refrigeration unit experiment

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LAB Report - Refrigeration

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The refrigerant unit experiment is carried out to observe how the thermodynamic

refrigeration unit works. The equipment that is used in the laboratory to perform the

experiment is the Refrigeration Laboratory Unit (Model: R713/24822). The varying

conditions that is involved are the condenser pressure, evaporator pressure and water flow

rate. The derived results would be compressor power, compressor suction and evaporating

temperature. Each experiment used the same starting step that is the system would run out for

2 minutes before taking the necessary reading. The water flow rate is adjusted depending on

the required condenser pressure.

INTRODUCTION

The Refrigeration Laboratory Unit (Model: R713/24822) has been designed to provide

students with a practical and quantitative demonstration of a vapour compression cycle, and

is suitable for all course levels (intermediate and undergraduate). Refrigerators applies the

vapour compression cycle. The experiment shows that a large freely available energy source,

such as the atmosphere is to be upgraded for refrigeration. The unit will be of particularly

interest to those studying Mechanical Engineering, Energy Conservation,

Thermodynamics, Building Services, Chemical Engineering, Plant and Process Engineering,

Refrigeration and Air Conditioning.

• To measure the pressure and temperature of the refrigerant in the evaporator and condenser.
• To obtain the temperature of the water entering and leaving the coils in the evaporator and condenser.
• To know the water flow rate through the coils.
• To know the temperature of the refrigerant at the compressor inlet and outlet.

Ideal Refrigeration cycle 1. An Ideal Refrigeration unit is modelled by a reversed Carnot cycle, which is internally and externally reversible. Steps of the cycle: 2. The dry saturated refrigerant vapour enters to a compressor. The compressor compressed the vapour by increased its pressure and temperature. 3. Hot vapour was cooled at constant pressure. The heat is rejected by the Refrigerant. 4. Simple throttle valve expansion from high to low pressure. The enthalpy is constant during the process. 5. Heating the liquid at constant pressure. The heat supply is from the cold sources.

USEFUL DATA (See also Specification) Compressor (Two cylinder, single acting.) Bore = 38mm; Stroke = 19mm. Swept volume 43 cm 3 per revolution. = (nominally) 0.

Belt Pulley Ratio, = = 1.

Hence, if Compressor RPM nc = 719 Motor RPM = 718x 1. nm = 1423 RPM Typical Compressor Friction Force (at spring balance) = 5N Dynamometer Torque arm radius O Condenser Mean heat transfer area, 0 2 Specific heat of water = 4 kJ kg-

Digital Temperature Indicator - This becomes operative as soon as the electrical supply is turned on. The temperature may be measured at six points in the circuit by selecting stations 2 to 6 on the switch

Compressor Motor and Dynamometer - The motor shaft power is the product of torque and angular velocity in radians. The motor speed is determined from the compressor speed and pulley ratio

Condenser Pressure - This is controlled by the cooling water flow rate and its inlet temperature. Reduce the cooling water flow rate to increase the condenser pressure

Wattmeter - The wattmeter is an instrument for measuring the electric power and control the input work of refrigerator.

Figure-1: T-s diagram for a simple vapour compression cycle

1-2 Isentropic compression in the compressor 2-3 Constant pressure heat rejection in the condenser 3-4 isentropic expansion 4-1 Constant pressure heat addition in the evaporator.

ENERGY BALANCE FOR CYCLE

Process 1-2: compression Compressor (q1, 2= 0): w1, 2 = (h 1 - h 2 ) w = (h 2 - h 1 )

Process 2-3: Condensation Condenser (wE=0): q3, 2=q out =T2or3 (s 3 -s 2 ) qe = (h 3 - h 2 )

Process 3-4: Expansion Expansion valve (q3,4= 0, w3,4=0):

h 4 = h 3 and s 3 =s 4

REFRIGRATION LABORATORY UNIT TEST OBSERVATIONS Atm pressure 101 kpa.

SERIES TEST No. Unit 1 2 3 4 5 6

Condenser pressure (abs.)

KN/M 2 1200 750 850 1050 950 1300

Evaporator pressure (abs.)

KN/M 2 240 140 270 300 350 360

Compressor suction oC -0 -5 -1 5 6 9.

Compressor delivery

oC 40 40 48 47 49 56.

Liquid leaving condenser

oC 23 21 21 22 22 22.

Evaporator inlet oC -1 2 6 8 10 13.

Water inlet oC 19 13 13 13 12 12.

Water outlet oC 22 17 18 18 19 2.

Water flow rate gm/sec 50 40 40 40 40 40

R134a flow gm/sec 8 9 9 10 11 11

Evaporator load Watt 400 300 450 600 750 900

Motor input Watt 577 515 589 580 628 643

Spring balance N 7 12 13 14 15 15.

Compressor speed Rpm 822 818 812 812 808 807

Motor speed Rpm 1612 1612 1610 1610 1600 1600

TEST No. Unit 1 2 3 4 5 6 h 1 kJ/kg 338 335 341 344 415 415. h 2 kJ/kg 438 434 442 440 442 448. h 3 = h 4 kJ/kg 230 229 229 231 231 231. Work done on compressor

kJ/kg 100 99 100 96 26 33

Heat transfer at condenser

Qc/kJ 1669 1947 2015 2202 2317 2382.

Heat transfer at evaporator

Qe/kJ 862 1000 1061 1191 2019 2019.

CoPref - 4 1 1 1 6 5. Shaft power, Ps W 235 207 221 249 275 288. Friction power, Pf W 235 207 221 249 275 288. Indicated power, Pi W 235 207 221 249 275 288. Heat losses, Qrad kJ 1042 1155 1174 1260 568 651.

SAMPLE CALCULATIONS

Work done on compressor w = h 1 – h 2 = 338-448= -100 kJ/kg

Heat transfer at condenser Qc = m (h 3 – h 2 ) = 8 (230– 448) = - 1749 kJ

Heat transfer at evaporator Qe = m (h 1 – h 4 ) = 8 (1195 –234) = 961 kJ

CoPref CoPref

Shaft power Ps = 0 x F x = 0 x 13 x = 365 W

Friction power

Pf = 0 x Ff x = 0 x 5 x = 140 W

Indicated power Pi = Ps - Pf 365 – 140 = 224 W

Heat losses

Qrad = m (h 2 – h 1 ) + Ps = 1(424 – 1195) = 365 = -406 kJ

Similarly, temperature at the condenser will increase. As load increase on evaporator cause refrigerant to carry more heat to the condenser. This heat must be removed to the surrounding. The condenser acts as a device that release heat carried by refrigerant to the surrounding. More heat needs to be transfer will affect the temperature and performance of the condenser. Load increase on evaporator will also affect the expansion valve. As discussed earlier, increase in load requires more refrigerant to flow through expansion valve. Expansion valve function as a throttling device. As such, it will force more refrigerant to flow to the inlet of evaporator.

It should be noted however that there are discrepancies between the experimental results and theoretical results. The main cause for this is to keep in mind that all the components used are not 100% efficient. This affect p-h diagram and COP. Besides that, a few errors and bad experiment practice was present when conducting the experiment. Firstly, it is required to wait for the unit to stable. However, due to time constraints, we had to rush the experiment and did not give sufficient time for the unit to stabilize. Some of the values were taken while the reading fluctuates. Secondly, there might be a few parallax errors such as when taking measurement from the gauges. Thirdly, is should be noted that the force reading were obtained from the spring balance. This makes it highly inaccurate as we were required to pull the spring balance, a better alternative would be to use a dynamometer.

The objectives of the experiment were achieved and were stated as in the results section. It can be concluded that with the increase of load, all components of the refrigeration cycle were affected. The parameters such as pressure, enthalpy and temperature at the four states of the refrigeration cycle were determined and calculated. This enables us to calculate the COP and efficiency of the refrigeration cycle. As discussed in the section before, there were errors that jeopardize our results.

• Multiple Choice

Course : mechanical engineering

University : politeknik sultan azlan shah.

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