past exam paper & memo n3 - ekurhuleni tech …...compressor has a bore of 60 mm and a stroke...

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T1660(E)(D1)T NOVEMBER EXAMINATION

NATIONAL CERTIFICATE

REFRIGERATION TRADE THEORY N3

(11041583)

1 December 2016 (X-Paper) 09:00–12:00

Nonprogrammable calculators may be used.

This question paper consists of 6 pages and 1 formula sheet of 8 pages.

(11041583) -2- T1660(E)(D1)T


DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA

NATIONAL CERTIFICATE REFRIGERATION TRADE THEORY N3

TIME: 3 HOURS MARKS: 100

INSTRUCTIONS AND INFORMATION 1. 2. 3. 4.

Answer ALL the questions. Read ALL the questions carefully. Number the answers according to the numbering system used in this question paper. Write neatly and legibly.

(11041583) -3- T1660(E)(D1)T


QUESTION 1 1.1 Various options are provided as possible answers to the following questions.

Choose the answer and write only the letter (A–D) next to the question number (1.1.1–1.1.10) in the ANSWER BOOK.

1.1.1 Which one of the following is correct ? In a domestic refrigerator, periodic defrosting is required because frosting

A B C D

Causes corrosion of materials Reduces heat extraction Overcools food stuff Partially block refrigerator flow

1.1.2 Consider the following statements: In domestic refrigeration, the coefficient of performance is a function of:

1. Electrical conductivity of materials 2. Peltier coefficient 3. Seebeck coefficient 4. Temperature at hot and cold junctions 5. Thermal conductivity of materials

A B C D

1,2,3,4 and 5 are correct 1,2,3 and 5 are correct 1,2, 4 and 5 are correct 2,3,4 and 5 are correct

1.1.3 In a vapour compression refrigeration system, liquid to suction heat exchange is used for

A B C D

Keep the COP constant Prevent the liquid refrigerant from exiting the compressor Sub cool the vapour refrigerant leaving the compressor Sub cool the vapour refrigerant from the evaporator

1.1.4 If refrigerant circulation rate is 0.025kg/s and WD = 84 kJ/kg. the refrigerant effect is equal to:…

A

B C D

2.1 KW 2.5 KW 3.0 KW 4.0 KW

1.1.5 Air cooling is used for Freon compressors whereas water jacketing is adopted for cooling ammonia compressors. This is because …

A B C D

Latent heat of ammonia is higher than that of Freon Thermal conductivity of water is higher than that of air. Specific heat of water is higher than that of air. Of the larger superheat horn of ammonia compression cycle.

(11041583) -4- T1660(E)(D1)T


1.1.6 Excessive pressure drop in liquid line in a refrigeration system

causes

A

B C D

High condenser pressure Flashing of the liquid refrigerant Higher evaporator pressure Under cooling of the liquid the refrigerant

1.1.7 Which of the following statements are true for ammonia as a

refrigerant? 1. It has a higher compressor discharge temperature

compared to fluorocarbons 2. It is toxic to mucous membranes 3. It requires large displacement per Tr compared to

fluorocarbons 4. It reacts with copper and its alloys 5. It is a colourless gas

A

B C D

1,2,3,4 and 5 are correct 1,2,3 and 5 are correct 1,2, 4 and 5 are correct 2,3,4 and 5 are correct

1.1.8 Environmental friendly refrigerant R134a is used in the new

generation domestic refrigerators. Its chemical formula is

A

B C D

CH ClF2 C2Cl2F3 C2Cl2F4 C2H2F4

1.1.9 In a vapour absorption refrigerator, heat is rejected in :… A

B C D

Condenser only Generator only Absorber only Condenser and absorber

1.1.10 A centrifugal compressor is suitable for which of the following…

A

B C D

High pressure ratio, low mass flow low pressure ratio, low mass flow High pressure ratio, high mass flow low pressure ratio, high mass flow

[10]

(11041583) -5- T1660(E)(D1)T


QUESTION 2

2.1 At constant temperature, 120 m³ of gas exerts a pressure of 150 kPa on an ordinary pressure gauge and is compressed to 350 kPa? Calculate the new volume.

(3)

2.2 What is the pressure of a quantity of confined gas in a closed container when raised to 15 °C, if the original pressure was 60 kPa absolute and the temperature at 5 °C at constant volume?

(3)

2.3 A volume of 40 m³ of gas at 5 °C is raised to 40 °C at constant pressure. Calculate the new volume.

(3)

2.4 A six-cylinder compressor rotating at 25 revolutions per second delivers refrigerant vapour at a rate of 0,0225 m³/s under these specifications. The compressor has a bore of 60 mm and a stroke length of 60 mm.

Calculate the volumetric efficiency of the compressor.

(3)

2.5

The cooling capacity of a refrigeration plant is 50 kW. The refrigerant-enthalpy gain in the evaporator is 130 kJ/kg. The coefficient of performance of the plant is 5.

Calculate the following:

2.5.1 The refrigerant mass-flow rate (2)

2.5.2 The work of compression (2)

2.5.3 The power required by the compressor motor if the efficiency is 90%.

(2)

2.5.4 The total amount of heat rejected by the condenser

(2) [20]

QUESTION 3

3.1 List THREE advantages of a centrifugal compressor. (3)

3.2 Explain the process of capacity control on compressors by hot gas bypass to evaporator inlet in direct expansion (DX) applications.

(5) [8]

QUESTION 4

4.1 Name TWO types of water-cooled condensers. (2)

4.2 Give FOUR consequences which may occur if insufficient or incorrect insulation methods are used

(4)

4.3 Define the following terms as used in refrigeration lubrication systems:

4.3.1 Floc test

4.3.2 Dielectric strength

4.3.3 Viscosity (1 x 3)

(3)

(11041583) -6- T1660(E)(D1)T


4.4 List THREE advantages of soft drawn copper tubing as used in refrigeration

applications

(3) 4.5 Describe the principle operation of the evaporative condenser. (5) 4.6 Name any FOUR different rooms as used in a locker plant (4)

[21] QUESTION 5 5.1 List four specific areas which are targeted during an inspection or

maintenance visit to compressor installations

(4) 5.2 Complete the following paragraph by using the word(s) in the list below. Write

only the word(s) next to the question number (5.2.1 – 5.2.7) in the ANSWER BOOK.

horizontal; vertical; vibration; flexible connectors; rubber hose; bellows connectors; vibration eliminator

Compressors and pumps will transmit (5.2.1) to their connecting pipe work.

Where (5.2.2) are inserted in the suction and discharge lines, they should be in the same plane as the compressor shaft i.e. (5.2.3) for semi- hermitic and open reciprocating types and (5.2.4) for all scrolls. Water or brine pumps may be insulated with (5.2.5). For small- bore pipes, these can be ordinary reinforced (5.2.6), suitably fastened at each end. For larger pipes, corrugated or (5.2.7) of various types can be obtained.

(7)

5.3 State FOUR characteristics required for the selection of an air-cooled

condenser.

(4) [15]

QUESTION 6 6.1 Explain the operating principle of a solid absorption system (10) 6.2 Describe the operation of the following components as found in a continuous

absorption system:

6.2.1 thermostat (5)

6.2.2 burner (4)

6.2.3 Burner safety valve (5) 6.3 What are the two basic causes of insufficient refrigerant in a continuous

system? (2)

[26]

TOTAL: 100

(11041583) -7- T1660(E)(D1)T


FORMULA SHEET


1. Density = v

m

volume

mass kg/m

3

also

1

kg/m3 ( = specific volume)

2. Specific volume

11

density

m3/kg

3. Greek symbols: = Efficiency (neta)

= Density (rho)

4. Mass flow rate:

m

Vm kg/s

OR

V

mVm1

kg/s

5. Quantity of heat (Q) (for a change in temperature)

TCmQ kJ

(For a change of phase)

Q = m.hfg

(hfg = latent heat required)

When the heat transfer rate is required, the formula changes to:

hfgmQ .

kJ/s or kW

6. Specific enthalpy (h)

To calculate the total energy transfer

hmQ

kJ/s or kW

7. Boyle's law:

P1V1 = P2V2 (at constant temperature)

-1-

(11041583) -2- T1660(E)(D1)T


8. Charles's law:

2

2

1

1

T

V

T

V (at constant pressure)

9. Ideal gas law:

2

22

1

11

T

VP

T

VP

10. The gas constant (R)

T

PR

J/kg.K

11. Compressor efficiency = )(

)(

VpntdisplacemelTheoretica

VntdisplacemeActual

%1

100

Vp

V

12. Theoretical displacement:

4

2 NnLDVp

m3/min

4

2 RNsDOR

m3/min

NOTE: ( = 3,142)

13. Evaporator capacity:

(a) The quantity of sensible heat transferred in watts

(evaporator sensible capacity)

Q

tAUQ

.. watts

(b) Overall heat transfer coefficient ('U'-factor)

R

U1

W/m2.°C

(c) Resistance of surface to heat flow (R)

11

111

hA

Ao

k

x

hoUR m

2.°C/W

(11041583) -3- T1660(E)(D1)T


(d) Log mean temperature difference (LMTD):

LMTD =

trt

trtenLog

trttrte

°C

14. Condenser load:

(a) Total heat rejection of condenser in kW (THR)

THR =

QiQe kW

Also THR = refrigeration capacity × heat rejection factor

(b) Condenser capacity

Q

tAUQ

.. kW

(c) LMTD =

tetr

ttrnLog

tetrttr

15. The pressure enthalpy diagram:

(a) Refrigerating effect (Re)

Re = h1 - h4 kJ/kg

(b) Heat of compression (WD)

WD = h2 - h1 kJ/kg

(c) Heat rejection (THR)

THR = h2 - h3 kJ/kg

(d) Coefficient of performance (COP):

COP = WD

Re (cooling)

COP = WD

THR (heat pump)

(e) Refrigerant mass flow

m :

Re

Loadm

kg/s

(11041583) -4- T1660(E)(D1)T


(f) Plant capacity = Re

m kW

(g) Theoretical power = WDm

kW

(h) Actual power =

WDm

kW

(i) Total heat rejection in kW (THR):

THR in kW =

m × THR in kJ/kg kW

(j) The efficiency of a compressor:

Theoretical power input % = × 100 Actual power input

OR

% = 100

112

12

hh

hh

16. Cooling load calculations:

(a) Basic heat transfer formula

TAUQ

.. watts (J/s)

(b) To calculate total resistance against heat flow

io hk

d

k

d

hRt

1...

1

2

2

1

1 m2.°C/W

(c) Heat transfer coefficient (U)

U = tR

1 W/m

2.°C

(d) Resistance against heat flow (R)

R = k

d m

2.°C/W

(e) Total air change in a cold room

V

40086

fVV

m3/s

(11041583) -5- T1660(E)(D1)T


(f) Total air change load

Q

(i) 000112

hhVQ watts

or when using the table for the number of air

changes:

(ii) 4,86

hfVQ

watts

The above formula is the combination of the

2 formulae given in (e) and (f)(i).

(g) The product load:

(i) Sensible heat:

4,86

.. tCmQ

watts

(ii) Latent heat:

4,86

hmQ

watts

(iii) Heat respiration:

hmQ

watts

(h) Miscellaneous loads:

Equipment output × hours of operation (i) Equipment load = watts 24

NB: Load and output must be given in watts.

Wattage of lighting × hours of operation (ii) Lighting load = watts 24

NB: When lighting is by means of fluorescent lights, add 25% to the lighting

load.

(iii) Occupancy load:

° Heat equivalent per person × number of persons × hours Q = watts 24

(11041583) -6- T1660(E)(D1)T


17. Ventilation and air conditioning:

(a) Removal of latent heat:

hfggmQ

kW

V

m kg/s

hfggVQ

kW

(b) Room sensible heat factor (RSHF):

RSHF = RTH

RSH

RLHRSH

RSH

(c) Efficiency of an evaporative cooler:

tbta

tcta

(d) Cooling and dehumidification:

ghfghmQ

kW

(e)

V of the return air

smfaVraVraV /3

(f) Air temperature of mixture in front of coil

saV

tdbraravtdbfafav

T

°C

(g) Quantity of air:

tC

QV

m

3/s

(h) Amount of air forced through an opening by wind:

AEV m3/s

(i) Volume flow due to temperature difference:

thkAV

m3/s

(11041583) -7- T1660(E)(D1)T


(j) Air changes per hour (f):

f = V

V 6003

changes/hour

(k) Volume air flow per m2 opening:

A

VV

m3/s/m

2

(l) Free area of an opening (A):

A = total area × m2

(m) Louvre size for given air speed:

AspeedinletinletsofNumber

V

%

100

m2

18. Fans:

(a) Fan total pressure (Pt)

Pt = Ps + Pv kPa

(b) Theoretical power of the fan:

= PtV

kW

(c) True power of the fan:

=

PtV

kW

(d) Air power (Ap)

Ap = N

PtQ

kW

(e) Brake power (Bp)

s

PsQ

t

PtQBp

s

Ps

t

PtBp

(for the same volume)

(11041583) -8- T1660(E)(D1)T

Copyright reserved

19. Fan laws:

(a)

32

31

2

1

2

1

DIA

DIA

N

N

VOL

VOL

(b)

2

1

22

21

22

21

2

1

DIA

DIA

N

N

P

P

(c)

2

15

2

51

32

31

2

1

DIA

DIA

N

N

kW

kW

1

2

2

1

N

N

DIA

DIA


NATIONAL CERTIFICATE

NOVEMBER EXAMINATION


1 DECEMBER 2016

This marking guideline consists of 5 pages.

MARKING GUIDELINE

MARKING GUIDELINE -2- T1660(E)(D1)T REFRIGERATION TRADE THEORY N3


QUESTION 1 1.1 1.1.1

1.1.2 1.1.3 1.1.4 1.1.5 1.1.6 1.1.7 1.1.8 1.1.9 1.1.10

B C C A A B C D D D (10 × 1)

(10) QUESTION 2 2.1

(3) 2.2

kPa

To

TnPoPn

ToVnTnPo

16,62

278

17280

2735

2731560

(3)



2.3

278

28370

2735

2734040

To

TnVoVn

Tn

Vn

To

Vo

(3) 2.4 Volumetric Efficiency of compressor

VP 4

2 RNSD

VP 4

25606,0)06,0(142,3 2 xxxx

VP sm /0424,0 3 →

VP

V

1

100

0424,0

0225,0x

%066,53 →

(3)

2.5 2.5.1 Refrigerant mass flow rate

Mass flow = Re

Load

= 130

50

= 385,0 kg/s →

(2)

2.5.2 Work of compression.

(2) kgkJ

COPWD

/26

5

130

Re



2.5.3 Power required.

Power

MxWD

9,0

385,026x

= 11,122 kW →

(2)

2.5.4 Total heat rejected.

(2) [20]

QUESTION 3

3.1 Simple operation

Low cost and maintenance

Insignificant excessive pressure built up in casing

Impeller and shaft are the only moving parts

Quiet operations

Wide range of pressure, flow and capacities

Utilize small floors pace in different positions (Any applicable 3 × 1)

(3)

3.2 In ( DX) applications, refrigerant vapour is bypassed from the compressor discharge line to the inlet of the evaporator coil. Sensing a decrease in suction pressure, a pressure actuated valve opens to bypass hot refrigerant vapour from the compressor discharge line to the inlet of the evaporator coil, between the expansion valve and the liquid distributor. This increases the rate at which liquid refrigerant is boiled off within the evaporator coil and causes the temperature of the refrigerant leaving the coil to rise. Sensing this increased temperature, the expansion valve feeds additional refrigerant to the coil, increasing the suction pressure and temperature.

(Any applicable 5 × 1)

(5)

QUESTION 4 4.1 Shell and tube

Tube within a tube type

(2)

4.2 Inconsistent temperature

Overloaded equipment

High running costs

High maintenance costs

Inefficient energy consumption

High condensation in the system

Increased noise of equipment (Any applicable 4 × 1)

(4)

kgkJ

WDTHR

/156

26130

Re


Copyright reserved

4.3.1 It determines the measurable temperature at which wax separates from an oil (1) 4.3.2 Dielectric strength is a measure of the resistance of an oil to an electric

current (1)

4.3.3 Viscosity means how thick or thin an oil is at a given temperature and how

readily it flows at that temperature

(1) 4.4 Can easily be bent around obstacles and for direction changes using a

tube bender

It is easy to join by brazing or soldering

It is easy to flare or swag (Any applicable 3 × 1)

(3)

4.5 An evaporative condenser consists of an air-cooled condenser over which

water is allowed to flow.The air blast from the fan causes the water to evaporate from the surface of the condenser fins, absorbing heat from the refrigerant in the evaporation process. The water, which is exposed to the air flow, evaporates very rapidly.The latent heat required for the evaporating process is obtained by a reduction in sensible heat, therefore a reduction in the temperature of the remaining water. The hot refrigerant vapour is piped through a spray chamber where it is cooled by the evaporation of the water coming in contact with the refrigerant tubing. Since the cooling is accomplished by the evaporation of water, the water is only a fraction of what is used in conventional water-cooled

condensers. (Any applicable 5 × 1)

(5)

4.6 Chill room

Processing room

Quick-freezer room

Locker room

(4) [20]

QUESTION 5 5.1 Mechanical vibration analysis for rotating machinery

Periodic lubricant testing for refrigerating compressors

Vessel inspection with ultrasonic thickness gauges

Infrared inspection of all equipment (Any applicable 4 × 1)

(4)

5.2 5.2.1 Vibration

5.2.2 Vibration eliminators 5.2.3 Horizontal 5.2.4 Vertical 5.2.5 Flexible connectors 5.2.6 Rubber hose 5.2.7 Bellows connectors

(7)


Copyright reserved

5.3 • The capacity required in kW T.H.R

• The entering of dry bulb temperature • The condensing temperature in °C • The altitude where the condenser is to operate • Type of refrigerant • Amount of sub-cooling required in °C • Saturated suction temperature of the plant

• Type of compressor (Any applicable 4 × 1)

(4) [20]

QUESTION 6 6.1 Heat is applied to the generator or absorber, which will liberate ammonia gas

from its absorbent and will increase the pressure to the point where the air or water cooled condenser will remove sufficient heat from the high pressure gaseous ammonia to reduce it to a liquid. The liquid refrigerant is forced from the condenser to the receiver or storage tank by pressure of the vapour entering the condenser. The liquid refrigerant is forced from the condenser to the receiver or storage tank by the pressure of the vapour entering the condenser. After sufficient ammonia is driven into the condenser and receiver, the heat is discontinued and the generator or absorber cools. When the temperature of the absorber is lowered sufficiently, it begins taking back or reabsorbing the ammonia gas. As the absorber cools it attracts the gas ammonia molecules and as they enter the absorber are changed to a liquid two things happen. First, the absorber heats up and this heat must be removed. Second, the p[pressure in the container is reduced to the level where the liquid ammonia can convert again to a gas at such a low heat level that refrigeration temperatures are produced. The apparatus is so constructed that it can obtain the ammonia only from the cooling unit, which necessitates an evaporation taking place. This will cause removal of heat from the cooling unit and its surroundings which results in refrigeration. As the ammonia evaporates from the cooling unit, it is replaced by liquid ammonia from the receiver. This operation continues until the proper amount of ammonia is reabsorbed by the generator, when heat is again applied and ammonia is again moved to the receiver


(10)

6.2 6.2.1 Thermostat

This mechanism varies the amount of cooling by varying the amount of heat. The more that is fed to the generator the cooler the evaporator will become. A thermal bulb clamped to the cooling coil will create pressure if the cooling coil becomes warmer. This pressure is carried to a diaphragm by a capillary tube. An increase of pressure in the diaphragm opens the gas valve and more gas flows to the burner. The increased cooling resulting from the added heat will cool the thermal bulb and the diaphragm pressure will decrease, closing the valve


(5)


Copyright reserved

6.2.2 Burner The burner is used to carefully mix the air and gas in the proper proportions and burn the mixture to provide the most efficient heat. The gas enters the top, travels past the safety valve and passes through the turbulator, through the carefully sized orifice sperd, mixes with the primary air, burns at the end of the mixing tube where the secondary air and the heat enters the generator tube and flue. The burner must be exactly the right distance from the generator flue to obtain this correct amount of secondary

air. (Any applicable 4 × 1)

(4)

6.2.3 Burner safety valve The purpose of the burner safety valve is to prevent unburnt gas from escaping if the flame is accidentally snuffed out. The safety valve is operated by the flame. A metal plate, touches this flame and becomes quite hot. This heat is transmitted to a bi-metallic disc which will move with a snap and open the gas valve when hot. If this disc cools somewhat the bi-metallic disc will dish air the other way pulling the valve shut and stopping all gas flow. It is important that the heat conductor strip be kept up against the

outer edge of the flame (Any applicable 5 × 1)

(5)

6.3 Overloaded cabinet

Improper condensing temperatures

Little or no heating of the generating unit (Any applicable 2 × 1)

(2)

TOTAL: 100

Website: www.ekurhulenitech.co.za Email: [email protected]

REGISTERING WITH OUR COLLEGE

CONSIDER REGISTERING WITH OUR COLLEGE AND WE HAVE THE

FOLLOWING TYPES OF LEARNING:

ONLINE OR CORRESPONDENCE LEARNING. THIS IS THE MOST

PREFARED ONE BY THOUSANDS

PART-TIME CLASSES DURING SATURDAYS IF NEAR OUR

OFFICES IN KRUGERSDORP

FULL-TIME CLASSES IF NEAR OUR CLASSES IN

KRUGERSDORP

ABOUT EXTRA TEXTBOOKS

IF LOOKING FOR TEXBOOKS FOR CERTAIN SUBJECTS I N1-N6 ENGINEERING

STUDIES PLEASE SEND US AN EMAIL ON [email protected]

ABOUT VIDEO MATERIAL

WE HAVE VIDEO MATERIAL FOR EXTRA UNDERSTANDING ON CERTAIN

ENGINEERING FOR A FEE. REQUEST A QUOTE. SEND US AN EMAIL ON

[email protected]

EKURHULENI TECH COLLEGE. No. 3 Mogale Square, Krugersdorp.

Website: www. ekurhulenitech.co.za

Email: [email protected] TEL: 011 040 7343 CELL: 073 770 3028/060 715 4529



past exam paper & memo n3 - ekurhuleni tech …...compressor has a bore of 60 mm and a stroke...

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