## Lesson 21. Numerical on multi-evaporation and compression systems

Module 5. Multi-evaporator and compressor systems

Lesson 21
NUMERICAL ON MULTI-EVAPORATION AND COMPRESSION SYSTEM

21.1 Numericals

a). A 10 ton ammonia vapour compression refrigeration system consists of one evaporator and two stage compression. The suction temperature is -30 oC and condensing temperature of 35 oC. Flash inter-cooling is done between two stages of compression. Find theoretic kW of each compressor and COP of the plant.

Solution: The working cycle is represented on P-H diagram.

Fig. 21.1 Cycle on P-H diagram

From R-717 (ammonia) Table data,

And Pc = 13.5 bar at +35 oC

Pe = 1.2 bar at -30 oC

The flash refrigerant required (mi) at intermediate pressure (4 bar at -2 oC) for inter-cooling can be calculated as under. This refrigerant is also compressed by higher stage compressor.

m (h2 - h3) = mi (h3 - h5)

∴ 0.0331(1557-1442.5) = mi (1442.5-347.5)

∴ mi = 3.46 × 10-3 kg/s

The total amount of refrigerant compressed by higher stage compressor is as under.

= 0.0331 kg/s + 0.00346 kg/s

=0.03656 kg/s

"kW of higher stage compressor = 0.03656 × (h4- h3)

= 0.03656 (1617.0 - 1442.5)

= 6.38 kW

COP= (3.5 ×10)/(5.01+6.38)

∴ COP = 3.07

b) Multi-evaporation and multi-compression ammonia vapour compression refrigeration system is used in commercial plant (Figure given below). The condensing temperature of the plant is 30 oC. The intermediate pressure between two stages of compression is 3.8 bar (absolute). Find the theoretical C.O.P. and work of compression of each compressor.

Fig. 21.2 Multievaporator refrigeration system

Solution: The working cycle of the above system is indicated on P-H diagram

Fig. 21.3 Multievaporator refrigeration on P-H digram

From the properties tables & chart

h10 = 1405.6 kJ/kg

h8 = 1433.1 kJ/kg

h3 = 1441.0 kJ/kg

h2 = 1450.0 kJ/kg

h4 = 1595.0 kJ/kg

h5 = 323.1 kJ/kg

Enthalpy h1 can be calculated by heat and mass balance of refrigerants of E1 and E2 as under.

m1 h10 + m2 h7 = (m1 + m2) h1

(0.1617 × 1405.6) + (0.2523 × 1433.1) = (0.1617 + 0.2523) h1

227.29 + 361.57 = 0.414 h1

h1 = 1422.37 kJ/kg

The h1 point can be located on P-H diagram corresponding to enthalpy 1422.37 kJ/kg.

kW of first stage compressor = (m1 + m2) × (h2 - h1)

= (0.1617 + 0.2523) × (1450.0-1422.37)

= 11.44 kW

The refrigerant used for inter-cooling between two stages of compression can be estimated as under.

(m1 + m2) × (h2 - h3) = mi (h3 - h5)

(0.1617 + 0.2523) × (1450.0 - 1441.0) = mi (1441.0 - 323.1)

0.414 × 9 = mi (1117.9)

mi =0.00333 kg/s

Second stage compressor compresses m1 + m2 + mi quantity of refrigerant.

kW for second compressor = ( m1 + m2 + mi) × (h4 - h3)

= (0.1617 + 0.2523 + 0.00333) (1595.0 - 1441.0)

= 0.4173 × 154

= 64.26 kW

"COP = " "3.5 (50 +80)" /"(11.44 +64.26)"

= 6.0