics:
Length = 250 feet, width = 200 feet, ceiling height = 35 feet
Average thermal resistance of walls and ceilings = 10 ft2-F-hr/Btu
Average ventilation rate = 10,000 cfm
Outdoor air temperature = 30 F
Indoor air temperature = 70 F
Average electrical power = 100 kW
Efficiency of heating system = 80%
Explicit Method:
A = (250 ft + 200 ft + 250 ft + 200 ft) x 35 ft + (250 ft x 200 ft) = 81,500 ft2
Qenv = A / R (Tia – Toa) = 81,500 ft2 / 10 ft2-F-hr/Btu x (70 – 30) F = 326,000 Btu/h
Qair = V pcp (Tia – Toa)
Qair = 10,000 cfm x 60 min/hr x 0.018 Btu/F-ft3 x (70 – 30) F = 432,000 Btu/h
Qi = 100 kW x 3,413 Btu/kWh = 341,300 Btu/h
Qh = [ Qenv + Qair – Qi ]+
Qh = [ 326,000 Btu/hr + 432,000 Btu/hr – 341,300 Btu/hr ]+ = 416,700 Btu/h
Qf = Qh / Effh = 416,700 Btu/hr / 0.80 = 520,875 Btu/h
Balance-Temperature Method:
A = (250 ft + 200 ft + 250 ft + 200 ft) x 35 ft + (250 ft x 200 ft) = 81,500 ft2
A / R = 81,500 ft2 / 10 ft2-F-h/Btu = 8,150 Btu/h-F
V pcp = 10,000 cfm x 60 min/hr x 0.018 Btu/F-ft3 = 10,800 Btu/h-F
HC = A / R + V pcp = 8,150 Btu/h-F + 10,800 Btu/h-F = 18,950 Btu/h-F
Tb = Tia – Qi / HC = 70 F – 341,300 Btu/h / 18,950 Btu/h-F = 52.0 F
Qh = HC (Tb – Toa)+ = 18,950 Btu/h-F (52.0 F – 30 F) = 416,700 Btu/h
Qf = Qh / Effh = 416,700 Btu/h / 0.80 = 520,875 Btu/hr
Calculating Annual Fuel Use with the Degree-Day Method
A useful method for calculating annual heating and cooling energy use is called the degree-day method. As before, the balance temperature, Tb, is defined as the outside air temperature when the building needs no heating. This means, whenever Toa < Tbal, the building needs heating and whenever Toa > Tbal the building needs cooling.
In the past, average daily temperatures were readily available but hourly temperatures were not. Thus, it was common to use average daily temperatures to determine the heating and cooling temperature drivers. To calculate annual heating degree days (HDD), sum the difference between the balance and average daily outdoor air temperatures for all days during the year when Toa < Tbal.
HDD (Tbal) = (Tbal – Toa)+ (11)
To calculate annual cooling degree days (CDD), sum the difference between the balance and average daily outdoor air temperatures for all days during the year when Toa > Tbal.
CDD (Tbal) = (Toa - Tbal)+ (12)
Annual heating and cooling equipment energy use are then:
Qf = [HC / Effh] HDD (13)
Qe = [CC / Effc] CDD (14)
Thus, heating and cooling degree days provide an effective way to summarize the heating and cooling driving potentials of a given climate. For example, consider the following degree days data, calculated with different balance temperatures from TMY3 data for Dayton, Ohio and Phoenix, Arizona. At a 65 F balance temperature, a the heating load for facility in Dayton, OH would be 4.62 times greater than a similar facility in Phoenix Arizona. Similarly, the cooling load for a facility in Phoenix, Arizona would be 4.17 times greater than for a similar facility in Dayton, OH.
Fraction Difference = HDDTb=65,Dayton – HDDTb=55,Phoenix / HDDTb=55,Phoenix
Fraction Difference = (5,902 – 1,050) / 1,050 = 462%
Fraction Difference = CDDTb=65, Phoenix – CDDTb=55, Dayton / CDDTb=55, Dayton
Fraction Difference = (4,643 – 898) / 4,643 = 417%
Annual heating and cooling degree days (F-day/year) for Dayton, Ohio and Phoenix, AZ based on TMY3 data.
In many residences, the heating and cooling balance temperatures are about 65 F. Thus, it became common practice to assume the balance temperature is 65 F and calculate degree days based on this assumption. Base 65-F heating and cooling degree days for a given time period are often posted on the internet and reported by weather services. However, in manufacturing facilities with a wide range of internal temperatures and widely varying amounts of internal heat gain, the assumption of 65-F balance temperature is often incorrect. The loss of accuracy from using base 65 F degree days to calculate energy use depends on the difference between the actual building balance temperature and the assumed 65 F balance temperature. For example, using HDD from Dayton, Ohio calculated with a balance temperature of 65 F when the actual building balance temperature is 55 F would overestimate annual heating energy use by 62%.
Fraction Change = HDDTb=65 – HDDTb=55 / HDDTb=55 = (5,902 - 3,659) / 3,659 = 61%
Thus, the use of nominal base 65 F heating and cooling degree days to calculate heating and cooling energy use is not recommended. However, the degree-day method is accurate and useful when balance temperatures are calculated and heating or cooling degree days to the appropriate balance temperature are used.
Example
Calculate annual fuel use using HDDs for a facility in Dayton, Ohio with the following characteristics:
Length = 200 feet, width = 200 feet, ceiling height = 30 feet
Average thermal resistance of walls and ceilings = 10 ft2-F-hr/Btu
Average ventilation rate = 10,000 cfm
Average indoor air temperature during heating season = 65 F
Average internal electrical power = 75 kW
Efficiency of heating system = 80%
Solution:
A = (200 ft + 200 ft + 200 ft + 200 ft) x 30 ft + (200 ft x 200 ft) = 64,000 ft2
A / R = 64,000 ft2 / 10 ft2-F-hr/Btu = 6,400 Btu/hr-F
V pcp = 10,000 cfm x 60 min/hr x 0.018 Btu/F-ft3 = 10,800 Btu/hr-F
HC = A/R + V pcp = 6,400 Btu/hr-F + 10,800 Btu/hr-F = 17,200 Btu/hr-F
Qint = 75 kW x 3,413 Btu/kWh = 255,975 Btu/hr
Tb = Tsp – Qint / HC = 65 F - 255,975 Btu/hr / 17,200 Btu/hr-F = 50.1 F
(from table) HDD(50) = 2,757 F-day/yr
Qf = HC / Effh HDD(50)
Qf = 17,200 Btu/hr-F / 0.80 x 2,757 F-day/yr x 24 hr/dy x 106 Btu/mmBtu
Qf = 1,423 mmBtu/yr
A graph showing monthly space-heating fuel use versus outdoor air temperature for this facility is plotted below.