Refrigeration & Air-Conditioning

Module 1. Fundamentals of refrigeration

Lesson 7
ELEMENTARY NUMERICAL ON REFRIGERATION

7.1 Problems

7.2 A 50 ton ammonia vapor compression refrigeration system operates an evaporating temperature of -10 ºC and condensing temperature of 30 ºC. There is no sub-cooling of liquid refrigerant and superheating of suction vapour. Determine (i) theoretical C.O.P. of the plant and (ii) operating cost of the plant, if the cost of electricity is Rs. 5.00 per kWh.

Solution

Given data

Capacity of the plant : 50 ton
Refrigerant used : Ammonia (R717)
Evaporating Temperature : -10 ºC
Condensing Temperature: 30 ºC

Fig. 7.1 Charts

From the property table and chart

h1 = 1433.05 kJ/kg (From table)
h2 = 1620.00 kJ/kg (From P-H chart)
h3 = h4 = 323.08 kJ/kg (From table)

7.3 Find the capacity of vapor compression refrigeration system to be used for the bulk milk cooler from the following data.

i. Capacity of the bulk cooler: 1000 liters

ii. Initial temperature of supply milk: 37 ºC

iii. Temperature to which milk is to be cooled: 2 ºC

iv. Time for cooling: 3.5 hour

(Make necessary assumptions and indicate them clearly.)

Solution:

Given data :

Capacity of the Bulk cooler (BC) = 1000 liters
Initial temperature of supply milk = 37 ºC
Temperature to which milk is to be cooled = 2 ºC
Time for cooling = 3.5 h

Assumptions made:

Specific gravity of milk = 1.032

Specific heat of milk = 3.89 kJ/kg ºC

1 TR = 12600 kJ/h

Total heat to be removed from the milk = Q = m s ΔT

= (1000 ×1.032) × 3.89 × (37 - 2)

= 140506.8 k J

Heat removed / hour = 140506.8 / 3.5

= 40144.8 kJ/h

Hence, Capacity of refrigeration system required = 40144.8 / 1260

= 3.186 TR

7.4 A water cooler using R-22 as refrigerant works between -5 ºC evaporating temperature and 45 ºC condensing temperature. The output of cold water is 200 kg per hour cooled from 40 ºC to 10 ºC. Assuming 30 per cent loss of useful cooling and 80 per cent volumetric efficiency of the compressor, calculate compressor power in kW and volumetric displacement of the compressor.

Solution: Given data

Refrigerant used = R-22

Evaporating Temperature = - 5 ºC

Condensing Temperature = 45 ºC

Output of cold water = 200 kg/h

Initial temperature of water = 40 ºC

Final temperature of the water = 10 ºC

Cooling loss = 30 %

Volumetric efficiency of compressor = 80 %

Fig. 7.2 Chart

From the property table and chart

h1 = 249.88 kJ/kg (From table)
h2 = 290 kJ/kg (From P-H chart)
h3 = h4 = 105.58 kJ/kg (From table)

Specific volume of R-22 at -5 ºC is 0.05545 m³/kg

Considering 80% volumetric efficiency,

The volumetric displacement = 3.76×0.05545× (100/80)

= 0.2606 m³/min

7.5 An ammonia vapour compression plant is to be selected for a milk chilling centre for the following requirements. Estimate the capacity, theoretical C.O.P. and kW of the compressor for the refrigeration plant. (Make appropriate assumptions and specify them clearly.)

(i) Milk handling capacity of chilling centre = 60,000 litres/day.

(ii) Condensing pressure = 12 bar

(iii) Evaporating pressure = 2 bar

(iv) Temperature of supply milk = 35 ºC

(v) Working hours of the plant = 18 hours/day

Solution:

Assuming the milk is cooled to 2 ºC

TR required for the chilling centre = m s ΔT

= (60000×1.032)×3.89×(35-2)

=7948670.4 kJ/day

TR of the plant = 7948670.4 / 18 X 12600

=35.04 TR

Considering 20% refrigeration losses,

Capacity of the plant = 35.04+ (35.04×0.2)

= 42 TR

Fig. 7.3 Charts

From the property table and chart

h1 = 1422.72 kJ/kg (From table)
h2 = 1660 kJ/kg (From P-H chart)
h3 = h4 = 327.89 kJ/kg (From table)

7.6 An ammonia refrigeration plant is working between the temperature limits of 30 ^oC and -15 ^oC. The load on the unit is 20 ton. Find the (i) Theoretical C.O.P. and (ii) kW of the compressor. If the temperature required in the evaporator is -30 ^oC, then find out the change in theoretical C.O.P. and kW of the compressor. There is no change in condensing temperature.

Solution

Given data:

Refrigerant used: Ammonia (R-717)
Evaporating Temperature: -15 ^oC and -30 ^oC
Condensing Temperature:30 ^oC
Capacity of the plant: 20 TR

At operating conditions of -15 ^oC and 30 ^oC

Fig. 7.4 Charts

h₁ =1426.58 kJ/kg (From table)

h₂ =1640 kJ/kg (From P-H chart)

h₃ =323.08 kJ/kg (From table)

At operating conditions of -30 ^oC and 30 ^oC:

Fig. 7.5 Charts

h₁ =1405.6 kJ/kg (From table)

h₂ =1742 kJ/kg (From P-H chart)

h₃ =323.08 kJ/kg (From table)

7.7 A single cylinder single acting R-22 compressor works between -10 ºC and 40 ºC. The vapour leaves the evaporator dry and saturated. Find (i) capacity (ii) kW of motor from the following data.

(i) Diameter of the compressor cylinder = 15 cm.

(ii) Stroke of the compressor = 15 cm.

(iii)R.P.M. of the compressor =850.

(iv)Volumetric efficiency of the compressor =75 %

(v)Mechanical efficiency of the compressor = 96%

(vi)Motor efficiency = 97%

Solution:

Fig. 7.6 Charts

h₁ = 247.72 kJ/kg (From table)

h₂ = 285 kJ/kg (From P-H chart)

h₃ = 98.44 kJ/kg (From table)

7.8 A 10 ton Ammonia vapour compression refrigeration system operates at suction temperature of -10 ºC and condensing temperature of 35 ºC. There is no sub cooling and superheating of suction vapour. Calculate the theoretical COP, kW and temperature of gas leaving the compressor (Uses only refrigerant property data).

Solution:

Data given:

Refrigerant used: Ammonia(R-717)

Capacity of the plant: 10 TR

Evaporating temperature: -10 ^oC

Condensing temperature: 35 ^oC

Fig. 7.7 Charts

h₁ = 1433.05 kJ/kg (From table)

h₃ = 347.5 kJ/kg (From table)

h_2’ = 1471.43 kJ/kg (From table)

Fig. 7.8 Charts

S₁ = 5.477 kJ/kg K (From table)

S_2’ = 4.93 kJ/kg K (From table)

Taking specific heat of ammonia as 2.8 kJ/kg K, the enthalpy h₂ can be obtained as under: