Tài liệu Handbook of Mechanical Engineering Calculations P12 pdf

SECTION 12 REFRIGERATIONRefrigeration Required to Cool an Computing Refrigerating Capacity Needed for Air-Conditioning Loads 12.12 Water-Vapor Refrigeration-System Analysis 12.15 Analyzi

SECTION 12 REFRIGERATION

Refrigeration Required to Cool an

Computing Refrigerating Capacity

Needed for Air-Conditioning Loads

12.12

Water-Vapor Refrigeration-System

Analysis 12.15

Analyzing a Steam-Jet Refrigeration

System for Chilled-Water Service

12.17

Heat Pump and Cogeneration

Combination for Energy Savings 12.21

Comprehensive Design Analysis of an

Absorption Refrigerating System

12.28

Cycle Computation for a Conventional

Compression Refrigeration Plant 12.41

Design of a Compound

Compression-Refrigeration Plant with Water-Cooled

Intercooler 12.45

Analysis of a Compound Refrigeration Plant with a Water- Cooled Intercooler and Liquid Flash Cooler 12.47

Compression-Computation of Key Variables in a Compression Refrigeration Cycle with Both Water- and Flash-Intercooling 12.50

Refrigeration System Selection 12.54 Selection of a Refrigeration Unit for Product Cooling 12.56

Energy Required for Steam-Jet Refrigeration 12.62

Refrigeration Compressor Cycle Analysis 12.64

Reciprocating Refrigeration Compressor Selection 12.68

Centrifugal Refrigeration Machine Load Analysis 12.71

Heat Pump Cycle Analysis and Comparison 12.72

Central Chilled-Water System Design to Meet Chlorofluorocarbon (CFC) Issues 12.76

REFRIGERATION REQUIRED TO COOL AN

OCCUPIED BUILDING

The building in Fig 1 is to be maintained at 75⬚F (23.9⬚C) dry bulb and 64.4⬚F(18.0⬚C) wet-bulb temperatures This building is situated between two similar unitswhich are not cooled There is a second-floor office above and a basement below.The south wall, containing 45 ft2(4.18 m2) of glass area, has a southern exposure

On the north side of the building there are two show windows which are ventilated

to the outside and are at outside temperature conditions Between the show windows

is a doorway This doorway is normally closed but it is frequently opened andallows an average of 600 ft3/ min (16.98 m3) of outside air to be admitted Opening

of the door by customers will cause slightly more than two air changes per hour

in the building The number of persons in the building is 35; lighting is 1100 watts

FIGURE 1 Plan of building cooled by refrigeration.

on a sunny day Basement temperature is 80⬚F (26.7⬚C) The maximum outsideconditions for design purposes are 95⬚F (35.0⬚C) dry bulb and 78⬚F (25.6⬚C) wet-bulb temperature There is a 0.5 hp (373 W) fan motor in the interior of the building.What is the refrigeration load for cooling this building?

Calculation Procedure:

1. Assemble the overall coefficients of heat transfer for the building materials

Using the ASHRAE handbook, find the U values, Btu / ft2䡠h䡠 ⬚F as follows: Eastand west walls (24-in [60.96-cm] brick, plaster one side), 0.16; North partition(1.25-in [3.18-cm] tongue-and-groove wood), 0.60; Plate-glass door, 1.0; South wall(13-in [33-cm] brick, plaster one side), 0.25; Windows (single-thickness glass),1.13; Floor (1-in [2.54-cm] wood, paper, 1-in [2.54-cm] wood over joists), 0.21;Ceiling (2-in [5.08-cm] wood on joists, lath and plaster), 0.14 To determine the SIoverall coefficient, multiply the given value above by 5.68 to obtain the W / m2䡠

⬚C Thus, the values are: 0.908; 3.4; 5.68; 1.42; 6.42; 1.19; 0.79, respectively

2. Compute the temperature differences for the walls, ceiling, and floor

For the walls and ceiling the temperature difference⫽ outside design temperature

⫺indoor design temperature⫽95⫺75⫽20⬚F (36⬚C) For the floor, the ature difference⫽ 80⫺75 ⫽5⬚F (9⬚C)

temper-3. Calculate the heat flow into the building

The heat leakage for any surface⫽U(ft2[m2] surface area)(temperature difference).For each of the surfaces in this building the heat leakage is computed thus: Eastand west walls ⫽(0.16)(1440)(20) ⫽ 4610 Btu / h (1350.7 W); North partition⫽

REFRIGERATION 12.3

(0.6)(299)(20)⫽3590 Btu / h (1051.9 W); Glass door⫽(1.0)(35)(20)⫽700 Btu/ h (205.1 W); South wall⫽ (0.25)(300)(20)⫽ 1500 Btu / h (439.5 W); Windows

⫽ (1.13)(91)(20)⫽ 2060 (603.5 W); Floor ⫽(0.21)(1104)(5) ⫽1160 (339.9W);Ceiling ⫽ (0.14)(1104)(20) ⫽ 3090 (905.4 W) Summing these individual heatleakages gives 16710 Btu / h (4896 W)

4. Determine the sensible heat load

Using the conventional heat loads for people, lights, motor hp, the sensible heatload is: Occupants ⫽ (35)(300) ⫽ 10,500 Btu / h (3076.5 W); Lights ⫽(1100)(3.413)⫽3760 Btu / h (1101.7 W); Motor horsepower⫽(0.5)(2546)⫽1275Btu / h (373.6 W)

5. Find the air leakage heat load

Use the relation: Air leakage heat load, Btu / h (W) ⫽(air change, ft3/ h)(specificheat of air)(temperature difference) / (specific volume of air, ft3/ lb) Or(36,000)(0.24)(95⫺75) / 13.70⫽12,600 Btu / h (rounded off) (3691.8 W)

6. Calculate the sun effect heat load

For the glass on the south wall, the sun effect ⫽ (45 ft2)(30 Btu / h ft2) ⫽ 1350Btu / h (399.6 W) The sun effect on the south wall⫽(300 ft2)(0.25)(120⫺95)⫽

1875 Btu / h (549.4 W)

7. Find the dry tons (W) of refrigeration required

Sum the heat gains computed above thus: 16,710 ⫹ 10,500 ⫹ 3760 ⫹ 1275 ⫹12,600⫹1350⫹1875⫽48,070 Btu / h (14,084.5 W)⫽grand total heat loss Thedry tons (W) of refrigeration is then found from 48,070 / 12,000 Btu / ton ⫽ 4.01tons (14.1 kW)

8. Evaluate the moisture latent heat load

List the air leakage conditions thus:

The, the air latent heat ⫽(lb / h)(latent heat of air, Btu / lb) ⫽(15.95)(1040) ⫽16,588 Btu / h (4860.3 W) Person load ⫽ (35)(100) ⫽ 3500 Btu / h (1025.5 W).Total latent heat⫽16,588⫹3500⫽20,088 Btu / h (5885.8 W), or 20,088 / 12,000

⫽1.67 tons Then, the total refrigeration load ⫽ 4.01⫹1.67 ⫽5.68 tons (19.98kW)

Related Calculations. As you can see, if you are to perform repeated lations for buildings and rooms, a form listing both the equations and items to becomputed will be helpful in saving you time The procedure given here is usefulfor the occasional computation of buildings of all types: residential, commercial,

calcu-industrial, etc The same general procedure can be used for trucks, ships, aircraftand other mobile applications.

DETERMINING THE DISPLACEMENT OF A

RECIPROCATING REFRIGERATION COMPRESSOR

What is the needed displacement of a reciprocating refrigeration compressor rated

at 50 tons (45.4 t) when operating with a refrigerant at 0⬚F (⫺17.8⬚C) in the orator expansion coils? At this temperature, the heat absorbed by the evaporation

evap-of 1 lb (0.45 kg) evap-of the refrigerant is 500 Btu (527.5 kJ) refrigerating effect, andthe specific volume is 9 ft3/ lb (0.56 m3/ kg) The vapor enters the compressor inthe saturated state If the compressor speed of rotation is 180 r / min, and the stroke

is 1.2⫻bore, what is the bore and stroke of this single-acting compressor?

Calculation Procedure:

1. Determine heat absorbed and compressor displacement

The heat to be absorbed (tons of refrigeration)(heat equivalent of 1 ton of eration, Btu / min) Or the heat absorbed ⫽ 50(200 Btu / min / ton refrigeration) ⫽10,000 Btu / min (10,550 kJ / min)

refrig-The compressor displacement ⫽ (heat absorbed)(specific volume of therefrigerant) / (refrigerating effect) Substituting for this refrigerant, displacement⫽10,000(9) / (500)⫽ 180 ft3(5.09 m3)

2. Find the compressor bore

Let N ⫽the compressor speed, rpm; D ⫽compressor cylinder bore, ft (m); L ⫽

compressor stroke, ft (m); V⫽piston displacement, ft3(m3) per stroke Then:

V⫽0.785D L N⫻V⫽180 ft / min

3 180

3. Find the compressor piston stroke

Using the relation in step 2, L ⫽ 1.2D ⫽ 1.2 ⫻ 1.02 ⫽ 1.224 ft (0.373 m), or14.69 in The bore and stroke as computed here are typical for a compressor ofthis capacity

Related Calculations. With the phasing out of chlorofluorcarbons (CFC) cause of environmental restrictions, engineers must be able to evaluate the perform-ance of alternative refrigerants The procedure given above shows exactly how toperform this evaluation for any refrigerant whose thermodynamic and physical char-acteristics are known by the engineer, or can be obtained from standard data ref-erence

be-While many liquids boil at temperatures low enough for refrigeration, few aresuitable for refrigeration purposes Those liquids suitable for practical refrigeration

Superheat pickup Controlled superheat

Expansion valve

Low side High side

20 psi 5.5 °F

185 psi

235 °F

Compressor (heat pump)

C B

A

F F'

B

A D

Fig 2a.

When the boiling refrigerant removes sensible heat from environment at a rateequivalent to the melting of one ton (2000 lb; 980 kg) of water ice in 24 h, therate of heat removal is a ton (3.516 kW) of refrigeration Since the heat of melting(sublimation) of 1 lb (0.454 kg) of water ice is 144 Btu (151.9 kJ), a ton ofrefrigeration is equivalent to 2000 lb (144 Btu / 24h)⫽ 12,000 Btu / h (3516 W, or3,516 kW).

By comparison, the heat of sublimation (melting) of dry ice (CO2) is 275 Btu / lb(640.8 kJ / kg) To say that a refrigeration machine has a capacity of 10 tons (35.16kW) is to say that the rate of refrigeration is 10⫻ 200⫽ 2000 Btu / min (35.16kW) Note that 1 ton of refrigeration equals a rate of 200 Btu / min (3.516 kW)

To determine the amount of refrigerant that must be circulated, divide the frigerating effect of the refrigerant in But / lb (kg) into 200 Btu / min (W) Thus,with a refrigerating effect 25 Btu / lb (58.3 kJ / kg), the quantity of refrigerant to becirculated is 200 / 24⫽4 lb / min (1.82 kg / min)

re-The work of compression is the amount of heat added to the refrigerant duringcompression in the cylinder or rotary compressor It is measured by subtracting theheat content of 1 lb (0.454 kg) of refrigerant at the compressor suction conditions,

point F⬘, F, or A in Fig 2b from the heat content of the same pound (kg) at the compressor discharge conditions, point B or B⬘in Fig 2b.

The theoretical horsepower (kW) requirements of a refrigeration compressor can

be found by multiplying the work compression in Btu / lb (kJ / kg) by the pounds(kg) of refrigerant circulated in one hour, and dividing this product by 2545 Btu /hp-h, hpt⫽(work of compression)(refrigerant circulated) / 2545 Multiply by 0.746

to obtain theoretical kW input

A good example of the practical value of this calculation is in the recent life example of the upgrading of the HVAC system in a 400-unit apartment com-plex Two older refrigerating machines using CFC-refrigerants were replaced bytwo new chillers using HCFC-123 refrigerant

real-The older machines required an input of 0.81 kW per ton of cooling capacity(refrigeration) while the newer machines require only 0.55 kW per ton Annualsavings of more than 30 percent in energy costs for refrigeration are expected withthe new machines and refrigerant The new machines also reduce greenhouse gasemissions because of reduced electrical power needed to run them Payback timefor the new machines will be less than 2.5 years because the energy savings are sosignificant Several examples of typical piping arrangements for reciprocating re-frigeration compressors and chillers are shown in Fig 3 through Fig 6

Another example of using this procedure is substitution of natural-gas fueledengine drives for refrigeration chillers using new refrigerants In one departmentstore installation of such a chiller the estimated annual energy cost savings are

$54,875 Such drives reduce electric demand charges, are compact in size, areenvironmentally friendly, and are used in hospitals, nursing homes, schools, col-leges, office buildings, retail, and industrial / process facilities Some of the newestcentrifugal chillers on the market report a required input of just 0.20 kW / ton whenoperating at 60 percent load with entering condenser water at 55⬚F (12.8⬚C)

HEAT-RECOVERY WATER-HEATING FROM

REFRIGERATION UNITS

How much heat can be obtained from heating water for an apartment house having

150 apartment units served by two 200-ton (180 t) air-conditioning units if a 70⬚F

REFRIGERATION 12.7

FIGURE 3 Layout of suction and hot-gas lines for multiple-compressor

operation (Carrier Corporation.)

FIGURE 4 Interconnecting piping for multiple condensing reciprocating refrigeration units.

(Carrier Corporation.)

FIGURE 5 Piping arrangement for two centrifugal coolers in series arrangement (Carrier poration.)

Cor-(38.9⬚C) temperature rise of the incoming cold water is required? Determine thenumber of gallons (L) of water that can be heated per hour and the total gallonage(L) of heated water that can be delivered with the air-conditioning units operating

8, 12, and 16 hours per day

Calculation Procedure:

1. Determine the quantity of heat available

Heat is available from the high-pressure gas at the refrigeration compressor charge This is valid regardless of the type of compressor used: reciprocating, rotary,

dis-or centrifugal The quantity of heat available from a specific compressdis-or depends

on the outlet-gas temperature, gas flow rate, and the efficiency of the heat exchangerused

To recover heat from the hot gas, a heat exchanger, Fig 7, is placed in thecompressor discharge line, ahead of the regularly used condenser Cold water fromeither the building’s outside water supply line, or from the building’s heated-waterstorage tank, is pumped through the heat exchanger in the compressor dischargeline Leaving the heat exchanger, the heated water returns to the hot-water storagetank

Experience shows that a typical well-designed heat exchanger, such as a ventional water-cooled condenser, can transfer 25 to 35 percent of the Btu (kJ)rating of the refrigeration compressor, i.e., the air-conditioning unit’s rating Usingthe lower value in this range for these units gives, Heat Available ⫽ 0.25 (200)(12,000 Btu / h / ton)⫽0.25 (200)(12,000)⫽600,000 Btu / h (633,000 kJ / h) With

con-35 percent, Heat Available ⫽ 0.35 (200)(12,000) ⫽ 840,000 Btu / h (886,200 kJ /h) With two refrigerating units the heat available would be double the computedamount, or 1,200,000 Btu / h (1,266,000 kJ / h) and 1,680,000 Btu / h (1,772,400 kJ/ h)

2. Find the hourly water heating rate for the system

Water weighs 8.34 lb / gal (1.02 kg / L) To raise the temperature of one pound ofwater (0.454 kg) 1⬚F (0.55⬚C) requires a heat input of 1 Btu (1.055 kJ) With thespecified 70⬚F (38.9⬚C) water-temperature rise required in this building, the rate ofwater heating will be: gal / h (L / h)⫽(heat available) / (lb / gal)(temperature rise re-quired)

With 25 percent heat transfer, we have, gph heated ⫽(600,000) / (8.33)(70) ⫽

1029 gal / h (3900 L / h) And with 35 percent heat transfer, gph ⫽ (840,000) /

FIGURE 7 Heat-recovery heat-exchanger location in a ation cycle.

refriger-(8.33)(70)⫽1440 gal / h (5458 L / h) Again, with two units the hourly heating ratewill be doubled, or 2058 gal / h (7800 L / h) and 2880 (10,915 L / h)

3. Compute the daily total gallonage of hot water produced

Since air-conditioning refrigeration units operate varying numbers of hours per day,depending on the outside weather conditions, the gallonage of hot water availablefrom heat recovery will vary For the range of operating hours specified, we have,per 200-ton (180-t) unit:

Gallonage available (L) per hours of operation

25 percent 8232 (30,458) 12,348 (45,688) 16,464 (62,399)

35 percent 11,520 (42,624) 17,280 (65,491) 23,040 (87,322)

Again, we double these numbers for two units in the building

Related Calculations. The normal discharge temperature of modern ants makes heat recovery for domestic and / or process water heating an attractiveoption, especially in an environmentally conscious world Further, heat recovery is

refriger-REFRIGERATION 12.11

FIGURE 8 Layout of heat-recovery heat-exchanger piping for refrigeration cycle.

such a simple design challenge that the investment is often recovered in fuel savings

in less than two years

For maximum efficiency, the designer must try to match hot-water needs withthe operating time of the refrigeration unit If there is a disparity between thesetwo variables, a sufficiently large water storage tank can be designed into the system

to store hot water during times when the refrigeration unit is not operating Such

a storage tank is not expensive and it will have a long life if properly maintained.Heat recovery for heating incoming cold water can be used in office buildings,apartment houses, hotels, motels, factories, and commercial buildings wherever aneed for hot water exists and a refrigeration unit of some kind operates in thestructure Energy savings can range from 25 to 100 percent of the cost of heatingwater for either domestic or process uses A pump, Fig 8, is used to force-circulate

FIGURE 9 Basic dehumidifier consisting of cooler followed by heater.

water through the hot-gas heat exchanger while the refrigeration unit is operating.This action recovers heat when it is available and stores the heated water for futureuse

A regulating valve, Fig 8, controls water flow and the temperature of the waterleaving the heat exchanger The valve ensures that maximum heat recovery willoccur When designing the piping for heat recovery, be certain to arrange to havethe coldest water enter the heat exchanger Then maximum benefit will be obtainedfrom the heat recovery

COMPUTING REFRIGERATING CAPACITY

NEEDED FOR AIR-CONDITIONING LOADS

An air-conditioned space is supplied 10,000 ft3/ min (283 m3/ min) of air at 65⬚F(18.3⬚C) and 60 percent relative humidity Air is supplied to the dehumidifier at

85⬚F (29.4⬚C); wet-bulb temperature is 70⬚F (21.1⬚C) Condensate leaves the humidifier at the same temperature as the outgoing air, which is at standard at-mospheric pressure throughout the system Air leaves the dehumidifier in the sat-urated state Determine the refrigerating load for this air-conditioned space

flow rate of air, M a⫽ flow rate, ft3/ min / specific volume ⫽10,000 / 13.39 ⫽746

lb dry air per min (338.7 kg / min)

2. Compute the quantity of moisture condensed and the air dew point

Moisture condensed ⫽ (moisture content of incoming air ⫺ moisture content ofleaving air) / 7000 gr / lb ⫽ (86 ⫺ 55.5) / 7000 ⫽ 0.00435 lb / lb dry air (0.00197

kg / kg) Figure 11 shows the psychrometric process The dew point, from the chart,

t2⫽51⬚F (10.6⬚C); enthalpy h ƒc⫽51 ⫺32⫽19 Btu / lb of water (44.2 kJ / kg)

3. Set up an energy balance about the dehumidifier

The energy balance for an adiabatic mixing process is given by M1(h m1)⫹M2(h m2)

⫽M3(h m3 ), where the M values are the weight flow rates in the respective air-vapor

REFRIGERATION 12.13

FIGURE 10 Dehumidification process on skeletonized psychrometric chart.

mixtures, and the h values are the respective enthalpies Then, the energy relation for the cooler is Q R⫽ (h m1⫺ h m2)⫺ (W v1 ⫺ W v2 )h ƒc where the symbols are asgiven above

Substituting, Q R ⫽ (33.96 ⫺ 20.85) ⫺ (0.00436)(19) ⫽ 13.03 Btu / lb (30.36

kJ / kg) Then, the total heat removed at the dehumidifier⫽ (13.03)(746)⫽ 9740Btu / min (171.2 kW)

4. Find the refrigeration capacity required

Since 200 Btu / min⫽ 1 ton of refrigeration, the refrigeration capacity required⫽

9740 / 200⫽ 48.7 tons (43.6 t)

Related Calculations. This procedure could have been performed using thepressure and humidity relations of the air streams Approximately the same resultwould have been obtained By considering the enthalpy of the condensate the es-timated refrigeration load is decreased by about 0.2 ton (0.18 t), which for thisinstallation is insignificant The tonnage reduction will always be small when thefinal dew point is low, and in most cases it can be neglected Further, the true state

of which the condensate is removed is usually difficult to establish Neglecting theenergy of the condensate will give a refrigeration capacity requirement that is onthe safe side, i.e., a larger value than would be arrived at with the exact solution.Depending on the number of banks of spray nozzles, the direction of the spray,and the air velocity, the percentage of untreated air passing through an air washer,Fig 12, may range from 5 to 35 percent for two banks and one bank of nozzles,respectively, according to ASHRAE For a coil-type dehumidifier the percentage ofuntreated air may range from 2 to 39 percent for coils eight rows to one row deepoperating with a face velocity of 300 ft / min (91.4 m / min) In general, the refrig-

FIGURE 11 Values for the dehumidification process.

FIGURE 12 Schematic of complete air-conditioning system for evaluating variables in design.

eration power requirements of an air-conditioning system are dependent upon theheat-transfer characteristics of the dehumidifier used

This procedure, and Figs 9 through 12, are the work of Norman R Sparks,Professor and Head, Department of Mechanical Engineering, The PennsylvaniaState University, and Charles C DiLio, Associate Professor of Mechanical Engi-neering, The Pennsylvania State University SI values were added by the handbookeditor

pressor is 0.97 Find: (a) the capacity of the machine, tons; (b) the pounds (kg) of vapor to be removed from the evaporator per minute; (c) the volume of vapor to

be removed from the evaporator, ft3/ min (m3/ min) and ft3/ (min䡠ton) (m3/ t) ther, consider that the machine in Fig 13 is equipped with a centrifugal compressorand mechanical vacuum pumps The condenser pressure is 2 in Hg (5.1 cm) abs.Compression efficiency is 0.65 Mechanical efficiency of the compressor is 98 per-cent The condensate leaves the condenser at 90⬚F (32.2⬚C) Power required to drive

Fur-the condensate and air pumps is 6 percent of Fur-the total power input Find: (d ) Fur-the compressor hp (kW) and hp / ton (kW / t); (e) the heat rejected in the condenser;

Btu / min (kJ / min); (ƒ) the coefficient of performance

Calculation Procedure:

1. Draw a sketch of the system showing the important state points

Figure 13 shows a schematic of the system and the state points

2. Compute the rate of refrigerant flow

cu3/ min) or 387 ft3/ min / ton (12.2 m3/ t)

4. Find the actual change in enthalpy

h3⫽1049.4; s3⫽2.0794; h5⫽58 (135.1 kJ / kg)

After isentropic compression to 2 in Hg,h⬘4⫽1168 (2721.4 kJ / kg)

Isentropic⌬h⫽1168⫺1049.4⫽118.6 Btu / lb⫽ h⬘4⫺h3(276.3 kJ / kg)Actual⌬h⫽118.6⫽ 182.5 Btu / lb⫽h4⫺h3(425.2 kJ / kg)

0.65

h4⫽ 1049.4⫹182.5 ⫽1231.9 (2 in Hg and 380⬚F) (2870.3 kJ / kg) (5.08 cm

Hg and 193.3⬚C)

5. Compute the compressor and total power input

⫽30.66(182.5)⫽5595 Btu / min (5902.7 kJ / min)

W i

J

Compressor hp⫽ 5595 ⫽134.6, or 0.859 hp / ton (0.71 kW / t)

42.4(0.98)Total hp⫽134.6⫽143.2 or 0.914 hp / ton (0.758 kW / t)

0.94

6. Determine the condenser heat rejection and COP

Q⫽ 30.66(1231.9⫺ 58)⫽35,992 Btu / min, or 230 Btu / min / ton (269.6 kJ / t)Coefficient of performance⫽ 200 ⫽5.17 (based on total hp)

0.914(42.4)

See Fig 14 for a typical water-vapor refrigeration system using steam ejectorsand a surface condenser Figure 15 shows a water-vapor refrigeration system using

a steam-jet (barometric condenser)

Related Calculations. The rather high coefficient of performance (COP) of thismachine is caused by the comparatively favorable temperature range through whichthe machine operates With today’s environmental concern over safer refrigerants,water has a 0.00 ozone-depletion potential Further, water also has a 0.00 immediateglobal warming potential, an a 0.00 100-year global warming potential And sincesuch refrigeration systems can be gas-fired, they have less potential atmosphericpollution compared to coal or oil Likewise, water has zero flammability

Air cooling is one industrial application of refrigeration that does not ordinarilyrequire, in the refrigerating sense, low temperatures Hence, the water-vapor refrig-erating system is ideal for such applications Further, water is a cheap refrigerant,and it is truly a safe one, and as a result, it is finding greater use in industry today,especially in view of the new, stricter environmental regulations that seem to beimposed every year

This procedure, and Figs 13 through 15, are the work of Norman R Sparks,Professor and Head, Department of Mechanical Engineering, The PennsylvaniaState University, and Charles C DiLio, Associate Professor of Mechanical Engi-neering, The Pennsylvania State University SI values were added by the handbookeditor

REFRIGERATION 12.17

FIGURE 14 Typical water-vapor refrigeration system using steam ejectors and surface condenser.

(Ingersoll-Rand Co.)

ANALYZING A STEAM-JET REFRIGERATION

SYSTEM FOR CHILLED-WATER SERVICE

A steam-jet water-vapor refrigerating machine is to produce 250 gal / min (15.8 L /s) of chilled water at 45⬚F (7.2⬚C) Makeup and recirculated water are at 60⬚F(15.6⬚C), and the quality of the vapor leaving the evaporator and entering the ejector

is 0.97 Condenser pressure is 2 in (5.08 cm) Hg abs Condensate leaves the denser at 90⬚F (32.2⬚C) The motive steam is supplied at 140 lb / in2(abs) (946 kPa),

con-370⬚F (187.8⬚C) System manufacturer supplies the following efficiencies: Nozzleefficiency ⫽ 90 percent Entrainment efficiency⫽ 65 percent Diffuser efficiency

⫽75 percent Steam consumption of the auxiliary ejectors is 6 percent of the total

steam requirement Neglecting the power demands of the water pumps, find: (a) Steam consumption of the main ejector; (b) the total steam consumption; (c) heat

rejected in the primary condenser

Calculation Procedure:

1. Determine the pertinent steam enthalpies and quality

Referring to Fig 16 for symbols, we have

h A⫽1203.5 at 140 lb / in2(abs), 370⬚F (2804 kJ / kg at 964.6 kPa, 187.8⬚C)

FIGURE 15 Steam-jet (barometric condenser), water-vapor refrigeration system.

(Ingersoll-Rand Co.)

REFRIGERATION 12.19

FIGURE 16 Steam-jet compressor.

⫽803 (after isentropic expansion to 0.3 in Hg) (1871 kJ / kg) (0.76 cm Hg)

Assume x c⫽0.87, for which h c⫽943 (2197.2 kJ / kg)

Thenh⬘d⫽1043 (after isentropic compression to 2 in Hg) (2430.2 kJ / kg at 5.08

M a is too low Therefore, h cmust be increased appreciably

3. Make a second computation of M a

M a⫽ ⫽1.59 lb / lb evaporator vapor (0.72 kg / kg)

1205.5⫺11440.65(1.59) (360.5)⫽(2.59) (144)

373⫽373Since the entrainment energy equation shows a perfect check, the assumed

h c and the value of M aderived from this second trial are considered correct

4. Calculate the system variables

(a) Data, as necessary, have been taken from the previous procedure.

Evaporator vapor⫽30.66 lb / min

Ejector steam consumption ⫽ 1.59(30.66) ⫽ 48.7 lb / in, or 2920 lb / h(1325.7 kg / h)

Steam rate⫽ 48.7 ⫽0.311 lb / (min䡠ton), or 18.7 lb / (h䡠ton) [9.43 kg / (h

or 550 Btu / (min䡠ton) [644.7 Btu / (min䡠t)]

The heat required for the generation of the motive steam will be h A⫺h5, or, inthis instance, 1203.5⫺58⫽ 1145.5 Btu / lb (2669 kJ / kg) The ratio of the refrig-erating effect to the heat supplied to produce that effect is therefore 12,000 /19.9(1145.5)⫽0.526

Related Calculations. Like the absorption system, the steam-jet water-vaporcycle requires but a very small portion of mechanical energy for its operation Acoefficient of performance (COP) may not, therefore, be truly expressed for thistype of machine But a pseudocoefficient can be arbitrarily devised for the purpose

of comparison with other types of systems

While there is no standard procedure for setting up a pseudocoefficient, if themotive steam were used in, say, a turbine to drive a compressor, about 50 percent

of the energy available in expanding to condenser pressure might be considered asthe turbine work output and the compressor input For this procedure, the energyavailable per pound (kg) of motive steam between 140 lb / in2 (abs) (964.6 kPa),

370⬚F (187.8⬚C) and 2 in (5.08 cm) Hg abs is 317.5 Btu / lb (739.8 kJ / kg) Thecoefficient of performance would thus be, according to this standard, 12,000 / [19.9(0.5)(317.5)]⫽3.80, based on actual steam consumption The corresponding horse-power per ton⫽ 1.237 (1.025 kW / t)

As a practical problem, the determination of the minimum available energy ofthe motive steam at which the ejector will operate is not of particular importancesince excessive steam consumption will render the operation unsatisfactory longbefore this point is reached However, the condition of the motive steam to besupplied to an ejector for any given maximum allowable steam consumption, asdictated, for example, by cooling-water considerations, may frequently be of inter-

FIGURE 17 Schematic of heat pump used to heat a building.

est This may be closely approximated by the application of the principles given

in this, and the preceding, calculation procedures

This procedure is the work of Norman R Sparks, Professor and Head, ment of Mechanical Engineering, The Pennsylvania State University, and Charles

Depart-C DiLio, Associate Professor of Mechanical Engineering, The Pennsylvania StateUniversity SI values were added by the handbook editor

HEAT PUMP AND COGENERATION

COMBINATION FOR ENERGY SAVINGS

In the heat-pump system shown in Fig 17, the building requires 100,000 Btu / h

(29.3 kW) for heating Calculate for this system: (a) The capacity of the heat pump, tons (t) as a refrigerating machine; (b) the motor horsepower (kW); (c) the hourly energy input to the motor; (d ) the overall coefficient of performance; (e) the heat

received from the outside air; (ƒ) the electrical consumption per hour for direct

heating; (g) the coal consumption for direct heating with a 50 percent furnace efficiency; (h) the oil consumption for direct heating with a 65 percent furnace

efficiency; (i) the natural-gas consumption for direct heating with a 70 percent furnace efficiency; ( j ) the manufactured-gas consumption for direct heating with

70 percent furnace efficiency; (k) the energy-saving potential with a conventional

diesel engine driving the compressor in a cogeneration mode

Calculation Procedure:

1. Specify the power and fuel constants to use in the design

For this system the following power and fuel constants will be sued; such constantsare readily available in standard reference works: Thus, the power consumption ofmotor for the unit, considered as a standard refrigerating machine, will be 1.7 hp /ton (1.49 kW / t) Motor efficiency is 87 percent Heat of combustion of the fuels:coal, 14,000 Btu / lb (32,620 kJ / kg); oil, 19,000 Btu / lb (44,270 kJ / kg); natural gas,

1100 Btu / ft3(41,007 kJ / m3); manufactured gas, 600 Btu / ft3(22,367 kJ / m3) cific gravity of the fuel oil⫽0.88

Spe-2. Find the total heat output of the heat pump

1.7 hp⫽ 1.7(2544)⫽4325 Btu / h (1.27 kW)

Motor input⫽ 4325⫽ 4970 Btu / (h䡠ton) (1.62 kW / t)

0.87The motor loss is therefore 4970 ⫺4325 ⫽ 645 Btu / (h䡠ton) [0.21 kW / h䡠t)]which may be considered as applied to direct heating

Heat output of condenser (and cylinder jacket) per ton ⫽ 12,000 ⫹ 4325 ⫽16,325 Btu / h (4.8 kW)

Total heat output of machine per ton ⫽ 16,325⫹ 645 ⫽ 16,970 Btu / h (4.97kW)

In this instance, this is equivalent to 12,000⫹4970

3. Compute the system characteristics

5.89 tons (6.5 t)

1.1 gal / h (4.2 L / h)

m3/ h)

REFRIGERATION 12.23

4. Determine the cogeneration potential for the installation

With a Diesel-engine drive for the compressor, the specific fuel consumption willtypically be 0.45 lb / bhp䡠h (0.27 kg / kW䡠h) Then, the fuel required per ton (t)refrigerating capacity will be 1.7 (0.45)⫽ 0.765 lb / h (0.347 kg / h), representing0.765 (19,000)⫽ 14,535 Btu / h (4.26 kW) Of this, 4325 Btu / h (1.26 kW) leavethe engine in the form of work The remainder, 10,210 Btu / h䡠ton (10,722 kJ / h䡠t), is rejected in cooling water, exhaust, and by radiation Much of this loss can berecovered and applied to direct heating

Assuming that 80 percent of the waste can be recovered, a typically safe sumption, the heat supplied to the building per ton (t) capacity is 16,325⫹ 0.80(10,210)⫽ 24,495 Btu / h (7.17 kW), using data from the steps above in this pro-cedure Then:

as-Capacity required⫽100,000 / 24,495 ⫽4.08 tons (3.67 t)

Power required⫽4.08(1.7)⫽6.94 hp (5.2 kW)

Fuel required⫽ 6.94(0.45)⫽ 3.12 lb / h or 0.425 gal / h (1.6 L / h)

Coefficient of performance, based on shaft work⫽100,000 / 6.94(2545)⫽5.65

Related Calculations. Relative fuel and electrical energy costs can be obtainedfor any given locality from the utility serving the area, and with the deregulationtaking place throughout the electrical power industry at this writing, competitivecosts may be obtained from two, or more, utilities This means that even morefavorable rates, in general, will be obtained

Cost calculations may indicate that, on the basis of fuel consumption alone, theDiesel-driven heating unit would be more economical for the system above thanany other heating system, including direct heating with coal, in many localities As

a heating unit alone, however, the initial cost of the heating cycle is greater thanthat of any other system But when a substantial proportion of the capacity of thesystem may be applied to cooling in the summer, the combined heating and air-conditioning first cost should compare favorably with that of a conventional heatingplant plus air-conditioning equipment For this reason, air conditioning should pref-erably be incorporated with the heating cycle

Recently developed rare-earth additives for Diesel engines simultaneously cut

NOx emissions and eliminate more than 90 percent of particulates in the engineexhaust The system combines a cerium-based organic fluid and a conventionalceramic particulate trap Using such emission controls allows Diesel engines inheavily populated areas where they formerly might not have been welcome Thus,Diesel-engine-drive of heat pumps in a cogeneration mode could be more popular

in the future The additive and trap for Diesel engines results in NOxof 2.0 gramsper brake-horsepower-hour (2.68 g / kWh and 0.013 g / kWh), compared to 0.1 g /bhp䡠h (0.13g / kWh) for best traps as of this writing EPA has set emission limits

of 2.5 g / bhp䡠h NOx and 0.1 g / bhp䡠h, to go into effect by 2004 The rare-earthadditive will increase fuel cost by 2 to 4 percent in the U.S

For winter operation of this heat pump, a humidifier would be used Provisionwould be made for ventilation by admission of the required amount of fresh air to

be conditioned and mixed with that which is recirculated For summer use, thecondenser and evaporator in Fig 17 would be so connected that they wouldexchange functions, the summer evaporator acting as an air cooler and dehumidifier.Heat removed from the inside air would be rejected, along with the heat equivalent

of the compressor work, to the atmosphere or to the water which acts as the source

of heat for cold-weather operation, and as the cooling medium in warm weather

FIGURE 18 Conventional unitary heat-pump system heats, cools, and dehumidifies at terminal

unitary heat pumps (UHPs) (Gershon Meckler, P.E., and HPAC magazine.) See procedure for

SI values.

Unitary heat pumps (UHPs) using closed-water loops are used extensively today

in office buildings because of their flexibility As described by Gershon Meckler,

P.E., in HPAC magazine,* UHPs are small, individual air conditioners located and

controlled within each building zone, that dehumidify and cool or heat at the minal without activating a central chiller or other UHPs Best of all, they can beswitched back and forth between heating and cooling, and some can be heatingwhile others are cooling Figure 18 shows a conventional unitary heat-pump system

ter-To retain the benefits of UHPs, while making them more responsive to the newutility rate structures and more energy efficient, Gershon Meckler developed a UHP

system that reduces electric demand in several ways, amongst which are: (a) Shifts

the dehumidification load from the UHPs to a small, central ice plant that operatesoff-peak (or, in an alternative version, to an efficient two-stage desiccant system);

(b) distributes dehumidification via a small quantity of very dry ventilation air (no increase in primary air quantity over a conventional UHP system); (c) uses UHPs for sensible heating / cooling only, with no condensation at terminals; (d ) operates

UHPs in cooling mode at 55⬚F (12.8⬚C) refrigerant coil temperature (instead of at

40 to 45⬚F [4.4 to 7.2⬚C]), resulting in a 35 to 40 percent reduction in compressor

horsepower (kW); (e) deactivates the UHP compressors when winter cooling and

night heating are required

As described by Gershon Meckler, P.E., Fig 19 depicts schematically and Fig

20 charts psychrometrically the ice dehumidification / unitary heat-pump system

Desiccant enthalpy exchange wheel

Run-around coil Exhaust fan-1

FIGURE 20 Psychrometric process for office building UHP system with central plant ice

dehumidification and UHP sensible cooling (Gershon Meckler, P.E and HPAC magazine.) See

procedure for SI values.

signed for a six-story, 159,000 ft2 (14,771 m2) office building in northern NewJersey As shown, three techniques are employed in this patented system in thecentral plant to limit the ice load or system demand:

(1) Incoming outside air passes through a desiccant-impregnated enthalpy exchange

wheel that handles 30 to 50 percent of the building’s dehumidification taskwithout adding to the refrigeration / ice load The rotating wheel absorbs bothheat and moisture from the incoming air stream and transfers them to the drierexhaust air stream

(2) A run-around coil further limits the load on the ice plant by precooling /

re-heating the air entering / leaving the ice cooling coil in the primary air handler.The very dry primary / ventilation leaves the air handler air at 60⬚F (15.6⬚C) and

36 gr / lb (79.3 gr / kg) (grains of moisture per lb or kg of dry air)—dry enoughfor the 0.18 ft3/ min / ft2(0.05 m3/ m2) distributed to handle 100 percent of thebuilding’s dehumidification load

water that dehumidifies via the primary air handler cooling coil The ification or latent cooling load that is shifted off peak typically represents 20

dehumid-to 40 percent of the dehumid-total cooling load (24 percent of the Fig 18 building)

REFRIGERATION 12.27

Recent research and development on heat pumps has focused on absorption heatpumps and ground-coupled geothermal heat pumps.** Much of the stimulus forthis R & D results from the ban of CFC refrigerants to protect the ozone layer.The year 2040 is the absolute deadline for complete banning of these refrigerantsaround the world It is thought that heat-actuated absorption systems for mobileair-conditioning / heat-pump use powered by automotive waste heat could save up

to 70 percent of the energy consumed

New controls for air-, water-, or ground-source heat pumps reduce the annualenergy used by reducing the number of defrosts a system must undergo during aheating season Heat-transfer enhancements have been made on both the air andthe refrigerant sides of the heat-exchanger equipment for heat pumps Materialrequirements have been reduced by about 15 to 20 percent while maintaining thesame overall heat transfer as earlier designs.**

Absorption heat pumps and chillers are energy-conversion machines driven byheat.** Fuel cost for an absorption machine is usually lower than for an electricmachine but the first cost of an absorption machine is larger The absorption heat-pump industry is based on lithium bromide / water technology A significant number

of large tonnage absorption chillers serve building-cooling uses Today single-,

dou-ble-, and triple-effect absorption machines are in use The effect term refers to the

number of times the input heat transfer is recycled internally to produce additionalvapor and additional capacity There is a tradeoff—each effect requires more equip-ment

Absorption-cycle improvements are possible primarily through advanced-cycledesign which incorporate higher heat input temperatures.** Examples of currentdevelopments include triple-effect cycles and generator-absorber heat-exchange(GAX) cycles The GAX cycle is another method for increasing the performance

of an ammonia-water absorption cycle For low-temperature lift applications (such

as air conditioning), the properties of ammonia-water allow the absorption cycle to

be arranged so the high-temperature end of the absorber (heat-rejection device)overlaps the low-temperature end of the generator (heat-input device) Due to thistemperature overlap, a portion of the absorber heat rejection can be used to provideheat to the generator This cycle holds promise as a gas-fired residential heat pump.Ground-coupled heat-pump systems use the ground as the thermal source or sinkfor the heat-pumping process.** Such systems include those in which heating /cooling coils are placed in horizontal trenches, vertical boreholes, under a building

or parking lot, or in bodies of water, such as a pond They may also use ground water directly in what is usually called a groundwater system The primaryadvantage of this technology is that the earth provides a relatively constant tem-perature for heat transfer, improving the energy efficiency (COP) over that of con-ventional air systems and reducing electric-utility peak loads Several electricalutilities have estimated that the installation of a single residential ground-coupledsystem reduces summer peak loads by 1 to 2 kW and winter peak loads by 4 to 8

under-kW Because of the reduced peak demands, it is often more economical to age home owners and others to install heat-pump systems than to provide additionalgenerating capacity

encour-A number of advances in heat-pump and ground heat-transfer technology haveoccurred: (1) System reliability has improved with new materials and system in-stallation techniques; (2) New configurations for the in-ground heat exchanger helpreduce material requirements and system cost Today’s in-ground piping systemstypically consist of thermally fused polyethylene or polybutylene piping with anexpected lifetime of at least 50 years Installation techniques have also been im-proved, requiring less space and excavation for the in-ground piping.**

This procedure is the work of Norman R Sparks, Professor and Head, ment of Mechanical Engineering, The Pennsylvania State University, and Charles

Depart-C DiLio, Associate Professor of Mechanical Engineering, The Pennsylvania StateUniversity SI values were added by the handbook editor

Additional information in this procedure comes from *Gershon Meckler, P.E.,

‘‘Unitary Heat Pumps Plus Ice Storage,’’ HPAC magazine, August, 1989, and

**Karen Den Braven, Keith Herold, Viung Mei, Dennis O’Neal, and Steve

Pen-oncello, ‘‘Improving Heat Pumps and Air Conditioning,’’ Mechanical Engineering,

September, 1993

COMPREHENSIVE DESIGN ANALYSIS OF AN

ABSORPTION REFRIGERATING SYSTEM

An ammonia absorption refrigerating system is to have a capacity of 100 tons (90t) The cooling-water temperature is such that a condenser pressure of 170 lb / in2(abs) (1171 kPa) exists The brine temperature requires an evaporator pressure of

30 lb / in2 (abs) (206.7 kPa) Pressures throughout the system are: generator, 175

lb / in2 (abs) (1206 kPa); rectifier entrance, 174 lb / in2 (abs) (1199 kPa); rectifierexit, 171 lb / in2(abs) (1178 kPa); absorber, 29 lb / in2(abs) (199.8 kPa); aqua pumpsuction, 25 lb / in2(abs) (172.3 kPa); aqua pump discharge, 185 lb / in2(abs) (1274.7kPa) Temperatures are: generator, 220⬚F (104⬚C); vapors and drip from rectifier,

110⬚F (43⬚C); liquid from condenser and to expansion valve, 76⬚F (24⬚C); rator exit, 10⬚F (⫺12⬚C); weak aqua from exchanger and to absorber, 100⬚F (38⬚C);strong aqua from absorber, 80⬚F (27⬚C) The aqua pump handles 570 lb (259 kg)

evapo-of strong ammonia per minute Steam is supplied to the generator coils at 20 lb /

in2(abs) (137.8 kPa) dry saturated Find: (a) the heat supplied to the generator, and

the heat removed from the absorber, rectifier, and condenser per lb (kg) of ammonia

circulated; (b) the pump horsepower (kW) required per ton (t), and total, if the combined hydraulic and mechanical efficiency is 65 percent; (c) the pump piston displacement with 20 percent slip; (d ) pounds (kg) of steam required per minute per ton (t); (e) an energy balance of the system on the basis of Btu / min (kJ / minute)

per ton (t) Show an alternate arrangement that might be used for this design

Calculation Procedure:

1. Compute the absorber variables

Three equations may be written that deal with the mass flow through the absorber,

Fig 21a They are

the data given By solving two of the above equations simultaneously, either ofthese cases may be satisfied, or the following analysis may be used.

The weight ratio of ammonia to water in terms of the weight concentration X

may be expressed for the absorber

the average concentration is used

Now, for the absorber:

Enthalpy of liquid entering the expansion valve at 76⬚F⫽127.4 (296.8 kJ / kg)Enthalpy of vapor leaving evaporator at 30 lb / in2 (abs) and 10⬚F ⫽ 617.8(1439.5 kJ / kg)

Refrigerating effect per lb ammonia⫽ 617.8⫺127.4 ⫽490.4 Btu (1142.6 kJ/ kg)

Pound anhydrous ammonia per (min䡠ton)⫽200 / 490.4⫽0.408 (0.21 kg / min

䡠t)

Pound anhydrous ammonia per (min䡠ton)⫽0.408(100)⫽40.8 [20.6 kg / (min

䡠t)]

For the following, refer to Fig 21a for notation.

M B⫽40.8 lb / min, or 0.408 lb / (min䡠ton)

M C ⫽ 570 lb / min, or 5.7 lb / (min䡠ton), or 14 lb / lb (6.4 kg / kg) anhydrousammonia

M A⫽570⫺ 40.8⫽529.2 lb / min, or 5.29 lb / (min䡠ton), or 13 lb / lb (5.9 kg /kg) anhydrous ammonia

t A⫽100⬚F; h⬘A⫽155.2; hⴖA⫽ 68; t C⫽80⬚F; h⬘C ⫽132; hⴖC⫽48; h Bat 30 lb /

in2(abs), 10⬚F⫽617.8

From generator conditions (175 lb / in2(abs), 220⬚F), (1205.8 kPa, 104⬚C)

REFRIGERATION 12.31

X A⫽0.325

X C⫽0.325(13)⫹1⫽ 0.373

14Average concentration⫽0.325⫹0.373⫽0.3495

2

q ⫽345⫺ 250(0.3495)⫺ 2550(0.34953⫽ 148.8 Btu / lb (345.7 kJ / kg) drous ammonia absorbed

anhy-M⬘A⫽0.325(13)⫽4.23 lb / lb anhydrous ammonia (1.9 kg / kg)

MⴖA⫽13 ⫺4.23⫽8.77 lb / lb anhydrous ammonia (3.9 kg / kg)

M⬘C⫽4.23 ⫹1⫽5.23 lb / lb anhydrous ammonia (2.4 kg / kg)

Q R⫽ 4.23(155.2) ⫹8.77(68) ⫹617.8 ⫹ 148.8 ⫺5.23 ⫺ 8.77⫽ 908 Btu / lbanhydrous ammonia, or 908(0.408) ⫽ 370 Btu / (min䡠ton), or 908(40.8) ⫽37,000 Btu / min (39,035 kJ / min)

2. Determine the aqua pump power input

The power representing the useful work actually done on the fluid may be termedthe hydraulic hp, and the ratio of this to the power input to the pump is the combinehydraulic and mechanical efficiency The hydraulic work, determined by a steady-flow analysis applied to a fluid, the pressure of which is increased under reversibleadiabatic action, may be stated mathematically as:

integrated between the pressure limits P d and P s, the discharge and suction pressure,respectively, as follows:

In reciprocating pumps, the liquid delivered is theoretically equivalent in volume

to the piston or plunger displacement, but actually a lesser quantity is discharged.The difference between theoretical and actual volumetric discharge expressed in

terms of the piston displacement is called slip, which is thus the proportion of the

displacement that is unproductive in so far as the output of the pump is concerned.The weight of strong aqua to be pumped is given as 570 lb / min (258.8 kg / min)

X⫽0.373; P d⫽185 lb / in2(abs) (1274.7 kPa); P a⫽25 lb / in2(abs) (172 kPa)

v⬘ƒat 80⬚F⫽0.0267;v⬘ƒat 80⬚F⫽0.0161

With the usual 16 percent absorption of the ammonia liquid in these systems,

vsol⫽0.84(0.373) (0.0267)⫹ 0.627(0.0161)⫽0.0185 ft3/ lb (0.0012 m3/ kg)Hydraulic hp⫽144(570) (185⫺25) (0.0185)⫽7.37 (5.5 kW)

33,000

Hp input⫽7.37 ⫽11.33 or 0.1133 hp / ton (0.09 kW / t)

0.65Piston displacement ⫽ 570(0.0185) ⫽ 13.2 ft3/ min, or 0.132 ft3/ (min䡠ton)

0.8(0.004 m3/ t)

The work input to the aqua pump,

W / J ⫽ 11.33(42.4) ⫽ 480 Btu / min, or 4.80 Btu / (min䡠ton) [5.6 kJ / min䡠t)] or11.8 Btu / lb (27.4 kJ / kg) anhydrous ammonia

Of energy supplied to drive the pump, all but a fraction of that converted to heat

by mechanical friction is represented in the strong aqua by an increase in enthalpy.Although this increase is small per pound of solution, it becomes noticeable whenexpressed in energy units per minute or per minute per ton, or when appearing in

an energy balance for the system In the exchanger calculations which follow inthe next section, it will be noted that the enthalpy of the strong aqua leaving theabsorber has been augmented by the pump work before entering the exchanger.This is assuming that all of the energy input to the pump is taken up by the liquid

3. Calculate the exchanger heat transfer

The weight relationships for the exchanger are simple since there is no change in

concentration of either fluid The weights at A and at D are equal for both the total fluid and components, and the same relation holds for the weights at B and C, Fig 21b.

The process for the weak solution is that of simple cooling The heating of thestrong aqua, however, may be accompanied by vaporization of a small portion ofthe constituents if the temperature is carried high enough This will tend to repressthe temperature rise of the strong aqua and will make the final temperature some-what difficult to predict When this temperature is desired, it may be found by trialand error, but in problems involving energy quantities alone the temperature isunimportant so long as the energy of the solution leaving the exchanger may bedetermined for application to the generator This may be done, irrespective of anyvaporization that may occur, by adding to the energy of the strong aqua enteringthe exchanger the heat removed from the weak aqua Since in the strong aqua therelative quantities of ammonia and water have not changed during the heating, theheat of absorption need not be considered until the generator calculations are carriedout

The energy equation is

M⬘C C h⬘ ⫹MⴖC C hⴖ ⫹M⬘A A h⬘ ⫹MⴖA A h⬘ ⫽M⬘B B h⬘ ⫹MⴖB B hⴖ ⫹M⬘D D h⬘ ⫹MⴖD D hⴖ

or

M⬘D D h⬘ ⫹MⴖD D hⴖ ⫽M⬘C C h⬘ ⫹MⴖC C hⴖ ⫹M A A⬘h⬘ ⫹MⴖA A hⴖ ⫺M⬘B B h⬘ ⫺MⴖB B hⴖ

REFRIGERATION 12.33

where M h⬘ ⬘D D ⫹ M hⴖ ⴖD D may be considered as the energy of the strong solutionentering the analyzer if the differential heat of absorption is subsequently taken intoaccount The heat exchanged through the surface is written:

Q⫽M⬘C C h⬘ ⫹MⴖC C hⴖ ⫺M⬘B B h⬘ ⫺MⴖB B hⴖ ⫽M⬘CB (h⬘ ⫺C h⬘B)⫹MⴖCB (hⴖ ⫺C h Bⴖ)For every pound of anhydrous ammonia circulated in the system,

The work input to the aqua pump,

W / J (from pump computation)⫽11.8 Btu / lb (27.4 kJ / kg) anhydrous ammonia

M h⬘ ⬘A A ⫹ M hⴖ ⴖA A ⫽ 1111 ⫹ 11.8 ⫽ 1122.8 Btu / lb (2609.1 kJ / kg) anhydrousammonia

(This is determined solely for use in generator calculations.)

The heat transferred through the exchanger surface from weak to strong aqua:

Q⫽ 4.23(313⫺155.2) ⫹8.77(188.1⫺68)⫽ 1723 Btu / lb anhydrous ammonia,

or 702 Btu / (min䡠ton), or 70,200 Btu / min (74,061 kJ / kg)

4. Find the analyzer vapor properties

The analyzer is a direct-contact heat exchanger that serves to further preheat thestrong aqua before entrance to the generator by cooling the vapors leaving thegenerator and thus dehydrating them to a certain extent The analyzer may beintegral with the generator, i.e., built into the generator structure, or it may beexternal to or separate from the generator Irrespective of type, the analyzer consists

of a series of trays over which the strong aqua and the rectifier drip, introduced atthe top, flow by gravity counter to the flow of vapors In this manner, by exposing

a large solution surface to the vapors, a good transfer of heat is effected Afterpassing through the analyzer, the vapors go to the rectifier for more complete de-hydration while the strong aqua is introduced to the generator for further heating.Because the weight of strong aqua is several times that of the vapors and becausethe specific heat of the solution is more than double that of the vapors, the coolingeffect on the latter is considerably greater, measured in temperature, than is theheating effect on the solution Depending upon the efficiency of the analyzer, atemperature difference between incoming solution and outgoing vapors of from 10

to 20⬚may be had or, in cases where the exchanger is functioning very effectively,the temperature difference may be less than 10⬚ The vapors leaving will closelyapproximate equilibrium conditions for the pressure and temperature since there iscontinuous contact with the solution at nearly the same temperature The effect of

the analyzer on the vapors is to progressively increase the ammonia content andreduce the proportion of water vapor, while the concentration of the strong aqua

is, of course, also somewhat reduced at the same time

In system design, it is convenient and satisfactory to regard the generator andanalyzer as a single unit Viewed in this way, the analyzer affects the calculationsfor the generator simply by reducing the quantity of energy leaving in the vapors,decreasing by an equivalent amount the heat to be supplied

That part of this calculation procedure which concerns the analyzer alone isconfined to the determination of the properties of the vapors leaving

The temperature of the strong aqua leaving the exchanger and entering the alyzer may be estimated as 180 to 190⬚F (82 to 88⬚C) The temperature of thevapors at analyzer exit may therefore be taken as 200⬚F (93⬚C)

an-Properties of vapors:

Pressure⫽ 175 lb / in2(abs) (1205.8 kPa)

Partial pressure of water vapor⫽7.2 lb / in2(abs) (49.6 kPa)

Partial pressure of ammonia vapor⫽167.8 lb / in2(abs) (1156.1 kPa)

Temperature of both⫽200⬚F (93.3⬚C)

For the water vapor, h⫽1147.8,v⫽54.29 (126.1 kJ / kg)

For the ammonia vapor, h⫽707.1,v⫽2.423

Ammonia vapor per lb water vapor⫽54.29 / 2.423⫽22.4 lb (10.2 kg)

5. Solve for the rectifier variables

The rectifier flow diagram is shown in Fig 21c Three combined quantities are

involved: the vapors from the analyzer, the drip to the analyzer, and the vapors tothe condenser In the diagram, these quantities are shown as divided into their

respective components Thus, in the vapors entering, a represents ammonia and b water vapor; in the drip leaving, c is ammonia and d is water; and, at the vapor exit, ammonia vapor is at e and water vapor at ƒ The weights at each of these

points will be determined first

From analyzer data, the weight ratio of ammonia to water vapor in the mixture

entering the rectifier M a / M bis known The exit pressure and temperature, assuming

all of the drip to leave at this point, determine the concentration X of the solution, and the weight concentration may be expressed as M c / (M c ⫹M d) From this, the

weight ratio of ammonia to water, M c / M d, may be found by using the equation

erties to be found as well as, from the specific volumes, the weight ratio M e / Mƒ.*

The quantity of ammonia M e passing to the condenser must be the same as thatpassing through the evaporator and returning to the absorber, and it is therefore

known This permits Mƒto be immediately solved for The weight relationship forthe respective fluids are

*The volume of vapors to the condenser is equal to the volume of ammonia vapor at its partial pressure

or the volume of water vapor at its partial pressure Mathematically, V⫽M e v e⫽M f v f.

REFRIGERATION 12.35

M a⫽M c⫹ M e

for the ammonia, and

M b⫽ M d⫹ Mƒfor the water and water vapor, in which both M e and Mƒare known for any particular

case With the ratios M a / M b and M c / M dalso known, either of the foregoing tions can be written in the terms of the other, and, by solving simultaneously, anyone of the four unknown quantities may be found, with the remaining three theneasily determined

equa-The energy quantities for the rectifier may be handled in the customary manner,treating the various components individually The heat of absorption associated withthe formation of the solution must in this instance be used in its entirety becausethe solution is entirely formed within the rectifier The error involved in taking theaverage concentration over a wide range, as in this case, is relatively small and theaverage may be so taken for simplicity without vitally affecting the results Whenaqua ammonia concentrations exceed 0.45, the ammonia absorbed above this con-centration supposedly produces no heat of absorption Therefore, in dealing withthe formation of a solution from the pure constituents, the concentration cannotexceed 0.45 for that portion of the ammonia that produces heat of absorption; andfor the remainder of the ammonia, with the solution above a weight concentration

of 0.45, there will be no heat absorption It thus becomes necessary, in calculatingthe heat of absorption for a strong solution, to apply the integrated mean heat of

absorption for a concentration from zero to 0.45 to that quantity of ammonia M x

that will give a solution strength of 0.45 It is also necessary to ignore the heat ofabsorption for any additional ammonia that may go into solution above this con-centration

The rectifier energy equation is

M a h a⫹M b h b⫹m x q⫽M c h c⫹M d h d⫹M e h e⫹Mƒhƒ⫹ Q R

or

Q R⫽M a h a⫹ M b h b⫹M x q⫺M c h c⫺M d h d⫺M e h e⫺Mƒhƒ

The known data are: At rectifier exit, p⫽171 lb / in2(abs); t⫽110⬚F; X (drip)

⫽0.74, partial pressure of water vapor⫽0.18 lb / in2(abs), M e⫽40.8 lb / min; and

M a / M b⫽ 22.5, or M a⫽22.4M b

M c 0.74

⫽ ⫽2.84, or M c⫽2.84

M d 0.26Partial pressure of ammonia vapor at exit⫽170.83 lb / in2(abs) (1176.9 kPa)Specific volume of ammonia vapor at exit,v e ⫽1.879

Specific volume of water vapor at exit⫽ 1545(570) ⫽1886*

Expressing in terms of M b and M d,

22.4M b⫽2.84M d⫹40.8

M b⫽M d⫹0.0407Solving simultaneously,

M d⫽2.04 lb / min

M b ⫽ 2.04 ⫹ 0.0407 ⫽ 2.08 lb / min; M a ⫽ 1.08(22.4) ⫽ 46.6 lb / min (21.2

kg / min); M c⫽46.6⫺ 40.8⫽5.8 lb / min (2.6 kg / min)

For the energy quantities:

h a⫽707.8; h b⫽1147.8; h c⫽167; h d⫽ 77.94; h e⫽ 648.9; hƒ⫽1,108.5

q ⫽ 345 ⫺ 125(0.45) ⫺ 637.5(0.45)3 ⫽ 230.6 Btu / lb (536 kJ / kg) ammoniaabsorbed up to a concentration of 0.45

Ammonia absorbed to bring concentration to 0.45⫽0.45(2.04) / 0.55⫽1.67

lb / min (0.76 kg / min)⫽M x Remaining ammonia has no heat of absorption Q R

⫽ 46.6(707.1) ⫹ 2.08(1146.8) ⫹ 1.67(230.6) ⫺ 5.8(167) ⫺ 2.04(77.9) ⫺40.8(648.9) ⫺ 0.0407(1109.5) ⫽ 8068 Btu / min, or 80.7 Btu / (min䡠ton), or197.8 Btu / lb (460 kJ / kg) anhydrous ammonia circulated

6. Compute the generator variables

The generator is the still, or boiler, for the system It consists of a shell with suitableconnections for the introduction and withdrawal of fluids and, when in operation,

is partially filled with aqua ammonia solution Heat is usually supplied by pressure steam that is passed into coils or tubes built into the generator beneath theliquid level At the outlet, the steam coil is provided with a trap that permits thepassage of condensate only

low-As stated earlier, the analyzer will be considered in calculations as a part of thegenerator, and the two will be treated as an integral unit The strong aqua and thedrip enter the generator by way of the analyzer from the exchanger and the rectifier,respectively, and the vapors depart for the rectifier by way of the analyzer Theweak aqua is usually taken from the bottom of the generator as far as possible fromthe point where the strong aqua is introduced The generator action, like that in theabsorber but in the reverse direction, is simply a process for the restoration ofequilibrium in the solution The aqua ammonia from the exchanger, being too strongfor the generator conditions, is rapidly reduced in ammonia content until an equi-librium concentration is established This is as far as the process can be carried,and the solution is then removed to be subjected to the reverse action in the ab-sorber

From exchanger data:

M g h g⫹M h h h⫽ 116,100 Btu / min

M e h e⫽ 40.8(4.23)(313)⫽54,000 Btu / min

Mƒhƒ⫽40.8(8.77)(188.1)⫽67,300 Btu / min (71,001 kJ / min)

From rectifier data:

M a h a⫽46.6(707.1)⫽32,950 Btu / min

M x q rect⫽385 Btu / min (406.2 kJ / min)

Average generator concentration X⫽0.373⫹0.325 / 2⫽0.3495

qgen⫽345⫺250(0.3495)⫺ 2550(0.3495)3⫽ 148.8 Btu / lb (345.7 kJ / kg) hydrous ammonia circulated

an-Mq⫽385⫹40.8(148.8)⫽6455 Btu / min (6810 kJ / min)

Qs⫽32,950⫹2385⫹54,000⫹67,300⫹ 6455⫺ 116,100⫺ 968⫺159⫽45,863 Btu / min, or 458.6 Btu / (min䡠ton), or 1125 Btu / lb (2614.2 kJ / kg) an-hydrous ammonia

7. Show the use of an alternate generator-energy-balance solution

Figure 22 shows the analysis for the first law of thermodynamics applied to theseveral components of the absorption system When energy quantities have beenestablished for the rectifier, absorber, and aqua pump, the steady-flow analysis forthe portion of the system bounded by the imaginary control surface of Fig 22materially simplifies the calculation of the heat quantity to be transferred at thegenerator Referring to Fig 22 and the values previously established, the generator

energy balance is as shown in the table below (Note: The water vapor into the

condenser and out of the evaporator has not been included because of its icance.)

insignif-Energy into system, Btu / min Energy out of system, Btu / min 40.8 (617.8) ⫽ 25,206

(26,592 kJ / min)

40.8 (648.9) ⫽ 26,475 (27,931

kJ / min) Pump work ⫽ 480 (506.4 kJ

/ min)

Qrectifier⫽ 8,068 (8512 kJ/min)

min) Sum of energy into system ⫽

The energy balance then consists simply of a balance between the summation

of the energies into and out of the system from or to an outside source The balancemay be made on any basis desired In the present case, the basis is to be Btu /minute per ton capacity

Solution (Condenser, and system energy balance.)

Condenser:

For the ammonia entering, h (from rectifier)⫽648.9

For the ammonia leaving, hƒat 76⬚F⫽127.4

For the water vapor entering, h (from rectifier)⫽1,109.5

For the water leaving, hƒat 76⬚F⫽44

Q R⫽ 40.8(648.9 ⫺127.4)⫹ (0.0407(1109.5⫺44) ⫽21,313 Btu / min, or 213.1Btu / (min䡠ton) [249.8 kJ / min䡠t)]

REFRIGERATION 12.39

Energy balance:

Part of system

Energy into system

Btu / (min 䡠 ton) [kJ / (mint)] 䡠

Energy out of system

Btu / (min 䡠 ton) (kJ / (mint)] 䡠

Related Calculations. Reviewing the ammonia absorption system in the light

of this calculation procedure, it may be seen that this type of machine requiresapproximately seven times the energy input* for a given refrigerating capacity thatthe compression machine needs The quantity of energy alone, however, is hardly

a reasonable criterion for comparison because the compression machine consumeshigh-grade energy and the absorption system may utilize almost entirely relativelylow-grade energy It would obviously be uneconomical to operate an absorptionmachine on an input of electrical energy, but it might prove the more economicalsystem when the energy is obtainable from a low-cost source such as low-pressureexhaust or process steam, or cheap fuel

The cooling-water load, measured in heat units to be disposed of, is about 21⁄2times as heavy in the absorption as in the compression machine But because ofthe possibility of using the water through a considerably greater temperature range

in the absorption system, there is not such a disparity in actual cooling-water sumption, although this will always be heavier than for a compression system undercomparable conditions

con-Concerning the actual operation, the absorption machine undoubtedly demandsmore careful attention than the compression system This applies to the regulation

of load and to the general care required in more frequent purging, in the prevention

of corrosion, and in checking solution strength and vapor rectification

Absorption refrigerating systems, once thought to be on a declining usage curve,are returning with great vigor Applications include gas-turbine intake-air cooling,Fig 23, and combined-cycle cogeneration, Fig 24 In the plant shown in Fig 24,

a gas turbine provides electricity, steam, and chilled water for a health center inCanada The 1500-ton (1350-t) absorption chiller keeps the gas-turbine intake air

at the best temperature for this turbine, namely 50⬚F (10⬚C) The absorption chiller

is powered by 15-lb / in2(gage) (103.4-kPa) steam extracted from the steam turbine

in the cycle This unit was reported to be the world’s first gas turbine to exceed 40percent thermal efficiency It is part of a natural-gas fired combined-cycle cogen-eration plant, a type which is becoming more popular every year

Another cogeneration plant, Fig 25, provides steam generated by gas-turbineexhaust gas to power an ammonia absorption refrigerating system producing 550

*Absorption refrigeration systems are often compared on the basis of the ratio of refrigerating effect to the heat input.

Air cooling coil

Chilled water pump Combustion air

Gas turbine Heat

Head tank

Fuel

Cooling

tower

FIGURE 23 Integrated plant system with absorption-refrigerating-system

chiller gas-turbine intake air (Power.)

tons (495 t) of ice per day Using this gas-turbine-exhaust generated steam allowsthe shutdown of electrically driven 300-ton (270 t) reciprocating refrigeration com-pressors which are expensive to operate during summer months when ice-makingdemand peaks This cogeneration plant, as the one in Fig 24, uses an HRSG torecover heat from the gas turbine exhaust Electricity from both these plants is sold

to local utilities Inlet air for the cogeneration plant in Fig 25 is cooled to within

2⬚F (1.1⬚C) of the prevailing wet-bulb temperature

Lithium-bromide absorption refrigeration systems are popular for chilled-waterservice in large air-conditioning units Developed by Carrier Corporation, theseunits find wide use in large office, residential, and commercial buildings Advan-tages of absorption refrigerating systems ass detailed by Carrier Corporation includecompactness, vibrationless operation, ease of installation anywhere in a building,from the basement to the roof, where space and a heat source are available Suchmachines use the cheapest, safest, and most reliable refrigerant available, ordinarytap water

Favorable situations outlined by Carrier Corporation for lithium-bromide sorption systems include: (1) where low-cost fuel is available; (2) where electricrates are high; (3) where steam or gas utilities desire to promote summer loads; (4)where low-pressure heating boiler capacity is largely, or wholly, unused during thecooling season; (5) where waste steam is available; (6) where there is a lack ofadequate electric facilities for installing a conventional compression machine; theabsorption machine uses only 2 to 9 percent of the electric power required bycompression equipment Absorption machines can be used with gas or diesel en-gines, gas and steam turbines

ab-Ammonia absorption refrigerating systems are popular for large and heavy dustrial applications in a variety of industries, ice-making, oil refining, chemicalmanufacturing, food chilling and storage, etc While the gas does have some neg-ative characteristics, it has a zero ozone depletion potential Handled in industrialsituations, ammonia has proven to be relatively safe and its cost is considerablyless than the conventional Freon refrigerants

in-This procedure is the work of Norman R Sparks, Professor and Head, ment of Mechanical Engineering, The Pennsylvania State University SI values wereadded by the handbook editor Illustrations of cogeneration applications came from