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

*Module 5. Multi-evaporator and compressor systems*

**Lesson 21NUMERICAL 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^{ o}C and condensing temperature of 35^{ o}C. 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 P_{c} = 13.5 bar at +35 ^{o}C

P_{e} = 1.2 bar at -30 ^{o}C

The flash refrigerant required (m

_{i}) at intermediate pressure (4 bar at -2

^{o}C) for inter-cooling can be calculated as under. This refrigerant is also compressed by higher stage compressor.

m (h_{2} - h_{3}) = m_{i} (h_{3} - h_{5})

∴ 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 × (h_{4}- h_{3})

= 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^{ o}C. 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

h_{10 }= 1405.6 kJ/kg

h_{8 }= 1433.1 kJ/kg

h_{3 }= 1441.0 kJ/kg

h_{2 }= 1450.0 kJ/kg

h_{4 }= 1595.0 kJ/kg

h_{5 }= 323.1 kJ/kg

Enthalpy h_{1} can be calculated by heat and mass balance of refrigerants of E_{1} and E_{2 }as under.

m_{1} h_{10} + m_{2} h_{7} = (m_{1} + m_{2}) h_{1}

(0.1617 × 1405.6) + (0.2523 × 1433.1) = (0.1617 + 0.2523) h_{1}

227.29 + 361.57 = 0.414 h_{1}

h_{1} = 1422.37 kJ/kg

The h_{1} point can be located on P-H diagram corresponding to enthalpy 1422.37 kJ/kg.

kW of first stage compressor = (m_{1} + m_{2}) × (h_{2} - h_{1})

= (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.

(m_{1} + m_{2}) × (h_{2} - h_{3}) = m_{i} (h_{3 }- h_{5})

(0.1617 + 0.2523) × (1450.0 - 1441.0) = m_{i} (1441.0 - 323.1)

0.414 × 9 = m_{i} (1117.9)

m_{i }=0.00333 kg/s

Second stage compressor compresses m_{1} + m_{2} + m_{i} quantity of refrigerant.

kW for second compressor = ( m_{1} + m_{2} + m_{i}) × (h_{4} - h_{3})

= (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**