Refrigeration Lab Report

Refrigeration and Air Conditioning Laboratory Report Name: Leejat Kumar Pradhan Roll No : 09ME1055

4/18/2012

EXPERIMENT No. 1a Studies on refrigerant compressors Objective: Study of reciprocating compressors used in refrigeration and air conditioning applications.

Compressors: The compressor is the most critical component of a mechanical vapor compression refrigeration system. Classification: Compressors: 1. Positive Displacement Type: i) Reciprocating ii) Vane iii) Screw iv) Scroll 2. Rotodynamic Type: i)Centrifugal Compressor The Reciprocating type of compressor is the mainstay of the refrigeration and air conditioning industry. These compressors are further classified as: i) Open Type compressor ii) Hermetically sealed type compressor Main parts of a reciprocating compressor include: i) Cylinder Head ii) Valve Plate iii) Inlet and outlet valves iv) Crank case v) Crank shaft vi) Connecting rods vii) Piston viii) Oil return path ix) Seals

Discussion: 1)Define stroke, bore, clearance volume and displacement volume of the compressor? The distance traveled by the piston from bottom dead center (BDC) to top dead center (TDC) within the cylinder is known as the stroke of the compressor. The bore is the diameter of the compressor cylinder. The volume between the piston and the cylinder head when it is at it’s minimum, i.e. when the piston is at the position of Inner Dead Centre is called the clearance volume. The volume of the compressor may be defined as the volume swept by the piston in its stroke between the bottom and the top dead centers respectively. 2) Do the piston rings have a taper? While mounting the ring the sharp edge should be on the top or the bottom? Yes, the piston rings have a taper. While mounting, the sharp edge should be on the bottom. 3) In which direction should the piston rings scrap the lubricating oil? If more than one piston ring is used what precautions should be used? Ans>The piston rings should scrap the oil towards the crankcase. The piston ring which is to be farthest from the TDC has to be mounted first. 4) What are the methods used for capacity control of compressors? Ans> Increasing the amount of clearance volume is a way of decreasing the capacity of compressor. Capacity control of compressor is achieved in reciprocating type compressor by varying speed in engine driven units through fuel flow control. But speed reduction is limited to 80% of the top speed to provide constant torque. Capacity control for screw type compressor is done by variable speed and variable compressor displacement. For the later technique, a slide valve is positioned in the casing that directs a portion of the compressed air to the suction valve when the capacity is reduced. 5) What are the safety devices used with refrigerant compressors? Ans>Some of the safety devices used are as follows: 1. Low and high suction pressure cut-off devices 2. Freeze protectors 3. Proofing switches 4. Exhaust fan interlocks 5. Temperature and pressure pop-off valves 6. Low gas pressure cut-off valves 6) What type of motors are generally used with refrigerant compressors? Ans>Fordomectic refrigerators, generally a single phase motor is used to drive a hermitic sealed compressor. For industrial compressors, generally 3 phase motor of higher capacity is used. 7) Describe various types of seals used in compressors. Ans> Open type compressors require seals to minimize leakage. These may be of stationary or rotary types. The stationary type seal is basically a metallic bellow and a hardened shaft. The rotary type seal employs a synthetic seal tightly fitted to the shaft which prevents leakage using a carbon nose. 8) What are the advantages/disadvantages and applications of open type and hermetic compressors? Ans> Open type compressors are generally installed in higher capacity refrigeration units. The motor and the compressor are separate units and there are more possibilities of refrigerant leakage so they have to be serviced frequently. But it has got more efficiency than hermitic sealed compressors. Open compressors are flexible in terms of speed as they use belt drives. The belt drives also provide safety because if the motor attains an excessive high speed the belt snaps. Heat is rejected to the surroundings

by both the motor and the compressor. In case of hermitic compressor, the motor and the compressor are locked in a sealed case. They are used generally for low capacity domestic purposes. The refrigerant serves to be the cooling agent in this case as it comes in contact with the entire body, before it enters the compressor. As the refrigerant absorbs some of the heat, these kind of compressors have low efficiency than the open type. Servicing is not frequently required in these type of compressors. Also, hermetically sealed compressors give satisfactory and safe purpose only over a range of deasign temperature. 9) Discuss briefly the advantages, disadvantages and applications of other type of positive displacement (screw, rolling piston, rotary vane and scroll types) and dynamic (centrifugal) compressors. Ans> Centrifugal compressors are preferred over the reciprocating compressors for higher efficiency over a large range of load and a large volume of the suction vapour and hence have a larger capacity to size ratio. However, the capacity of reciprocating machines is not affected much by an increase in the condensing temperature followed by adverse ambient conditions. Rotary compressors have high volumetric efficiencies due to negligible clearance. Screw compressor combines many advantageous features of both centrifugal and reciprocating systems. As it is a positive displacement machine, high pressure refrigerants may be used and due to high speeds large volumes may be handled. It has no surging problems. It has small pipe dimensions and positive pressure due to the use of high pressure refrigerants. It has high compression efficiency, continuous capacity control, unloaded starting and no balancing problems. It is hence suitable for large capacity installations. 10> Discuss the various material and manufacturing issues involved in the design of refrigerant compressors. Ans>The cylinder head, valve plate and crankcase are generally made of cast iron. The valves are generally made of high speed, spring steel as they have to open and close properly for years together. The crankshaft is made of forged steel, with hardened bearing surfaces. Current practice is to use cast shafts of either alloy or nodular iron or high tensile grey iron. Connecting rods are either eccentric strap or scotchyoke type or blade and cap-type. The first two types are made of bronze or aluminium, whereas the last one may be made of forged steel or cast iron. The piston is usually made of cast iron or aluminium. Cast iron gives a tighter fit. The clearance varies from 0.003 mm per cm dia to 0.03 mm per cm.

Experiment no. 1b Studies on Refrigeration Test Rig cum Demonstration Unit Objective:Studies on refrigeration test rig-cum-demonstration unit Observation:

Result Evaporator water flow rate [g/s] Condenser water flow rate [g/s] Evaporator Gauge Pressure [kN m-2] Evaporator Temperature [oC] evaporator water inlet temperature [C] evaporator water outlet temperature[C] Condenser gauge pressure [] Condenser temperature condenser water inlet temperature [C] condenser water outlet temperature [C] Compressor discharge temperature heat transfer rate in evaporator [J/s] heat transfer rate in condenser [J/s] COP

50

30

10

50

30

10

50

30

10

50

50

50

30

30

30

10

10

10

50

50

50

50

50

50

50

50

50

17

17.5

16

18

18

16.5

18.5

18.5

17

23.5

24.5

25

25

25.5

26

25.5

26

26.5

23

23.5

22.5

24

24

23

25

25

24

140

140

140

140

140

140

150

150

150

33

32.5

33

34

34.5

35

37

38

38.5

25

24.5

25

22

25.5

26

26

26.5

26.5

26

25.5

26

25

27

27.5

31

31.5

31.5

81

81

81

81.5

81.5

81.5

82.5

82.5

83

104.6 125.52 104.6 209.2 1

209.2 188.28 125.52

209.2 209.2 376.56 188.28 188.28 NA

1

1.25

NA

NA

104.6 125.52 104.6 209.2 NA

209.2 209.2 NA

NA

Variation of COP with flow rate of water at evaporator with different flow rate of water at condenser.

Discussion   

The COP varied with the heat rejection rate from the condenser and the cooling load. Condenser water flow rate remaining constant, remaining constant, the CoP decreases asflow rate of water at the evaporator increases. Evaporator water flow rate remaining constant, the CoP decreases as the flow rate ofwater at the condenser increases.

EXPERIMENT No. 2 Objective: To carry-out steady-state measurements on the test-rig of a vapor compression refrigeration system in order to determine: (i) Carnot COP, Cycle COP and Actual COP of the refrigeration system (ii) Overall heat transfer coefficients for the evaporator and the condenser (iii) Overall volumetric efficiency of the compressor

Components in the test-rig of a vapor compression refrigeration system: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

Hermetic compressor Air cooled condenser Capillary tube / Thermostatic expansion valve Water cooled evaporator Shut-off valves Sight glass Drier Filter Solenoid valve Pressure gauges Thermometric wells Rota meter

Refrigerant used: R-12

Observations:

Calculations: 1. CoP: Carnot CoP: CoP = Tevp / (Tcond – Tevp) = (279.5) / (329.5-279.5) = 5.59 Cycle CoP: CoP = (h1 – h3) / (h2 – h1) Where h1 = enthalpy of refrigerant vapor at evaporator outlet = 192.7 kJ/kg h 2 = enthalpy of refrigerant vapor at compressor outlet = 220.4 kJ/k h3 = enthalpy of refrigerant liquid at condenser outlet =80.81 kJ/kg CoP = 4.03 Actual CoP: CoP = actual refrigerant effect / actual energy input to compressor = energy meter reading of heater/ energy meter reading of compressor = 750/ 400 = 1.875 Overall heat transfer coefficients of evaporator and condenser (U evp, Ucond): Uevp = actual refrigerant effect / Acoil * (Twater – Te) Actual refrigeration effect = Energy meter reading of heater Acoil = π⋅de⋅π⋅De⋅Nte

2

= 0.3865 m 0 Twater = 14.5 0 Te = 6.5 C

C

2

Uevp = 242.56 w/m /K Ucond = condenser heat rejection rate / A cond * LMTDcond Condenser heat rejection rate = (m * Cp)*( Tair,out – Tair,in) = (0.109*1)( 39 -31) = 0.872 W A

=

cond

𝜋

𝑏 × ℎ × 𝑁𝑓 − 𝑁𝑓 × 𝑁𝑡𝑐 ×

× 𝑑𝑐2

4

2

= 3.27m =21.276 K

LMTD

cond

U

cond

1.

𝜋 × 𝐿 × 𝑑𝑐 × 𝑁𝑡𝑐 − 𝜋 × 𝑒 × 𝑑𝑐 × 𝑁𝑓 × 𝑁𝑡𝑐 + 2 ×

2

= .011 W/m /K

Overall volumetric efficiency of compressor, η ov:

η ov=actual refrigerant mass flow rate / maximum possible mass flow rate Actual refrigerant mass flow rate = Qliq / vliq = 32.27 kg/h Maximum possible mass flow rate = PD / vsuction = 60.22 kg/h

η

= 53.43% All thermo-physical properties are calculated using REFPROP 7.

Error analysis: 1.Carnot COP: Absolute error in T = 1 K evp

Absolute error in T

=1K

cond

Relative error in Tevp = .00357 Relative error on Tcond – Tevp = .00305 Relative error in CoP = .0067 2. Error in Cycle CoP: [cycle CoP- act CoP]/ cycle CoP : (4.03-1.875)/4.03= 0.535

Discussion What are the errors in measurement of refrigerant temperature? Suggest a better method of measuring the temperature. The following errors may creep into refrigerant temperature measurement:  Faulty calibration of thermometer.  Due to heat losses in the thermometer contacts  Ineffective wetting of thermometer bulb by the connecting liquid. These may be remedied by using solid state thermoelectric sensors. What are the errors in energy meter reading? The following are some sources of error in an energy meter reading:  Human error in measurements  Inaccurate Calibration  Unsteady or unknown power factor for AC source. What are the assumptions made in calculating the Carnot, cycle and actual COPs and overall heat transfer coefficients? Carnot COP: Actual COP:

A saturation cycle has been assumed between the measured temperatures. We directly calculated the values from the energy meter readings. These may have errors and also do not account for electrical losses.

How does one find cooling capacity and condenser heat rejection rate from the refrigerant side? The flow rate is determined from the Rota meter and the thermodynamic state functions and variables of the refrigerant at inlet and outlet of the Condenser and Evaporator are calculated from the property tables knowing the pressure and temperature etc. The same is used to calculate the cooling capacity and condenser heat rejection rate. Why is thermostat used in the water tank? A thermostat is used to keep the water temperature constant in the tank. This is useful to maintain a steady rate of heat rejection by the water to the refrigerant, and find the cooling capacity from the power consumption of the water heater.

What are the safety devices used in the set-up? The following are the safety devices used in the system:  Filter, drier absorb moisture and dust.  High and Low pressure cut off systems for the compressor. The experiment gives COP at a fixed temperature. Suggest a method to find the COP at different condenser and evaporator temperatures. COP at different temperatures of Evaporator and Condenser can be found out by carrying out the experiment at various Ambient Temperatures (Room Temp control) and loads (varying Heater output) What is the significance of volumetric efficiency of compressor?

Volumetric efficiency determines the critical pressure ratio for a particular compressor. It gives us a limit for the Condenser pressure, for a given Evaporator pressure.

EXPERIMENT No. 3 Objective: To study the pull down and cycling characteristics of a cold storage

Components of a cold storage: i) Open compressor 3TR ii) Condensing unit iii) Air cooled condenser iv) Hand valve v) Dryer vi) Solenoid valve vii) Electric bulb viii) Sight glass ix) Sub-cooling heat exchanger x) Thermostatic expansion valve xi) Direct expansion coil xii) Thermostat xiii) HP/LP cutout

Observation: o

Initial temperature: 73 F Cut out pressure = -8 inches of Hg Cut in pressure = 18 inches of Hg Refrigerant used: R22

Time (in seconds) Temperature vs. time indicating the solenoid and compressor condition

Pull down time = 16:35 min Cycle time = 13.7 min On time = min Off time = 13 min

Discussion: Describe various methods of capacity control of refrigeration systems. “On and Off Control”, “Holding the Valves Open” ,”Hot Gas Bypass” and “UsingMultiple Units” are some of the control methods used.

Describe a method for precise control of temperature.

A thermostat with a solenoid coil connection is used in order to control the temperature in the cold storage. The thermostat has its probe at the farthest corner of the cold storage so that it it is ensured that the entire cold storage has reached that temperature. When the temperature falls below the cut out point, the solenoid valve closes and stops the flow of fluid to the evaporator. The compressor sucks in the vapor refrigerant that is left behind in the evaporator. This causes a low pressure to develop in the evaporator which in turn switches off the compressor motor with the help of HP/LP switch. Again, due to heat transfer from the surroundings, the temperature inside the cold storage rises and when it goes above the cut in point, the thermostat opens the solenoid valve and the liquid at condenser being at higher pressure flows to the evaporator. When the pressure in the evaporator becomes higher than the LP pressure setting of the HP/LP switch, the compressor is again switched on by the HP/LP switch. In this way the cycle continues. Describe methods for humidity control. The humidity of the ambient air can be measured by comparing the wet bulb temperature with the dry bulb temperature. Humidity can be controlled by humidifiers and vaporizers. Vaporizers are of many types like impeller, ultrasonic, evaporative and steam humidifiers. Sometimes, even hygroscopic substances like calcium chloride and lithium chloride are used for controlling the humidity of the air. Some old types of cold storage have the cooling coil mounted near the ceiling. These are called bunker coils. These do not have any fan for air circulation. Howdoes the air circulate in such a cold storage? In such systems, the air circulates by natural convection. In some cold storages the chilled air is injected at high velocity into the coldstorage. What are the disadvantages and advantages of this method?

The high velocity results in high heat transfer coefficient and hence faster cooling.

What is the advantage of buying a condensing unit rather buying separatecomponents and assembling them? In any vapor based assembly, slight mismatches in the components may lead toleakage, besides loss of efficiency.

Where and why is TEV with external pressure equalizer used? The purpose of the thermal expansion valve is to maintain a constant temperature in between the temperature of the liquid refrigerant leaving the condenser and the temperature of the vapor refrigerant leaving the evaporator. This is to ensure that even when there is change in load requirements the temperature and pressure at all points of the system remain constant even though the mass flow changes. Pressure equalizer is used when the superheat in the evaporator is very high.The equalizer

equates the pressure of the evaporator outlet to that in the bellows through which the fluid enters the evaporator.

Why are cross charge, liquid charge and vapor charge used in TEV? Cross charge uses “power fluids” which have a more flat pressure-Temperature curvethan the refrigerant hence helping maintain a constant degree of superheat fordifferentevaporator temperatures. Describe pull down characteristics of a compressor and outline the procedure tobypass the power peak while starting a compressor. In order to bypass the power peak while starting a compressor the condenser and evaporator should be at the same pressure at the time of starting or the pressuredifference should be minimum.

What decides the cut-in pressure? The pressure of the evaporator at which the compressor; through the HP LP setupstarts working is called the cut in temperature. The design evaporator pressure and thestarting torque capacity of the motor determine it’s magnitude.

What happens if instead of using the LP cut-out/cut-in, the thermostat itselfsimultaneously closes the solenoid valve and switches off compressor? The solenoid valve will prevent compressor slogging. However on turning on againthe compressor will have to work against a high pressure difference so a high starting torque will be required.

What differences did you notice between the components of the refrigerationsystem of the cold storage and the vapor compression refrigeration test rig(Experiment 1)? Use of an open type compressor instead of a hermetic compressor, the capacity of the compressor, the use of HP/LP cut off switches (as opposed to the critically charged VCRS system), use of a thermostatic expansion valve as opposed to the float valve used in the test rig, and the choice of the refrigerant.

EXPERIMENT No.4 Performance evaluation of a summer air conditioning system Objective: To determine: i) ii) iii)

The actual COP of an air-conditioning system Condensate rate (moisture removal rate) The apparatus dew point and the by-pass factor of the cooling coil

Components in an air-conditioning system: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) Refrigerant used: R134a

Observations:

hermetic compressor of 1 TR capacity air-cooled condenser finned tube direct expansion coil as an evaporator capillary tube/thermostatic expansion valve orifice meter pre-heat coil re-heat coil humidification components

Case1: Without preheating or reheating (Reading 2) Fan supply voltage (Vfan) = 150V 2 Evaporator pressure (P1) = 250 kN/m 2 Condenser inlet pressure (P2) = 1320 kN/m 2 Condenser outlet pressure (P3) = 1310 kN/m Temperature In degree Celsius T1 32.3 T2 21 T3 32.5 T4 20 T5 21.3 T6 17.7 T7 21.3 T8 17.7 T13 13.3 T14 87.6 T15 52.1 Manometer reading = 8.6 mm of H2O Mass flow rate = 12 gm/ sec Radius of orifice or outer duct = 0.075 m Actual COP = actual cooling capacity / power input to compressor Actual Cooling capacity=Ma(ha,i-ha,o) – mwhw Ma = ρa Aoutlet Vavg 3 3 ρa = P / RT = 250*10 / 287.06*(273.15 + 13.3) = 3.0403 gm/m 2 2 Aoutlet = Π *(0.075) = 0.0177 m

Vavg = 28 ft/sec = 8.5344 m/sec Ma = 3.0403*0.0177*8.5344 = 0.4593 kg/sec Now, from psychometric chart, o o ha,i (at DBT 32.3 C and WBT 21 C) = 60.45 kJ/kg wi = 0.010930 kg/kg dry air o o ha,o (at DBT 21.3 C and WBT 17.7 C) = 49.83 kJ/kg wo = 0.011187 kg/kg dry air hw= 34.2 kJ/kg Now, mw = Ma*(wi - wo) = 0.4593 x (0.010930 - 0.011187) -3 = 0.1180 x 10 kg/sec Qe = Ma(ha,i-ha,o) – mwhw -3 = 0.4593*(60.45-49.83) – 0.1180x10 *34.2 = 4.874 kJ/sec Power input to compressor = MREF*(enthalpy of compressor inlet-evaporator outlet) = 0.012*(466.01-411.43) = 0.65496 kW Hence, COP = Qe / Work input to compressor = 7.4416 By pass factor = [Toutlet,DBT - TADP] / [Tinlet,DBT - TADP] From psychometric chart TADP = 15.7°C Thus, By-pass factor = (21.3 – 15.7) / (32.3 – 15.7) = 0.3373

Case 2 : Considering preheating affect (Reading 5) Fan supply voltage (Vfan) = 150V 2 Evaporator pressure (P1) = 320 kN/m 2 Condenser inlet pressure (P2) = 1475 kN/m 2 Condenser outlet pressure (P3) = 1450 kN/m Temperature In degree Celsius T1 32.8 T2 21.4 T3 43.5 T4 22.5 T5 26.9 T6 19.9 T7 26.9 T8 19.9 T13 18.9 T14 89 T15 55.7 Manometer reading = 9.8 mm of H2O Mass flow rate = 16.5 gm/ sec Radius of orifice or outer duct = 0.075 m Actual COP = actual cooling capacity / power input to compressor Actual Cooling capacity=Ma(ha,i-ha,o) – mwhw Ma = ρa Aoutlet Vavg 3 3 ρa = P / RT = 320*10 / 287.06*(273.15 + 18.9) = 3.817 gm/m 2 2 Aoutlet = Π *(0.075) = 0.0177 m Vavg = 28 ft/sec = 8.5344 m/sec Ma = 3.817*0.0177*8.5344 = 0.5766 kg/sec Now, from psychometric chart, o o ha,i (at DBT 32.8 C and WBT 21.4 C) = 61.8620 kJ/kg wi = 0.011283 kg/kg dry air o o ha,o (at DBT 26.9 C and WBT 19.9 C) = 56.8145 kJ/kg

wo = 0.011676 kg/kg dry air hw= 34.2 kJ/kg Now, mw = Ma*(wi - wo) = 0.5766 x (0.011283 - 0.011676) -3 = 0.2266 x 10 kg/sec Qe = Ma(ha,i-ha,o) – mwhw -3 = 0.5766*(61.86-56.81) – 0.2266x10 *34.2 = 2.90 kJ/sec Power input to compressor = MREF*(enthalpy of compressor inlet-evaporator outlet) = 0.0165*(465.49-415.22) = 0.829455 kW Hence, COP = Qe / Work input to compressor = 3.496 By pass factor = [Toutlet,DBT - TADP] / [Tinlet,DBT - TADP] From psychometric chart TADP = 18°C Thus, By-pass factor = (26.9 – 18) / (32.8 – 18) = 0.6014

Case 3 : Considering both preheating and reheating affect (Reading 8) Fan supply voltage (Vfan) = 150V 2 Evaporator pressure (P1) = 250 kN/m 2 Condenser inlet pressure (P2) = 1350 kN/m 2 Condenser outlet pressure (P3) = 1340 kN/m Temperature In degree Celsius T1 32.9 T2 21.6 T3 33.3 T4 20.2 T5 30.4 T6 19 T7 32.4 T8 21 T13 15.3 T14 87.6 T15 52.4 Manometer reading = 10.5 mm of H2O Mass flow rate = 13.5 gm/ sec Radius of orifice or outer duct = 0.075 m Actual COP = actual cooling capacity / power input to compressor Actual Cooling capacity=Ma(ha,i-ha,o) – mwhw Ma = ρa Aoutlet Vavg 3 3 ρa = P / RT = 250*10 / 287.06*(273.15 + 15.3) = 3.0192 gm/m 2 2 Aoutlet = Π *(0.075) = 0.0177 m Vavg = 28 ft/sec = 8.5344 m/sec Ma = 3.0192*0.0177*8.5344 = 0.4560 kg/sec Now, from psychometric chart, o o ha,i (at DBT 32.9 C and WBT 21.6 C) = 62.58 kJ/kg wi = 0.011525 kg/kg dry air o o ha,o (at DBT 30.4 C and WBT 19 C) = 53.69 kJ/kg wo = 0.009049 kg/kg dry air hw= 34.2 kJ/kg Now, mw = Ma*(wi - wo) = 0.4560 x (0.010930 - 0.013985) -3 = 1.129 x 10 kg/sec Qe = Ma(ha,i-ha,o) – mwhw

-3

= 0.4560*(62.58-53.69) – 1.129x10 *34.2 = 4.02 kJ/sec Power input to compressor = MREF*(enthalpy of compressor inlet-evaporator outlet) = 0.0135*(465.59-413.18) = 0.7075 kW Hence, COP = Qe / Work input to compressor = 5.682 By pass factor = [Toutlet,DBT - TADP] / [Tinlet,DBT - TADP] From psychometric chart TADP = 15°C Thus, By-pass factor = (30.4– 15) / (32.9 – 15) = 0.8603

Psychometric chart for reading 2:

Psychometric chart for reading 5:

Psychometric chart for reading 8:

Error analysis: Absolute error in T = 0.1 K Error in TADP=0.5 K Error in enthalpy of input=.33 kJ/kg Error in enthalpy of air after cooling =0.26 kJ/kg Error in enthalpy of water at evaporator temperature= 0.42 kJ /kg Error in Qe=ma(Δha,I –Δha,o) - mwΔhw=0.101 KJ/kg Relative error in Qe=4.2% Error in enthalpy of refrigerant at compressor outlet=0.11 kJ/kg Error in enthalpy of refrigerant at compressor inlet=.05 kJ/kg Error in compressor work=1.6*10-3kJ/kg Relative error in compressor work=0.3% Relative Error In COP=4.6% Relative Error in bypass factor=(0.1+0.5)(1/(To – TADP)+1/(Ti – TADP))=1.54 Discussion: How does the bypass factor vary with air velocity, answer with reason? Ans.More the air velocity, more will be the bypass factor. This is because as the velocity of air increases, the air gets less time to be in contact with the coolant and higher is the outlet air temperature. What are the effects of fin spacing and number of rows on bypass factor? Ans. Increase in the number of fins causes increase in heat transfer rate between the refrigerant and the air and hence less is the bypass factor. However, as the number of fins is further increased, the bypass factor increases due to decrease in flow rate of air. What is the relationship between by-pass factor and the performance of the system? Ans. The bypass factor indicates what portion of the enthalpy of the air is absorbed by the refrigerant. A higher bypass number means a less efficient system. Describe methods for independent control of DBT and RH. Ans. For independent control of dry bulb temperature, pre-heaters and reheaters are used. And for independent control of relative humidity, air washers are used.

What are the errors in the measurements of condensate rate? Ans. The condensate formation may not occur only at the cooling coil surface and the entire condensate is not possible to collect. Also there may be some impurities in the air which may alter the condensate formation rate. What precautions should be taken in measurement of WBT? Ans. The sling psychrometer used to measure the WBT should be swung enough number of times to ensure that the water totally evaporates from the bulb surface and does not stagnate. Also the WBT should be noted within a short interval of time of swinging it. Also water must be sufficiently present in the storage of the psychrometer. Suggest alternate methods (other than measuring WBT) for finding state points of moist air? Ans.Other means of finding state points include moisture-absorbing polymer films, chilled gold mirrors with optical sensors etc. How do you design the thermometer walls used for measurement of temperature? Ans.It should be of some non conducting material, and transparent. They should also be thick enough so that the change in reading is considerate. Suggest methods for estimating the system performance more accurately Ans. The system must be in a controlled environment where the ambient temperature and humidity is constant at all time. Also it should be perfectly insulated and all the condensate form should be collected mechanism.