Energy-Efficient Process Cooling

Introduction The cooling of equipment and products is an integral part of many manufacturing processes. This chapter begins by discussing guiding principles for reducing cooling energy use. The chapter then describes typical process cooling systems used in manufacturing and the approximate cost of cooling for each system. The body of the chapter discusses common methods, organized according to the inside-out approach, for improving the cooling system energy efficiency. For each method, the fundamental equations for estimating savings are presented, and the method is illustrated with an example.

Principles of Energy-Efficient Process CoolingHeat Exchange Effectiveness Approach Cooling systems remove the required quantity of heat at the required temperature. An energy balance on a system shows that the rate of heat removed, Q, is a function of heat transfer effectiveness and temperature:

Q = UA (Tp – Tc) = m cp Tc (1)

where UA is the overall heat transfer coefficient, Tp is the temperature of the process and Tc is the temperature of the cooling fluid, m is mass flow rate and cp is the specific heat.

This relation demonstrates that cooling energy use can be reduced by either:

Increasing the required temperature of cooling, represented by Tc Increasing heat transfer effectiveness, represented by UA, and subsequently

deceasing m or increasing Tc

Exergy Balance Approach Exergy is a thermodynamic property derived from the combination of the First and Second Laws of Thermodynamics. An exergy balance shows that useful work (exergy) is destroyed in all real processes, and provides a method for quantifying the losses.

Exin – Exout – Edestroyed = d Exsystem /dtEdestroyed = Exin – Exout – d Exsystem /dt

In cooling processes, important mechanisms of exergy destruction are:

Friction Turbulence Mixing Heat transfer through large temperature differences

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Mismatch between the quality of energy supplied and that needed by the end-use.

Minimizing these losses inevitably improves system efficiency. Thus, seeking to identify and reduce these losses is a useful guide to improving cooling system energy efficiency.

Opportunities for Improving The Energy-Efficiency of Process Cooling SystemsMany energy systems, including process cooling systems, can be organized into energy conversion, distribution and end use components. In process cooling systems, the distribution system includes the pumps/fans, piping and tanks necessary to transport heat from the process loads. Reducing required flow rate, m, leads to reduced friction losses and reduced pump/fan energy costs. Similarly, increasing cooling fluid temperature reduces heat gain into the cooling fluid during distribution. Reduced end-use and distribution loads substantially reduce the total cooling load on the conversion equipment. Moreover, cooling equipment energy use is decreased even more since most cooling equipment runs more efficiently at part load and when delivering higher temperatures than at full load with low cooling fluid delivery temperature. Thus, inside-out savings are substantial. Finally, most plants that employ process cooling also employ process heating. Looking across the entire plant for opportunities to simultaneously reduce process cooling and heating loads can yield significant savings opportunities.

Combining the heat exchange effectiveness, exergy balance and inside-out approaches, common opportunities to improve the energy efficiency of process cooling systems include:

Reduce end use loadso Add insulation to cold surfaceso Add heat exchangers between heated and cooled processeso Improve heat exchange effectiveness

Improve efficiency of distribution systemo Reduce friction and flow in piping systemso Avoid mixingo Avoid unnecessary heat exchange

Improve efficiency of energy conversiono Use cooling towers in place of chillers when possibleo Use VFD drives on cooling tower fanso Stage chillers to optimize part-load efficiencyo Use high efficiency chillers

Typical Process Cooling Systems and Cooling CostsMany industrial processes use water to transport heat from the process or equipment back to a primary cooling unit. The most common types of primary cooling units are cooling

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towers, water-cooled chillers, air-cooled chillers and absorption chillers. In addition, water is sometimes used in an open loop to cool processes or equipment and then discharged to sewer. Finally, compressed air is sometimes expanded through vortex tubes to produce cooling. Diagrams of these systems, and the approximate costs of cooling are shown below. In all cases, the cost of electricity is assumed to be $0.10 /kWh, the cost of natural gas is $10 /mmBtu and the cost of water is $6.00 per 1,000 gallons.

Cooling TowerCooling towers cool water by evaporating about 1% of the water passing through the tower. A typical cooling tower cooling system is shown in Figure 1. The system uses an open tank as a well for return water from the process and cooling tower. In the cooling tower loop, water is pumped from the chilled water tank to the top of the cooling tower, where it gravity feeds back to the chilled water tank. The process loop shown below includes a bypass loop to accommodate flow from a constant speed pump if the water required by the process loads varies.

ProcessLoad 1

ProcessLoad 2

Chilled Water Tank

Cooling Tower

BypassValve

Process PumpCooling Tower Pump

Figure 1. Cooling tower system.

The approximate cost of cooling with a cooling tower can be estimated by considering a cooling tower with a rating of 500 cooling-tower tons. A cooling-tower ton is 15,000 Btu/hr. Water flow through most cooling towers is 3 gpm per cooling-tower ton. Total pressure rise through a cooling tower pump is frequently about 40 ft-H20, and pumps are about 70% efficient. A cooling tower with a rating of 500 cooling-tower tons typically uses a 30-hp cooling tower fan that is about 80% loaded. Both pump and fan motors are 90% efficient. Thus, the pump power use, Pp, and fan power use, Pf, are about:

Pp = 1,500 gpm x 40 ft-H20 / (3,960 gpm-ft-H20/hp x 70% x 90%) x 0.75 kW/hp = 18 kWPf = 30 hp x 80% x 0.75 kW/hp / 90% = 20 kW

The rate of cooling provided, Q, electricity per unit cooling, E/Q, and electricity cost per unit cooling, EC/Q, are:

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Q = 500 cttons x 15,000 Btu/hr-ctton = 7.5 mmBtu/hrE/Q = (18 kW + 20 kW) / 7.5 mmBtu/hr = 5 kWh/mmBtuEC/Q = 5 kWh/mmBtu x $0.10 /kWh = $0.50 /mmBtu

In addition, cooling towers evaporate about 1% of water flow. Assuming the total of the water and sewer charges for water is $6.00 per 1,000 gallons, the quantity of makeup water per unit cooling, W/Q, and water cost per unit of cooling, WC/Q, are about:

W/Q = (1,500 gal/min x 60 min/hr x 1%) / 7.5 mmBtu/hr = 120 gal/mmBtu WC/Q = 120 gal/mmBtu x $6.00 / 1,000 gallons = $0.72 /mmBtu

The total unit cost is of cooling, C/Q, with a cooling tower is about:

C/Q = EC/Q + WC/Q = $0.50 /mmBtu + $0.72 /mmBtu = $1.22 /mmBtu

Water-Cooled ChillerChillers can provide lower temperature water than cooling towers. Water-cooled chillers reject process and compressor heat by passing water from a cooling tower over condensor coils of the chiller. A cooling system with a water-cooled chiller is shown in Figure 2. Water-cooled chillers are slightly more energy efficient than air-cooled chillers, but require the additional expense of a cooling tower. Water-cooled chillers require about 0.8 kW per ton of cooling, including the cooling tower fan and pump. Thus, the electricity use per unit of cooling, E/Q, and cost per unit cooling, C/Q, to provide 1 mmBtu of cooling are about:

E/Q = 0.8 kW/ton / 12,000 Btu/ton x 1,000,000 Btu/mmBtu = 67 kWh/mmBtuC/Q = 67 kWh/mmBtu x $0.10 /kWh = $6.70 /mmBtu

Process Pump

ProcessLoad 1

ProcessLoad 2

Chiller

Cooling Tower

Cooling Tower Pump

BypassValve

Figure 2. Water-cooled chiller system.

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Air-Cooled ChillerAir-cooled chillers reject process and compressor heat by blowing ambient air over condensor coils of the chiller. A cooling system with an air-cooled chiller is shown in Figure 3. Air-cooled chillers are slightly less energy efficient than water-cooled chillers, but are generally less expensive to purchase and easier to maintain. Air-cooled chillers require about 1 kW per ton of cooling. Thus, the electricity use per unit cooling, E/Q, and cost per unit cooling, C/Q, to provide 1 mmBtu of cooling are about:

E/Q = 1 kW/ton / 12,000 Btu/ton x 1,000,000 Btu/mmBtu = 83 kWh/mmBtuC/Q = 83 kWh/mmBtu x $0.10 /kWh = $8.30 /mmBtu

Process Pump

ProcessLoad 1

ProcessLoad 2

Chiller

BypassValve

Air

Figure 3. Air-cooled chiller system.

Absorption ChillerAbsorption chillers use heat rather than electricity as the primary source of energy. Thus, absorption chillers can be powered with waste heat from other processes, or with a dedicated source of heat such as a boiler. A cooling system with an absorption chiller is shown in Figure 4.

Process Pump

ProcessLoad 1

ProcessLoad 2

AbsorptionChiller

BypassValve

Boiler

Steam

Figure 4. Absorption chiller cooling system.

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The efficiency of the absorption chillers increases with increasing temperature heat. The coefficient of performance for absorption chillers powered with steam is about 1, thus about 1 Btu of heat is required to generate a Btu of cooling. Assuming the steam is generated by an 80% efficient boiler, the electricity use per unit cooling, E/Q, and cost per unit cooling, C/Q, to generate 1 mmBtu of cooling are about:

E/Q = 1 Btu-heat / Btu-cooling / 80% x 1,000,000 Btu/mmBtu = 1.25 mmBtu-heat/mmBtu-coolingC/Q = 1.25 mmBtu-heat/mmBtu-cooling x $10.00 /mmBtu = $12.50 /mmBtu-cooling

Open-Loop Water CoolingIn open-loop cooling, cooling water is discharged to the sewer after cooling a process. An open-loop cooling system is shown in Figure 5.

ProcessLoad 1

ProcessLoad 2

From City Water Supply

To Sewer

Figure 5. Open-loop cooling system.

Assuming the temperature of the water increases by 10 F during the cooling process, the quantity of water, V, needed to provide 1 mmBtu of cooling is about:

V = 1 mmBtu / (8.32 lb/gal x 1 Btu/lb-F x 10 F) = 12,000 gallons

Assuming the total water and sewer charge for water is $6.00 / 1,000 gallons, the cost per unit cooling, C/Q, providing 1 mmBtu of cooling is about:

C/Q = 12,000 gallons/mmBtu x ($6.00 / 1,000 gallons) = $72 /mmBtu

Compressed Air CoolingCompressed air cools when discharged to the atmosphere. This cooling effect can be enhanced by a vortex tube. A schematic of a vortex tube is shown in Figure 6. Compressed air enters through the top port, hot air is rejected through the right port and cool air is supplied through the left port.

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Figure 6. Vortex tube (EXAIR Corporation, 2007)

Performance specifications report that a typical vortex tube uses 150 scfm of compressed air at 100 psig to produce 10,200 Btu/hr of cooling (EXAIR Corporation, 2007). Centrifugal air compressors produce about 4.5 scfm of compressed air at 100 psig per hp of work applied to the compressor. Assuming the air compressor motor is 90% efficient, the electrical energy required to generate 1 mmBtu of cooling, E/Q, from a vortex tube is about:

E/Q = [150 scfm / 4.5 scfm/hp x 0.75 kW/hp / 90%] / [10,200 Btu/hr x 1 mmBtu / 1,000,000 Btu] E/Q = 2,723 kWh/mmBtu

The cost per unit cooling, C/Q, for generating 1 mmBtu of cooling with a vortex tube is about:

C/Q = 2,723 kWh/mmBtu x $0.10 /kWh = $272 /mmBtu

Relative Costs of Process Cooling SystemsBased on these results, the cost of cooling varies from about $1 per mmBtu for cooling towers, to about $10 per mmBtu for chillers, to about $70 per mmBtu for open-loop cooling to about $270 per mmBtu for compressed air (Figure 7). These near order-of-magnitude cost differences underscore the importance of avoiding compressed air and open-loop water cooling, and using cooling towers instead of chillers whenever possible.

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0

50

100

150

200

250

300

Compressed air Open loopcooling

Chillers Cooling towers

$/m

mBt

u co

olin

g

Figure 7. Comparative costs of cooling.

End Use: Add InsulationInsulation reduces heat transfer into cooled tanks and piping, and decreases the likelihood of condensation forming on the outside of the tanks and piping. Heat is transferred to cooling equipment by radiation and convection. At large temperature differences, radiation heat transfer becomes the dominant. At small temperature differences, convection is dominant. Thus, in cooling applications, radiation heat gain can be neglected with minimal error. The cooling energy savings, Qsav, from insulating a cooled surface are:

Qsav = A (1/R1 – 1/R2) (Ta – Tc) dt (2)

Where A is the area of the cold surface, R1 and R2 are the thermal resistances between the surface and air before and after adding insulation, Tc and Ta are the temperatures of the cooling medium and ambient air, and dt is the time period considered. Even at small temperature differences between cooling medium and ambient air, insulating cold surfaces is generally cost effective.

ExampleConsider an uninsulated tank at 40 F surrounded by plant air at 80 F. The total thermal resistance of the metal tank and air film is 1 hr-ft2-F/Btu. Calculate the reduction in cooling load and cooling energy cost from adding R-13 insulation to the tank if the cooling was provided by an air cooled chiller at a cooling cost of $8.30 /mmBtu and the installed cost of the insulation is $1.00 /ft2.

The reduction in cooling load from adding R-13 insulation to the tank, Qsav, would be:

Qsav = (1/1 – 1/14) (Btu/hr-ft2-F) x (80 F – 40 F) x 8,760 (hr/year) = 0.33 mmBtu/ft2-yr

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The cost savings, Csav, would be:

Csav = 0.33 mmBtu/ft2-yr x $8.30/mmBtu = $2.80 /ft2-yr

The simple payback, SP, would be:

SP = Implementation Cost / Annual SavingsSP = [($1.00 /ft2) / ($2.80 /ft2-yr)] x 12 months/year = 4 months

End Use: Reduce Cooling Load with Heat Exchangers Continuous ProcessesMany manufacturing processes require heating at one stage of the process and cooling at another. In many cases, the strategic use of heat exchangers can reduce both the heating and cooling loads. Consider, for example, a process where a fluid is heated to some high temperature and then cooled for packaging or further processing (Figure 9a). Both the heating and cooling loads can be decreased by adding a heat exchanger to transfer heat from the hot fluid before cooling (Figure 9b).

T1 T2 T3

Qh1 Qc1

T1 T2 T3

Qh2 Qc2

New HX

T2B

T2A

A.

B.

Figure 9. A. Original process. B. Process with additional heat exchanger.

To quantify the savings for installing an additional heat exchanger in this system, consider the following relations. The original heating and cooling loads, Qh1 and Qc1, are the product of the mass flow rate, m, specific heat, cp, and temperature, T, differences across the heat exchangers. Defining the product of the mass flow rate and specific heat as the mass capacitance, mcp, gives the following relations for the heating and cooling loads in the original configuration.

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Qh1 = mcph (T2 – T1) (3)Qc1 = mcpc (T2 – T3) (4)

Heat exchanger effectiveness, e, is defined as the ratio of actual heat transfer, Qhx, to the maximum heat transfer, Qm. The maximum heat transfer is the product of the minimum mass capacitance of the hot and cold streams, mcpmin, and the difference of the incoming temperatures. Thus, the rate of heat exchange is:

Qhx = e mcpmin (T2 – T1) (5)

After the heat exchanger has been added, the new temperatures T2A and T2B can be calculated from energy balances on the fluid as it passes through the heat exchanger.

Qhx = mcph (T2A – T1) = mcpc (T2 – T2B) (6)

The total heat transfer can also be represented in terms of the log mean temperature difference Tlm where U is overall conductance of the heat exchanger and A is the heat exchange surface area:

Qhx = U A Tlm (7)

The log mean temperature difference Tlm is:

Tlm = (TB – TA) / ln(TB / TA) if mcph <> mcpc Tlm = TB = TA if mcph = mcpc (8)

If the heat exchanger is a counter-flow design, the temperature differences are:

TB = T1 – T2B and TA = T2A – T2 (9)

ExampleConsider a system that heats 20 gpm of soup from 70 to 200 F then cools it to 40 F for packaging. The density of the soup is 8.32 lb/gal and the specific heat is 1 Btu/lb-F. The overall efficiency of the boiler system delivering the required heat is 70% and the electrical power requirement of the air cooled chiller that delivers the required cooling is 1 kW/ton. Calculate the reduction in heating and cooling energy use if a 50% effective heat exchanger where added between the heating and cooling operations. If the overall conductance of the heat exchanger is 10 Btu/hr-ft2-F, calculate the required surface area of the heat exchanger.

The product of the mass flow rate and specific heat, mcp, is:

mcp = 20 gpm x 60 min/hr x 8.32 lb/gal x 1 Btu/lb-F = 9,984 Btu/hr-F

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The current heating and cooling loads are:

Qh = mcp (T2 – T1) = 9,984 Btu/hr-F x (200 F - 70 F) = 1,297,920 Btu/hrQc = mcp (T2 – T3) = 9,984 Btu/hr-F x (200 F - 40 F) = 1,597,440 Btu/hr

The heat transferred, Qhx, by a 50% effective heat exchanger would be:

Qhx = e mcpmin (T2 – T1) Qhx = 0.50 x 9,984 Btu/hr-F x (200 F - 70 F) = 648,960 Btu/hr

The reduction in heating and cooling energy use would be:

Qhsav = 648,960 Btu/hr / 0.70 = 927,086 Btu/hrQcsav = 648,960 Btu/hr x 1 ton / 12,000 Btu/hr x 1 kW/ton = 54.08 kW

From energy balances on the fluid streams, the exit temperatures from the new heat exchanger are:

T2A = T1 + Qhx / mcph = 70 F + 648,960 Btu/hr / 9,984 Btu/hr-F = 135 FT2B = T2 - Qhx / mcph = 200 F - 648,960 Btu/hr / 9,984 Btu/hr-F = 135 F

If the heat exchanger is a counter-flow design, the temperature differences and log mean temperature are:

TB = T1 – T2B = 70 F – 135 F = -65 FTA = T2A – T2 = 135 F – 200 F = -65 FTlm (when mcph = mcpc) = TB = TA = 65 F

If the heat exchange coefficient is 10 Btu/hr-ft2-F, the required heat exchanger surface area is:

A = Qhx / [U Tlm] = 648,960 Btu/hr / (10 Btu/hr-ft2-F x 65 F) = 998 ft2

Determining Heat Exchanger Effectiveness for Maximum Energy SavingsThe maximum heat transferred between the two streams is limited by the smaller of Qh1 and Qc1, where:

Qh1 = m cp (T2 – T1)Qc1 = m cp (T2 – T3)

Thus, if Qh1 is smaller, then (T2 – T1) < (T2 – T3) and energy savings are maximized by eliminating Qh1. In this case, T2A should approach T2 and

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Qhx = emax mcp (T2 – T1) = mcp (T2A – T1) = mcp (T2 – T1)emax = 1.0

If Qc1 is smaller, then (T2 – T1) > (T2 – T3) and energy savings are maximized by eliminating Qc1. In this case, T2B should approach T3 and

Qhx = emax mcp (T2 – T1) = mcp (T2 – T2B) = mcp (T2 – T3)emax = (T2 – T3) / (T2 – T1)

The rule can be summarized in Boolean logic as:

IF (T2 – T1) < (T2 – T3) THEN (10)emax = 1.0

ELSE emax = (T2 – T3) / (T2 – T1)

END IF

ExampleDetermine the heat exchanger effectiveness that maximizes energy savings if a) T1 = 70 F, T2 = 200 F and T3 = 40 F and b) T1 = 40 F, T2 = 200 F and T3 = 70 F.

a) For T1 = 70 F, T2 = 200 F and T3 = 40 F:

(T2 – T1) = 200 F – 70 F = 130 F(T2 – T3) = 200 F – 40 F = 160 F

Thus, (T2 – T1) < (T2 – T3) and emax = 1.0

b) For T1 = 40 F, T2 = 200 F and T3 = 70 F:

(T2 – T1) = 200 F – 40 F = 160 F(T2 – T3) = 200 F – 70 F = 130 F

Thus, (T2 – T1) > (T2 – T3) and emax = (T2 – T3) / (T2 – T1) = (130) / (160) = 0.813

Batch ProcessesThe same principal of transferring heat from processes requiring cooling to processes requiring heating is also applicable for processes with multiple tanks. In general, it is cost effective to transfer heat between processes, whenever the processes that need cooling are 10 F higher than the process that need heating; transferring heat across smaller temperature difference requires very large heat exchangers and the cost of the heat

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exchanger can outweigh the possible savings. When multiple options for transferring heat exist, the “pinch” approach can determine the configuration which maximizes savings.

Consider for example, the following case study. The maximum temperatures of the processes that require cooling are shown in the Table 1. The table also shows those design values for the associated heat exchangers and pumps. The largest cooling load is for the three 35,000-gallon work tanks.

Table 1. Maximum temperatures of tanks which require cooling

Tanks That Need CoolingName Tmax (F) Q (mmBtu/hr) V (gpm)

5.6 115 1 4005.1 130 1 400

Work 145 3 x 15 3 x 1,720

The minimum temperatures for several heated rinse tanks are shown in the Table 2. Table 2 also shows those design values for the associated heat exchangers and pumps. Inspection of the Tables 1 and 2 indicates that heat could be cost effectively transferred from the work tanks at 145 F to all rinse tanks with temperatures of 135 F or lower in order to decrease the steam heating load in the plant.

Table 2. Minimum temperatures of tanks that require heating.

Tanks That Need HeatingName Tmin (F) Q (mmBtu/hr) V (gpm)

5.1 165 5.2 1,0605.2 165 5.2 1,0605.3 165 2.2 5305.4 160 2.4 5305.5 165 2.4 5305.8 125 2.0 5305.9 130 2.1 530

5.11 130 2.1 5305.13 130 2.1 5305.15 145 2.2 5305.16 120 2.0 5305.17 120 2.0 5305.18 145 2.2 530

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Heat Exchanger Network AnalysisOne way to visualize the potential for heat recovery is by plotting composite curves for the cooling and heating processes on a temperature versus enthalpy diagram. To construct a composite cooling curve, the process lines showing the temperature drop versus energy loss for all processes that require cooling are linked. Similarly, a composite heating curve is constructed by linking the process lines showing the temperature gain versus energy gain for all processes that require heating. The composite heating curve can then be translated horizontally until the minimum temperature difference between the two curves is 10 F. The overlapping section where the cooling curve is greater than the heating curve represents processes where heat can be cost-effectively transferred from cooled processes to heated processes. Utility heating is still necessary on the non-overlapping portion of the heating curve. Similarly, utility cooling is still necessary on the non-overlapping portion of the cooling curve.

The original and shifted composite cooling and heating curves for this case study are shown in Figure 10. The curves show that the work tanks could transfer heat to rinse tanks 5.8, 5.16 and 5.17 since an adequate temperature difference exists. This would simultaneously reduce the heating load and cooling loads.

Original Composite Curves

100110120130140150160170180

0 5 10 15 20 25

Q (mmBtu/hr)

T (F

) Tc

Th

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Shifted Composite Curves

100110120130140150160170180

0 10 20 30 40

Q (mmBtu/hr)

T (F

) Tc

Th

Figure 10. Original (A) and shifted (B) composite curves.

End Use: Reduce Cooling By Improving Heat TransferWith improved heat transfer effectiveness, the same quantity of heat can be transferred with smaller flow rates or smaller temperature differences. Reducing flow rates and temperatures reduces energy losses and improves system efficiency.

Consider the following graphs of heat transfer effectiveness for counter flow, cross flow and parallel flow heat exchanger configuration. The effectiveness of cross flow heat exchange is always greater than the effectiveness of the other configurations

Figure 11. Heat transfer effectiveness for counter flow, cross flow and parallel flow heat exchange (Incropera and DeWitt, 1996).

The equations for heat exchanger effectiveness are:

Ch = mh * cph Cc = mc * cpc Cmin = min(Ch, Cc)

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Cmax = max(Ch, Cc)Cr = Cmin / CmaxNTU = UA/Cmin

Cross flow: e = 1-exp[(1/Cr)*(NTU0.22)*{exp((-Cr)*(NTU0.78))-1}]Counter flow: e = [1 - exp(-NTU (1-Cr))] / [1 – Cr*exp(-NTU (1 - Cr))] (use Cr = 0.999 when Cr = 1.0)Q = e*Cmin*(Th1-Tc1)

ExampleConsider cross flow heat transfer between extruded plastic and cooling water. Currently mcpmin = 83.2 Btu/min-F, NTU = UA/Cmin = 3, Cmin/Cmax = 1.0, the temperature of the extruded plastic is 300 F and the entering temperature of chilled water is 50 F. Calculate the rate of heat transfer, and the required entering water temperature if the mode of heat transfer were converted from cross flow to counter flow.

The effectiveness of cross flow heat transfer for NTU = 3, Cmin/Cmax = 1.0 is:

e = 0.69

The rate of heat transfer is:

Q = e mcpmin (Tp – Tw1) = 0.69 83.2 (300 – 50) = 14,352 Btu/min

The effectiveness of counter flow heat transfer for NTU = 3, Cmin/Cmax = 1.0 (use Cmin/Cmax = 0.999) is:

e = 0.78

An energy balance on the process gives:

Q = e mcpmin (Tp – Tw1) 4,352 Btu/min = 0.78 83.2 (300 – Tw1)Tw1 = 79 F

A cooling tower, which use about 1/10 as much energy per unit cooling as a chiller could supply 79 F water much of the year. Thus, improving the effectiveness of heat transfer and running a cooling tower instead of a chiller would reduce cooling energy costs by about 10 times.

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Distribution System: Avoid MixingUnnecessary mixing of hot and cold streams inevitably leads to losses. Consider for example, the case of a common hot and cold water tank, as shown in Figure 1. The system uses an open tank as a well for return water from the process and cooling tower. In the cooling tower loop, water is pumped from the chilled water tank to the top of the cooling tower, where it gravity feeds back to the chilled water tank.

Adding a wall inside the chilled water tank to separate the hot and cold sides of the tank, as shown in Figure 10, would minimize mixing. This would decrease the temperature of water to the process loads and increase the temperature of the water to the cooling tower.

ProcessLoad 1

ProcessLoad 2

Chilled Water Tank

Cooling Tower

BypassValve

Tp2

Tc2 Process Pump

Tp1

Cooling Tower Pump

Tc1

Figure 12. Process with separate hot and cold water tanks.

Cooling tower water outlet temperature is a function of the temperature drop of the water through the cooling tower, called the range, and the ambient wet bulb temperature. In this process shown in Figures 1 and 12, the range is set by the process load. Thus, simple energy balances show that the temperature of the water to the process loop is reduced by an amount equal to the temperature range by splitting the tanks. This reduced supply temperature could enable more production if cooling was a constraint. Alternately, the cooling tower could be set to run at the lower fan speeds if set so the temperature of water supplied to the process was the same as with a single tank.

Distribution System: Eliminate Unnecessary Heat ExchangersThe efficiency of a well-insulated heat exchanger approaches 100%, meaning that all of the heat extracted from the hot fluid is transferred to the cold fluid with no energy loss. This does not mean, however, that use of heat exchangers does not incur an energy use penalty. Heat exchangers require temperature difference to transfer heat, and exergy is always destroyed during heat transfer through a finite temperature difference. In heat exchangers, this exergy destruction manifests itself by requiring hotter temperatures on the hot side and/or colder temperatures on the cold side. Generating these excess temperatures requires more energy. In addition, friction losses through heat exchangers require more energy from pumps and fans to push the fluids. Thus, eliminating unnecessary heat exchangers improves the efficiency of process cooling systems.

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Example: Eliminate HX Between Backup Chiller and Process Cooling Loop

The current piping system is shown below. In this system:

the flow rates through each side of the current HX are the same the effectiveness of the current heat exchanger is 0.50 the pressure drop through the current HX is 15 psig. the temperatures of the supply water to the process and return water from the process

are 55 F and 60 F. the process cooling load is 150 tons the system operates 1,000 hours per year

Calculate the chiller electricity savings (kWh/yr) and pump electricity savings (kWh/yr) from removing the heat exchanger so that chilled water from the backup chiller flows directly to the process as shown below.

To calculate savings, assume the heat added to the water by the pumps is negligible after HX is removed and the pressure drop at Ts connecting process and chiller loops is negligible.

Chiller Electricity SavingsHeat added by the process, Qp, is:

Qp = (m cp) (Th2– Th1)

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Heat added by the process, Qp, is also transferred by the heat exchanger to the backup chiller. According heat exchanger effectiveness theory, the heat is:

Qp = e (m cp)min (Th1 – Tc1)

If (m cp) is the same on both sides of the heat exchanger, combing equations and solving for Tc1 gives:

Tc1 = Th1 – (Th2 - Th1) / e

If the heat exchanger were removed, the leaving water temperature of the chiller could be increased to the minimum temperature required by the process:

Tc1 = Th1

The following relation gives chiller EER as a function of leaving water temperature, LWT, when the condenser air temperature is 75 F.

EER (Btu/Wh) = 4.688564 + 0.20999 * LWT (F) -.000518143 * LWT (F) ^ 2

Chiller electricity use is given by:

Welec (kWh/yr) = Qevap (kBtu/hr) / EER (Btu/Wh) x HPY (hr/yr)

Using the equations from above and these values, the savings would be about:

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CHILLERCurrentTh1 (F) 60Th2 (F) 55e 0.5Tc1 (F) = Th1 – (Th1 - Th2) / e 50EER (Btu/Wh) 13.89Qp (tons) 150HPY (hr/yr) 1000We (kWh/yr) = Qp (kBtu/hr) / EER (Btu/Wh) x HPY (hr/yr) 129,564ProposedTc1 (F) = Th2 55EER (Btu/Wh) 14.67Qp (tons) 150HPY (hr/yr) 1000We (kWh/yr) = Qp (kBtu/hr) / EER (Btu/Wh) x HPY (hr/yr) 122,694SavingsWs (kWh/yr) = We,current - We,proposed 6,870

Pump SavingsFrom an energy balance, the water flow rate on the process side is:

V = Qp / (p cp (Th1- Th2))

The conversion between pressure drop P psi and elevation head h (ft-H20) is:

h = P (psi) x 2.31 (ft-H20/psi)

The fluid power, Wf, and electrical energy, We, to push water through each side of the heat exchanger is:

Wf (hp) = V (gal/min) htotal (ft-H20) / 3,960 (gal-ft-H20/min-hp) We (kWh/yr) = Wf (hp) x 0.75 (kW/hp) / (Epump Emotor) x HPY (hr/yr)

If the heat exchanger were removed, and the flow rates were reduced to the current values by slowing the pumps or trimming the impellors, the total pump energy savings, Ws, would be:

Ws (kWh/yr) = 2 We (kWh/yr)

Plate and frame heat exchangers are typically designed with a pressure drop of 15 psi on each side. We assume the pumps are 70% efficient, the pump motors are 92% efficient. If so, the total pump energy savings would be about:

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PUMPQp (tons) 150(Th1- Th2) (F) 5V (gpm) = Qp / (p cp (Th1- Th2)) 721dP (psi) 15dh (ft-H20) = psi x 2.31 ft-H20/psi 34.65Wf (hp) = V (gal/min) dh (ft-H20) / 3,960 (gal-ft-H20/min-hp) 6.310Epump 0.7Emotor 0.92HPY (hr/yr) 1000Ws (kWh/yr) = 2 x Wf / (Epump Emotor) * HPY 14,697

Distribution System: Employ Variable Speed Pumping on Process LoopsMany chilled-water cooling systems are candidates for variable frequency drive (VFD) pumping retrofits. For example, consider process cooling system shown in Figure 14. The cooling system consists of two pumping loops: a cooling tower loop and a process loop. VFDs are generally not applied to the cooling tower loop since cooling tower manufacturers generally recommend that cooling tower water flow be kept constant. On the other hand, varying production schedules and parts often result in highly variable process cooling loads. Thus, process cooling loops are often candidates for variable speed pumping retrofits.

warm water

cool water

cooling tower

city water make-up

7.5 hp pump

25 hp pump

reservoir

process water return

bypass / pressure

relief valve

cooling water to process loads

dP

VSD

Figure14. VFD pumping retrofit to process cooling loop.

VFD pumping retrofits typically require making three changes to the existing pumping system:

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Install a VFD on the power supply to the pump motor. In parallel pumping configurations, one VFD is generally needed for each operational pump, but not for the backup pump. If some flow will always occur through process loads, close all valves on the bypass piping. If all flow through process loads could be shut off, then it is necessary to maintain some flow through the bypass loops to ensure proper pump operation. Minimum flow through the pumps can be assured by installing a pressure-controlled valve on the bypass loop that opens as the pressure difference across the valve increases.

Install a differential-pressure sensor between the supply and return headers at the process load located the farthest distance from the pump. Determine the pressure drop needed to guarantee sufficient flow through the farthest process load at this point. Control the speed of the VFD to maintain this differential pressure.

Pumping power savings from VFD retrofits can be significant, because pump affinity laws show that fluid power, Wf, varies with the cube of flow, V,. Wf2 = Wf1 (V2 / V1)3 (12)

However, after accounting for the reduced efficiency of the motor, pump and VFD at low loads, electrical power to the VFD, We, generally varies with the 2.5 power of flow.

We2 = We1 (V2 / V1)2.5 (13)

Moreover, all energy added to the cooling water by pumps must be removed by the cooling tower or chiller. Hence, reducing pumping power also reduces cooling energy requirements.

Example: Add VFD to Process Pump

Currently, the process pump runs continuously and draws 32 kW of power. Each process load runs 14 hours per day. The valves that control the flow of water to the process loads are two way valves; hence, excess water is diverted through the primary bypass valve. The total specific power requirement of the chiller is 1 kW/ton, and the chiller can accommodate variable chilled-water flow through the evaporator. Calculate the pump electricity savings (kWh/yr) and chiller electricity savings (kWh/yr) from adding a VFD and differential-pressure sensor between the supply and return headers to control the VFD.

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Current system Proposed System

Assuming the current pump is sized for peak chilled water demand, the ratio of required to peak flow is about:

V2/V1 = 14/ 24 = 0.583

Based on these observations, the savings would be about:

CurrentHPY (hr/yr) 8,760Pump power P1 (kW) 32.00Pump electricity use E1 (kWh/yr) = P1 HPY 280,320

ProposedFrac flow rate required (V2/V1) 0.5833333Pump power P2 (kW) = P1 x (V2/V1)^2.5 8.32Pump electricity use E2 (kWh/yr) = P2 HPY 72,853

Pump SavingsDemand savings DS (kW) = P1 - P2 23.68Electricity savings ES (kWh/yr) = E1 - E2 207,467

Chiller SavingsCooling Load Reduction (Btu/hr) = DS 3,412 / 12,000 6.73Specific Chiller Power (kW/ton) 1.00Electricity savings ES (kWh/yr) = SPC CLR HPY 58,990

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Distribution System: Eliminate Throttling of Cooling Tower WaterMany cooling tower water supply pumps are oversized. After installation, throttling valves are partially closed to deliver the required flow to the cooling tower. This results in a continuous waste of pumping energy (see figure below). More efficient methods of controlling flow are to slow the pump with a VFD or trim the pump impellor.

0%

20%

40%

60%

80%

100%

0% 20% 40% 60% 80% 100%

Volume Flow Rate (%)

Pow

er (%

)

Outlet Damper Variable Inlet Vane Variable Frequency Drive

Example

Currently, the throttling valve on the water supply to a cooling tower is 50% open. The cooling tower operates continuously. The 30-hp cooling tower pump was designed to be 80% loaded at full flow. The pump motor is 92% efficient. Calculate the savings from opening the balancing valves and adding a VSD to the cooling tower pump.

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According to the figure above, pump power is reduced to about 82% by the balancing valves at 50% of full flow. If the balancing valves were fully open, and a VFD were installed to slow the pump, pump power would be reduced to 28% at the same flow. Based on these observations, the savings would be about:

CurrentHPY (hr/yr) 8,760Pump rated power Pr (hp) 30Pump motor effi ciency Em 0.92Frac Loaded at full flow FL0 0.80Pump power at full flow Pf (kW) = Pr FL0 (.75kW/hp) / Em 19.57Frac loaded after throttling FLt 0.82Pump power P1 (kW) = Pf FLt 16.04Pump electricity use E1 (kWh/yr) = P1 HPY 140,541

ProposedFrac loaded after VFD FLvfd 0.28Pump power P2 (kW) = Pf FLvfd 5.48Pump electricity use E2 (kWh/yr) = P2 HPY 47,990

SavingsDemand savings DS (kW) = P1 - P2 10.57Electricity savings ES (kWh/yr) = E1 - E2 92,551

Cooling Tower PerformanceThe performance of a typical cooling tower at water flow rates of 3 gpm/ton of refrigeration and 5 gpm/ton of refrigeration is shown below. The actual heat rejected by the cooling tower is sum of the refrigeration load and power input to the chiller. Because chillers typically operate at a COP of about 4.0, one ton of refrigeration is equivalent to

Actual cooling load = 1 ton refrigeration + 1 ton refrigeration / COPActual cooling load = 12,000 Btu/hr + 12,000 Btu/hr / 4 = 15,000 Btu/hr

Hence cooling tower tons are defined to be 15,000 Btu/hr. Similar performance data for specific cooling towers can be obtained from the manufacturer.

These curves predict the temperature of the cold water leaving the cooling tower as a function of the water temperature range (Th-Tc) and entering air web bulb temperature. Temperature range is generally known and can be used as an input value in these charts, since the temperature range is set by the water flow rate and heat rejection rate of the condenser.

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Source: ASHRAE Handbook, HVAC Systems and Equipment, 2004.

Relations for the temperature of cooling water leaving the tower, Tc, can be derived from regressing data from the 3 gpm/ton and 5 gpm/ton curves shown above. The relation and regression coefficients are shown below. The R2 for these relations exceeds 0.995 and the average error, [abs(Tc – Tc,pred)], is less than 0.8 F. Tc = a + b Twb + c Tr + d Twb2 + e Tr2 + f Tr Twb

Coef 3 gpm/ton 5 gpm/tona 16.790751 24.6299229b 0.6464308 0.45007792c 2.2221763 3.32229591d 0.0016061 0.00261818e -0.0159268 -0.0324886f -0.015954 -0.0190476

Use Cooling Towers Instead of Chiller When PossibleBecause cooling towers use about 1/10 as much electrical energy as chillers to deliver the same amount of cooling, it is advantageous to use cooling towers in place of chillers whenever possible. This can be done by installing a heat exchanger between the cooling towers and the process load, and directing flow through the heat exchanger instead of the chillers whenever possible.

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Cooling systems with and without heat exchanger to enable direct cooling by cooling tower.

ExampleA continuous cooling process requires 100 tons of 70 F chilled water and returns the water at 80 F. The water is supplied by a water-cooled chiller using 0.60 kW/ton of process cooling. Using binned weather data for Dayton, Ohio, determine energy savings from operating the cooling tower instead of the chiller whenever possible.

The current power requirement of the chiller, P1, is:

P1 = 100 tons x 0.60 kW/ton = 60 kW

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The first five columns of the table below show binned weather data for Dayton, Ohio. The temperature of water delivered by the cooling tower, Tc, is calculated using the following relation with coefficients for 3 gpm/ton and is shown in column 7:

Tc = a + b Twb + c Tr + d Twb2 + e Tr2 + f Tr Twb

StrTemp EndTemp T(F) Twb(F) hrs1-24 Tr (F) Tc (F)======== ======== ======== ======== ======== ======== ======

90 94 91 75.2 1 10 83.185 89 87 72.5 87 10 81.280 84 82.2 69.4 338 10 78.975 79 76.8 66.7 502 10 77.070 74 72.4 64.8 661 10 75.765 69 67.9 61.4 865 10 73.460 64 62.4 56.4 943 10 70.055 59 57 51.6 735 10 66.850 54 52 47.1 644 10 63.945 49 47.4 42.9 505 10 61.340 44 43.2 39.8 644 10 59.335 39 37.5 35.6 713 10 56.830 34 32.1 31.4 726 10 54.325 29 27.5 27.9 438 10 52.320 24 23.2 24.9 343 10 50.515 19 17.7 21.1 292 10 48.410 14 12.2 17.5 156 10 46.45 9 7.3 14.3 87 10 44.70 4 2.7 11.3 50 10 43.1-5 -1 -1.3 8.7 29 10 41.8

-10 -6 -5.1 6.4 1 10 40.6Total hours when Tc < 70 F = 6,306

Summing the number of hours in all temperature bins when Twb < 70 F, gives 6,306 hours per year. Thus, the energy savings, Es, from turning off the chiller whenever the cooling tower could deliver 70 F water would be:

Es = P1 x dt = 60 kW x 6,306 hours/year = 378,360 kWh/yr

The power requirement of cooling tower pumps and fans is about 0.06 kW/ton. Thus, the fraction savings, fs, from running the cooling tower instead of the chiller is:

fs = 0.06 kW/ton / (0.06 kW/ton + 0.60 kW/ton) = 91%

Cooling Tower Performance at Reduced Air Flow RatesCooling tower cooling capacity decreases at reduced air flow rates. The relation between fraction cooling capacity and fraction fan speed for a typical cooling tower is shown below.

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Fan Speed vs Cooling Tower Cooling

0.0

0.2

0.4

0.6

0.8

1.0

0.0 0.2 0.4 0.6 0.8 1.0

Fraction Cooling Capacity

Frac

tion

Fan

Spe

ed

Regressing this data gives the following relation between Fraction Fan Speed, FFS, and Fraction Cooling Capacity, FQ:

FFS = -0.08173 + 1.185565 FQ - 1.13455 FQ2 + 1.02769 FQ3

Cooling tower fan power varies with air flow rate according to fan affinity laws. Thus, fan power savings from VFD retrofits can be significant, because fan affinity laws show that fluid power, Wf, varies with the cube of flow, V,. Wf2 = Wf1 (V2 / V1)3 (15)

However, after accounting for the reduced efficiency of the motor, fan and VFD at low loads, electrical power to the VFD, We, generally varies with the 2.5 power of flow.

We2 = We1 (V2 / V1)2.5 (16)

ExampleA cooling tower has a single speed 20 hp fan that cycles on and off to deliver cooling water at a set-point temperature of 80 F. When running, the fan is 95% loaded and 90% efficient. The wet-bulb temperature is 60 F and the temperature range of the cooling water is 10 F. Determine the fan energy savings from adding a variable frequency drive to the cooling tower fan and modulating fan speed to achieve the desired set-point temperature.

A cooling tower with a cycling fan can be modeled as:

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An energy balance on Tm, in terms of the fraction of time the cooling tower fan is on, fon, gives:

1) Tm = fon Tc + (1-fon) Th

The temperature range across the cooling tower, Tr, is:

2) Tr = Th – Tc

The regression for Tc in terms of Tr and Twb is:

3) Tc = a + b Twb + c Tr + d Twb2 + e Tr2 + f Tr Twb

An energy balance on the load gives:

4) Th - Tm = dTl

This system of 4 equations has 4 unknowns (fon, Tc, Th, Tr). Solving the system when Tm = 80 F, Twb = 60 F and dTl = 10 F gives:

fon = 0.7128

The average power of the fan, P1, is:

P1 = 0.7128 x 20 hp x 0.95 / 0.90 x 0.75 kW/hp = 11.29 kW

For a variable speed fan, the fraction fan speed, FFS, to generate the required cooling at fon is: FFS = -0.08173 + 1.185565 FQ - 1.13455 FQ2 + 1.02769 FQ3

FFS = -0.08173 + 1.185565 (0.7128) - 1.13455 (0.7128)2 + 1.02769 (0.7128)3 = 0.5591

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Assuming power varies with the 2.5 power of flow, the electrical power by a VFD fan, P2, would be:

P2 = P1 (V2 / V1)2.5 = 11.29 kW x 0.55912.5 = 2.64 kW

The power savings, PS, would be:

PS = P1 – P2 = 11.29 kW - 2.64 kW = 8.65 kW

The fraction savings, fs, from replacing the single-speed fan with the VFD fan is:

fs = 8.65 kW / 11.29 kW = 77%

Cooling Tower Evaporation RateAs discussed in the previous section, cooling in cooling towers is dominated by evaporation. The evaporation rate can be calculated from the psychrometric relations in the previous section, if the inlet and exit conditions of the air are known. For example, consider the case in which the cooling load, Ql, mass flow rate of air, ma, (which can be calculated based on the fan cfm and specific volume of the inlet air), and inlet conditions of air are known. The enthalpy of the exit air, ha2, can be calculated from an energy balance.

Ql = ma (ha2 – ha1)ha2 = ha1+ Ql / ma

The state of the exit air can be fixed by assuming that it is 100% saturated with an enthalpy ha2. The evaporation rate, mwe, can be determined by a water mass balance on the air.

mwe = ma (wa2- wa1)

The fraction of water evaporated is:

mwe / mw

Using this method for entering air temperatures from 50 F to 90 F, we determined that the fraction of water evaporated typically ranges from about 0.5% to 1%, with an average value of about 0.75%.

Another way to estimate the fraction of water evaporated is to assume that all cooling, Ql, is from evaporation, Qevap. The cooling load Ql, is the product of the water flow rate, mw, specific heat, cp, and temperature difference, dT. The evaporative cooling rate is the product of the water evaporated, mwe, and the latent heat of cooling, hfg.

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Ql = Qevapmw cp dT = mwe hfg

Assuming the latent heat of evaporation of water, hfg, is 1,000 Btu/lb, and the temperature difference of water through the tower, dT, is 10 F, the fraction of water evaporated is:

mwe / mw = cp dT / hfg = 1 (Btu/lb-F) x 10 (F) / 1000 (Btu/lb) = 1%

If on average, 75% of the cooling were from evaporation and 25% from sensible cooling, then the evaporation rate would be:

75% x 1% = 0.75%

Thus, both methods suggest that 0.75% is a good estimate of the rate of evaporation. However, we have seen manufacturer data indicating average evaporation rates as low as 0.30%. Water lost to evaporation should not be subjected to sewer charges.

ExampleA multi-stage cooling tower is sized to cool condenser water from a 1,000 ton chiller plant, with a design flow rate of 3 gpm/ton-chiller capacity. The cooling tower operates 8,000 hours per year. The water utility charges $3.00 per ccf for water discharged to sewer, and assumes that all water entering the plant is discharged to sewer. Assuming the evaporation rate is 0.75%, estimate the savings from applying for a “sewer exemption” for water evaporated from the cooling tower.

The quantity of water evaporated from cooling tower is:

1,000 tons x 3 gpm/ton x 0.0075 x 60 minutes/hour x 8,000 hours/year / [7.481 gal/ft3 x 100 ft3/ccf] = 14,437 ccf/year

The savings from exempting evaporated water from the sewer charge would be:

14,437 ccf/year x $3 /ccf = $43,310 /year

Air-Cooled ChillersAir-cooled chillers reject heat to the atmosphere by blowing air over a fin-tube heat exchanger through which flows the refrigerant. Air condensers are less expensive and easier to maintain than the cooling towers required by water-cooled chillers. However, the evaporative effect of cooling towers enables water cooled chillers to operate at lower condenser temperatures and pressures, which increases chiller efficiency. Thus, air-cooled

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chillers are recommended for applications in which low first cost and maintenance costs outweighs the increased energy costs compared to water-cooled chillers.

The energy efficiency of air-cooled chillers is rated in terms of Energy Efficiency Ratio (EER). EER is a dimensional measure of efficiency. It is the ratio of the rate of cooling to electrical power consumption by the evaporator and the condenser fans.

EER (Btu/Wh) = Qevap (Btu/hr) / (Wcomp + W condfans) (W) @ ARI Std 590-92 conditions

The performance of a typical air cooled chiller is shown below. As predicted by a Carnot analysis, efficiency improves when the temperature of air entering the condenser declines or the leaving water temperature increases.

Air-Cooled Chiller Evaporator and Condenser Temperature Performance Chart

Using the data in the preceding table, the following relation gives chiller EER as a function of leaving water temperature when the condenser air temperature is 75 F. The R2 of the regression is 0.9998.

EER = 4.688564 + 0.20999 * LWT -.000518143 * LWT ^ 2

The performance of another air cooled chiller, as functions of outdoor air temperature and part-load ratio is shown below.

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Air-Cooled Chiller Part-Load and Condenser Temperature Performance Map. Source: Chan, K. and Yu, F., 2004, “How Chillers React to Building Loads”, ASHRAE Journal, August, pp 52-58.

Water – Cooled Chillers The energy efficiency of water-cooled chillers is typically described in terms of specific power:

kW/ton = Wcomp / Qevap

The nominal kW/ton rating of a chiller is reported at 100% load and ARI standard conditions of 44°F leaving chilled water and 85°F inlet condenser water. Typical nominal kW/ton ratings are shown below. In general, centrifugal chillers are the most energy efficient, followed by screw compressors and then reciprocating compressors.

New (kW/ton) Older (kW/ton)Reciprocating .78 to .85 .90 to 1.2Screw .62 to .75 .75 to .85Centrifugal .40 to .70 .70 to .80

Chiller efficiency is dependent on the temperature of chilled water leaving the chiller and the temperature of condenser water entering the chiller. For example, chiller efficiency improves by about 1.5%, for every degree F increase in leaving chilled water temperature.

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This means increasing chilled water supply temperature from 44°F to 54°F can cut chiller power use by about 15%. Similarly, chiller efficiency improves by about 1.5%, for every degree F decrease in entering condenser water. This means decreasing the temperature of entering condenser water from the cooling tower by 10 F, can cut chiller power use by about 15%.

In addition, chillers seldom operate at full load since design conditions rarely occur. A performance map for a standard-efficiency constant-speed chiller as a function of entering condenser water temperature and part load is shown below.

Standard-Efficiency Water-Cooled Chiller Performance Map

A relation for the specific power of this standard-efficiency chiller, KWPT kW/ton, as a function of part load, PL, and condensing water temperature CDWT can be derived from regressing data from the curves shown above. The relation and regression coefficients are shown below. The R2 for this relation is 0.98.

KWPT = a + b PL + c PL2 + d CDWT + e CDWT2 + f PL CDWT

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a 0.573417156b -1.202312827c 0.794812922d 0.005196469e 2.29267E-05f -0.000805732

A performance map for high-efficiency chillers showing constant speed (red) and variable speed (blue) chiller performance as a function of entering condenser water temperature and is shown below. At equal condenser water temperature, the constant speed chiller uses less kW/ton at full load than the variable speed chiller. However, variable-speed chiller performance improves significantly at part load, and approaches 0.15 kW/ton at some operating conditions. The declining specific power at low loads indicates excellent control efficiency.

High-Efficiency Water-Cooled Constant-Speed And Variable-Speed Chiller Performance Map for 1,000 Ton Chiller with entering condensing water temperatures from 62 F to 85 F. Source: http://ateam.lbl.gov/cleanroom/doc/Applied_Final.pdf. Lawrence Berkeley National Laboratory, Applications Team.

A relation for the specific power, KWPT kW/ton, as a function of part load, PL, and condensing water temperature CDWT for the VSD chiller can be derived from regressing data from the curves shown above. The relation and regression coefficients are shown below. The R2 for this relation is 0.98.

KWPT = a + b PL + c PL2 + d PL3 + e CDWT + f CDWT2

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a -0.18972b -1.4381c 2.5595d -1.2500e 0.010935f -2.1739 E-05

The performance of screw chillers at part load is shown below.

(Reindl, Douglas T. and Jekel, T. B., 2003, “Selection Of Screw Compressors For Energy Efficient Operation”, International Congress of Refrigeration 2003, Washington, D.C.)

Follow Load by Staging Multiple Chillers or Using Variable Speed ChillerBased on the preceding discussion, it is apparent that the energy efficiency of constant-speed chillers decreases at load cooling loads. To avoid poor part-load performance, it is possible to stage multiple chillers using differential set point temperatures, so that one chiller operates at full load before additional chillers turns on. To do so, the chilled water set point temperature of the base-loaded chiller should be about 5 F lower than the secondary chiller. If so, the secondary chiller will not turn on unless the base-load chiller cannot maintain the desired chilled water temperature.

Example

A chiller system utilizes two 150-ton screw chillers which are 50% loaded. The leaving water set point temperatures for the compressors are both set at 55 F. Calculate the electricity savings (kWh/yr) from setting the leaving water set point temperatures at 58 F and 53 F, to stage the compressors so that one runs fully loaded before the other chiller turns on.

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The figure above shows than screw chiller energy use per cooling produced increases from about 0.25 kW/kWt at full load to about 0.31 kW/kWt at 50% load. If the compressors were fully staged the power would be reduced to:

P2 = P1 .25/.32

Thus, the savings would be:CurrentTerm ValueHPY (hr/yr) 8,760Chiller amps I (A) 153Chiller volts V (V) 480Chiller power factor PF 0.80Compressor effi ciency Eff1 (kW/kWt) 0.31Total chiller power P1 (kW) = I V (3^.5) PF /1000 101.76Chiller electricity use E1 (kWh/yr) = P1 HPY 891,430

ProposedTerm ValueCompressor effi ciency Eff2 (kW/kWt) 0.25Eff2/Eff1 0.806Chiller power P2 (kW) = P1*Eff2/Eff1 82.07Chiller electricity use E2 (kWh/yr) = P2 HPY 718,895

SavingsTerm ValueDemand savings DS (kW) = P1 - P2 19.70Electricity savings ES (kWh/yr) = E1 - E2 172,535

Summary and ConclusionThis chapter calculates the comparative costs of typical types of industrial process cooling. The costs indicate that cooling towers should be used for cooling whenever possible. The chapter also describes several methods for cost effectively improving the energy efficiency of process cooling systems in manufacturing. Together, an awareness of the relative costs of cooling and an understanding of these widely applicable methods for reducing cooling loads, reducing distribution losses and improving the efficiency of primary cooling units can promote cost-effective and energy-efficient process cooling systems.

ReferencesASHRAE, 2004, ASHRAE Handbook, HVAC Systems and Equipment

EXAIR Corporation, 2007, http://www.exair.com/vortextube/vt_page.htm

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Kissock, K., 1997, CoolSim Cooling Simulation Software, University of Dayton Industrial Assessment Center (www.engr.udayton.edu/udiac)

Mott, R., 2000, Applied Fluid Mechanics, Prentice Hall, New Jersey.

Incropera and DeWitt, 1996, Fundamentals of Heat and Mass Transfer, John Wiley and Sons

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