The Absorption Refrigeration Cycle andChilled Water Plant Tour – 2015

Objectives To review the compression refrigeration cycle, introduce the absorption refrig-eration cycle and tour an industrial chilled water facility.

Tour The tour replaces the lecture on March 3rd. Go to the chilled waterfacility on the northeast corner of Bogue and Service Rds. at your regular lecturetime. Mr. Mike Crouch, who is the lead refrigeration technician on campus, will give thetour. Wear comfortable clothes and shoes. The entrance is a rather plain door on the eastside of building. NOTE: you may get ticketed for parking in the lot directly adjacent to thechiller plant. The worksheet is attached.

BACKGROUND

A refrigeration system extracts energy from a cold space (i.e., low temperature reservoir) andrejects energy to a warm space (i.e., a high temperature reservoir), typically the surroundingatmosphere. So, a refrigeration system may be thought of as pumping energy from a region ofcolder temperature to a region of warmer temperature. The second law of thermodynamicsstates that this never happens spontaneously at the continuum scale. Thus, this pumpingof energy requires an input of either work or heat.

Most refrigeration systems run on either an ideal vapor compression cycle or an absorptioncycle. Other refrigeration systems include gas cycles (e.g., Stirling), the Hilsch-Ranquevortex tube cooler, and thermoelectric cooling. Large refrigeration plants often run on oneof the first two mentioned. The vapor compression cycle resembles an Ideal Rankine Cyclewith Superheat operating in the reverse direction, that is, the refrigeration cycle operatesunder the steam dome and in the superheated domain between two isobars and can bethought of consisting of four steps. The working fluid or refrigerant may be one of a numberof liquids, including ammonia (a.k.a. R-717), fluorocarbons (e.g., dichlorodifluoromethane,a.k.a. R-12 or FC-12), or hydrofluorocarbons (e.g., 1,1,1,2-tetrafluoroethane, a.k.a., R-134aor HFC-134a).

1. Saturated vapor (i.e., x = 1) is compressed isentropically from a low-pressure isobarat 1 to a high-pressure isobar at 2 . The specific work done on the system is

wcompression = h1 − h2,

which is less than 0.

2. The now superheated vapor at 2 cools isobarically to saturated liquid (i.e., x = 0) at

3 . The heat transfer from the system to the high temperature reservoir is

qhot = h3 − h2,

which is again less than 0.

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Figure 1: A typical vaporcompression refrigeration cy-cle defined by two isobars,one isentropic process, andone isenthalpic process. Workis added to system in thecompression from states 1

to 2 . Heat transfer to theevaporator from cold reser-voir occurs from states 4

to 1 . Heat rejection occursfrom the condenser to the hotreservoir from states 2 to

3 .

3. The saturated liquid 3 expands isenthalpicly to the low-pressure isobar at state 4 .There is no work or heat transfer because

h3 = h4.

4. The saturated mixture 4 is heated isobarically at the low pressure to a saturated

vapor 1 . The specific heat transfer from the cold reservoir is

qcold = h4 − h1,

which is greater than 0. This is the cooling rate of the cycle.

The desired effect is the heating of the saturated mixture, between states 4 and 1 ,which corresponds to cooling of the cold reservoir (either an ice box, a building space, orautomobile interior). The cost is the power required to run the compressor. Appropriatingthe concept that the efficiency of a power cycle is ratio of what we want to what we pay, wecan define a metric of performance, called the coefficient of performance, and define it as

β =ql

−wc

=h1 − h4h2 − h1

(1)

Example Consider a compression refrigeration cycle using ammonia (NH3). The evapo-rator pressure is 108.37 kPa (Tsat = −32 ◦C) and the condenser pressure is 1166.49 kPa(Tsat = 30 ◦C). Referring to Figure 1, h1 = 401.6 kJ kg-1, s1 = 5.8156 kJ kg-1 K-1. Interpo-lating for s2 = 5.8156 kJ kg-1 K-1 at p2 = 1166.49 kPa, h2 = 1760 kJ kg-1. Condensing alongthe 1166.49 kPa isobar to the saturated liquid yields h3 = 322.9 kJ kg-1. Since h3 = h4,h4 = 322.9 kJ kg-1.

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Given these properties, the energy absorbed from the cold reservoir per mass of refrigerantflowing is

qc = h4 − h1 = 1410.6 − 322.9 = 1090 kJ kg−1,

the work of the compressor is

wcomp = h1 − h2 = 1410.6 − 1760 = −350 kJ kg−1.

The coefficient of performance is then

β =1090

350= 3.11.

Note that β is usually greater than 1. Furthermore, the energy rejected to the environmentis

qh = h3 − h2 = 322.9 − 1760 = −1440 kJ kg−1,

which is, of course the difference between qc and wcomp.

Absorption Cycle The work done on the system in the isentropic process of getting fromstates 1 to 2 of a compression cycle results in a considerable cost. In a compressionrefrigeration system, which is most common for domestic and automobile use, this is donewith a compressor. The compressor is driven either electrically, in the case of a home refrig-erator or air-conditioner, or driven by taking power from the automobile engine. Reducingthis cost could have a great benefit.

Consider the circumstance for which there exists a large heat source that might otherwisego to waste. Consider also that the cost of pumping a liquid from low to high pressure istrivial compared to the change in enthalpy of vapor between the same two isobars. (Yousaw an example of this when calculating the Simon power plant performance with its 93psig steam.) Is there a means to harness this energy and take advantage of the small cost ofpumping liquid? Yes the absorption refrigeration cycle.

In the absorption refrigeration cycle, two devices, the generator and the absorber, replacethe compressor of the vapor compression refrigeration cycle. The essential idea is that a liquidmixture of an absorbent and a refrigerant is pumped to the generator. In the generator, therefrigerant is boiled from this high-pressure solution, creating a pure refrigerant vapor atthat high pressure, and a concentrated absorbent solution. The solution is expanded toreturn to the absorber, while the refrigerant runs through the condenser, expansion valve,and evaporator.

There are several absorbent–refrigerant combinations. If the cold reservoir is to be frozen,Ammonia–Water is a common combination (ammonia is the refrigerant). If only chilled wateris needed (no freezing), water–lithium bromide is a good choice (water is the refrigerant).By controlling the LiBr concentration, the pressures of the condenser and evaporator can becontrolled in much the same way that a compressor controls the pressures in a conventionalvapor compression refrigeration system. Water from the plants cooling tower provides boththis cooling and the high temperature heat reservoir in the condenser. The system is shownin schematically in Figure 2.

Table 1 lists the capacities of some of the chilled water units arounds campus.

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CONDENSERWATER OUTTO TOWER

CONDENSERGENERATOR

ABSORBER

EVAPORATOR

HEATEXCHANGER

LOW PRESSUREWATER VAPOR

STEAM IN

STEAM OUT

CHILLEDWATER IN

CHILLEDWATER OUT

REFRIGERANTPUMP

GENERATORPUMP

ABSORBERPUMP

CONDENSERWATER INFROM TOWER

Figure 2: Schema of an absorption refrigeration system. The components on the left handside replace the compressor.

Table 1: Chiller capacities at the MSU main chiller plant.

Chiller # Capacity Pump Flow Rate Pump Flow Rate(tons) (gpm) (gpm)

at design ∆T at ∆T = −10 ◦C

1-Carrier 16JA054 475 1150 at -12 ◦C 8142-Carrier 16JA054 475 1150 at -12 ◦C 8143-Trane C12A-2 1170 2900 at -12 ◦C 20064-Trane C12A-2 1170 2900 at -12 ◦C 20065-Trane ABSC-12A4 1250 2143 at -10 ◦C 21436-Trane ABSC-12A4 1250 2143 at -10 ◦C 21437-Trane ABSC-12A4 1250 2143 at -10 ◦C 21438-Trane ABSC-12A4 1250 2143 at -10 ◦C 21439-Trane ABSC-12A4 1250 2143 at -10 ◦C 214310-Trane ABSC-12A4 1250 2143 at -10 ◦C 2143

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Figure 3: Picture of the outside of a typical industrial absorption unit.

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Figure 4: Schema of the Trane absorption refrigeration system.

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MSU Chilled Water Plant Tour– Spring 2015 Worksheet

Student Name:

Signature of TA to Verify Attendance:

1. What devices in an absorption refrigeration system replace the compressor of a con-ventional vapor compression refrigeration system?

2. Describe the four (4) fluid streams within a Trane absorption unit.

3. How does a cooling tower operate?

4. What is the rated capacity of the chilled water plant in tons of refrigeration, Btu/hr,and kW?

5. Given the chilled water flow rate (from the pump) and the entering and exiting temper-atures of the chilled water stream for one of the absorption units, calculate the actualcooling load (in tons of refrigeration) the unit is providing.

6. Determine the Carnot cycle COP for the absorption cycle used at the MSU chilledwater plant.

7. Using the Carnot cycle COP and assuming a cooling load of 1250 tons, determine therequired mass flow rate of steam.

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