Chapter 19 Vapour Compression and Vapour AbsorptionSystems – Thermal Engineering

Chapter 19

Vapour Compression and Vapour Absorption Systems

19.1 ❐ INTRODUCTION

The vapour compression refrigeration system is based upon the fact that fluids absorb heat while changing from a liquid phase to vapour phase and reject heat while changing from a vapour phase to a liquid phase. The temperature during the phase change remains constant. However, the temperature varies with the pressure and the fluid. In vapour compression system, the refrigerant condenses and evaporates at temperatures and pressures close to the atmospheric conditions. The closed refrigeration cycle is used in which refrigerant does not leave the system. The refrigerant alternately condenses and evaporates. During evaporating the refrigerant absorbs its latent heat from the brine (salt water), which is used for circulating it around the cold chamber. While condensing, it gives out its latent heat to the circulating water of the cooler. The vapour compression system is used in a small domestic refrigerator to big industrial refrigeration systems.

19.2 ❐ COMPARISON OF VAPOUR COMPRESSION SYSTEM WITH AIR REFRIGERATION SYSTEM

Following are the advantages and disadvantages of the vapour compression refrigeration system over air refrigeration system:

Advantages

  1. It has smaller size for the given capacity of refrigeration.
  2. It has less running cost.
  3. It can be employed over a large range of temperatures.
  4. The coefficient of performance is quite high.

Disadvantages

  1. The initial cost is high.
  2. The prevention of leakage of the refrigerant is the major problem in vapour compression system.
19.3 ❐ SIMPLE VAPOUR COMPRESSION REFRIGERATION SYSTEM

The schematic diagram of a simple vapour compression refrigeration system is shown in Fig. 19.1. It consists of the following parts:

  1. Compressor.
  2. Condenser
  3. Throttle valve
  4. Evaporators.
  1. Compressor: The low pressure and temperature vapour refrigerant from the evaporator is drawn into the compressor through the suction valve. It is compressed to a high pressure and temperatures in the compressor. The high pressure and temperature vapours are discharged into the condenser through the delivery valve.
  2. Condenser: The condenser or cooler consists of coils of pipes in which the high pressure and temperature vapour refrigerant is cooled and condensed. The refrigerant, while passing through the condenser coils gives up its latent heat to the condensing medium. The condensing medium is normally air or water.
  3. Throttle Valve: Its function is to allow the liquid refrigerant under high pressure and temperature to pass at a controlled rate after reducing its pressure and temperature. Some of the liquid refrigerant evaporates. The enthalpy of the refrigerant remains unaltered in passing through the throttle valve.
  4. Evaporator: It consists of coils of pipe in which the liquid-vapour refrigerant at low pressure and temperature is evaporated and converted into vapour refrigerant at low pressure and temperature. During evaporation, the liquid-vapour refrigerant absorbs its latent heat of vaporisation from the medium (air, water or brine), which is to be cooled.

Figure 19.1 Schematic diagram of simple vapour compression refrigeration system

19.4 ❐ VAPOUR COMPRESSION REFRIGERATION SYSTEM

A schematic diagram of a vapour compression refrigeration system is shown in Fig. 19.2. It consists of a compressor, condenser, expansion device for throttling and an evaporator. In plants with a large amount of refrigerant charge, a receiver and a drier is installed in the liquid line. The p-v, T-s, and p-h diagrams for the system are shown in Fig. 19.3. The sequence of operations is as given below:

  1. The process 1-2 is isentropic compression.

    s2 = s1, Q = 0. Work done by the compressor per kg of refrigerant

    W1-2 = −∫ v dp = −∫ dh = −(h2 h1)

  2. The process 2-3 is constant pressure condensation.

    Heat rejected, Qr = h2h3

  3. The process 3-4 is isenthalpic expansion.

    h3 = h4 = hf4 + x(h1 hf4)

    or Dryness fraction,

  4. The process 4-1 is constant pressure evaporation. Refrigeration effect or Heat absorbed,

    Qa = h1 h4

    Figure 19.2 Vapour compression system

Figure 19.3 Vapour compression cycle: (a) p-v diagram, (b) T-s diagram, (c) p-h diagram

  1. From Fig. 19.3(b), we have for vapours dry and saturated at the beginning of compression and superheated at the end of compression.

    Refrigerating effect, Qa = area 1–4−6−5−1

    Heat rejected, Qr = area 2−2′−3−7−5−2

    Work done, W = Qr Qa

    = area 1−2−2′−3−7−6−4−1
    Area 4−4′−7−6−4 = area 3−0−4′−3
  2. When vapours are dry and saturated at the end of compression, as shown in Fig. 19.4.
    Work done = area 1−2−3−0−1 = Heat at 2 − Heat at 1 = h2h1
    Refrigeration effect = area 1−4−6−5 = Heat at 1 − Heat at 4 = h1h4
    Area 3−0−4′−3 = area 4−4′−7−6−4

    ∴ Work done, W = area 1−2−3−7−6−4−1

    Dryness fraction at suction,

Figure 19.4 Vapour compression cycle: (a) T-s diagram, (b) p-h diagram

The COP of vapour compression cycle is lower than that of reversed Carnot cycle. Power per ton of refrigeration,

where v1 = specific volume of the refrigerant at the inlet of compressor

D = dia of compressor piston or cylinder

L = stroke length

N = rpm

n = number of cylinders in compressor

ηv = volumetric efficiency

c = clearance ratio

Mass flow rate of cooling medium in the condenser:

19.5 ❐ USE OF T-s AND p-h CHARTS

The T-s (Temperature-entropy) diagram for the vapour compression cycle is shown in Fig. 19.5.

Figure 19.5 T-s diagram for vapour compression cycle

Figure 19.6 p-h diagram for vapour compression cycle

Figure 19.7 Pressure-enthalpy (p-h) chart

Process 1-2: Isentropic compression. State 1 represents saturated vapour and state 2 is super heated vapour.

Process 2-3: Condensation. State 2 represents saturated vapour and state 3 is saturated liquid. Heat is rejected to the condensing medium.

Process 3-4: Throttling process. The enthalpy of saturated liquid at state 3 is equal to the enthalpy at state 4 or 1.

Process 4-1: Evaporation. Heat is absorbed from the medium and refrigeration effect is produced.

The p-h (pressure-enthalpy) diagram is shown in Fig. 19.6. The various processes are also shown in this figure. The p-h chart is shown in Fig. 19.7. It is the most convenient chart for studying the behaviour of a refrigerant. The vertical ordinates represent pressure and horizontal ordinates represent enthalpy. The saturated liquid line and the saturated vapour line merge into one another at the critical point. The space to the left of the saturated liquid line is the sub-cooled liquid region. The space between the liquid and vapour lines is called the wet region. The space to the right of the saturated vapour line is the superheated vapour region.

Example 19.1

An R-12 vapour compression system is operating at a condenser temperature of 40°C and an evaporator temperature of −5°C. Calculate the Carnot COP and actual COP of the cycle.

Solution

From p-h diagram of R-12, we have

h1 = 185.4 kJ/kg, h2 = 208.0 kJ/kg, h3 = h4 = 74.6 kJ/kg

Carnot

Actual

Example 19.2

An ammonia refrigerating machine has working temperature of 35°C in the condenser and −15°C in the evaporator. Calculate the COP for (a) dry Compression, and (b) wet compression. The data for ammonia is given below:

Solution

  1. By interpolation for entropy,

    Degree of superheat at discharge = 50 + 27.1 = 77.1°C

    Discharge temperature = 35 + 77.1 − 112.1°C

    Refrigerating effect, Qa = h1 h4 = 1426 − 347.5 = 1078.5kJ/kg

    Work done, W = h2 h1 = 1685.4 −1426 = 259.4 kJ/kg

  2. h1 = 112.3 + 0.88(1426 −112.3) = 1268.4 kJ/kg
    v1 = vf1 + x1 (vg1vf1)
    = 0.00152 + 0.88(0.509 − 0.00152) = 0.448m3/kg
    Qa = h1 h4 = 1268.4 − 347.5 = 920.9 kJ/kg
    W = h2 h1 = 1471−1268.4 = 202.6 kJ/kg

Example 19.3

A refrigerating plant using CO2 as refrigerant works between 25°C and −5°C. The dryness fraction of CO2 is 0.6 at the entry of compressor. Find the ice formed per day if the relative efficiency is 50%. Ice is formed at 0°C from water at 10° C. The quantity of CO2 circulated is 6 kg/min. For water cp = 4.187 kJ/kg, L = 335 kJ/kg.

Properties of CO2

Temperature °C hf kJ/kg hfg kJ/kg sf kJ/kg.K
25 81.25 121.5 0.2513
−5 −7.53 254.8 −0.04187

Solution

The T-s cycle is shown in Fig. 19.8.

T1 = 273 − 5 = 268K,
T2 = 273+ 25 = 298K
h1 = hf1 + x1hfg1 = −7.53 + 0.6 245.8
= 140 kJ/kg
s1 = s2
x2 = 0.63
h2 = hf2 + x2hfg2 = 81.25 + 0.63 × 121.5 = 157kJ/kg

Work done, w1−2 = h2h1 = 157−140 = 17kJ/kg

Figure 19.8 T-s diagram

Refrigerating effect, qa = h1h4 = h1h3 = h1hf3

Actual COP = 3.45 × 0.5 = 1.725

Work done per second =

Actual cooling effect = 1.725 × 1.7 = 2.94 kJ/s

Heat carried to form 1 kg of ice

= cp (t2t1) + L = 4.187 ×10 + 335 = 376.87 kJ

Ice formed per hour =

Example 19.4

A refrigerator operating on a standard vapour compression cycle has a coefficient of performance of 6.5 and is driven by a 50 kW compressor. The enthalpies of saturated liquid and saturated vapour refrigerant at the operating condensing temperature of 35°C are 69.55 kJ/kg and 201.45kJ/kg respectively. The saturated refrigerant vapour leaving the evaporator has an enthalpy of 187.53kJ/kg. Find the refrigerant temperature at the compressor discharge. The cp of refrigerant vapour may be taken to be 0.6155kJ/kg.K.

Solution

COP = 6.5, Compressor power = 50 kW

Refrigerating capacity = 50 × 6.5 = 325 kW

Heat extracted per kg of refrigerant = 187.53 − 69.55 = 117.98 kJ/kg

Enthalpy of vapour after compression = 187.53 + 18.15 = 205.68 kJ/kg

Superheat = 205.68 − 201.45 = 4.23 kJ/kg

Temperature of refrigerant at compressor discharge = 35° + 6.87° = 41.87°C

Example 19.5

A refrigeration system of 15 tons capacity operates on standard simple vapour compression cycle using Refrigerant -22 to an evaporator temperature of 5°C and condensing temperature 50°C. Draw the p-h diagram for the cycle. Calculate (a) the refrigerant mass flow rate, and (b)the compressor intake volume flow rate if the compressor volumetric efficiency is 0.72. Use the refrigerant property data given in Table below.

Figure 19.9 (a) Simple vapour compression system, (b) p-h diagram

Solution

Refer to Fig. 19.9.

  1. Capacity of plant
    h1 = 407.1 kJ/kg,
    h4 = h3 − 263.3kJ/kg
  2. Volume flow rate

Example 19.6

A refrigeration cycle uses Freon −12 as the working fluid. The temperature of the refrigerant in the evaporator is −10°C. The condensing temperature is 40°C. The cooling load is 150 W and the volumetric efficiency of the compressor is 80%. The speed of compressor is 720 rpm. Calculate the mass flow rate of the refrigerant and the displacement volume of the compressor.

Properties of Freon − 12

Solution

Cooling load Q = 150 W

 

Refrigeration effect = h1h4 (Fig.19.10)
= 183 − 74.5
= 108.5kj/kg
(h1h4) = Q

Figure 19.10 p-h diagram

Mass flow rate of refrigerant,

Displacement volume of compressor,

Example 19.7

In a simple vapour compression cycle, following are the properties of the refrigerant R-12 at various points.

Compressor inlet: h1 = 183.2 kJ/kg, v1 = 0.0767 m3/kg
Compressor discharge: h2 = 222.6 kJ/kg, v2 = 0.0164 m3/kg
Condenser exit: h3 = 84.9 kJ/kg, v3 = 0.0083 m3/kg

The piston displacement volume for the compressor is 1.5 litres per stroke and its volumetric efficiency is 80%. The speed of the compressor is 1600 rpm.

Find (a) the power rating of the compressor (kW), (b) the refrigerating effect (kW).

Solution

Piston displacement volume (Fig. 19.11),

Figure 19.11 p-h diagram

Compressor discharge,

Mass flow rate of compressor,

= 39.12kg/min

Power of compressor = (h2h1)

Refrigerating effect = (h1h4) = 39.12(183.2 − 84.9) = 3845.5 kJ/min or 64.1kW

Example 19.8

In a standard vapour compression refrigeration cycle operating between an evaporator temperature of −10°C and condenser temperature of 40°C, the enthalpy of the refrigerant, Freon-12, at the end of compression is 220 kJ/kg. Show the cycle diagram on T-s plane.

Calculate (a) the COP of the cycle, (b) the refrigerating capacity and the compressor power assuming a refrigerant flow rate of 1 kg/min. You may use the extract of the Freon-12 property table given below:

t(°C) p(MPa) hf(kJ/kg) hg(kJ/kg)
−10 0.2191 26.85 183.1
40 0.9607 74.53 203.1

Solution

The T-s diagram is shown in Fig. 19.12

Refrigerating capacity

= (h1h4)
= 1(183.1 − 26.85)
= 156.25 kJ/min or 2.6 kW
Compressor power = (h2h1)
= 1 (220 − 183.1)
= 36.9 kJ/min or 0.615 kW

Figure 19.12 T-s diagram

Example 19.9

An R-717 (ammonia) system operates on the basic vapour compression refrigeration cycle. The evaporator and condenser pressures are 0.119 MPa and 1.389 MPa respectively. The mass flow rate of refrigerant is 0.1 kg/s. The volumetric efficiency of the compressor is 84%. Determine the compressor displacement rate. If the COP of the cycle is 2, determine the power input to the compressor.

Saturation Properties of R-717 (ammonia)

Solution

Compressor displacement (Fig. 19.13)

Refrigerating effect = h1h4

= 1423.6 − 371.4
= 1052.2 kJ/kg

Figure 19.13 T-s diagram

Power input to compressor = 526.1 × 0.1 = 52.61 kW

Example 19.10

In a 5 kW cooling capacity refrigeration system operating on a simple vapour compression cycle, the refrigerant enters the evaporator with an enthalpy of 75 kJ/kg and leaves with an enthalpy of 183 kJ/kg. The enthalpy of the refrigerant after compression is 210 kJ/kg. Show the cycle on T-s and p-h diagrams. Calculate the following: (a) COP, (b) power input to compressor, and (c) rate of heat transfer at the condenser.

Solution

Refer to Fig. 19.14.

Mass flow rate of refrigerant,

Power input to compressor,

= (h2h1)= 0.046 (210 − 183) = 1.24 kW

Rate of heat transfer to condenser,

 

Qr = (h2h4)
h4 = h3
Qr = 0.046 (210 − 75) = 6.21 kW

Figure 19.14 (a) T-s diagram, (b) p-h diagram

Example 19.11

An ice making plant using refrigerant R-12 is having an evaporator saturation temperature of −25°C and the condenser saturation temperature of 35°C. The vapour is leaving the compressor at 65°C. Following table shows the properties of the refrigerant:

Temperature Pressure Saturation enthalpy, kJ/kg
°C kPa Liquid Vapour
−25 123.7 13.3 176.5
35 850.0 69.6 201.5

Enthalpy of superheated refrigerant at 850 kPa and 65°C = 225.5 kJ/kg

  1. Calculate the coefficient of performance (COP) of this system.
  2. If the capacity of the plant is 5 kW, calculate mass flow rate of refrigerant and power consumption.

Solution

Refer to Fig. 19.15.

  1. (h1h1) = 5

Power consumption = (h2h1)

= 0.0467 × 49 = 2.288 kW

Figure 19.15 p-h diagram

19.6 ❐ EFFECT OF SUCTION PRESSURE

The suction or evaporator pressure decreases due to the frictional resistance to the flow of refrigerant. Let the suction pressure ps decrease to ps′ as shown in p-h diagram of Fig. 19.16. The effect of decrease in suction pressure are:

  1. The refrigerating effect decreases from (h1h4 ) to (h1′h4′)
  2. The compression work increases from (h2h1) to (h2′h1′ )

Therefore, the COP decreases for the same amount of refrigerant flow. Hence, the refrigerating capacity of the system decreases and the refrigeration cost increases.

Figure 19.16 p-h diagram

19.7 ❐ EFFECT OF DISCHARGE PRESSURE

The discharge (or condenser) pressure pd increases to pd’ due to frictional resistance to flow or refrigerant. The effects of increase in discharge pressure are (Fig. 19.17):

  1. The refrigerating effect decreases from (h1h4) to (h1h4’).
  2. The compression work increases from (h2h1) to (h2h1).

Figure 19.17 p-h diagram

Thus the COP decreases. The effect of increase in discharge pressure is similar to that of decrease in suction pressure.

19.8 ❐ EFFECT OF SUPERHEATING OF REFRIGERANT VAPOUR

The vapour compression refrigeration cycle with superheated vapour before compression on T-s and p-h diagrams are shown in Fig. 19.18. The evaporation starts at state 4 and continues upto state 1′, when it is dry and saturated. The vapours are now superheated from state 1′ to state 1, where they enter the compressor. Its effect is to increase the COP.

Figure 19.18 Superheated vapour compression cycle: (a) T-s diagram, (b) p-h diagram

19.8.1 Superheat Horn

Consider the T-s diagram for a vapour compression cycle shown in Fig 19.19. The vapours of refrigerant are wet at state 1′ and dry saturated at state 2′. The process 1′−2′ represents wet compression. The vapours at state 1 are dry saturated and at state 2 superheated. The process 1−2 represents dry compression. The increased work of the cycle due to the replacement of wet compression by dry compression appears as the area 2−2′−2′′, generally known as superheat horn.

Figure 19.19 Wet and dry compression processes

19.9 ❐ EFFECT OF SUBCOOLING (OR UNDERCOOLING) OF REFRIGERANT VAPOUR

Consider the T-s and p-h diagrams for the vapour compression cycle shown in Fig. 19.20 in which the refrigerant after condensation process 2′−3′, is cooled below the saturation temperature T3before throttling process to temperature T3. Such a process is called undercooling or subcooling of the refrigerant. It is generally done along the saturated liquid line. The effect of undercooling is to increase the COP.

Refrigerating effect, RE = h1 h4 = h1 hf3, where hf3 = hf3′cpf(T3′ T3)

Work done, w = h2h1

Figure 19.20 Vapour compression cycle with subcooling of refrigerant: (a) T-s diagram, (b) p-h diagram

Example 19.12

A food storage locker requires a refrigeration system of 42 kW capacity at an evaporator temperature of −5°C and a condenser temperature of 40°C. The refrigerant, R-12, is sub-cooled 5°C before entering the expansion valve, and the vapour is superheated 6°C before leaving the evaporator coil. The compression of the refrigerant is reversible adiabatic. A two-cylinder vertical single acting compressor with stroke equal to 1.5 times the bore is to be used operating at 960 rpm. Determine (a) the refrigerating effect/kg, (b) the mass of refrigerant to be circulated per minute, (c) the theoretical piston displacement per minute, (d) the theoretical power, (e) the co-efficient of performance, (f) the heat removed through condenser/kg and (g) the theoretical bore and stroke of compressor.

Data for R−12 refrigerant is given below:

Solution

Given: Q = 42 kW, T1 = −5°C or 273−5= 268 K, T2′ = 40°C or 273 + 40 = 318K, T3′T3 = 5°C, T1T1′ = 6°C, n = 2, L = 1.5D, N = 960 rpm.

The T-s diagram is shown in Fig. 19.21.

Figure 19.21 T-s diagram

Process l′−1: superheating

h1 = hg1′ + cpgl′(T1T1′) = 350.44 + 0.6455 × 6
= 354.313 kJ/kg
= 1.5778 kJ/kg.K

Process 1−2: Isentropic compression

T2 = 326.38 K
h2 = hg2′ + cpg2′(T2T2′)
= 368.81 + 0.6455 (326.38 − 313)
= 379.00 kJ/kg

Process 3′−3: Subcooling

 

h2 = hg3′ + cpg3′(T3T3′) = 239.03 − 1.030 × 5 = 233.88kj/kg

Process 34: Throttling

h4 = h3 =233.88

Specific volume at suction to compressor

  1. Refrigeration effect/kg, q4−1 = h1h4 = 354.313 −233.88 = 120.433kJ/kg
  2. Mass of refrigerant required/min,
  3. Theoretical suction volume/min = × v1 = 20.924 × 0.06706 = 1.403 m3/min
  4. Power,
  5. Heat removed through the condenser/kg = h2 h3 = 379.00 − 233.88 = 145.12 kJ/kg
  6. Theoretical suction volume per cylinder per minute
19.10 ❐ VAPOUR ABSORPTION SYSTEM

The vapour absorption system uses heat energy instead of mechanical energy in order to change the conditions of the refrigerant required for the operation of the refrigeration cycle. In the vapour absorption system, the compressor is replaced by an absorber, a pump, a generator and a pressure reducing valve. The vapour refrigerant from the evaporator is drawn into the absorber where it is absorbed by the weak solution of the refrigerant forming a strong solution. This strong solution is pumped to the generator where it is heated by some external source like steam or heating oil. During the heating process, the vapour refrigerant is driven off by the solution and enters into the condenser where it is liquefied. The liquid refrigerant then flows into the evaporator, thereby completing the cycle. A line diagram of vapour absorption refrigeration system is shown in Fig. 19.22.

Figure 19.22 Vapour absorption refrigeration system

19.11 ❐ WORKING PRINCIPLE OF VAPOUR ABSORPTION REFRIGERATION SYSTEM

The schematic diagram of vapour absorption system is shown in Fig. 19.23. It consists of

  1. an absorber,
  2. a pump,
  3. a generator and
  4. a pressure reducing valve

    The function of the pressure reducing valve is to replace the compressor of vapour compression system. The other components of the system are

  5. condenser,
  6. receiver,
  7. expansion valve and
  8. evaporator as in the vapour compression system.

Figure 19.23 Schematic diagram for vapour absorption system

19.11.1 Working

In the vapour absorption system, the low pressure ammonia vapour leaving the evaporator, enters the absorber where it is absorbed by the cold water in the absorber. The solution of ammonia vapour in water is called aqua-ammonia. The absorption of ammonia vapour in water lowers the pressure in the absorber which in turn draws more ammonia vapour from the evaporator. As a result of this, the temperature of solution rises. Some cooling arrangement (usually water cooling) is employed in the absorber to remove the heat of solution evolved there. This increases the absorption capacity of water. The strong solution thus formed in the absorber is pumped to the generator by the liquid pump.

The strong ammonia solution in the generator is heated by some external source such as hot gas, steam or heating oil. During the heating process, the ammonia vapour is driven off the solution at high pressure leaving behind the hot weak solution of ammonia in the generator. This weak ammonia solution flows back to the absorber at low pressure after passing through the pressure reducing valve. The high pressure ammonia vapour from the generator is condensed in the condenser to a high pressure liquid ammonia. This liquid ammonia is passed to the expansion valve through the receiver and then to the evaporator. This completes the vapour absorption cycle.

19.12 ❐ ADVANTAGES OF VAPOUR ABSORPTION SYSTEM OVER VAPOUR COMPRESSION SYSTEM

The advantages of vapour absorption system over vapour compression system are listed in Table 19.1.

 

Table 19.1 Comparison of vapour absorption and vapour compression systems

19.13 ❐ COEFFICIENT OF PERFORMANCE OF AN IDEAL VAPOUR ABSORPTION SYSTEM

The ideal vapour absorption refrigeration system is shown in Fig. 19.22

Let Qg = heat given to the refrigerant in the generator.

Qrc = heat rejected from the condenses to the atmosphere.

Qra = heat rejected from the absorber to the atmosphere.

Qe = heat absorbed by the refrigerant in the evaporator.

Qp = heat added to the refrigerant due to pump work.

Qr = Qrc + Qra

= total heat rejected from the condenser and absorber to atmosphere

For heat balance, neglecting Qp according to the first law of thermodynamics, we have

 

Qr = Qrc + Qra = Qg + Qc

Let Tg = temperature at which heat is given to the generator.

Tr = temperature at which heat is rejected to atmosphere or cooling water from the condenser and absorber.

Te = temperature at which heat is absorbed in the evaporator.

The vapour absorption system can be considered as a perfectly reversible system. Therefore, the initial entropy of the system must be equal to the entropy of the system after the change in its conditions.

Maximum coefficient of performance,

Neglecting pump work,

If Ta = temperature at which heat is rejected from the absorber

Tc = temperature at which heat is rejected from the condenser.

Example 19.13

A geothermal well at 130°C supplies heat at a rate of 100,500 kJ/h to an absorption refrigeration system. The environment is at 30°C and the refrigerated space is maintained at −22°C. Determine the maximum possible heat removal from the refrigerated space.

Solution

Given: Tr = 273 + 35 = 308 K, Tg = 273 + 110 = 383 K, Te = 273 − 5 = 268 K

Example 19.14

In an absorption type refrigerator, the heat is supplied to ammonia generator by condensing steam at 2.5 bar and dryness fraction 0.9. The temperature in the refrigerator is to be maintained at −5°C. Calculate the maximum COP possible.

If the refrigeration load is 20 tonnes and actual COP is 75% of the maximum COP, find the mass of steam required per hour. Take ambient temperature as 30°C.

Solution

Given: p = 2.5 bar, x = 0.9, Te = 273 − 5 = 268 K, Q = 20TR, (COP)max = 0.75(COP)max, Tr = 273 + 30 = 303 K, From steam tables hf = 535.34 kJ/kg. hg = 2716.9 kJ/kg.

Saturation temperature of steam at p = 2.5 bar, Ts = 273 + 127.43 = 400.43 K = Tg

(COP)act = 0.75 × 1.863 = 1.397

Actual heat supplied =

Latent heat of steam, hfg = hghf = 2716.9 − 535.35 = 2181.56 kJ/kg

Mass of steam required per hour,

19.14 ❐ AMMONIA-WATER (OR PRACTICAL) VAPOUR ABSORPTION SYSTEM (nh3 – h2o)

In the NH3 − H2O vapour absorption system, ammonia is the refrigerant and water is the absorber. An analyser, a rectifier and two heat exchangers are added to the simple vapour absorption system. The complete schematic diagram of ammonia-water absorption system is shown in Fig. 19.24.

The functions of various accessories are:

  1. Analyser: In the simple vapour absorption system, some water is also varporised along with ammonia and flows into the condenser. These water vapours further move and enter into the expansion valve, where they freeze and choke the pipeline. The analyser is used to remove these water vapours. The analyser may be an integral part of the generator or as a separate piece of equipment. It consists of a series of trays mounted above the generator. The strong solution from the absorber and the aqua from the rectifier (or dehydrator) are introduced at the top of the analyser and flows downward on the trays and into the generator. In this way, considerable liquid surface area is exposed to the vapour rising from the generator. The vapours are cooled and most of the water vapours condense, so that mainly ammonia vapours leave the top of the analyser. This also reduces the external heat required in the generator since aqua is heated by the vapour.
  2. Rectifier or Dehydrators: The function of the rectifier is to completely remove the water vapours still left in the analyser. The rectifier is a closed type vapour cooler of the double pipe type, shell and coil, or shell and tube type. Its function is to further cool the ammonia vapours leaving the analyser so that the remaining water vapours are condensed. The condensate from the rectifier is returned to the top of the analyser by a drip return pipe.
  3. Heat exchangers: The heat exchanger-1 provided between the pump and the generator is used to cool the weak hot solution returning from the generator to the absorber. The heat removed from the weak solution raises the temperature of the strong solution leaving the pump and going to analyser and generator. This operation reduces the heat supplied to the generator and the amount of cooling required for the absorber.

    Figure 19.24 Ammonia-water vapour absorption system

    The heat exchanger-2 provided between the condenser and the evaporator may be called liquid sub-cooler. The liquid refrigerant leaving the condenser is sub cooled by the low temperature ammonia vapour from the evaporator. This sub-cooled liquid is now passed to the expansion valve and then to the evaporator.

    Net refrigerating effect, RE = Heat absorbed in the evaporator

    Total energy supplied to the system,

    Wt = Work done by the pump + Heat supplied in generator
19.15 ❐ LITHIUM BROMIDE-WATER VAPOUR ABSORPTION SYSTEM (LiBr-H2O)

In the LiBr-water system, water is the refrigerant and LiBr the absorber. The lithium bromide solution has a strong affinity for water vapour because of its very low vapour pressure. The lithium bromide-water vapour absorption system is shown in Fig. 19.25. The evaporator and absorber are placed in one shell which operates at the same low pressure of the system. The generator and condenser are placed in another shell which operates at the same high pressure of the system.

Figure 19.25 Lithium bromide-water vapour absorption system

19.15.1 Working Principle

The water for process requirements is chilled as it is pumped through the chilled-water tubes in the evaporator by giving up heat to the refrigerant water sprayed over the tubes. Since the pressure inside the evaporator is maintained very low, therefore, the refrigerant water evaporates. The water vapours thus formed are absorbed by the strong lithium bromide solution which is sprayed in the absorber. In absorbing the water vapour, the lithium bromide solution helps in maintaining very low pressure needed in the evaporator, and the solution becomes weak. This weak solution is pumped by a pump to the generator where it is heated up by using steam in the heating coils. A portion of water is evaporated by the heat and the solution becomes more strong. This strong solution is passed through the heat exchanger and then sprayed in the absorber. The weak solution of lithium bromide from the absorber to the generator is also passed through the heat exchanger. This weak solution gets heat from the strong solution in the heat exchanger, thus reducing the quantity of steam required to heat the weak solution in the generator.

The refrigerant water vapours formed in the generated due to healing of solution are passed to the condenser where they are cooled and condensed by the cooling water flowing through the condenser water tubes. The cooling water for condensing is pumped from the cooling water pond. This cooling water first enters the absorber where it takes away the heat of condensation and dilution. The condensate from the condenser is supplied to the evaporator to compensate the water vapor formed in the evaporator. The pressure reducing valve reduces the pressure of condensate from the condenser pressure to the evaporator pressure. The cooled water from the evaporator is pumped and sprayed in the evaporator in order to the cool the water flowing through the chilled tubes. This completes the cycle.

19.15.2 Lithium Bromide-Water System Equipment

A line diagram of two-shell lithium bromide-water vapour absorption refrigeration system is shown in Fig. 19.26. The various components are:

Figure 19.26 Lithium bromide-water vapour absorption system

  1. Generator: It consists of tube bundles submerged in the solution heated by steam or hot liquids.
  2. Condenser: It. consists of tube bundles located in the vapour space over the generator shielded from carry over of salt by eliminators. Cooling water to condenser first passes through the absorber.
  3. Absorber: It consists of bundles over which the strong absorbent is sprayed. Refrigerant vapours are condensed into the absorbent releasing heat to the cooling water passing through it.
  4. Evaporator: Evaporators are tube bundles over which the refrigerant water is sprayed and evaporated. The liquid to be cooled passes inside the tubes.
  5. Solution heat exchangers: Solution heat exchangers are all of steel shell and tube construction.
  6. Purgers: All units include a purger which is used to remove non-condensable gases. Non-condensable gases present in small quantities can raise the total pressure in the absorber sufficiently to significantly change the evaporator pressure. Very small pressure increases cause appreciable change in the refrigerant evaporating temperatures.
  7. Expansion Device: Mechanical expansion valves are not used in absorption units. The flow or refrigerant liquid to evaporator is controlled by an orifice or other fixed restriction between the condenser and the evaporators.
  8. Pump: Other details about pumps etc., are shown in Fig. 19.26.

Capacity control: All lithium bromide-water cycle absorption machines meet load variations and maintained chilled water temperature control by varying the rate of re-concentration of the absorbent solution.

At any given constant load, the chilled water temperature is maintained by a temperature difference between refrigerant and chilled water. The refrigerant temperature is maintained in turn by absorber being supplied with a flow rate and concentration of solution, and by absorber cooling water temperature.

Load changes are reflected by corresponding changes in chilled water temperature. Aload reduction, for example, results in less temperature difference being required in the evaporator and a reduced requirement for solution flow or concentration. The resultant chilled water temperature drop is met basically by adjusting the rate of re-concentration to match the reduced requirements of the absorber.

All units sense the chilled water temperature changes resulting from a load change with a thermostat in the leaving chilled water.

19.16 ❐ COMPARISON OF AMMONIA-WATER AND LITHIUM BROMIDE-WATER ABSORPTION SYSTEMS

The comparison of ammonia-water and lithium bromide-water systems are given m Table 19.2

 

Table 19.2 Comparison of ammonia-water and lithium bromide-water systems

Exercises

19.1. The evaporator and condenser temperatures in an NH3 refrigeration system are –10°C and 40°C respectively. Determine per TR basis: (a) mass flow rate; (b) compressor work, (e) condenser heat rejection; (d) C.O.P.; and (e) refrigerating efficiency. Use only the propertiesof NH3 given below:

For superheated NH3 at 15.55 bar, the following values may be taken

[Ans. 0.198 kg/min: 0.82 kW; −4. 32 kW; 4.27; 81.2%]

19.2. In a vapour compression refrigeration system using R-12, the evaporator pressure is 1.4 bar and the condenser pressure is 8 bar. The refrigerant leaves the condenser sub-cooled to 30°C. The vapour leaving the evaporator is dry and saturated. The compression process is isentropic. The amount of heat rejected in the condenser is 13.42 MJ/min. Determine: (a) refrigerating effect in kJ/kg; (b) refrigerating load in TR; (c) compressor input in kW; and (d) COP.

[Ans. 114 k//kg; 49 TR: 51.4 kW; 3.35]

19.3. A vapour compression refrigerator works between the temperature limits of −20°C and 25°C. The refrigerant leaves the compressor in dry saturated condition. If the liquid refrigerant is undercooled to 20°C before entering the throttle valve determine:

  1. work required to drive the compressor ;
  2. refrigerating effect produced per kg of the refrigerant; and
  3. theoretical COP.

Assume specific heat of the refrigerant as 4.8. The properties of the refrigerant are

[Ans. 189.7 kJ/kg; 990.2 kJ/kg; 5.01]

19.4. A food storage chamber requires a refrigeration system of 12 TR capacity with an evaporator temperature of −8°C and condenser temperature of 30°C. The refrigerant R-12 is sub-cooled by 5°C before entering the throttle valve, and the vapour is superheated by 6°C before entering the compressor. If the liquid and vapour specific heats are 1.235 and 0.733 kJ/kg.K respectively, find: (a) refrigerating effect per kg; (b) mass of refrigerant circulated per minute; and (c) coefficient of performance.

The relevant properties of the refrigerant R-12 are given below:

[Ans. 130.05 kJ/kg; 19.4 kg; 6.2]

19.5. An ammonia refrigerator produces 30 tonnes of ice form and at 0°C in 24 hours. The temperature range of the compressor is from 25°C to −15°C. The vapour is dry saturated at the end of compression and an expansion valve is used. Assume a coefficient of performance to be 60% of the theoretical value. Calculate the power required to drive the compressor. Latent heat of ice = 335 kJ/kg. Properties of ammonia are.

[Ans. 33.24 kW]

19.6. A freezer of 20 TR capacity has evaporator and condenser temperature of −30°C and 25°C respectively. The refrigerant R-12 is sub-cooled by 4°C before it enters the expansion valve and is superheated by 5°C before leaving the evaporator. The compression is isentropic and the valve throttling and clearance are to be neglected. If a six cylinder, single acting compressor with stroke equal to bore running at 1000 rpm is used, determine (a) COP, of the refrigerating system, (b) mass of refrigerant to be circulated per min, (c) theoretical piston displacement per minute, and (d) theoretical bore and stroke of the compressor. The specific heat of liquid R-12 is 1.235kJ/kg K and of vapour R-12 is 0.733 kJ/kg.K. The properties of R-12 are given below:

[Ans. 3.64; 34.12 kg/min; 5.56 m/min; 0.106 m]

19.7. A refrigeration plant of 8 TR capacity has its evaporation temperature of − 8°C and condenser temperature of 30°C. The refrigerant is sub-cooled by 5°C before entering into the expansion valve and vapour is superheated by 6°C before leaving the refrigerator. The suction pressure drop is 0.2 bar in the suction valve and discharge pressure drop is 0.1 bar in the discharge valve.

If the refrigerant used is R-12, find out the COP, of the plant and theoretical power required for the compressor. Assume compression is isentropic. Use p-h chart for calculation.

19.8. An ammonia refrigerator works between − 6.7°C and 26.7°C, the vapour being dry it the end of isentropic compression. There is no under cooling of liquid ammonia and the liquid is expanded through a throttle valve after leaving the condenser. Sketch the cycle on the T-s and p-h diagram and calculate the refrigeration effect per kg ammonia and the theoretical coefficient of performance of the unit with the help of the properties given below:

[Ans. 1028.3 kJ/kg, 7.2]

19.9. An ammonia refrigerating machine fitted with an expansion valve works between the temperature limits of −10°C and 30°C. The vapour is 95%, dry at the end of isentropic compression and the fluid leaving the condenser is at 30°C If the actual coefficient of performance is 60% of the theoretical, find the ice produced per kW hour at 0°C from water at 10°C. The latent heat of tee is 335 kJ/kg. The ammonia has the following properties:

[Ans. 33.24 kg/kWh]

19.10. A R-12 refrigerating machine works on vapour-compression cycle. The temperature of refrigerant in the evaporator is − 20°C. The vapour is dry saturated when it enters the compressor and leaves it in a superheated condition. The condenser temperature is 30°C. Assuming specific heat at constant pressure for R-12 in the superheated condition as 1,884 kJ/kg.K, determine:

  1. condition of vapour at the entrance to the condenser:
  2. condition of vapour at the entrance to the evaporator; and
  3. theoretical COP of the machine. The properties of R-12 are:

[Ans. 33.8°C;29% dry; 4.07]

19.11. A CO2 refrigerating plant fitted with an expansion valve, works between the pressure limits of 54.81 bar and 20.93 bar. The vapour is compressed isentropically and leaves the compressor cylinder at 32°C. The condensation takes place at 18°C in the condenser and there is no undercooling of the liquid. Determine the theoretical coefficient of performance of the plant. The properties of CO2 are:

[Ans. 4.92]

19.12. A single stage NH3 refrigeration system has cooling capacity of 500 kW. The evaporator and condenser temperatures are −10°C and 30°C respectively. Assuming saturation cycle, determine: (a) mass flow rate of refrigerant; (b) adiabatic discharge temperature, (c) compressor work in kW; (d) condenser heat rejection, (e) COP; and (f) compressor swept volume in m3/min, if volumetric efficiency is 70%.

The following values may be taken:

hg (−10°C) = 1431.6 kJ/kg; hf (30°C) = 322.8 k.J/kg;
vg (−10°C) = 0.4185 m3/kg; sg = (−10°C) = 5.4717 kJ/kg.K

The properties of superheated NH3 at condenser pressure of 1 1.66 bar (30°C) are as follows:

At 85°C, h = 1621.8 kJ/kg; s = 5.5484 kJ/kg.K.

At 90°C, h = 1634.5 kJ/kg;

s = 5.4838 kJ/kg.K;

[Ans. 0.45 kg/s; 88.3°C; 89.5 kW; 590 kW; 5.585; 16.2 m3/min]

19.13. A 15 TR Freon 22 vapour compression system operates between a condenser temperature of 40°C and an evaporator temperature of 5°C.

  1. Determine the compressor discharge temperature:
    1. Using the p-h diagram for Freon 22.
    2. Using saturation properties of Freon 22 and assuming the specific heat of its vapour as 0.8 kJ/kg.K.
    3. Using superheat tables for Freon 22.
  2. Calculate the theoretical piston displacement and power consumption of the compressor per ton of refrigeration.

19.14. A simple saturation ammonia compression system has a high pressure of 1.35 MN/m2 and a low pressure of 0.19 MN/m2. Find per 400,000 kJ/h of refrigerating capacity, the power consumption of the compressor and COP of the cycle.

19.15.

  1. A Freon 22 refrigerating machine operates between a condenser temperature of 40°C and an evaporator temperature of 5°C. Calculate the increase (per cent) in the theoretical piston displacement and the power consumption of the cycle:
    1. If the evaporator temperature is reduced to 0°C,
    2. If the condenser temperature is increased to 45°C.
  2. Why is the performance of a vapour compression machine more sensitive to change in evaporator temperature than to an equal change in the condenser temperature?

19.16. In a vapour compression cycle saturated liquid Refrigerant 22 leaving the condenser at 40°C is required to expand to the evaporator temperature of 0°C in a cold storage plant.

  1. Determine the percentage saving in network of the cycle per kg of the refrigerantif an isentropic expander could be used to expand the refrigerant in place of thethrottling device.
  2. Also determine the percentage increase in refrigerating effect per kg of refrigerant as a result of use of the expander. Assume that compression is isentropic from saturated vapour state at 0°C to the condenser pressure.