Air- conditioning

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TABLE OF CONTENT NO 1 2 3

ITEMS INTRODUCTION FUNDAMENTAL EQUATION

PAGES 2 3-6 7-19

STAGE 1 

QUESTION 1



QUESTION 2



QUESTION 3



QUESTION 4

4

20-34 STAGE 2

5 6

1



QUESTION 5



QUESTION 6



QUESTION 7



QUESTION 8

CONCLUSION REFERENCES

35 36

INTRODUCTION

Air- conditioning has a working principal as the refrigerator. But, instead of cooling small space area inside of the refrigerator, an air conditioner capable to cool a room, a whole house or an entire hall. The main purpose of air conditioning is as an equipment to ensure personal human comfort especially inside a house. Air-conditioner also being used in industry and not limited only for human comfort. For example, in manufacturing industry, airconditioner used to maintain correct operation of equipment that need to working in specific environmental conditions.

Air conditioner often used chemical such a refrigerant 134a because theses substance easily to change from a gas to liquid and reverted back. The purpose of this refrigerant in the system to transfer heat from air in home to outside. This system mainly combination of 4 main components which are evaporator, compressor, condenser and throttling valve. Mostly, air conditioning systems prepared for cooling using the principal of chilled water. The basic concept is air go through a coil of chilled liquid and the heat is delivered from the air to the chilled water. The chilled water is rotated through so that chilled water is always exist as air comes from.

A chiller performs on the concepts of vapor compression or vapor absorption. This system capable to provide a continuous flow of coolant to water cool side at about 10℃.

Refrigerant inside a chiller system will circulated through the four main devices. All the components in chiller system will undergoes thermodynamic process. Different components carry out different task and each task play vital role to ensure process running smoothly.

Evaporator through the process of coolant absorb heat and delivered to the refrigerant. This process resulting in evaporation of the refrigerant and changed into vapor state.

Compressor carry out multipurpose task by removes refrigerant from evaporator and increasing temperature refrigerant high enough to reached condenser. As the name condenser, function to condense refrigerant and change it back to liquid state. Cooling medium which is water/air will carry out the latent heat energy from the environment.

2

FUNDAMENTALS QUESTIONS

i)

Heat transfer through building mainly in three ways which is convection, conduction and radiation.

Convection is the heat flow in a fluid which the higher temperature fluid rise upward and the lower temperature move down. At homes, the air is fluid and in the ocean the water is the fluid. When the temperature becomes higher the buoyancy of air becomes higher and the molecule of the air move apart and hence floating upward. The result from the air moving upward, the cold air is being pulled to the below to replace the hot air. To take the details for air flow in the building, these are the parables which is difference in the temperature (ΔT). For the heat to flow in a building, it is crucial for temperature difference from one space to another. Then time (t). We observed the time during the air movement occurs. After that, the volume of air (V). The width, length and height of the space interior is being calculated in a building and the volume is constant whether the air change or not. Lastly the changes of air per hour (AC/hr). The air change is being calculate as the rate of the air movement.

Conduction is the heat transfer from a molecule into another molecular in a solid state. The molecule is vibrated within its own place and hence transfer the heat to another. Every building material has their own rates of conduction. The metals such as steel and copper have high conductivity which at a high potent rate.

There is also a low conductivity material such as rigid foam and fiberglass batts.

This serve as an insulator and is placed in between the high conductive material such as a wall or a roof. The rate is measured using U-value which is the heat transfer rate and the R-value which is the heat transfer resistance.

3

Radiation is an electromagnetic wave that move heat through space and not air. Sun emitted heat energy to the earth in form of solar radiation. Furthermore, the air does affect the radiation. There are several variables that affect the heat transfer rate. First, the two surfaces distance. If the earth were close enough to the sun, the earth would vaporize by the sun’s immerse energy output. The example is similar to campfire which is the further from the campfire, the less heat we would feel. Next, the surfaces optical properties. For example, dark surfaces absorb higher radiant energy while white or shiny surfaces reflect the radiant energy. In summer, roofs and windows are two main sources for heat gains in the building.

As to control the heat, there are a thin metal coating on the window surface to reflect the radiant heat. On roof, the manufacture use light coloured or install radiant barrier to reflect the heat. Lastly, the angle between the two surfaces with same optical properties. If both of the surfaces are facing each other, the energy is absorbed while if it is not facing each other, the radiant energy will bounce. The energy radiant moves in a straight line and if the surface is parallel to each other.

4

ii)

a) Heat transfer can be reduced in winter by installing an insulator in the building.

Insulators in building such as plastic foam and fibre glass is used to trap the air in the building and to reduce the heat energy transfer so that inside the building the heat stays. The insulators are being put into the wall’s cavity to trap heat. Next, close the gap between the timber flooring and skirting boards. The gaps have a big impact in holding the heat in the living space. Furthermore, use glazing in windows known as low e coating or low emissivity coating. The function is to reduce the heat transfer from inside and heat loss to the outside in winter. Lastly solar passive heating design is being used to trap heat from the sun in more lucrative manner. This means make sure that the window facing east and west as the heat energy that is gain from the sun are major contribution in heating the building. b) Heat transfer in summer can be reduced is by using window properties that is low in Solar Heat Gain Coefficient (SHGC). The SHGC is used to rate the amount of energy that is permissible through the window. The lower the SHGC, the less energy transfer that can be transmit through the window in solar energy. Next, face the window to south and north to avoid higher solar heat gain in summer.

Reduce the number of window and horizontal projecting shaded, and install interior shades.

Brick

Insulation Figure 1 Assume : Height of wall = 2.7m Length = 7m 5

T∞,1 = 38℃ T∞,2 = 30℃ Brick wall

Insulation layer

Thickness

0.23m

0.0575 m

Thermal Conductivity, k

0.77 W/m. K

0.077 W/m. K

Heat transfer coefficient, h

0.84 W/m2. K

BRICK WALL WITHOUT INSULATION

Rconv, 1 =

1 hA

Rconv, 1 =

1 ( 0.84 ) (2.7 x 7)

=

= 0.0630⁰C/W

1 kA

Rbrick wall =

BRICK WALL WITH INSULATION

= 0.0630⁰C/W

=

1 hA

1 kA

Rinsulation =

1 (0.077)(2.7 x 7)

=

= 0.06871⁰C/W Rconv,2 =

1 ( 0.84 ) (2.7 x 7)

=

1 (0.77)(2.7 x 7)

=

1 hA

= 0.6871⁰C/W

1 kA

Rbrick wall =

1 (0.84)(2.7 x 7)

= 0.0630⁰C/W

=

1 kA

=

1 (0.77)(2.7 x 7)

= 0.06871⁰C/W RTotal = 0.0630 + 0.06871 + 0.0630

Rconv,2 =

= 0.1947 ⁰C/W =

1 hA 1 (0.84)(2.7 x 7)

= 0.0630⁰C/W RTotal = 0.0630 + 0.6871 + 0.06871 + 0.6871 + 0.0630 = 1.5689⁰C/W 6

QTotal =

∆T Rtotal

QTotal =

∆T Rtotal

=

T ∞ ,1−T ∞ ,2 RTotal

=

T ∞ ,1−T ∞ ,2 RTotal

=

38−30 0.1947

=

38−30 1.5689

= 41.089 W

= 5.099 W

Problem Statement To finalize the design and operation of an air-conditioning system for the WISMA TUNWAN building, as shown in Figure 1.

System Analysis

Figure2: System schematic of the centralized chiller system for the WISMA TUNWAN building.

Stage 1 Question 1 Component Windows



Detail 20 windows at both North and South side of the building.



10 windows at West side and no window on the East side of the building

Wall

7



2ft. width and 4ft. height for each window.

 

The glass is single, no shading and no outside awnings.

120ft.width and 20ft. height at North and South sides.



100ft. width and 20ft. height at East and West sides.



The East and West side is exposed to the sun.

Roof



Size of the roof is 120ft. x 100ft.



Roof is directly exposed to the sun.



It is flat with no vented air and it is non-insulated.

Table 1: Specification of the WISMA TUNWAN building.

8

5 possible heat load causes and its standard quantities when hall is used for seminar.

Head Load Direct Solar Heat Gain by the Windows

Standard Quantities Size: 2 x 4 = 8 sq ft.

The rounded factor for window in West is 150 while for window in North and South is 100.

10 x 8 x 150 = 12000 BTU/HR. 40 x 8 x 100 = 32000 BTU/HR. So, the highest value which is 32000 BTU/HR has been selected. Thus, the total solar heat gained by the window for this building is

Heat Gained by the Walls

32000 BTU/HR. Size (S&N): 120 x 20 = 2400 sq ft.

Size (E&W): 100 x 20 = 2000 sq ft.

The total area of walls that exposed to the sun is 2000 sq ft. Since the designed temperature difference is 22F and there is no insulation, the factor associated with it is 7.

2000 x 7 = 14000 BTU/HR. So, the total heat

Heat Gained by the Roof

gained by the wall is 14000 BTU/HR. Size: 100 x 120 = 12000 sq ft.

For 22F of design temperature difference, the factor associated with it is 34. Thus, the total heat gained by the roof is 12000 x 34 =

Heat Gained by the Electrical Appliances

408000 BTU/HR. 3 units of PA system with 150 W head load.

150 units of lighting with 3.4 W head load.

2 units of Laptop with 70 W head load.

2 units of projector with 616 W head load.

3 x 74 = 222 W 150 x 3.4 = 510 W

9

2 x 70 = 140 W 2 x 616 = 1232 W Total = 222 + 510 + 140 + 1232 = 2104 W Thus, the total heat gained by electrical appliances is 2104 x 3.412 = 7179.15 Heat gained by Human

BTU/HR. Assuming 220 participants.

30 members of Seminar Facilitator.

Heat gained by each participants and facilitator that seated and light worked is 245.

(220 + 30) x 245 = 61250 W Thus, heat gained by the human is 61250 x 3.412 = 208993.68 BTU/HR.

Table 2: Calculation of Heat Gained by 5 Possible Heat Load Causes

10

Question 2 Sample Calculation Convert :13

L kg =0.217 min s

C P water =4.21 kJ /kg . k C P air =1.005 kj/kg . k

´ V́ =ḿ water =0.217 kg /s ρair =1.225 kg /m3 Qc ,chilled water =ḿ×C P ,water × ( ∆ T chilled water )

(

¿ 0.217

kg kJ × 4.21 × ( 10° C−4 ° C ) s kg .˚ C

)(

)

Qc ,chilled water =5481.42W Energy Balance ,Q c, chilled water=Qh , air

Qh ,air =0.9× Qc ,chilled water ¿ 0.9 ×5481.42W ¿ 4933.28 W CW =ḿ W C P W

¿ 0.217 ( 4.21 k ) ¿ 913.57 C air =ḿair C P air

¿ 0.204 ( 1.005 k ) ¿ 205.19 At T = 4˚C 3

v =0.001 m / kg

11

Sample Calculation At V = 10000 LPM ḿ=ρ V́

At V = 11000 LPM ∴ ḿ=ρ V́

¿ 1.225(0.167)

¿ 1.225 ( 0.183 )

¿ 0.204 kg /s

¿ 0.225 kg /s

∴ Q́=ḿ C P ∆ T

∴ Q́=ḿ C P ∆ T

4933.28=0.204 ( 1.005 k ) (38−T 2)

4933.28=0.225 ( 1.005 k ) (38−T 2 )

T 2 =13.97 ℃

4933.28=8592.75−226.125 T 2

At V = 12000 LPM

T 2 =16.18℃ At V = 13000 LPM

∴ ḿ=ρ V́

∴ ḿ=ρ V́

¿ 1.225 ( 0.2 )

¿ 1.225 ( 0.217 )

¿ 0.245 kg /s

¿ 0.265 kg /s

∴ Q́=ḿ C P ∆ T

∴ Q́=ḿ C P ∆ T

4933.28=0.245 ( 1.005 k ) (38−T 2 )

4933.28=0.265 ( 1.005 k ) (38−T 2 )

4933.28=9356.55−246.23T 2

4933.28=10120.35−266.33 T 2

T 2 =17.96℃ At V = 14000 LPM

T 2 =19.48℃ At V = 15000 LPM

∴ ḿ=ρ V́

∴ ḿ=ρ V́

¿ 1.225 ( 0.233 )

¿ 1.225 ( 0.25 )

¿ 0.286 kg /s

¿ 0.306 kg /s

∴ Q́=ḿ C P ∆ T

∴ Q́=ḿ C P ∆ T

4933.28=0.286 ( 1.005 k ) (38−T 2 )

4933.28=0.306 ( 1.005 k ) (38−T 2 )

4933.28=10922.34−287.43T 2

4933.28=11686.14−307.53 T 2

T 2 =20.84 ℃ At V = 16000 LPM

T 2 =21.97 ℃ At V = 17000 LPM

∴ ḿ=ρ V́

∴ ḿ=ρ V́

¿ 1.225 ( 0.267 )

¿ 1.225 ( 0.283 )

¿ 0.327 kg /s

¿ 0.347 kg /s

∴ Q́=ḿ C P ∆ T

∴ Q́=ḿ C P ∆ T

4933.28=0.327 ( 1.005 k ) (38−T 2 )

4933.28=0.347 ( 1.005 k ) (38−T 2 )

4933.28=12488.13−328.64 T 2

4933.28=13251.93−348.74 T 2

T 2 =22.97 ℃

T 2 =23.86 ℃

12

At V = 18000 LPM

At V = 19000 LPM

∴ ḿ=ρ V́

∴ ḿ=ρ V́

¿ 1.225 ( 0.3 )

¿ 1.225 ( 0.317 )

¿ 0.368 kg /s

¿ 0.388 kg /s

∴ Q́=ḿ C P ∆ T

∴ Q́=ḿ C P ∆ T

4933.28=0.368 ( 1.005 k ) (38−T 2 )

4933.28=0.388 ( 1.005 k ) (38−T 2 )

4933.28=14053.93−369.84 T 2

4933.28=14817.72−389.94 T 2

T 2 =24.64 ℃ At V = 20000 LPM

T 2 =25.35℃

∴ ḿ=ρ V́ ¿ 1.225 ( 0.333 ) ¿ 0.408 kg /s

∴ Q́=ḿ C P ∆ T 4933.28=0.408 ( 1.005 k ) (38−T 2 ) 4933.28=15581.52−410.04 T 2 T 2 =25.98℃

Table 3: Calculation of Temperature out from the system.

Air Flow Rate (LPM) 10000 11000 12000 13000 14000 15000 16000 17000 18000 19000 20000

Mass Flow Rate, ḿ( 0.204 0.225 0.245 0.265 0.286 0.306 0.327 0.347 0.368 0.388 0.408

kg ) s

T 2 (℃) 13.96 16.14 17.96 19.51 20.83 21.97 22.97 23.86 24.64 25.35 25.98

Table 4: Data of Air Flow Rate, Mass Flow Rate and Temperature (T2)

13

Object 239

Figure 3: Graph of Temperature (˚C) vs Air Flow Rate (LPM)

According to ASHRAE standard 55-2013, Thermal Environmental Conditions for Human Occupancy, notes that suitable temperature for thermal purposes could range between approximately 19.44˚C (67˚F) and 27.78˚C (82˚F).

Based on the range air flow range from 10000L/min to 20000L/min given in the question after calculating the exit temperature, the operating range of air flow has been identified. The suitable operating range for the air flow into the AHU needed to meet the standard comfort temperature requirement is ranging from 13000 – 20000 L/min.

14

Question 3 Free Convection T film= T film=

( T s +T ∞ ) 2

( 10+30 ) =20 ℃ 2

∴ ¿ the value of T film , the value that have been obtained are: −5

2

L=19 m , D=0.25 m, V =1.516 × 10 m /s , Pr=0.7309

μ=1.825 ×1 0−5 , k =0.02514 W /m . k , ρ=1.204 kg / m3

β=

β=

1 T film 1 −3 =3.413 ×10 20+273

Ra = D

gβ ( T s +T ∞ ) D V

2

3

× Pr

0.25 (¿¿ 3) × 0.7309 −5 2 (1.516 ×1 0 ) Ra =¿

9.81(3.413 ×10−3) ( 30−10 ) D

Ra =33.27 ×10 6 D

15

2

0.387 Ra 1 /6

{ [( )]} { [( )]}

Nu D = 0.6+

D

0.559 9/ 16 1+ Pr

8/ 27

6 1 /6

Nu D = 0.6+

{

Nu D = 0.6+

0.387 ( 33.27 ×10 ) 0.559 1+ 0.7309

9 /16 8/ 27

2

6.9404 =40.64 1.2019

}

h=

k × Nu D

h=

0.02514 × 40.64 0.25

h=4.087 W /m2 ℃

A s =πDL

A s =π (0.25)(19) A s =14.92m2

Q́=h A s ( T s −T ∞ ) Q́=4.087(14.92) ( 10−30 ) Q́=1219.56 W

16

2

Forced Convection At V = 2 m/s ℜ=

VD μ

ℜ=

2(0.25) 1.825 ×10−5

∴ ℜ=2.7397× 1 04

0.618

Nu=0.193 ℜ

Pr

1 /3

2.7397 ×1 04 ¿ ¿ Nu=0.193 ¿ ∴ Nu=96.09

h=

k × Nu D

h=

0.02514 × 96.09 0.25 3

∴ h=9.663W /m ℃

A s =πDL

A s =π (0.25)(19) ∴ A s =14.92m 2

Q́=h A s ( T s −T ∞ ) Q́=9.663 (14.92) ( 30−10 ) ∴ Q́=2883.44 W 17

18

V = 4 m/s 4 ℜ=5.4795× 10 0.805 1 /3 Nu=0.027 ℜ Pr

V = 6 m/s 4 ℜ=8.2192× 10 0.805 1 /3 Nu=0.027 ℜ Pr

Nu=158.73 3 h=15.96W /m ℃ Q́=4762.46W

Nu=220 3 h=22.12W /m ℃ Q́=6600.61 W

Table 5: Data of Re, Nu, h and Q́ for velocity 4 m/s and 6 m/s using same equation as 2 m/s

Object 338

Figure 4: Graph of Heat Transfer Rate (W) vs Velocity (m/s)

19

Question 4 Assuming the heat transfer is steady. The energy balance become: Qconvection =Qwater From calculation before, at velocity = 0 m/s (free convection) Qconvection =1219.56 W

Q́=ḿ C p ∆ T Q́=ḿ C p ( T ¿−T out )

l 0.001 m3 1 min kg ḿ=13 × × ×1000 3 min 1l 60 s m ḿ=0.217 kg /s

Specific heat of water, Find Cp at T=10˚C (Interpolate from table A-3) 10−0 C p −4.22 = 25−10 4.18−C p 41.8−10 C p =15C p−63.3

−25 C p =105.1 C p =4.204 kJ /kg . k

20

At V = 0 m/s

At V = 2 m/s

1219.56= ( 0.217 ) ( 4.204 k ) ( T ¿−10 )

2883.44=( 0.217 ) ( 4.204 k ) ( T ¿ −10 )

1.34=T ¿ −10

3.16=T ¿ −10

∴ T ¿ =11.34 ℃ At V = 4 m/s

∴ T ¿ =13.16 ℃ At V = 6 m/s

4762.46=( 0.217 ) ( 4.204 k ) ( T ¿ −10 )

6600.61=( 0.217 )( 4.204 k ) ( T ¿ −10 )

5.22=T ¿−10

7.24=T ¿ −10

∴ T ¿ =15.22 ℃

∴ T ¿ =17.24 ℃

Table 6: Calculation of temperature out for 0, 2, 4 and 6 m/s

Object 384

Figure 6: Graph of Temperature (˚C) vs Velocity (m/s)

5. Select a commercial refrigerant (other than R134a) and provide a list of important 21

properties for the refrigerant. Then, assign the suitable operating pressures of the chiller system based on all known operating conditions of the chiller system. Explain the choice of pressures.

Refrigerant is type of compound that usually in either a fluid or gaseous state. It operates as medium to absorb heat from the environment and can provide refrigeration or air conditioning when using with other components such as compressors and evaporators. Most commonly used commercial refrigerant is R134a. There is another suitable refrigerant for air conditioning which is R-407c. Important properties of refrigerant R-404a are: 1) R-407c is a ternary mixture of non-azeotropic combined refrigerant contain of the 3 ingredients R-125, R-134a, and R-32a, and was created as a alternative refrigerant for R-502 and R-22.

2) R-407c commonly used in air conditioning and commercial refrigerator equipment especially for low and medium temperature ranges and provide outstanding performances against many other HFCs. 3) Enable recommended properties as a refrigerant for showcases, refrigerated vehicles, ice makers and etc.

4) It also exhibits low compressor discharge temperatures which makes it suitable for singlestage compression. This will be necessary to avoid inter-stage cooling.

5) It is non-flammable and certificate as A1 group L1 which popular in high security refrigerant.

Physical properties

22

Property Molecular Weight (kg/kmol) Boiling point (°C) Saturated liquid density (25°C),

Value 86.2 −43.8 1138

kg/m3 Saturated vapour density (25°C), 43.8 kg/m3 Critical temperature (°C) 86.4 Critical pressure, bar 46.3 Liquid heat capacity @ 25°C, 1.533 (Kj/(kg·K)) Vapour heat capacity @ 1.013 bar 1.107 (Kj/(kg·K))

For evaporator, lowest operating pressure must be higher than atmospheric pressure.

This is to ensure leaking in the system can be avoided. Refer to property table R-407c and value of T inlet from question 4, the suitable operating pressure for evaporator is between 0.7 MPa to 1.0 Mpa.

Inlet temperature, at evaporator (from Q4)

Suitable pressure (Refer to R-407c property

11-18℃

table) 700 kPa – 1000 kPa

Table 7: Operating pressure for evaporator

As for condenser, temperature of refrigerant cannot lower than temperature of cooling medium. Then for ensure that process of heat rejection become isothermal saturation pressure of refrigerant has to below it critical pressure. Based on from question surrounding temperature is 30 ℃. Also at condenser the operating pressure must be able to make the 23

refrigerant in superheated state. Refer to R-407c the suitable operating pressure range is between 1.2 Mpa to 3.6 MPa.

Temperature surrounding, 30℃ 28.30℃ = 1.1 Mpa

Suitable operating pressure Must be above 1.16 Mpa

31.30 ℃ = 1.2 Mpa

P suitable = 1.2 Mpa – 3.6 Mpa

Interpolation method, P30 = 1.1 +

30−28.30 31.30−28.30

= 1.16 Mpa

Table 8: Operating pressure for condenser

24

6. Identify a specific multi-pass heat exchanger for the system. For one chosen case of chilled water inlet temperature from Q4, as well as using the LMTD method, discuss the overall heat transfer coefficient, U, profile of the evaporator if the effective surface area changes from 2m2 to 20m2. Choose the plot intervals accordingly.

Figure 7: System flow Heat exchanger used is 2-shell passes and 8-tubes.

Advantages over single passes, 1) Can double side velocity and improving heat transfer coefficient 2) Reduce probability effect of fouling in tube side 3) Reducing area required for the operation Given mass flow rate, 13L/m ṁ = (13𝑙/𝑚𝑖𝑛)×(1 𝑘𝑔)×(1 𝑚𝑖𝑛/60𝑠) = 0.217 𝑘𝑔/𝑠 Cp air, kJ/kg.K 1.005 Cp water, kJ/kg.K 4.186 Table 9: Specific heat for mass and water Based on Q4, T inlet chiller = Tout AHU = 13.16℃ at v = 2 m/s Q water = Q air 𝑄 𝑤𝑎𝑡𝑒� = 𝑚̇Cp,𝑤𝑎𝑡𝑒� Δ� = (0.217) (4.186) (10 – 4) = 5.46 kW 25

Assuming that the heat loss to the surrounding during the heat exchange process in the AHU is 10% from the emitted heat transfer rate, Qh ,air =0.9× Qc ,chilled water ¿ 0.9 ×5481.42W ¿ 4933.28 W Assume that heat exchanger in evaporator occur between hot air and the chilled water for the process rejecting heat to the air. Air at the initial condition is 38℃.

4.933 kW = (0.217 x 1.005) x (38 – Th out) 38 – Th out = 22.62 Th out = 15.38 ℃ Obtain Tc out, Cp = 4.204 kJ/kg.K (from Q2) 4.933 kW = (0.217) (4.204) (Tc out-13.16) Tc out – 13.16 = 5.407 Tc out = 18.58 ℃ Heat exchanger occurs between air and chilled water at the evaporator, Th in = 38℃

Tc in = 13.16 ℃ (from Q4)

Th out = 15.38 ℃

Tc out = 18.58 ℃

By using LMTD method, find log mean temperature, Δ�m and correction factor, F. Using counter-flow heat exchanger, 38 – 18.58 = 19.42 ℃ Δ�2= Th in – Tc out 15.38- 13.16 = 2.22 ℃ Δ�1= Th out – Tc in Table 10: Log mean temperature

26

ΔT 2−ΔT 1 ΔT 2 ln ⁡( ) ΔT 1

Δ�𝐿𝑀�� =

19.42−2.22 Δ�𝐿𝑀�� = ln ⁡( 19.42 ) 2.22

= 7.93 ℃

Obtain correction factor for 2- shell passes and 8-tubes from the graph,

Figure 8: Graph for 2-shell passes and 8 tubes P= P=

t 2−t 1 T 1−t 1

18.58−13.16 38−13.16

P = 0.22 (red line)

R= R=

T 1−T 2 t 2−t 1 38−15.38 18.58−13.16

R = 4.17 (orange line)

t = refer to tube side T = refer to shell side Table 11: Correction factor agent

27

Hence the correction factor obtained from graph, F = 0.85 (green line). 𝑄=𝑈𝐴�Δ�𝐿𝑀�� � Arrange the equation, U=

Q ASΔTLMTD F

U=

4.933 kW 2( 7.93)(0.85)

= 365.92 W/m2℃ for area 2m2

Area, m2 Overall heat transfer coefficient, W/m2℃ 2 365.92 4 182.96 6 121.97 8 91.48 10 73.18 12 60.99 14 52.27 16 45.74 18 40.66 20 36.59 Table 12: Area versus overall heat transfer coefficient

28

Figure 8: Graph area versus overall heat transfer coefficient From the graph, result indicate that when area decrease overall heat transfer coefficient will decrease. Based on newton law of cooling, Q =h.A.ΔT indicate that increase in heat transfer rate at higher transfer coefficient value. Convection can be described as bulk motion of fluid where it can occur naturally and forcefully. When the surface area bigger it will affect the velocity of fluid by decreasing it. Fluid move slower on bigger area and this proved by Bernoulli’s principle where ṁ = ρVA and it explain velocity and density inversely proportional with area. Hence heat transfer coefficient decreases if the velocity of fluid also decreases.

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7) For one chosen case of chilled water inlet temperature from Q4, determine the mass flow rate (in kg/s) of the refrigerant and calculate the Coefficient of Performance for the chiller system. Also assume the refrigerant exits the condenser as saturated liquid.

T inlet, T = 15.22 ℃ at V = 4 m/s and Q = 4762.46 Watt

Figure 9: T-s diagram for vapor compression cycle

At state 1 in saturated vapor, T, ℃ 15 15.22 16

Sg, kJ/kg.k 1.7785 S1 1.7776 Table 13: Interpolation of entropy

From property table R-407c and interpolation method and T = 15.22℃, P = 839.98 kPa S1 = 1.7785 +

15.22−15 16−15

(1.776-1.7785)

= 1.7780 kJ/kg.K.

T, ℃ 15 15.22 16

Hg, kJ/kg 421.8 h1 422.3 Table 14: Interpolation of enthalpy

h1 = 421.8 +

15.22−15 16−15

= 421.91 kJ/kg

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(422.3-421.8)

At state 2 in superheated and S1=S2 due to the isentropic compression, Assume pressure P2 = 3.6 MPa Sg, kJ/kg.K 1.7776 1.7780 1.7960

Hg, kJ/kg 459.5 h2s 466.4 Table 15: Interpolation of entropy

1.7780−1.7776 1.7960−1.7776

h2s = 459.5 +

(466.4-459.5)

= 459.65 kJ/kg When isentropic efficiency of the compressor is 88%, ŋ=

h2 s−h1 h 2−h 1

0.88 =

459.65−421.91 h 2−421.91

h2 = 464.80 kJ/kg

At state 3 saturated liquid and P3 = 3600 kPa, P 3570.8 3600.0 3642.0

Hf, kJ/kg 333.1 h3 335.9 Table 16: Interpolation of enthalpy

h3 = 333.1 +

3600−3570.8 3642−3570.8

(335.9-333.1)

= 334.25 kJ/kg

At state 4 h4 = h3 due to throttling device, h4 = 334.25 kJ/kg

Given equation for mass flow rate, QH = ṁ (h2-h3) 4762.46 = ṁ (464.80-334.25) ṁ = 0.0365 kg/s

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Then for COP the formula given, COP =

QL Win

QL Win ṁ (h1-h4) ṁ (h2-h1) 0.0365 (421.91-334.25) = 3.2 kW 0.0365(464.80-421.91) = 1.57 kW Table 17: Power load and power in

COP =

3.2kW 1.57 kW

= 2.04

Coefficient of performance air conditioning system indicates that effectiveness of the system to transfer heat against the amount of electrical power consumes. This means higher COP value, more efficient the air conditioning system. High COP also equate to lower operating cost. This question based on refrigeration cycle with saturated liquid instead of subcool liquid. Cycle with saturated liquid required smaller work input compare to sub-cooled liquid. The evaporator also designed with shell and multi pass heat exchanger. Hence, that are the reasons why the value of COP is higher.

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8. The cooling water temperature exiting the condenser was measured at 38℃ and returns to the condenser at 28℃ with a flow rate of 10 litres per minute. At the cooling tower, the inlet air humidity is 70%. Plot the saturated air exit temperature profile from the cooling tower if the air flow rate changes from 20,000 to 100,000 litres per minute.

Given, Mass flow rate, Kg/s 0.167 Cp Water, kJ/kg.K 1.8723 (Table A-2) Cp Air, kJ/kg.K 1.005 (Table A-2) Temperature air surrounding, ℃ 30 Relative humidity 0.7 Table 18: Data information for calculation

ṁw = 10 L/m = (10 𝐿𝑚) ( 1𝑘𝑔1𝐿 )( 1𝑚60𝑠 ) = 0.167 kg/s Q water = Q air Q water = ṁw Cpw (Th in – Th out) Q water = (0.167) (1.8723) (38-28) =3.127kW �1= 0.622 𝑃�/𝑃− 𝑃� 𝑚= ̇ �/̇ � 𝑃�=𝑅� �= 𝑅�/𝑃𝑎 𝑃𝑎=𝑃−𝑃�

At T = 30℃, P sat = 4.2469 kPa from table A-4 𝑃�= ∅ 𝑃𝑔 𝑃� = (0.7) x (4.2469 kPa) =2.9728 kPa �1 =

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0.622 x 2.9728 101.325−2.9728

= 0.0188 kg H2O/kg dry air

At T1 = 30℃ Pa = P-Pv Pa = 101.325 – 2.9728 kPa = 98.35 kPa �=

0.2870 x (30+273) 98.35

= 0.884 m3/kg

-Volume are flow rate are given by range of 20000-100000 L/m. Hence, to calculated mass V flow rate of air, the formula used is ṁ = v V air, L/m 20000 40000 60000 80000 100000

V air, m3/s v, m3/kg 0.3333 0.884 0.6667 0.884 1.0000 0.884 1.3333 0.884 1.6667 0.884 Table 19: Calculated data

At ṁ air = 0.3770 kg/s Q air = ṁ air Cpair (Tc out – Tc in) 3.127 Kw = 0.3770(1.005) x (Tc out – 30) Tc out = 38.25℃

At ṁ air = 0.7541 kg/s Q air = ṁ air Cpair (Tc out – Tc in) 3.127 Kw = 0.7541(1.005) x (Tc out – 30) Tc out = 34.13℃

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ṁair, kg/s 0.3770 0.7541 1.1312 1.5083 1.8854

At ṁ air = 1.1312 kg/s Q air = ṁ air Cpair (Tc out – Tc in) 3.127 Kw = 1.1312(1.005) x (Tc out – 30) Tc out = 32.75℃

At ṁ air = 1.5083 kg/s Q air = ṁ air Cpair (Tc out – Tc in) 3.127 Kw = 1.5083(1.005) x (Tc out – 30) Tc out = 32.06℃

At ṁ air = 1.8854 kg/s Q air = ṁ air Cpair (Tc out – Tc in) 3.127 Kw = 1.8854(1.005) x (Tc out – 30) Tc out = 31.65℃

ṁair, kg/s V air, L/m T exit, ℃ 0.3770 20000 38.25 0.7541 40000 34.13 1.1312 60000 32.75 1.5083 80000 32.06 1.8854 100000 31.65 Table 20: Data for mass flow rate, volume flow rate and Texit

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T exit, ℃ vs Volume air fow rate, L/m 45 40 35 30 25 20 15 10 5 0 20000

40000

60000

80000

100000

Figure 11: Graph oh T exit versus volume air flow rate

Heat exchanger occur at cooling tower between air and cooling water. Cooling water exit condenser in steam state which 38 ℃ and enter condenser at inlet with 28 ℃. Air with relative humidity 0.7% will cool the cooling water. Principal of thermal energy and thermodynamic is heat energy transfer from high pressure temperature to a lower temperature. Thus, resulting in increasement of temperature at exit for air and decreasing of water exit temperature. Graph shows T exit will decrease as long as volume air flow rate increase. This is due to moisture of air. High value of moisture will speed up to reduce heat of substance.

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CONCLUSION In a nutshell, the objective achieved because the problem statement answered with correct theoretical and mathematical formula. Essential key to optimizing air conditioning system is by making correct analysis and fully understand how the system work. Air shows different properties and can be control to adapt into required environment. The application is quite obvious where almost every house having air conditioner now days. This to make sure human personal comfort can be achieved especially during hot weather. Every transportation such as car has its own conditioning to make sure passenger feel cooled inside the car. In food industry, where air conditioning system used to keep the raw substance at desired temperature for prevent food spoiled.

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REFERENCES 1) Lienhard, John H. IV; Lienhard, John H. V (2019). A Heat Transfer Textbook (5th ed.). Mineola, NY: Dover Pub. 2) Yunus A.Cengel and Boles, M. (1998), Thermodynamics, An Engineering Approach, McGraw Hill.

3) Evans. P. (July 22, 2017). Chiller Cooling Capacity – How to Calculate. Retrieved from http://theengineeringmindset.com/chiller-cooling-capacity-calculate/ 4) Woodford,

C.

(March

24,

2018).

Air

Conditioners.

Retrieved

from

https://www.explainthatstuff.com/airconditioner.html 5) Property table R-407c (5/7/2020). Property table at various pressure and temperature. Retrieved from https://www.freon.com/en-/media/files/freon/freon407c--thermodynamic-properties-si.pdf.

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