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Refrigeration and Air Conditioning [ Question No: 1 To 20 ] :
Direction for [ Question No: 1 To 1 ] :
Common Data for the next two questions:
A refrigerator operates between 120 kPa and 800 kPa in an ideal vapor compression cycle with R-134a as the refrigerant. The refrigerant enters the compressor as saturated vapor and leaves the condenser as saturated liquid. The mass flow rate of the refrigerant is 0.2 kg/s. Properties for R-134a are as follows:
Saturated R-134a
| P (kPa) | T (°C) | hf (kJ/kg) | hg (kJ/kg) | sf (kJ/kg.K) | sg (kJ/kg.K) |
|---|---|---|---|---|---|
| 120 | -22.32 | 22.5 | 237 | 0.093 | 0.95 |
| 800 | 31.31 | 95.5 | 267.3 | 0.354 | 0.918 |
Superheated R-134a
| P (kPa) | T (°C) | h (kJ/kg) | s (kJ/kg.K) |
|---|---|---|---|
| 800 | 40 | 276.45 | 0.95 |
The power required for the compressor in kW is
Direction for [ Question No: 2 To 3 ] :
Statement for Linked Answer Questions:
A 1 hp motor is used for running a dual cylinder reciprocating compressor of a refrigeration system based on R-134a refrigerant having 185 kJ kg−1 cooling capacity. COP of the system is 4.2 and overall efficiency of the compressor is 80%. Specific volume of the refrigerant vapour at suction temperature is 0.15 m3 kg−1. The compressor with bore diameters of 40 mm each runs at 1440 rpm.
The compressor stroke length in mm is
The mass flow rate of the refrigerant in kg min−1 is
Halogenated hydrocarbons are the preferred refrigerants for household refrigerators and air conditioning. From the following, select the correct matching of halogenated hydrocarbon and their example.
One ton of refrigeration heat is equivalent to the amount of heat required to melt 1 ton of ice at 0°C in-
An industrial heat pump operates between the temperature of 27°C and -13°C. The rates of heat addition and heat rejection are 750W and 1000W, respectively. The COP for the heat pump is:
The COP of a heat pump operating between 6°C and 37°C will be:
A refrigeration system based on reverse Carnot cycle operates between −30 °C and 20 °C. The refrigeration capacity of the system is 70 kW. Heat rejected at the condenser to atmosphere (hot reservoir), in kW, is nearest to
In a vapour compression refrigeration cycle, the refrigerant enters the compressor in saturated vapour state at evaporator pressure, with specific enthalpy equal to 250 kJ/kg. The exit of the compressor is superheated at condenser pressure with specific enthalpy equal to 300 kJ/kg. At the condenser exit, the refrigerant is throttled to the evaporator pressure. The coefficient of performance (COP) of the cycle is 3. If the specific enthalpy of the saturated liquid at evaporator pressure is 50 kJ/kg, then the dryness fraction of the refrigerant at entry to evaporator is _________.
At 60 °C dry bulb temperature, saturated water vapour pressure is 20 kPa and relative humidity is 20%. The corresponding absolute humidity of air at atmospheric pressure is nearest to
(Take molecular weight of water = 18.02 g.mol−1, molecular weight of air = 28.97 g.mol−1, and atmospheric pressure = 101.325 kPa)
Match the following psychrometric processes (Column I) with the respective changes in thermodynamic properties (Column II) of air-water vapor mixture:
| Column I | Column II | ||
|---|---|---|---|
| 1 | Sensible cooling | P | Dry bulb temperature increases and dew point temperature decreases |
| 2 | Cooling and dehumidification | Q | Dry bulb temperature increases and dew point temperature increases |
| 3 | Chemical dehumidification | R | Dry bulb temperature decreases and dew point temperature remains constant |
| 4 | Heating and humidification | S | Dry bulb temperature decreases and dew point temperature decreases |
Air at 70 °C and 0.015 humidity ratio is cooled adiabatically by spraying water. The final temperature of the air is 55 °C. Specific heat capacities of dry air and water vapour are 1.005 and 1.88 kJ kg−1 K−1, respectively and latent heat of vapourization of water at 0 °C is 2501.7 kJ kg−1. The absolute humidity of the outlet air, in kg water vapour per kg dry air is
In a variable refrigerant flow (VRE) system, what allows the system to simultaneously provide heating and cooling to different zones in a building?
Which type of air conditioning system is commonly used for large buildings or entire city blocks and utilizes a central plant for cooling?
Which is NOT any type of vapor compression cycle?
The line 3-4 in the given T-s diagram of Bell-Coleman cycle represents
Which point on the psychrometric chart represents the condition of the air coming out of the cooling coil in an air conditioning system?
Which is one of the non-conventional refrigeration methods used for producing low temperature on the basis of the reverse seebeck effect?
Determine the C.O.P. of a refrigeration system if the work input is 80 kJ/kg and refrigeration effect produced is 120 kJ/kg of refrigerant flowing?
Which of the following is NOT a synthetic refrigerant in the refrigeration cycle?
Total number of questions : 20
Number of answered questions : 0
Number of unanswered questions : 20
Refrigeration and Air Conditioning [ Question No: 1 To 20 ] :
Direction for [ Question No: 1 To 1 ] :
Common Data for the next two questions:
A refrigerator operates between 120 kPa and 800 kPa in an ideal vapor compression cycle with R-134a as the refrigerant. The refrigerant enters the compressor as saturated vapor and leaves the condenser as saturated liquid. The mass flow rate of the refrigerant is 0.2 kg/s. Properties for R-134a are as follows:
Saturated R-134a
| P (kPa) | T (°C) | hf (kJ/kg) | hg (kJ/kg) | sf (kJ/kg.K) | sg (kJ/kg.K) |
|---|---|---|---|---|---|
| 120 | -22.32 | 22.5 | 237 | 0.093 | 0.95 |
| 800 | 31.31 | 95.5 | 267.3 | 0.354 | 0.918 |
Superheated R-134a
| P (kPa) | T (°C) | h (kJ/kg) | s (kJ/kg.K) |
|---|---|---|---|
| 800 | 40 | 276.45 | 0.95 |
The power required for the compressor in kW is
Your Answer : (Not Answered)
Correct Answer : C
The compressor needs about 7.9 kW.
Compression is isentropic, so the exit entropy equals the inlet entropy sg at 120 kPa, 0.95 kJ/kg.K. At 800 kPa this matches the superheated entry, where h = 276.45 kJ/kg:
\[ \begin{aligned} W &= \dot{m}(h_{2}-h_{1}) \\ &= 0.2(276.45-237) = 7.89\ \text{kW} \end{aligned} \]
Taking the exit as saturated vapour at 800 kPa (267.3) gives 6.1 kW, which is not offered and ignores the superheat. 39.5 forgets the mass flow rate of 0.2 kg/s (39.45 kJ/kg is the specific work), and 1.83 and 5.94 do not follow from the table.
Direction for [ Question No: 2 To 3 ] :
Statement for Linked Answer Questions:
A 1 hp motor is used for running a dual cylinder reciprocating compressor of a refrigeration system based on R-134a refrigerant having 185 kJ kg−1 cooling capacity. COP of the system is 4.2 and overall efficiency of the compressor is 80%. Specific volume of the refrigerant vapour at suction temperature is 0.15 m3 kg−1. The compressor with bore diameters of 40 mm each runs at 1440 rpm.
The compressor stroke length in mm is
Your Answer : (Not Answered)
Correct Answer : B
The stroke is about 33.7 mm.
The suction volume flow is \(0.813 \times 0.15 = 0.122\) m3/min. At 1440 rpm, with two single acting cylinders each making one suction stroke per revolution:
\[ \begin{aligned} V_{cyl} &= \frac{0.122}{1440 \times 2} = 4.23 \times 10^{-5}\ \text{m}^{3} \\ A &= \tfrac{\pi}{4}(0.04)^{2} = 1.257 \times 10^{-3}\ \text{m}^{2} \\ L &= V_{cyl}/A = 0.0337\ \text{m} \end{aligned} \]
That is 33.7 mm, taking the volumetric efficiency as 100%. Putting the whole flow through one cylinder doubles the stroke to 67.4 mm, and halving it to 16.8 mm would treat the cylinders as double acting.
The mass flow rate of the refrigerant in kg min−1 is
Your Answer : (Not Answered)
Correct Answer : C
The refrigerant mass flow is about 0.813 kg min−1.
The 1 hp motor supplies 0.746 kW. With the compressor 80% efficient, the work delivered to the refrigerant is 0.597 kW, and the COP gives the refrigerating effect:
\[ \begin{aligned} W &= 0.746 \times 0.80 = 0.597\ \text{kW} \\ Q_{e} &= 4.2 \times 0.597 = 2.507\ \text{kW} \\ \dot{m} &= \frac{2.507}{185} = 0.01355\ \text{kg/s} \\ &= 0.813\ \text{kg/min} \end{aligned} \]
Ignoring the compressor efficiency gives about 1.02 kg/min, which is not offered; 1.634 and 1.090 overstate the flow, and 0.240 understates it.
Halogenated hydrocarbons are the preferred refrigerants for household refrigerators and air conditioning. From the following, select the correct matching of halogenated hydrocarbon and their example.
Your Answer : (Not Answered)
Correct Answer : C
The correct match is \(HFC_{s}\)- R-134.
R-134a (tetrafluoroethane) is a hydrofluorocarbon (HFC): it contains hydrogen, fluorine and carbon but no chlorine, so it does not deplete ozone, and it replaced R-12 in household refrigerators and car air conditioners.
The other pairs are mismatched. R-22 is an HCFC (chlorodifluoromethane), not a CFC, and R-12 is a CFC (dichlorodifluoromethane), not an HCFC. 'HCFs' is not a refrigerant class, and R-18 is not a standard refrigerant number.
One ton of refrigeration heat is equivalent to the amount of heat required to melt 1 ton of ice at 0°C in-
Your Answer : (Not Answered)
Correct Answer : C
One ton of refrigeration is the heat needed to melt 1 ton of ice at 0°C in 24 hours.
The unit comes from the days when cooling was done with ice. Melting one short ton (2,000 lb) of ice absorbs 2,000 times its latent heat of 144 Btu/lb, spread over a day:
\[ \begin{aligned} \text{TR} &= \frac{2000 \times 144}{24} \\ &= 12{,}000\ \text{Btu/h} \approx 3.517\ \text{kW} \end{aligned} \]
Taking 1 hour instead would make the unit 24 times larger, 12 hours twice as large, and 1 minute 1,440 times as large. The day-long rate is what defines the ton.
An industrial heat pump operates between the temperature of 27°C and -13°C. The rates of heat addition and heat rejection are 750W and 1000W, respectively. The COP for the heat pump is:
Your Answer : (Not Answered)
Correct Answer : C
The COP of the heat pump is 4.0.
A heat pump's COP is the heat delivered to the hot side divided by the work input. The work is the difference between heat rejected and heat added:
\[ \begin{aligned} W &= 1000 - 750 = 250\ \text{W} \\ COP_{HP} &= \frac{1000}{250} = 4.0 \end{aligned} \]
7.5 is the Carnot COP, \(300/40\), for these temperatures, an upper limit rather than the actual value. 3.0 is \(750/250\), the COP if the machine were judged as a refrigerator. 6.5 does not follow from the data.
The COP of a heat pump operating between 6°C and 37°C will be:
Your Answer : (Not Answered)
Correct Answer : B
The COP is 10.
For a reversible heat pump, with temperatures in kelvin, \(T_{H} = 37 + 273 = 310\) K and \(T_{L} = 6 + 273 = 279\) K:
\[ \begin{aligned} COP_{HP} &= \frac{T_{H}}{T_{H} - T_{L}} \\ &= \frac{310}{31} = 10 \end{aligned} \]
Using the refrigerator formula \(T_{L}/(T_{H} - T_{L}) = 279/31 = 9\) gives one less, because \(COP_{HP} = COP_{R} + 1\). Using degrees Celsius instead of kelvin gives nonsense values. 8, 12 and 14 do not come from the correct temperatures.
A refrigeration system based on reverse Carnot cycle operates between −30 °C and 20 °C. The refrigeration capacity of the system is 70 kW. Heat rejected at the condenser to atmosphere (hot reservoir), in kW, is nearest to
Your Answer : (Not Answered)
Correct Answer : A
The heat rejected is about 84.40 kW.
A reversed Carnot cycle must use absolute temperatures: \(T_{L} = 243\) K and \(T_{H} = 293\) K.
\[ \begin{aligned} COP &= \frac{T_{L}}{T_{H} - T_{L}} = \frac{243}{50} = 4.86 \\ W &= \frac{70}{4.86} = 14.40\ \text{kW} \\ Q_{H} &= Q_{L} + W = 84.40\ \text{kW} \end{aligned} \]
14.41 kW is the compressor work, not the heat rejected. 116.67 kW is what the work comes to if the temperatures are left in degrees Celsius (COP = 30/50 = 0.6), and 186.67 kW is 70 plus that wrong work. Heat rejected must always be the refrigeration load plus the work input.
In a vapour compression refrigeration cycle, the refrigerant enters the compressor in saturated vapour state at evaporator pressure, with specific enthalpy equal to 250 kJ/kg. The exit of the compressor is superheated at condenser pressure with specific enthalpy equal to 300 kJ/kg. At the condenser exit, the refrigerant is throttled to the evaporator pressure. The coefficient of performance (COP) of the cycle is 3. If the specific enthalpy of the saturated liquid at evaporator pressure is 50 kJ/kg, then the dryness fraction of the refrigerant at entry to evaporator is _________.
Your Answer : (Not Answered)
Correct Answer : B
The dryness fraction at entry to the evaporator is 0.25.
Number the states: 1 evaporator exit (saturated vapour, \(h_{1} = 250\) kJ/kg), 2 compressor exit (\(h_{2} = 300\) kJ/kg), 3 condenser exit, 4 after throttling. Throttling keeps enthalpy constant, so \(h_{4} = h_{3}\).
\[ \begin{aligned} COP &= \frac{h_{1} - h_{4}}{h_{2} - h_{1}} = \frac{250 - h_{4}}{50} = 3 \\ h_{4} &= 250 - 150 = 100\ \text{kJ/kg} \end{aligned} \]
State 4 is a wet mixture at evaporator pressure, between saturated liquid (\(h_{f} = 50\)) and saturated vapour (\(h_{g} = 250\)):
\[ \begin{aligned} x_{4} &= \frac{h_{4} - h_{f}}{h_{g} - h_{f}} \\ &= \frac{100 - 50}{250 - 50} = 0.25 \end{aligned} \]
0.2 divides by 250 instead of by the latent enthalpy 200 (50/250). 0.3 and 0.35 do not follow from the given enthalpies; any error in the COP step changes \(h_{4}\) and every later figure.
At 60 °C dry bulb temperature, saturated water vapour pressure is 20 kPa and relative humidity is 20%. The corresponding absolute humidity of air at atmospheric pressure is nearest to
(Take molecular weight of water = 18.02 g.mol−1, molecular weight of air = 28.97 g.mol−1, and atmospheric pressure = 101.325 kPa)
Your Answer : (Not Answered)
Correct Answer : A
The absolute humidity is about 0.026 kg water per kg dry air.
The partial pressure of the vapour is the relative humidity times the saturation pressure, and the humidity ratio follows from the molar mass ratio:
\[ \begin{aligned} p_{v} &= 0.20 \times 20 = 4\ \text{kPa} \\ H &= \frac{18.02}{28.97} \cdot \frac{p_{v}}{P - p_{v}} \\ &= 0.622 \times \frac{4}{101.325 - 4} \\ &= 0.622 \times 0.0411 = 0.0256 \end{aligned} \]
That rounds to 0.026. 0.041 is the pressure ratio \(p_{v}/(P - p_{v})\) without the molar-mass factor 0.622. 0.004 and 0.032 do not come from the correct vapour pressure of 4 kPa; using the saturation pressure of 20 kPa instead would give a much larger 0.153.
Match the following psychrometric processes (Column I) with the respective changes in thermodynamic properties (Column II) of air-water vapor mixture:
| Column I | Column II | ||
|---|---|---|---|
| 1 | Sensible cooling | P | Dry bulb temperature increases and dew point temperature decreases |
| 2 | Cooling and dehumidification | Q | Dry bulb temperature increases and dew point temperature increases |
| 3 | Chemical dehumidification | R | Dry bulb temperature decreases and dew point temperature remains constant |
| 4 | Heating and humidification | S | Dry bulb temperature decreases and dew point temperature decreases |
Your Answer : (Not Answered)
Correct Answer : C
The correct matching is 1-R, 2-S, 3-P, 4-Q.
Dew point temperature depends only on the moisture content (humidity ratio) of the air, so it moves with moisture, while the dry bulb temperature follows heating or cooling.
- Sensible cooling removes heat at constant moisture, so dry bulb falls and dew point stays constant: R.
- Cooling and dehumidification lowers the temperature and condenses moisture out, so both fall: S.
- Chemical dehumidification by an adsorbent removes moisture and releases the heat of adsorption, so dew point falls while dry bulb rises: P.
- Heating and humidification adds both heat and moisture, so both rise: Q.
The first option swaps the pairings for chemical dehumidification and heating with humidification. The second and fourth options pair sensible cooling with a rising dry bulb temperature, which is impossible for a cooling process.
Air at 70 °C and 0.015 humidity ratio is cooled adiabatically by spraying water. The final temperature of the air is 55 °C. Specific heat capacities of dry air and water vapour are 1.005 and 1.88 kJ kg−1 K−1, respectively and latent heat of vapourization of water at 0 °C is 2501.7 kJ kg−1. The absolute humidity of the outlet air, in kg water vapour per kg dry air is
Your Answer : (Not Answered)
Correct Answer : C
The outlet humidity is about 0.021 kg/kg dry air.
Adiabatic cooling by spraying water keeps the enthalpy of the moist air essentially constant, with \(h = 1.005\,t + W(2501.7 + 1.88\,t)\) in kJ per kg dry air:
\[ \begin{aligned} h_{1} &= 1.005(70) + 0.015(2501.7 + 131.6) \\ &= 70.35 + 39.50 = 109.85 \\ 109.85 &= 1.005(55) + W_{2}(2501.7 + 103.4) \\ W_{2} &= \frac{109.85 - 55.28}{2605.1} = 0.0209 \end{aligned} \]
The air gains about 0.006 kg of water per kg dry air as it cools by 15 °C. 0.017 and 0.019 add too little moisture to balance the sensible heat given up, and 0.023 adds too much.
In a variable refrigerant flow (VRE) system, what allows the system to simultaneously provide heating and cooling to different zones in a building?
Your Answer : (Not Answered)
Correct Answer : A
Heat recovery technology allows a variable refrigerant flow system to give heating and cooling to different zones at the same time. In a heat recovery VRF system the heat removed from the zones that need cooling is moved to the zones that need heating, and this is done through refrigerant circuits, with three pipes in many models. Split type compressors, variable speed fans and dual thermostat control do not give the simultaneous heating and cooling by themselves.
Which type of air conditioning system is commonly used for large buildings or entire city blocks and utilizes a central plant for cooling?
Your Answer : (Not Answered)
Correct Answer : B
A district cooling system is used for large buildings or entire city blocks, with a central plant for cooling. Chilled water from the central plant is piped to the buildings. Portable units, window units and split systems are for single rooms or small spaces and have their own compressors, so they do not use a central plant.
Which is NOT any type of vapor compression cycle?
Your Answer : (Not Answered)
Correct Answer : D
The cycle with wet vapour before compression is not a type of vapour compression cycle in the key. In the vapour compression cycle the vapour is compressed in the compressor, and it enters either dry saturated or superheated, and may leave dry saturated or superheated. Compression of wet vapour is avoided because liquid droplets damage the compressor, so it is not a standard cycle, and the other three options are standard types.
Slugging, the entry of liquid into the cylinder, is the damage that wet compression risks, so practical machines are designed to take only dry or superheated vapour.
The line 3-4 in the given T-s diagram of Bell-Coleman cycle represents
Your Answer : (Not Answered)
Correct Answer : A
The line 3-4 of the Bell-Coleman cycle on the T-s diagram represents isentropic expansion. The line is vertical in the diagram, so the entropy is constant, and the temperature falls from \(T_3\) to \(T_4\), which means the air expands in the expander. The line 1-2 is isentropic compression, the line 2-3 is cooling at constant pressure, and the line 4-1 is heat absorption at constant pressure, which are the other three options.
This is why the air leaves the expander colder than the space being cooled, and that temperature fall is what produces the refrigeration effect.
Which point on the psychrometric chart represents the condition of the air coming out of the cooling coil in an air conditioning system?
Your Answer : (Not Answered)
Correct Answer : C
The condition of air coming out of the cooling coil is the apparatus dew point (ADP) on the psychrometric chart, which is the temperature of the coil surface when the air is cooled below its dew point. The wet bulb and dry bulb temperatures are properties of the air, not specific to the cooling coil's outlet, and 'dry point' is not a term used on the chart.
Which is one of the non-conventional refrigeration methods used for producing low temperature on the basis of the reverse seebeck effect?
Your Answer : (Not Answered)
Correct Answer : B
Thermoelectric refrigeration produces low temperature by the reverse of the Seebeck effect, which is the Peltier effect: a current passed through a junction of two dissimilar conductors absorbs heat at one junction. Magnetic refrigeration works on the magnetocaloric effect, liquid gas refrigeration on the evaporation of a liquefied gas, and dry ice refrigeration on the sublimation of solid carbon dioxide, so only the thermoelectric method matches the effect named.
Determine the C.O.P. of a refrigeration system if the work input is 80 kJ/kg and refrigeration effect produced is 120 kJ/kg of refrigerant flowing?
Your Answer : (Not Answered)
Correct Answer : B
The COP is 1.5. The coefficient of performance of a refrigerator is the refrigeration effect divided by the work input: \[ COP = \frac{120}{80} = 1.5 \] The value 1.33 would result from dividing 80 by 60 and 2.0 and 3.0 do not follow from dividing 120 by 80, so only 1.5 is right.
Which of the following is NOT a synthetic refrigerant in the refrigeration cycle?
Your Answer : (Not Answered)
Correct Answer : C
Propane (R290) is not a synthetic refrigerant. It is a natural hydrocarbon found in natural gas and petroleum, and it is used as a natural refrigerant. Hydrofluorocarbons (HFCs), perfluorocarbons (PFCs) and chlorofluorocarbons (CFCs) are all man-made halogenated compounds produced for refrigeration and other uses, so they are synthetic refrigerants.
Marks
Total number of questions : 20
Number of answered questions : 0
Number of unanswered questions : 20