Refrigeration & Air-Conditioning

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 P_c = 13.5 bar at +35 ^oC

P_e = 1.2 bar at -30 ^oC

The flash refrigerant required (m_i) 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) = m_i (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) = m_i (h₃ - h₅)

(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 m₁ + m₂ + m_i quantity of refrigerant.

kW for second compressor = ( m₁ + m₂ + m_i) × (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