possible heat load causes

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1. List 5 possible heat load causes (and its standard quantities) from the hall when it is being used for a specific function, ie. exhibition, wedding, seminar etc. [CO1]

Basically, there are many sources of heat load that happens during the ceremony in hall the when it using. Large number of occupants, each of occupants will comes with their own gadget, lighting and computers for PA system contributes to heat load causes which can make uncomfortable condition throughout the ceremony, especially for the important guests. Below is the description for each of the heat load causes from the hall when it is used for any ceremony during sunny day; a) Heat from human beings Occupants are major source of heat in building interiors. Considering that a human consumes hundreds of calories each day in the form of food, and part of this energy is released as heat during metabolic processes. The heat released by humans is even higher during intense physical activity, through perspiration. The heating effect of humans also increases depending on occupant density. As a result, the human contribution to the total heat load can be especially high in the hall. b) Heat from electrical and electronic appliances Indoor space in the hall are filled with electrical and electronic appliances such as lighting fixtures, smartphones for each occupants, laptop and LCD monitor. These appliances consume electricity and release some heat in air-conditioned spaces. c) Solar heat gain Three different ways in which heat from the sun can reach interior space of the hall which are conduction and convection. Conduction occurs across walls and roofs of the hall since they are exposed to a temperature difference between the hall interiors and the warmer outdoor environment. Convection refers to heat transfer from surfaces hot outdoor air or indoor air movement to surfaces at different temperatures.

d) Heat gained by the windows Windows of the hall are exposed directly to the surrounding and the heat from the sun enters the hall by radiation. As in the case of the walls, the heat gained by the hall through windows depends on their alignment. If there are enough curtains on the windows and the external awning the amount of heat gained by radiation reduces. The type of glass doors on the windows also affects the amount of heat gained through the windows by radiation. e) Outdoor air heat The warmer air outside of conditioned spaces of the hall is called atmospheric air. Due to higher temperature, atmospheric air tends to increase the average room temperature when it enters indoor spaces of the hall. Outdoor air can also enter the conditioned spaces of the hall through leaks around doors, windows. The heat held from outdoor air comes in great part from the sun, but it can also come from vehicles or from other buildings.

2. Assuming that the heat loss to the surrounding during the heat exchange process in the AHU is 10% from the emitted heat transfer rate, plot the changes in air exit temperature if the air flow rate into the AHU ranges from 10,000 to 20,000 litres per minute. Then, identify the suitable operating range for the air flow into the AHU to meet the standard comfort temperature requirement. [CO2] 10 °C

AHU ṁ = 13 L/m = 0.217 kg/s

Air Tair,1 = 38°C

4°C

ṁ (air) Convert 10000

L kg =166.667 min s

ṁ (water) Convert 13

L kg =0.217 min s

Qc ,chilled water =ḿ¿ C p ,water∗( ∆ T chilled water )

(

¿ 0.217

kg kJ ∗ 4.18 ∗( 10 ° C−4 ° C ) s kg . ° C

)(

¿ 5.442 kW

Energy Balance ,Q c, chilled water=Qh , air

Qh ,air =0.9∗Q c , chilled water ( 90 % )

)

Tair,2 ṁ=10,000 until 20,000 L/m

Qh ,air =0.9∗5.442 kW =4.898 kW Identifying the suitable operating range for the air flowinto AHU unit

(at ḿ=10000 minL ∨166.667 kgs ) Qh ,air = ḿ¿ C p ,air∗( ∆ T air )

(

4.898 kW = 166.667

kg kJ ∗ 1.007 ∗( 38 ° C−T 2 ,air ) s kg . ° C

)(

)

T 2 , air =37. 971° C

(at ḿ=20000 minL ∨333.333 kgs ) Qh ,air = ḿ¿ C p ,air∗( ∆ T air )

(

4.898 kW = 333.333

kg kJ ∗ 1.007 ∗ ( 38° C−T 2 ,air ) s kg . ° C

)(

)

T 2 , air =37.985 °C

(at ḿ=40000 minL ∨666.667 kgs ) Qh ,air = ḿ¿ C p ,air∗( ∆ T air )

(

4.898 kW = 666.667

kg kJ ∗ 1.007 ∗( 38° C−T 2 ,air ) s kg . ° C

)(

)

T 2 , air =37.993 °C

(at ḿ=60000 minL ∨1000 kgs ) Qh ,air = ḿ¿ C p ,air∗( ∆ T air )

(

4.898 kW = 1000 T 2 , air =37.995 °C

kg kJ ∗ 1.007 ∗ ( 38° C−T 2 ,air ) s kg . ° C

)(

)

(at ḿ=80000 minL ∨1333.333 kgs ) Qh ,air = ḿ¿ C p ,air∗( ∆ T air )

(

4.898 kW = 1333.333

kg kJ ∗ 1.007 ∗( 38° C−T 2 ,air ) s kg . ° C

)(

)

T 2 , air =37.996 ° C

(at ḿ=100000 minL ∨1666.667 kgs ) Qh ,air = ḿ¿ C p ,air∗( ∆ T air )

(

4.898 kW = 1666.667

kg kJ ∗ 1.007 ∗( 38 °C−T 2 ,air ) s kg . ° C

)(

)

T 2 , air =37.997 °C

Based from calculation for T2 at 10000L/m gives 37. 971 ℃ which emitted less heat to surrounding. To achieve standard comfort temperature (20℃−22℃) the new value of air flow rate must take into consideration.

(Air flow rate for temperature = 22° C ) Qh ,air = ḿ¿ C p ,air∗( ∆ T air )

(

4.898 kW =( ḿ )∗ 1.007 ḿ=0.304

kg s

kJ ∗ ( 38° C−22° C ) kg . ° C

)

= 18.24

L min

(Air flow rate for temperature = 20 ° C ) Qh ,air = ḿ¿ C p ,air∗( ∆ T air )

(

4.898 kW =( ḿ )∗ 1.007

kJ ∗ ( 38° C−20 ° C ) kg . ° C

)

ḿ=0.27

kg s

= 16.2

L min

Hence, to meet standard comfort temperature requirement, air flow rate into AHU are range from 16.2 L/min to 18.24 L/min must be taken into consideration.

Qh

ḿ (l/ min)

4.9212

10,000

4.9212 4.9212

11,000 12,000

4.9212

13,000

4.9212 4.9212

14,000 15,000

4.9212

16,000

4.9212 4.9212

17,000 18,000

4.9212

19,000

4.9212

20,000

ḿ (kg/s) 166.666 7 183.333 3 200 216.666 7 233.333 3 250 266.666 7 283.333 3 300 316.666 7 333.333 3

Cp (air)

T1

T2

1.005

38

37.971

1.005 1.005

38 38

37.973 37.976

1.005

38

37.977

1.005 1.005

38 38

37.979 37.980

1.005

38

37.982

1.005 1.005

38 38

37.983 37.984

1.005

38

37.985

1.005

38

37.985

Object 95

Graph 1: Air Exit Temperature versus Air Flow Rate Based from the graph, it shows that the air exit temperature increases as the air flow rate increase. Higher value of flow rate means the air moves faster, hence the air contact with the cold water with a minimum duration. Thus, the rate of heat transfer will become lower. Due to that, the amount of heat from air that did not transfer to cold water will carry along with the air to the exit (outlet) gives higher value of temperature as the flow rate is higher.

3. If the surface temperature of the pipe carrying the chilled water return (CWR) is assumed steady and uniform at 10oC along the pipe length, plot the heat transfer rate profile at the outer surface of the CWR pipe under surrounding conditions of natural convection current effect and wind effect of 2 ms -1, 4 ms-1 and 6 ms-1 velocities. The effective CWR pipe length is 19 meters with a pipe diameter of 25 cm.

[CO3]

�∞=10℃ �∞=2�/�

Ts = 10℃

Natural Convection V=0 Ra=

β = 1/(293 K) gβ ( T s−T ∞ ) L3c

( Ra=

Pr v2 m 1 9.81 2 (30−10 ) ( 0.25 )3 s 293 K

)( ) (1.516 ×10 ms )

2 2

( 0.7309 )=33274456.31

−5

From Table 9/1 (Heat and Mass Transfer by Cengel) 2

1 6 D

{ [ ( )]} 0.387 Ra

Nu= 0.6+

8

1+

0.559 169 27 Pr

33274456.31 ¿ ¿ 1 ¿ 6 0.6+0.387 ¿ 2=40.64 ¿ Nu=¿

Nu=

hl k

k 0.02514 W h= Nu= 40.64=4.09 l 0.25 m. K Q=h A s (T ∞−T s ) Q=( 4.09 ) (14.92 ) ( 30−10 )=1219.489 W Forced Convection At velocity, �=2 �/�

Properties of Air at T=20℃ �=1.204 𝑘�/� ³

�𝑓𝑖𝑙� =

T ∞+Ts 2

�=1.596×10−5 �2/� 𝑘=0.02154 �/�� 𝑃�=0.7309 �=1.825×10−5 𝑘�/��

30 ℃+10 ℃ 2 �𝑓𝑖𝑙� =20℃ �𝑓𝑖𝑙� =

ℜ=

( 2 )( 0.25 )

( 1.516 × 10−5 )

=32981.53

Using Churchill and Bernstein Equation; 1

u=0.3+

1

[ ( )] 1+

Nu=0.3+

Pr

3

)]

1 2

1 3

Q= 10.84

1

0.4 23 4 1+ 0.7309

k Nu= D

(

[ (

0.62 ( 32981.53 ) ( 0.7309 )

[ ( h=

)]

4

32981.53 58 5 1+ =107.76 282000

[ (

)]

W m.K W 107.76=10.84 0.25 m m. K

0.2514

W ( 14.92 m2) ( 30 ° C−10° C )=3234.66 W m. k

)

At velocity, �=4 �/� T ∞+Ts 2 30 ℃+10 ℃ �𝑓𝑖𝑙� = 2 �𝑓𝑖𝑙� =20℃ �𝑓𝑖𝑙� =

ℜ=

4

5 ℜ 8 5 1+ 1 282000 0.4 2 4

0.62 ℜ 2 Pr 3

( 4 ) ( 0.25 )

( 1.516 × 10−5 )

=65963.0607

Using Churchill and Bernstein Equation; 1

Nu=0.3+

1

0.62 ℜ2 Pr 3

[ ( )] 1+

5 4

[ (

)]

ℜ 8 5 1+ 1 2 282000 0.4 4 3

Pr

1

Nu=0.3+

1

0.62 ( 65963.0607 ) 2 ( 0.7309 )3

[ ( k h= Nu= D

1

0.4 23 4 1+ 0.7309

)]

(

W ( 14.92m2 ) ( 30° C−10 ° C )=4974.984 W m .k

)

T ∞+Ts 2 30 ℃+10 ℃ �𝑓𝑖𝑙� = 2 �𝑓𝑖𝑙� =20℃ �𝑓𝑖𝑙� =

( 6 ) ( 0.25 )

( 1.516 × 10−5 )

=98944.591

Using Churchill and Bernstein Equation; 1 2

u=0.3+

)]

W m.K W 165.793=16.6722 0.25 m m. K

At velocity, �=6 �/�

ℜ=

[ (

0.02514

Q= 16.6722

0.62 ℜ Pr

[ (

5 4

)]

ℜ 8 5 1+ 2 1 282000 0.4 4

[ ( )] 1+

1 3

Pr

3

4

65963.0607 58 5 1+ =165.793 282000

1 2

Nu=0.3+

0.62 ( 98944.591 ) ( 0.7309 )

[ ( 1+

h=

k Nu= D

(

Q= 21.74

1 3

1

0.4 23 4 0.7309

)]

4

98944.591 58 5 1+ =216.189 282000

[ (

)]

W m.K W 216.189=21.74 0.25 m m. K

0.02514

W ( 14.92 m2 ) ( 30 ° C−10° C )=6487.206 W m. k

)

At velocity, �=8 �/�

T ∞+Ts 2 30 ℃+10 ℃ �𝑓𝑖𝑙� = 2 �𝑓𝑖𝑙� =20℃ �𝑓𝑖𝑙� =

ℜ=

( 8 ) ( 0.25 )

( 1.516 × 10−5 )

=131926.121

Using Churchill and Bernstein Equation; 1

u=0.3+

1

[ ( )] 1+

Pr

3

[ (

)]

1 2

Nu=0.3+

4

5 ℜ 8 5 1+ 1 282000 0.4 2 4

0.62 ℜ 2 Pr 3

0.62 ( 131926.121) ( 0.7309 ) 1

0.4 23 4 1+ 0.7309

[ (

)]

1 3

4

131926.121 58 5 1+ =262.871 282000

[ (

)]

h=

k Nu= D

(

W m.K W 262.871=26.434 0.25 m m.K

0.02514

Q= 26.434

W ( 14.92 m2 ) ( 30 ° C−10° C )=7887.998 W m. k

)

4. Plot the prediction of the chilled water temperature at the inlet of the chiller under similar surrounding conditions. [CO2] At velocity, �=0 �/� �=�̇𝐶��̇𝐶�𝐶��Δ� 1219.489 =(0.217)(4.2K)(�2−�1) 1219.489 =(0.217)(4.2K)(�2−10) �2 =11.34℃

At velocity, �=2 �/� �=�̇𝐶��̇𝐶�𝐶��Δ� 3234.66 =(0.217)(4.2K)(�2−�1) 3234.66 =(0.217)(4.2K)(�2−10)

�� =13.54℃

At velocity, �=4 �/� �=�̇𝐶��̇𝐶�𝐶��Δ� 4974.984 =(0.217)(4.2K)(�2−�1)

Properties of Chilled water �̇=13 L per min = 0.217 𝑘�/� 𝐶�=4.28 KJ/kg.K

4974.984 =(0.217)(4.2K)(�2−10) ��=15.46 ℃

At velocity, �=6 �/� �=�̇𝐶��̇𝐶�𝐶��Δ� 6487.206=(0.217)(4.2K)(�2−�1) 6487.206 =(0.217)(4.2K)(�2−10)

��= 17.12℃

At velocity, �=8 �/� �=�̇𝐶��̇𝐶�𝐶��Δ� 7887.99 =(0.217)(4.2K)(�2−�1) 7887.99 =(0.217)(4.2K)(�2−10)

��= 18.65℃

Velocity, V (m/s) Heat Trasfer, Q (W) Temperature ,T2 (℃)

0

2

4

6

8

1219.489

3234.66

4974.984

6487.206

7887.998

11.34

13.54

15.46

17.12

18.65

Temperature ,T2 (℃) vs Heat Trasfer, Q (W) 20 18

Temperature ,T2 (℃)

16 14 12 10 8 6 4 2 0

0

1000

2000

3000

4000

5000

6000

7000

8000

9000

Heat Trasfer, Q (W)

Graph 2: Air Exit Temperature versus Heat Transfer Rate The cooling coil load is high and water flow rate increases to a certain level in the cooling coil, turbulent flow occurs in the tubes. As a result, both the water side and air side heat transfer coefficients will increase and lead to increase in overall heat transfer, which will have a positive increase of the chilled water temperatures