I85PREP
⌘ K
SOPEEC Refrigeration Operator A study materials
SOPEEC Refrigeration Operator A study resources

SOPEEC Refrigeration Operator A practice questions

One question at a time. Build your SOPEEC Refrigeration Operator A understanding.

First 50 of 190 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Refrigeration cycle, property references, and energy calculations

In a basic vapour-compression system, what is the evaporator's principal energy role?

  1. Absorb heat from the refrigerated load into the refrigerant Correct answer.
  2. Raise refrigerant pressure mechanically
  3. Reject the compressor's work directly to the room
  4. Reduce pressure without absorbing heat from the load

Explanation

The evaporator transfers heat from the load to low-pressure refrigerant, providing the useful cooling effect while other components perform compression, rejection, and expansion.

Question 2Refrigeration cycle, property references, and energy calculations

What is the compressor's main function in a vapour-compression cycle?

  1. Condense all refrigerant before it reaches the suction connection
  2. Create the useful cooling by throttling liquid
  3. Remove all noncondensable gases from every system automatically
  4. Raise refrigerant-vapour pressure so heat can be rejected at the high-side conditions Correct answer.

Explanation

Compression supplies work to the vapour and raises its pressure, enabling the cycle to reject heat at a temperature above the receiving sink.

Question 3Refrigeration cycle, property references, and energy calculations

For a simple steady refrigeration cycle with negligible other losses, what heat does the condenser reject?

  1. Evaporator heat absorbed plus compressor work input Correct answer.
  2. Evaporator heat divided by compressor work
  3. Compressor work minus evaporator heat
  4. Only the evaporator heat, with no work contribution

Explanation

Energy supplied as compressor work joins the heat absorbed at the evaporator, so both must leave through the condenser under the stated balance assumptions.

Question 4Refrigeration cycle, property references, and energy calculations

In the usual ideal throttling model for an expansion valve, which specific property remains approximately constant?

  1. Enthalpy Correct answer.
  2. Temperature
  3. Vapour quality
  4. Pressure

Explanation

Throttling is commonly modeled as approximately constant enthalpy when heat transfer, work, and kinetic effects are negligible, despite changes in pressure and phase condition.

Question 5Refrigeration cycle, property references, and energy calculations

A vapour line is at 12°C and its local saturation dew temperature is 5°C. What is the superheat?

  1. 17 K
  2. 7 K Correct answer.
  3. 5 K
  4. 12 K

Explanation

Superheat is the measured vapour temperature above the appropriate saturation temperature at the same pressure, so twelve minus five gives seven kelvins.

Question 6Refrigeration cycle, property references, and energy calculations

A liquid line is at 28°C and its local saturation bubble temperature is 35°C. What is the subcooling?

  1. 63 K
  2. 28 K
  3. 35 K
  4. 7 K Correct answer.

Explanation

Subcooling is the appropriate liquid saturation temperature minus the actual liquid temperature at that pressure, giving thirty-five minus twenty-eight, or seven kelvins.

Question 7Refrigeration cycle, property references, and energy calculations

For a pure refrigerant in saturation below its critical point, what generally happens to saturation temperature as pressure increases?

  1. It decreases for every pressure increase
  2. It remains fixed regardless of pressure
  3. It equals the surrounding air temperature automatically
  4. It increases Correct answer.

Explanation

The saturation pressure-temperature relationship links higher pressure with higher boiling or condensing temperature for a pure refrigerant in the relevant two-phase range.

Question 8Refrigeration cycle, property references, and energy calculations

A gauge reads 250 kPa and local atmospheric pressure is 100 kPa. What is the absolute pressure?

  1. 250 kPa absolute
  2. 350 kPa absolute Correct answer.
  3. 25 kPa absolute
  4. 150 kPa absolute

Explanation

For a positive gauge reading, absolute pressure equals gauge pressure plus local atmospheric pressure, so the supplied values give three hundred fifty kilopascals absolute.

Question 9Refrigeration cycle, property references, and energy calculations

A plant provides 90 kW of cooling using 30 kW of compressor input. What is its refrigeration COP for those stated boundaries?

  1. 3 Correct answer.
  2. 60
  3. 120
  4. 0.33

Explanation

Refrigeration coefficient of performance is useful cooling divided by the stated work input, so ninety kilowatts divided by thirty kilowatts equals three.

Question 10Refrigeration cycle, property references, and energy calculations

A simple cycle absorbs 120 kW at the evaporator and receives 40 kW of compressor work. What condenser duty follows from the stated energy balance?

  1. 80 kW
  2. 160 kW Correct answer.
  3. 120 kW
  4. 3 kW

Explanation

With other energy transfers neglected, the condenser rejects the absorbed heat plus the compressor input, giving one hundred twenty plus forty kilowatts.

Question 11Refrigeration cycle, property references, and energy calculations

A heat pump delivers 150 kW to the heated space using 50 kW of input. What is its heating COP?

  1. 3 Correct answer.
  2. 2
  3. 0.33
  4. 200

Explanation

Heating COP uses the useful heat delivered divided by the work input, so one hundred fifty divided by fifty gives three for the stated boundary.

Question 12Refrigeration cycle, property references, and energy calculations

A refrigerant enters an evaporator at h4 = 240 kJ/kg and leaves at h1 = 390 kJ/kg. What is the specific refrigerating effect?

  1. 240 kJ/kg
  2. 150 kJ/kg Correct answer.
  3. 1.625 kJ/kg
  4. 630 kJ/kg

Explanation

The evaporator's specific heat absorption is the outlet enthalpy minus the inlet enthalpy, so three hundred ninety minus two hundred forty gives one hundred fifty.

Question 13Refrigeration cycle, property references, and energy calculations

For an adiabatic idealized compressor, h1 = 400 kJ/kg and h2 = 450 kJ/kg. Neglecting kinetic changes, what is its specific work input?

  1. 0.89 kJ/kg
  2. 400 kJ/kg
  3. 850 kJ/kg
  4. 50 kJ/kg Correct answer.

Explanation

The idealized steady compressor work input per unit mass equals the enthalpy rise, giving four hundred fifty minus four hundred, or fifty kilojoules per kilogram.

Question 14Refrigeration cycle, property references, and energy calculations

A throttling valve has h3 = 260 kJ/kg. Under the stated isenthalpic model, what is h4?

  1. 520 kJ/kg
  2. 260 kJ/kg Correct answer.
  3. 0 kJ/kg
  4. 130 kJ/kg

Explanation

The ideal throttling assumption holds specific enthalpy constant across the device, so the outlet value equals the supplied inlet value of two hundred sixty.

Question 15Refrigeration cycle, property references, and energy calculations

A required cooling duty is 75 kW and the refrigerating effect is 150 kJ/kg. What refrigerant mass flow is needed?

  1. 0.5 kg/s Correct answer.
  2. 2 kg/s
  3. 5 kg/s
  4. 11 250 kg/s

Explanation

Cooling duty equals mass flow multiplied by specific refrigerating effect, so seventy-five kilojoules per second divided by one hundred fifty kilojoules per kilogram gives one-half kilogram per second.

Question 16Refrigeration cycle, property references, and energy calculations

An evaporator carries 0.8 kg/s of refrigerant with an enthalpy rise of 125 kJ/kg. What is its heat absorption rate?

  1. 100 kW Correct answer.
  2. 156.25 kW
  3. 0.0064 kW
  4. 250 kW

Explanation

Multiplying mass flow by enthalpy rise gives 0.8 times one hundred twenty-five kilojoules per second, or one hundred kilowatts of heat absorption.

Question 17Refrigeration cycle, property references, and energy calculations

Water flows at 2 kg/s and cools by 5 K. Using cp = 4.2 kJ/(kg·K), what cooling duty does that represent?

  1. 10 kW
  2. 42 kW Correct answer.
  3. 4.2 kW
  4. 21 kW

Explanation

The sensible cooling rate is mass flow times specific heat times temperature drop, so two times four point two times five equals forty-two kilowatts.

Question 18Refrigeration cycle, property references, and energy calculations

A 100 kg product cools by 10 K with constant cp = 3 kJ/(kg·K), without phase change. How much heat is removed?

  1. 3000 kJ Correct answer.
  2. 30 kJ
  3. 333 kJ
  4. 300 kJ

Explanation

Sensible heat removal is mass multiplied by specific heat and temperature change, giving one hundred times three times ten, or three thousand kilojoules.

Question 19Refrigeration cycle, property references, and energy calculations

Using a supplied latent heat of 330 kJ/kg, how much energy freezes 20 kg of water already at its freezing temperature?

  1. 350 kJ
  2. 16.5 kJ
  3. 330 kJ
  4. 6600 kJ Correct answer.

Explanation

For the stated phase-change step, multiply mass by latent heat: twenty kilograms times three hundred thirty kilojoules per kilogram gives six thousand six hundred kilojoules.

Question 20Refrigeration cycle, property references, and energy calculations

A 20 kW load operates steadily for 30 minutes. How much heat is removed during that interval?

  1. 36 kJ
  2. 1200 kJ
  3. 600 kJ
  4. 36 000 kJ Correct answer.

Explanation

Power is energy per second, so twenty kilojoules per second multiplied by eighteen hundred seconds gives thirty-six thousand kilojoules over thirty minutes.

Question 21Refrigeration cycle, property references, and energy calculations

Which unit is dimensionally equivalent to a kilowatt?

  1. Kilojoule per second Correct answer.
  2. Kilogram per second
  3. Kilopascal per second
  4. Kilojoule multiplied by second

Explanation

A kilowatt is a rate of energy transfer equal to one kilojoule each second, not an amount of energy or a mass-flow unit.

Question 22Refrigeration cycle, property references, and energy calculations

For a reversed Carnot refrigerator, use COP = Tc/(Th − Tc) with temperatures in kelvins. If Tc = 250 K and Th = 300 K, what is COP?

  1. 50
  2. 5 Correct answer.
  3. 6
  4. 0.2

Explanation

The absolute-temperature difference is fifty kelvins, so two hundred fifty divided by fifty gives an ideal reversed-Carnot refrigeration COP of five.

Question 23Refrigeration cycle, property references, and energy calculations

Using K = °C + 273.15, what absolute temperature corresponds to −10°C?

  1. −10 K
  2. 283.15 K
  3. 263.15 K Correct answer.
  4. −283.15 K

Explanation

Add two hundred seventy-three point one five to the Celsius value, including its negative sign, to obtain two hundred sixty-three point one five kelvins.

Question 24Refrigeration cycle, property references, and energy calculations

Suction is 200 kPa absolute and discharge is 1000 kPa absolute. What pressure ratio applies?

  1. 800
  2. 1200
  3. 5 Correct answer.
  4. 0.2

Explanation

Compressor pressure ratio uses discharge absolute pressure divided by suction absolute pressure, giving one thousand divided by two hundred, or five.

Question 25Refrigeration cycle, property references, and energy calculations

For this ideal single-acting compressor, displacement rate = piston area × stroke × cycles per second. With 0.01 m², 0.1 m, and 10 cycles/s, what is the rate?

  1. 0.1 m³/s
  2. 0.001 m³/s
  3. 1 m³/s
  4. 0.01 m³/s Correct answer.

Explanation

Multiplying the supplied area, stroke, and cycle rate gives 0.01 times 0.1 times ten, or one hundredth of a cubic metre per second.

Question 26Refrigeration cycle, property references, and energy calculations

A shaft rotates at 900 rpm. What is its rotational rate in revolutions per second?

  1. 1.5 rev/s
  2. 15 rev/s Correct answer.
  3. 900 rev/s
  4. 54 000 rev/s

Explanation

There are sixty seconds in one minute, so divide nine hundred revolutions per minute by sixty to obtain fifteen revolutions per second.

Question 27Refrigeration cycle, property references, and energy calculations

An idealized compressor displaces 0.05 m³/s but delivers 0.04 m³/s measured at suction conditions. What is volumetric efficiency?

  1. 125%
  2. 80% Correct answer.
  3. 0.8%
  4. 20%

Explanation

Volumetric efficiency compares delivered suction volume with displacement volume, so 0.04 divided by 0.05 is 0.8, or eighty percent.

Question 28Refrigeration cycle, property references, and energy calculations

Using the ideal incompressible relation power = Δp × volume flow, what is the hydraulic power at 200 kPa and 0.01 m³/s?

  1. 2 kW Correct answer.
  2. 20 kW
  3. 0.002 kW
  4. 200 kW

Explanation

A kilopascal times a cubic metre per second equals a kilowatt, so two hundred times 0.01 gives two kilowatts of ideal hydraulic power.

Question 29Refrigeration cycle, property references, and energy calculations

Use the supplied conversion 1 refrigeration tonne = 3.5 kW. What capacity corresponds to 70 kW?

  1. 73.5 refrigeration tonnes
  2. 2 refrigeration tonnes
  3. 245 refrigeration tonnes
  4. 20 refrigeration tonnes Correct answer.

Explanation

Divide the cooling duty by the stated kilowatts per refrigeration tonne, so seventy divided by three point five equals twenty under this supplied conversion.

Question 30Refrigeration cycle, property references, and energy calculations

For a wall, use heat gain = U × area × temperature difference. If U = 0.5 W/(m²·K), area = 100 m², and ΔT = 20 K, what is the heat gain?

  1. 250 W
  2. 100 W
  3. 10 000 W
  4. 1000 W Correct answer.

Explanation

Multiplying the supplied heat-transfer coefficient, area, and temperature difference gives one-half times one hundred times twenty, or one thousand watts.

Question 31Refrigeration cycle, property references, and energy calculations

Ten kilograms of water cool from 10°C to 0°C and then freeze. Use cp = 4 kJ/(kg·K) and latent heat = 330 kJ/kg. What total heat is removed?

  1. 400 kJ
  2. 3700 kJ Correct answer.
  3. 34 000 kJ
  4. 3300 kJ

Explanation

The sensible step removes four hundred kilojoules and freezing removes three thousand three hundred, so the two specified steps total three thousand seven hundred kilojoules.

Question 32Refrigeration cycle, property references, and energy calculations

For otherwise comparable refrigeration operation, why can reducing unnecessary condensing temperature improve COP?

  1. It can reduce the compressor lift and work needed for the useful cooling Correct answer.
  2. It makes throttling produce shaft work
  3. It requires the evaporator to reject the condenser heat
  4. It guarantees that all control limits can be ignored

Explanation

Reducing an unnecessary temperature and pressure lift can reduce compression demand, although safe minimums and equipment/control requirements still govern the operating range.

Question 33Compressor types, lubrication, capacity, and fault reasoning

What distinguishes a positive-displacement refrigeration compressor from a centrifugal compressor?

  1. It traps and reduces a volume of vapour rather than relying principally on dynamic velocity and pressure conversion Correct answer.
  2. It can safely compress unlimited liquid without damage
  3. It creates cooling by expanding refrigerant through its discharge valve
  4. It always operates without any lubrication requirement

Explanation

A positive-displacement machine compresses a trapped volume, while a centrifugal machine adds energy dynamically; each design still has specific operating and protection requirements.

Question 34Compressor types, lubrication, capacity, and fault reasoning

What is a main function of suction and discharge valves in a reciprocating compressor?

  1. Control the direction of vapour flow during the piston cycle Correct answer.
  2. Meter liquid directly into the evaporator
  3. Regulate condenser cooling-water chemistry
  4. Replace the piston rings as the only sealing elements

Explanation

The valves admit suction vapour and allow compressed vapour to leave at the appropriate part of the cycle, supporting one-way operation through the cylinder.

Question 35Compressor types, lubrication, capacity, and fault reasoning

Why can cylinder clearance reduce the volumetric efficiency of a reciprocating compressor?

  1. Clearance gas increases fresh suction volume without limit
  2. The piston displaces no volume when clearance exists
  3. Gas remaining in the clearance re-expands before fresh suction vapour enters Correct answer.
  4. Clearance removes the need for discharge valves

Explanation

Residual high-pressure vapour occupies the clearance space and re-expands on the return stroke, reducing the space available for new suction charge.

Question 36Compressor types, lubrication, capacity, and fault reasoning

Why is liquid slugging dangerous for a compressor designed to receive vapour?

  1. Liquid always reduces all cylinder pressures to zero
  2. Liquid makes the compressor's rated power irrelevant
  3. Liquid automatically repairs worn valves
  4. Liquid is much less compressible and can impose damaging mechanical loads Correct answer.

Explanation

A vapour compressor is not intended to compress a large liquid volume, which can create abnormal forces and damage valves, pistons, or other components.

Question 37Compressor types, lubrication, capacity, and fault reasoning

What is the primary purpose of a suction accumulator in a suitably designed system?

  1. Condense all discharge gas before the condenser
  2. Reduce the risk of liquid refrigerant reaching the compressor Correct answer.
  3. Replace the evaporator's heat-transfer surface
  4. Remove every noncondensable from the high side

Explanation

An accumulator can separate or retain liquid in the suction path according to its design, helping protect the compressor while allowing appropriate vapour return.

Question 38Compressor types, lubrication, capacity, and fault reasoning

What does a discharge-side oil separator principally do?

  1. Serve as the main liquid expansion valve
  2. Guarantee that no oil reaches any other component
  3. Separate entrained oil from refrigerant discharge flow for the specified return arrangement Correct answer.
  4. Remove every trace of moisture from the entire system

Explanation

An oil separator reduces oil carried onward in discharge gas and returns it through the designed oil-management system, without being a universal filter or complete separation guarantee.

Question 39Compressor types, lubrication, capacity, and fault reasoning

Why is net oil pressure often evaluated relative to compressor crankcase pressure?

  1. Crankcase pressure is always atmospheric
  2. A high gallery reading proves adequate differential at every crankcase pressure
  3. Bearing lubrication depends on an effective pressure differential, not only the gallery's absolute reading Correct answer.
  4. The oil gauge directly measures oil viscosity

Explanation

The useful driving pressure for oil delivery depends on both the oil-side indication and the pressure opposing it, so the relevant differential must be assessed.

Question 40Compressor types, lubrication, capacity, and fault reasoning

An oil gallery reads 450 kPa and the crankcase reads 300 kPa on the same pressure basis. What is the net differential?

  1. 150 kPa Correct answer.
  2. 750 kPa
  3. 1.5 kPa
  4. 450 kPa

Explanation

Subtract the crankcase pressure from the oil gallery pressure on the same reference basis, giving four hundred fifty minus three hundred, or one hundred fifty kilopascals.

Question 41Compressor types, lubrication, capacity, and fault reasoning

What should follow a low-oil-differential safety trip?

  1. Investigate and correct the cause under the approved procedure before restart Correct answer.
  2. Reset repeatedly without reviewing the oil system
  3. Increase the temperature thermostat setting as the only check
  4. Bypass the oil protection until the next service visit

Explanation

An oil-protection trip can indicate inadequate lubrication or a related fault, so repeated bypass or reset risks damage without establishing that the cause has been resolved.

Question 42Compressor types, lubrication, capacity, and fault reasoning

What is the intended purpose of a crankcase heater where the manufacturer specifies one?

  1. Limit refrigerant migration and dilution in the oil during relevant shutdown conditions Correct answer.
  2. Replace an oil pump during loaded operation
  3. Increase evaporator icing as quickly as possible
  4. Set the condenser pressure independently of the system

Explanation

A specified crankcase heater helps manage off-cycle refrigerant accumulation in the oil, but it must be used according to the equipment's actual design and sequence.

Question 43Compressor types, lubrication, capacity, and fault reasoning

Why may oil foam vigorously when a compressor starts after refrigerant has accumulated in the crankcase?

  1. A pressure change can cause dissolved refrigerant to boil out of the oil Correct answer.
  2. Foaming always means the oil level should be increased immediately
  3. The crankcase has necessarily become a condenser by design
  4. The foam proves the oil contains no refrigerant

Explanation

Refrigerant can dissolve in oil during shutdown and flash out as conditions change on startup, affecting oil level indication and lubrication quality.

Question 44Compressor types, lubrication, capacity, and fault reasoning

In a screw compressor with a slide-valve capacity control, what does unloading chiefly change?

  1. The effective compression capacity through the designed internal control arrangement Correct answer.
  2. The refrigerant's chemical identity
  3. The maximum pressure rating stamped on the vessel
  4. The required relief-valve certification interval

Explanation

A slide valve changes the screw compressor's effective loading according to its design; capacity and power behavior must be interpreted using the manufacturer's performance information.

Question 45Compressor types, lubrication, capacity, and fault reasoning

Why must part-load power not automatically be assumed proportional to compressor capacity?

  1. Power has no connection to compressor operation
  2. A half-loaded compressor always consumes zero power
  3. Unloading method and machine efficiency can make the relationship nonlinear Correct answer.
  4. A slide-valve position is a direct certified kilowatt reading

Explanation

Part-load performance depends on machine design and operating conditions, so actual curves or measurements are needed rather than a universal linear assumption.

Question 46Compressor types, lubrication, capacity, and fault reasoning

In an oil-injected screw compressor, what roles may the specified oil perform?

  1. Elimination of the need for an oil separator
  2. Permanent prevention of every high-discharge-temperature fault
  3. Replacement of all refrigerant flow through the evaporator
  4. Lubrication, sealing, and cooling within the designed machine Correct answer.

Explanation

Oil can serve several internal functions in this compressor design, so correct oil management and compatibility are important rather than treating it as only a bearing lubricant.

Question 47Compressor types, lubrication, capacity, and fault reasoning

What happens in the impeller and diffuser of a centrifugal compressor?

  1. The impeller adds energy and the diffuser converts part of velocity energy to pressure Correct answer.
  2. The diffuser throttles all vapour to evaporator pressure
  3. Both components operate as reciprocating inlet valves
  4. The impeller freezes the refrigerant before compression

Explanation

A centrifugal stage transfers energy dynamically, with velocity and pressure changes through its flow passages rather than trapping gas in a piston cylinder.

Question 48Compressor types, lubrication, capacity, and fault reasoning

What operating condition can contribute to centrifugal compressor surge?

  1. Any increase in stable refrigerant flow within the approved envelope
  2. The presence of a correctly installed oil filter alone
  3. Operation outside the stable flow range for the required lift Correct answer.
  4. A guarantee of adequate flow at zero load

Explanation

Surge is an unstable flow phenomenon related to the machine's operating point and pressure requirement, so the approved anti-surge and capacity controls must be respected.

Question 49Compressor types, lubrication, capacity, and fault reasoning

For otherwise similar suction conditions, why can a higher compression ratio increase discharge temperature?

  1. The expansion valve adds shaft work to the discharge line
  2. Higher ratio guarantees lower compressor work
  3. The vapour undergoes a greater compression demand Correct answer.
  4. The condenser must absorb heat from outside the system

Explanation

A greater pressure rise generally raises compression work and discharge heating under comparable conditions, although actual temperatures depend on refrigerant and machine behavior.

Question 50Compressor types, lubrication, capacity, and fault reasoning

Why can lower suction-vapour density reduce mass capacity for a given displacement rate?

  1. Mass capacity depends only on the number of cylinders
  2. Density changes only the liquid receiver level
  3. Less refrigerant mass is admitted per unit suction volume Correct answer.
  4. The same volume necessarily contains more mass at lower density

Explanation

For a displacement-based machine, mass flow relates to admitted suction volume multiplied by density, so a lower density can reduce the mass carried by the same volume.