System Analysis

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MEC551: ASSIGNMENT PHASE 1 System Analysis

Figure A. System schematic of the centralized chiller system for the DATC building

The DATC building houses a large hall for multi-purpose functions and its air-conditioning system is a centralized chiller system. Chilled water exits the chiller at 1oC, but the effects of pumping and ambient heating increases the water temperature to 4 oC as it enters the Air Handling Unit (AHU) of the DATC. The chilled water cools the air coming from the hall and then returns to the chiller system.

In the AHU of the DATC building, chilled water enters at 4 oC and exits at 10 oC at a flow rate of 10 litres per minute. Air from the building is to be cooled from its initial temperature of 38 oC. The chiller system operates based on the vapour compression refrigeration cycle with a single stage compression and a water-cooled condenser. The isentropic efficiency of the compressor is assumed at 90%. The evaporator design is based on the shell-and-tube multipass heat exchanger. The refrigerant enters and exits through the shell at its saturation temperature, while the chilled water flows through the tube and exits the evaporator at 1oC. For this air-conditioning system: 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 1000 to 5000 litres per minute. Then, identify the suitable operating range for the air flow into the AHU to meet the standard comfort temperature requirement. [CO2]

Question 2

Figure 1: The AHU system of DATC

Convert 10

L kg =0.1667 min s

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

(

¿ 0.1667

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

)(

)

¿ 4.1808 kW

Energy Balance ,Q c, chilled water=Qh , air Qh ,air =0.9∗Q c , chilled water ( 90 % ) Qh ,air =0.9∗4.1808 kW =3.7627 kW Identifying the suitable operating range for the air flowinto AHU unit

Referring ¿ Table A 15 ,Temp=30 ℃ , air density =1.1644

(at ḿ=1000 minL ∨0.0194 kgs ) Qh ,air = ḿ¿ C p ,air∗( ∆ T air )

(

3.7627 kW = 0.0194

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

)(

)

kg m3

T 2 , air =−154.61 ° C

(at ḿ=2000 minL ∨0.0388 kgs ) Qh ,air = ḿ¿ C p ,air∗( ∆ T air )

(

3.7627 kW = 0.0388

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

)(

)

T 2 , air =−58.30 ° C

(at ḿ=3000 minL ∨0.0582 kgs ) Qh ,air = ḿ¿ C p ,air∗( ∆ T air )

(

3.7627 kW = 0.0582

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

)(

)

T 2 , air =−26.20 ° C

(at ḿ=4000 minL ∨0.0776 kgs ) Qh ,air = ḿ¿ C p ,air∗( ∆ T air )

(

3.7627 kW = 0.0776

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

)(

)

T 2 , air =−10.15 ° C

(at ḿ=5000 minL ∨0.0970 kgs ) Qh ,air = ḿ¿ C p ,air∗( ∆ T air )

(

3.7627 kW = 0.0970

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

)(

)

T 2 , air =−0.52 ° C

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

Based on the range given in the question, after calculating the exit temperature, air flow range from 1000L/min to 2000L/min is not suitable for achieving the standard comfort temperature. We identify that the suitable operating range for the air flow into the AHU needed to be increase more so that it meets the standard comfort temperature requirement which is ranging from 12000 – 16000 L/min.

Question 3 Velocity, V (m/s) 0 2 4 6

Reynolds Number, Re n/a 32981.53034 65963.06069 98944.59103

Rayleigh Number, Ra 33274456.31 n/a n/a n/a

Nusselt Number, Nu 40.64 107.76 165.79 216.32

Heat transfer coefficient, h 4.09 10.84 16.67 21.75

Heat transfer rate, Q 2407.22 6383.11 9820.49 12813.38

Heat Transfer Rate vs Air Velocity 14000 12813.38 12000 9820.49

10000

Q (kW)

8000 6383.11 6000

4000

2407.22 2000

0

0

1

2

3

4

Air Velocity (m/s)

5

6

7

Calculations Assumptions iii-

The pipe is in horizontal orientation The ambient air temperature is 30 oC

T s +T ∞ 2 10+30 Film Temperature , T f = =20 ℃ 2 Film Temperature , T f =

Properties of air at 20 oC K = 0.02514 W/m.K V = 1.516 x10-5 m2/s Pr = 0.7309

As = 25m x 0.25 m = 6.25 m2

For Natural Convection V=0

β = 1/(293 K)

3

Ra=

gβ ( T s−T ∞ ) Lc 2

( Ra=

v m 9.81 2 s

Pr

)( 2931 K ) (30−10) ( 0.25 ) ( 0.7309 )=33274456.31 (1.516 ×10 ms ) 3

2 2

−5

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

1 6 D

{ [ ( )]}

Nu= 0.6+

0.387 Ra

9

8

0.559 16 27 1+ Pr

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

Nu=

hl k

W k m. K W h= Nu= 40.64=4.09 l 0.25 m m. K Q=h A s (T ∞−T s ) 0.2514

(

Q= 4.09

W ( 6.25 m2 ) ( 303 K −283 K )=2407.22 W m. k

)

Forced Convection At V = 2m/s

ℜ=

VD v

m 2 ) ( 0.25 m ) ( s ℜ= =32981.53

(

1.516 ×10−5

m2 s

)

From Churcill and Berstein Equation 1 2

Nu=0.3+

0.62 ℜ Pr

1 3

[ (

[ ( )] 1+

5 4

)]

ℜ 8 5 1+ 2 1 282000 0.4 4 3

Pr

1

Nu=0.3+

1

0.62 ( 32981.53 ) 2 ( 0.7309 ) 3 1

4

32981.53 58 5 1+ =107.76 282000

[ (

)]

0.4 23 4 1+ 0.7309 W 0.2514 k m.K W h= Nu= 107.76=10.84 D 0.25 m m. K W Q= 10.84 ( 6.25 m2 ) ( 303 K −283 K )=6383.11 W m. k

[ (

(

)]

)

At V = 4m/s

ℜ=

VD v

m 2 ) ( 0.25 m ) ( s ℜ= =32981.53

(

1.516 ×10−5

m2 s

)

From Churcill and Berstein Equation 1

Nu=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 ℜ2 Pr 3

0.62 ( 32981.53 ) ( 0.7309 ) 1

0.4 23 4 1+ 0.7309

[ (

)]

1 3

4

32981.53 58 5 1+ =107.76 282000

[ (

)]

h=

k Nu= D

(

Q= 10.84

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

0.2514

W ( 6.25 m 2 ) ( 303 K −283 K )=6383.11 W m. k

)

At V = 6m/s

ℜ=

VD v

m 2 ) ( 0.25 m ) ( s ℜ= =32981.53

(

1.516 ×10−5

m2 s

)

From Churcill and Berstein Equation 1

Nu=0.3+

1

[ ( )] 1+

3

Pr

[ (

)]

1

Nu=0.3+

4

5 ℜ 8 5 1+ 2 1 282000 0.4 4

0.62 ℜ2 Pr 3

1

0.62 ( 32981.53 ) 2 ( 0.7309 ) 3 1

[ (

)]

0.4 23 4 1+ 0.7309 W 0.2514 k m.K W h= Nu= 107.76=10.84 D 0.25 m m. K W 2 Q= 10.84 ( 6.25 m ) ( 303 K −283 K )=6383.11 W m. k

[ (

(

)]

)

At V = 8 m/s

ℜ=

VD v

m 2 ) ( 0.25 m ) ( s ℜ= =32981.53

(

1.516 ×10−5

m2 s

)

From Churcill and Berstein Equation 1 2

Nu=0.3+

0.62 ℜ Pr

1 3

[ ( )] 1+

[ (

5 4

)]

ℜ 8 5 1+ 2 1 282000 0.4 4 Pr

3

4

32981.53 58 5 1+ =107.76 282000

1 2

Nu=0.3+

0.62 ( 32981.53 ) ( 0.7309 ) 1

1 3

4

32981.53 58 5 1+ =107.76 282000

[ (

)]

0.4 23 4 1+ 0.7309 W 0.2514 k m.K W h= Nu= 107.76=10.84 D 0.25 m m. K W Q= 10.84 ( 6.25 m 2 ) ( 303 K −283 K )=6383.11 W m. k

[ (

(

)]

)

The calculation of the parameters for v= 4m/s and v=6m/s was done by replacing the value with the said velocity.

Question 4 Velocity, V (m/s) 0

Heat Transfer Rate Q (W) 2407.22

Mass flow rate (Kg/s) 0.17

ΔT (OC)

Tin (OC)

3.39

13.39

2

6383.11

0.17

8.98

18.98

4

9820.49

0.17

13.82

23.82

6

12813.38

0.17

18.03

28.03

Inlet Temperature vs Velocity 30

25

Temperature (oC)

20

15

10

5

0

0

1

2

3

Velocity (m/s)

4

5

6

7

Sample Calculation Assuming the heat transfer is steady. The energy balance become:

Qconvection =Qwater

From question 3, at velocity = 0 m/s (natural convection)

Qconvection =2407.22W Qwater =ṁC p Δ T =ṁ C p ( T inlet −10)

Mass flow rate of water

ṁ=10

l 0.001 m 3 1min kg × × ×1000 3 min 1l 60 s m

ṁ=0.17

kg s

Specific heat of water

C p , water =4.18

kJ kg . K

Q=ṁ C p (T inlet −10) T inlet = T inlet =

Q +10 ṁ Cp 2407.22W +10=13.39℃ kg kJ (0.17 )( 4.18 ) s kg . K

Calculation for others value of inlet temperature was done by substituting the values of Q respective to the velocity.