Considering building and air-handler loads in central-plant modeling
Editor's note: This is the fifth article in a five-article series.
The first four articles in this series1,2,3,4 considered only plant load in the modeling of central-chiller-plant performance. This article adds building and air-handler loads to the mix. A system serving a 1,672,000-sq-ft, one-shift facility is modeled over a 24-hr period for both design-day and fall-day weather conditions.
The system simulation modeling used in preparation of this article consists of a set of simultaneous equations that obey the laws of physics and thermodynamics and includes the nonlinear performance characteristics of plant equipment, air-side equipment, and buildings, always arriving at energy equilibrium after a change in steady-state condition.
Figure 1 illustrates the improvement in plant performance resulting from the control of distribution-pump power to provide air-handler leaving-water temperatures of 51°F and 54°F. The top chart gives primary/secondary- (P/S-) plant performance, while the bottom chart gives primary-only- (P-only-) plant performance. The chillers maintain a supply-water temperature of about 44°F. Implementing this control in a system experiencing low-load delta-T likely would result in complaints of excessive heat, requiring cleaning, parts replacement, and, possibly, water- and air-distribution-system modifications before 54°F air-handler leaving-water temperature could be maintained. Figure 1 shows little difference in plant performance between the two pumping approaches.
Figure 2 illustrates a P/S system operating at a peak plant design load of 4,902 tons. The lighting load of 1.7 w per square foot and plug load of 0.8 w per square foot are consistent with a 1980s-era system.
Equal load on each of the chillers in operation is assumed. Air-handler response is the average of the 75 units. Article 11 gives the performance characteristics of the chiller and tower. The plant in Figure 2 is the same one depicted in Figure 4 of Article 11.
The electricity demand of each of the plant's 75 fans at design load is:
cfm × (dh) ÷ [6,356 × (Efan-ASD)]
39,253 cfm × 3.91 in. ÷ (6,356 × 0.547) = 44.15 hp
0.746 kw per hp × 44.15 hp = 33 kw (Figure 2)
Each air handler includes a fan-powered terminal drawing 24.1 kw. Also included in Figure 2 are return-air and fresh-air fans. The chilled-water and tower pumps circulate water at a rate of 2,400 gpm and 3,000 gpm, respectively. The variable-speed secondary pump is at energy equilibrium at the conditions of Figure 2. The primary chiller pump is operating at 16 kw based on efficiency of 0.81 and head of 28.6 ft. The tower pump is operating at efficiency of 0.83 and head of 55.6 ft.
“Energy in equals energy out” is a condition of a real system, assuming change in internal energy is zero. Energy exits a system at the towers, exhausted to the atmosphere. For Figure 2, energy to the tower can be determined by the first law of thermodynamics:
Tons = gpm × delta-T ÷ 24 = 3,000 gpm × (81.87°F - 72.65°F) ÷ 24 = 1,152.5 tons
With the addition of tower-fan power (35 kw ÷ 3.517 = 9.95 tons), the total energy out is 1,163 tons for each tower, or 5,817 tons for all five. Note the building load of 3,513 tons is 5,817 tons when exhausted at the towers, a 65-percent increase attributed to the air handler and plant equipment.
Energy into the system in Figure 2 consists of the building loads plus the heat generated by the electricity required to drive the air-handler fans, pumps, and chillers/towers. Pump loading is modeled as a function of pump efficiency. All other electricity to the system, air handlers, chillers, and towers is modeled as fully loading the system.
Ein = (qs-ton) + (qsolar) + (FAton) + (qp-ton) + (light)ton + (plug)ton + (AHU)kw ÷ 3.517 + (Pheat)kw ÷ 3.517 + (FAheat)kw ÷ 3.517 + (Psec-kw × Esec) ÷ 3.517 + (Pc-kw × Ec × Pc#) ÷ 3.517 + (chillerkw × chiller#) ÷ 3.517 + (Pt-kw × Et × Pt#) ÷ 3.517 + (fant-kw × Tower#) ÷ 3.517
Ein = 1,422 tons + 501.8 tons + 178 tons + 222.4 tons + 808.7 tons + 380.5 tons + (4,883 kw ÷ 3.517) + 0 + 0 + (699.9 kw × 0.803 ÷ 3.517) + (16 kw × 0.81 × 5 ÷ 3.517) + (451.7 kw × 5 ÷ 3.517) + (37.8 kw × 0.83 × 5 ÷ 3.517) + (35 kw × 5 ÷ 3.517)
Ein = 3,513.4 + 1,388.4 + 159.8 + 18.42 + 642.2 + 44.6 + 49.8 = 5,817 tons
These relations are an integral part of the set of simultaneous equations and, therefore, hold for all figures and curves presented in the five articles in this series. The system always is at energy equilibrium. All data and curves meet the requirements of “energy in equals energy out” and the first law of thermodynamics. Any data point in any figure in any of the five articles in this series can be shown in a form similar to Figure 2.
The remaining figures in this article will consider the 24-hr operation of the system in Figure 2.
The top chart in Figure 3 shows the design-day ambient dry bulb and wet bulb, while the bottom chart shows the resulting building loads. The facility is assumed to be a one-shift operation; therefore, “on” control of lights and plug loads occurs from 8 a.m. to 6 p.m. The building-shell and fresh-air loads are set by outside temperature, while solar load is assumed as shown in the bottom chart.
The top chart in Figure 4 gives the total building load and the power required by the air handlers to transfer the building load to the coils. The bottom chart gives the site load, or load to the plant, and the load air-handler power adds to the total.
Figure 5 shows the 24-hr performance of the P/S and P-only plants, with performance as defined in Figure 1. Operating with 54°F air-handler leaving-water temperature significantly improves plant performance. Essentially, the performance of the P/S and P-only plants is the same when operated as defined in articles 22 and 33.
The top chart in Figure 6 shows the site power for the P/S and P-only plants, while the bottom chart shows the difference in site power. The biggest difference — 44 kw — occurs at noon and 4 p.m., but is less than 1 percent of the approximately 11,600-kw site power. Figures 4 and 5 of Article 11 show a P-only-plant advantage of 23 kw; the difference at 2 p.m. in Figure 6 of this article is about the same. In an attempt to equalize the operation of the two plants, evaporator leaving-water temperature is maintained at nearly identical levels, as shown by the secondary vertical axis in the bottom chart. Selecting a P-only plant over a P/S plant for slight potential power savings may not be wise; when ease of control and operation are considered, the P/S plant often is the preferred choice.
Figure 7 shows the cooling load and performance of the P/S plant. The top chart shows building load (Figure 3) is less than air-handler cooling load during off hours because of air-handler power. The bottom chart shows the power demand of the building, air handlers, and plant, with site power shown by the vertical axis.
The top chart in Figure 8 gives the dry bulb and wet bulb to be modeled, while the bottom chart gives the building loads over a 24-hr period. This set of building loads is much different than the one in Figure 3, with the shell load negative and perimeter heat required at night. These building-load components change only as the weather changes; the set of simultaneous equations iterates to this new condition of building equilibrium. The reduced building load causes the air handler to unload, which causes the plant to unload to the condition of energy equilibrium shown in Figure 9.
Figure 9 illustrates the system at energy equilibrium, with dry bulb 90°F, wet bulb 60°F, and the plant unloaded to four chillers/towers at 88-percent power. All changes from Figure 2 to Figure 9 are a result of the set of simultaneous equations iterating to a new condition of energy equilibrium as a result of a change in the weather.
The top chart in Figure 10 shows site power for fall-day operation similar to the site power for peak-day operation in Figure 7. The bottom chart shows that during off hours, the fall-day power is greater. This can be attributed primarily to the perimeter heat power required during fall-day night hours.
Actual site power in excess of modeled site power provides owners and operators an alarm for corrective action. The top chart in Figure 11 illustrates the variation in site power that could occur during fall-day operation, while the bottom chart illustrates the variation in site power that could occur during peak-day operation. Both charts illustrate the effect of no shutdown of lights and plug loads; partial shutdown would provide less difference between actual and expected performance.
The first article in this series1 showed that bypass control is a highly inefficient means of operating a P/S plant. The article also concluded that while overpumping reduces plant capacity at peak load, all lesser loads can be met by a plant. Overpumping by 43 percent results in a 3.5-percent reduction in plant capacity for the P/S system modeled. The article also shows that chiller capacity increases with a drop in wet-bulb temperature. Operating a P-only plant with the minimum number of chillers/towers on is more efficient than operating a P/S plant with bypass control.
Article 22 concluded that by operating chillers to full capacity before another chiller is turned on, the P/S plant could be operated more efficiently than the P-only plant of Article 11. This control strategy minimizes the number of chillers/towers on and results in a more efficient plant, even though return water mixes with supply water from the evaporator, requiring the chiller to provide sub-44°F water.
Article 33 showed that low-load delta-T for the P-only plant resulted in high evaporator velocities and increased distribution-pump power. Turning on another chiller/tower reduced the evaporator velocity and resulted in performance about the same as that of the P/S plant of Article 22.
Article 44 arrived at the general conclusions that load characteristics can cause low-load delta-T and achieving a load delta-T of 10°F is difficult, if not impossible, with a differential-pressure transmitter (DPT). Although installation of a high cutoff valve can have a positive effect for given secondary-pump power, the general trend is inefficient plant operation, if secondary-pump power is excessive. In general, control of a secondary pump with a DPT in a water-distribution system does not provide good control of load delta-T. A possible alternative is control of air-handler return-water temperature.
This article presented the concept of controlling secondary-pump power based on air-handler leaving-water temperature. A significant improvement in plant performance is possible with this control concept, which basically controls load delta-T at or near design value.
Conclusions regarding simulation modeling are:
A simulation model provides a tool for understanding the many operational and design characteristics of a system. A manual for operators could be developed with a simulation model.
A simulation model offers the ability to understand how a system should be operating at any hour and under any weather conditions. Comparing the site power of a real system with that of a modeled system gives a measure of how efficient the real system is operating.
Total site power demand generally is available to the owners and operators of a facility. A simulation model as part of a monitoring system could provide an energy alarm.
1980s systems still in operation represent a major opportunity for energy savings. A simulation model can define the effect of upgrades.
A simulation model provides the ability to plan the future of a facility, to evaluate modifications of or additions to the chilled-water system or expansion and modification of the site.
Nelson, K. (2011, April). Primary/secondary vs. primary-only pumping. HPAC Engineering, pp. 34-40. Available at http://bit.ly/Nelson_1
Nelson, K. (2011, May). Efficient control of a primary/secondary plant. HPAC Engineering, pp. 34, 37-41. Available at http://bit.ly/Nelson_2
Nelson, K. (2011, July). Efficient control of a primary-only plant. HPAC Engineering, pp. 32, 34, 36-39. Available at http://bit.ly/Nelson_3
Nelson, K. (2011, August). Anatomy of load delta-t. HPAC Engineering, pp. 34, 36-41. Available at http://bit.ly/Nelson_4
Did you find this article useful? Send comments and suggestions to Executive Editor Scott Arnold at scott.arnold@penton.com.
Kirby Nelson, PE, has been involved in the modeling of HVAC systems since the oil embargo of 1973 — first as corporate energy manager for Texas Instruments Inc., then as a consultant. Models he has used include DOE-2, E Cube, and models developed on an analog/digital computer, including models of cleanrooms. A life member of the American Society of Heating, Refrigerating and Air-Conditioning Engineers, he has presented numerous papers, led an energy engineering delegation to China, and more recently developed models for district cooling systems, thermal-storage systems, and central plants.
(Site)kw = (Bld)kw + (AHU)kw + (Plant)kw
(Site)ton = (CL) × (AH)#
(Bld)kw = (plug)kw + (light)kw + (Pheat)kw + (FAheat)kw
(Pheat)kw = perimeter heat, kilowatts
(FAheat)kw = fresh-air heat, kilowatts
(Bld)ton = (qs) + (qfa) + (qp) + (qsolar) + (Bld)kw ÷ 3.517
(qs) = shell dry-bulb weather load on campus, tons
(qfa) = fresh-air load on campus, tons
(qp) = people load on campus, tons
(plug)kw = building electric plug load, kilowatts
(plug)ton = building plug load, tons
(light)kw = building lighting electricity load, kilowatts
(light)ton = buildings lighting load, tons
(qsolar) = solar load on campus buildings, tons
(Bld)ft2 = total square footage, all buildings
(AHU)kw = total fan kw = (AH)# × (ahu)kw
(ahu)kw = (fan)VAV-kw + (fan)ter + (fan)ret + (fan)fa
(fan)VAV-kw = variable-air-volume-fan demand, kilowatts
(fan)ter = terminal-fan demand, kilowatts
(fan)ret = return-fan demand, kilowatts
(fan)fa = fresh-air-fan demand, kilowatts
(ahu)ton = (ahu)kw ÷ 3.52 = heat load
(Efan-ASD) = fan efficiency × motor-variable-speed-drive efficiency
(dh) = variable-air-volume-fan static pressure, inches of water
(AH)cfm = flow of air through coil, cubic feet per minute
(Tas) = temperature of air supplied to building
(Tar) = temperature of air returned to air handler
(UA) = heat-transfer coefficient, (U) × (A)
(LMTD) = coil log mean temperature difference, degrees Fahrenheit
(Ccap) = capacity, one coil, tons, (UA) × (LMTD)
(CL) = (Bld)ton ÷ (AH)# + (ahu)ton (load on one coil)
(ewt)AH = temperature of water entering air-handler coil, degrees Fahrenheit
(lwt)AH = temperature of water leaving air-handler coil, degrees Fahrenheit
(AH)# = number of air handlers serving buildings/campus
(P)sec-kw = secondary-pump kilowatt demand
(gpm)sec = secondary flow to/from campus, gallons per minute
(H)sec = secondary-pump head, feet
(E)sec = secondary-motor/pump efficiency
(chiller)kw per (evap)ton = (chiller)kw ÷ (evap)load
(Plant)kw = (P)sec-kw + [(chiller)kw + (P)t-kw + (fan)t + (P)c-kw] × C#
(Plant)kw per site ton = (Plant)kw per (Campus)ton
(CCWS)kw = (Plant)kw + (AHU)kw
(P)c-kw = chiller-pump demand, kilowatts
(gpm)c = chiller-pump flow, gallons per minute
(E)c = chiller-pump efficiency (0.81)
(gpm)b = flow in bypass, gallons per minute
(t)b = temperature of water in bypass, degrees Fahrenheit
(evap)load = (Site)ton ÷ Chiller# + ([(E)sec × (P)sec-kw] ÷ Chiller# + (E)c × (P)c-kw) ÷ 3.516
(lwt)evap = temperature of water leaving evaporator, degrees Fahrenheit
(TER) = evaporator refrigerant temperature, degrees Fahrenheit
(ewt)evap = temperature of water entering evaporator, degrees Fahrenheit
Evap# = number of evaporators on
(chiller)kw = chiller demand, kilowatts
(chiller)lift = (TCR) — (TER), degrees Fahrenheit
(chiller)% = motor load, percent
Chiller# = number of chillers on
(cond)load = (evap)load + [(chiller)kw + (P)t-kw × (E)t] ÷ 3.517
(TCR) = condenser refrigerant temperature, degrees Fahrenheit
Cond# = number of condensers
(Tower)# = number of towers on
(fan)t = tower-fan demand, kilowatts
(lwt)t = temperature of water leaving tower, degrees Fahrenheit
(ewt)t = temperature of water entering tower, degrees Fahrenheit
(gpm)t = tower water flow, gallons per minute
(P)t-kw = tower-pump demand, kilowatts
(E)t = tower-pump efficiency (0.83)
Tton-ex = tower exhaust, tons = (cond)load + (fan)t ÷ 3.52
OA = outside-air temperature, degrees Fahrenheit
(t)WB = wet-bulb temperature, degrees Fahrenheit
(fan)% = tower-fan-motor speed, percent
Eout = Tton-ex × (Tower)#
Ein = (qs) + (qfa) + (qp) + (light)ton + (plug)ton + (AHU)kw ÷ 3.52 + [(P)sec-kw × (E)sec] ÷ 3.52 + [(P)c-kw × (E)c + (P)t-kw × (E)t + (fan)t] × Chiller# ÷ 3.52
Ein = Eout
(CCWS)kw per ton = (CCWS)kw ÷ (Bld)ton
(Site)w per square foot = (Site)kw ÷ (Bld)ft2
Nomenclature.