1.0 RESULTS AND SAMPLE OF CALCULATION

7.1 Results

Experiment

Reading

No ProcessSensible Heating Steam

Humidification

Cooling and

Dehumidification

1 kW 1.5 kW 1 kW 1.5 kW

T1 (TA d) 26.9 26.8 27.3 27.9 29.1 29.5

T2 (TA w) 22.2 22.2 22.6 23.4 23.5 23.4

T3 (TB d) 27.0 35.8 36.9 31.4 29.6 29.7

T4 (TB w) 23.0 26.5 27.3 31.2 24.6 24.5

T5 (TC d) 23.7 32.2 36.2 30.8 18.2 18.2

T6 (TC w) 22.4 24.7 25.4 30.6 17.6 17.4

T7 (TD d) 23.7 31.1 38.9 30.9 18.8 18.6

T8 (TD w) 22.5 24.4 26.2 30.7 18.2 18.0

T9 (T 1) X X X 10.7 11.1

T10 (T 2) X X X 71.4 78.5

T11 (T 3) X X X 44.9 45.2

T12 (T 4) X X X 6.3 6.6

Qp (W) X 1009.7 1009.7 X X

Qr (W) X 0 536.9 X X

Qb (W) X X 2742.9 X

Pevap (P 1) (kN/m2) X X X 209.0 211.7

Pcond (P 3) (kN/m2) X X X 1121.8 1134.0

ma (g/s) 103.2 99.9 99.4 100.7 100.8 103.7

mr (g/s) X X X 12.3 12.1

Time (s) X X X X 486

Drain Water (ml) X X X 100 ml

Table 7.1.1 : Results

7.2 Sample of Calculation:

Test 1 : No Process

ma=103.2 g/ s@0.1032kg /s

Inlet Outlet

T dry=26.9℃ T dry=23.7℃

T wet=22.2℃ T wet=22.5℃

∅ i=76 % ∅ o=86 %

w i=15.5 x10−3kgH 2O/kgair wo=16.5 x10−3 kgH2O /kgair

Temperature drops at exit

∆T dry :26.9−23.7=3.2℃

Temperature rises at exit

∆T wet :22.5−22.2=0.3℃

Test 2 : Sensible Heating

1kW preheat

ma=99.9g /s@0.0999 kg/ s

Inlet Outlet

T dry=26.8℃ T dry=31.1℃

T wet=22.2℃ T wet=24.4℃

∅ i=67 % ∅ o=59 %

w i=15 x10−3kgH 2O/kgair wo=16.75 x10−3 kgH2O /kgair

Temperature rises at exit

∆T dry :31.1−26.8=4.3℃

∆T wet :24.4−22.2=2.2℃

1.5kW preheat

ma=99.9g /s@0.0999 kg/ s

Inlet Outlet

T dry=27.3℃ T dry=38.9℃

T wet=22.6℃ T wet=26.2℃

∅ i=68 % ∅ o=36 %

w i=15.25 x10−3 kgH 2O/kgair wo=15.75 x10−3 kgH2O /kgair

Temperature rises at exit

∆T dry :38.9−27.3=11.6℃

∆T wet :26.2−22.6=3.6℃

Test 3 : Steam Humidification

ma=100.7 g /s@0.1007 kg/ s

Inlet Outlet

T dry=27.9℃ T dry=30.9℃

T wet=23.4℃ T wet=30.7℃

∅ i=68% ∅ o=100%

w i=16.25 x10−3 kgH 2O/kgair wo=29 x10−3 kgH 2O /kgair

Temperature rises at exit

∆T dry :30.9−27.9=3℃

∆T wet :30.7−23.4=7.3℃

Dry air mass balance

ma1=ma2

=ma

∆ w=w2−w1

¿ (29 x10−3 )−(16.25 x10−3 )

¿12.75 x10−3kgH 2O/kgair

Water mass balance

ma2w2=ma1

w1+mw

mw=ma2w2−ma1

w1

¿ ma(w2−w1)

¿0.1007(12.75x 10−3)

¿1.2839 x10−3 kg /s

The relative humidity changes

∆∅=∅ o−∅ i

¿100 %−68 %

¿32%@0.32

Test 4 : Cooling and Dehumidification

ma=100.8 g/ s@0.1008 kg /s

Inlet Outlet

T dry=29.1℃ T dry=18.8℃

T wet=23.5℃ T wet=18.2℃

∅ i=63 % ∅ o=100%

w i=16 x10−3 kgH2O /kgair wo=13.75 x10−3 kgH2O /kgair

hi=70 kJ /kgair ho=54kJ /kgair

Temperature drops at exit

∆T dry :29.1−18.8=10.3℃

∆T wet :23.5−18.2=5.3℃

∆ w=w1−w2

¿ (16 x10−3 )−(13.75 x 10−3 )

¿2.25 x10−3kgH 2O/kgair

Water mass balance

ma1w1=ma2

w2+mw

mw=ma1w1−ma2

w2

¿ ma(w2−w1)

¿0.1008(2.25 x 10−3)

¿0.2268 x10−3kg /s

The relative humidity changes

∆∅=∅ o−∅ i

¿100 %−63 %

¿37%@0.37

Assumption:

T w=18.8℃

T h

15 62.982

18.8 hw

20 83.915

18.8−1520−15

=hw−62.982

83.815−62.982

hw=78.81508 kJ /kg

Energy balance

E¿=Eout

ma1h1=mw hw+ma2

h2+ Qout

Qout=ma (h1−h2 )−mwhw

¿0.1008 (70−54 )−0.0002(78.81508)

¿1.629kW

ma=103.7 g /s@0.1037 kg/ s

Inlet Outlet

T dry=29.5℃ T dry=18.6℃

T wet=23.4℃ T wet=18.0℃

∅ i=60% ∅ o=95%

w i=15.75 x10−3 kgH 2O/kgair wo=12.75 x10−3 kgH2O /kgair

hi=70kJ /kgair ho=51kJ /kgair

Temperature drops at exit

∆T dry :29.5−18.6=10.9℃

∆T wet :23.4−18.0=5.4℃

∆ w=w1−w2

¿ (15.75 x10−3 )−(12.75 x10−3 )

¿3 x10−3kgH 2O/kgair

Water mass balance

ma1w1=ma2

w2+mw

mw=ma1w1−ma2

w2

¿ ma(w2−w1)

¿0.1037(3x 10−3)

¿0.3111x 10−3 kg /s

The relative humidity changes

∆∅=∅ o−∅ i

¿95%−60 %

¿35 %@0.35

Assumption:

T w=18.6℃

T h

15 62.982

18.6 hw

20 83.915

18.6−1520−15

=hw−62.982

83.815−62.982

hw=77.98176 kJ /kg

Energy balance

E¿=Eout

ma1h1=mw hw+ma2

h2+ Qout

Qout=ma (h1−h2 )−mwhw

¿0.1037 (70−51 )−0.0003(77.98176)

¿1.947 kW

2.0 RESULTS AND SAMPLE OF CALCULATION

7.1 Results

Experiment

Reading

No ProcessSensible Heating Steam

Humidification

Cooling and

Dehumidification

1 kW 1.5 kW 1 kW 1.5 kW

T1 (TA d) 26.9 26.8 27.3 27.9 29.1 29.5

T2 (TA w) 22.2 22.2 22.6 23.4 23.5 23.4

T3 (TB d) 27.0 35.8 36.9 31.4 29.6 29.7

T4 (TB w) 23.0 26.5 27.3 31.2 24.6 24.5

T5 (TC d) 23.7 32.2 36.2 30.8 18.2 18.2

T6 (TC w) 22.4 24.7 25.4 30.6 17.6 17.4

T7 (TD d) 23.7 31.1 38.9 30.9 18.8 18.6

T8 (TD w) 22.5 24.4 26.2 30.7 18.2 18.0

T9 (T 1) X X X 10.7 11.1

T10 (T 2) X X X 71.4 78.5

T11 (T 3) X X X 44.9 45.2

T12 (T 4) X X X 6.3 6.6

Qp (W) X 1009.7 1009.7 X X

Qr (W) X 0 536.9 X X

Qb (W) X X 2742.9 X

Pevap (P 1) (kN/m2) X X X 209.0 211.7

Pcond (P 3) (kN/m2) X X X 1121.8 1134.0

ma (g/s) 103.2 99.9 99.4 100.7 100.8 103.7

mr (g/s) X X X 12.3 12.1

Time (s) X X X X 486

Drain Water (ml) X X X 100 ml

Table 7.1.1 : Results

7.2 Sample of Calculation:

Test 1 : No Process

ma=103.2 g/ s@0.1032kg /s

Inlet Outlet

T dry=26.9℃ T dry=23.7℃

T wet=22.2℃ T wet=22.5℃

∅ i=76 % ∅ o=86 %

w i=15.5 x10−3kgH 2O/kgair wo=16.5 x10−3 kgH2O /kgair

Temperature drops at exit

∆T dry :26.9−23.7=3.2℃

Temperature rises at exit

∆T wet :22.5−22.2=0.3℃

Test 2 : Sensible Heating

1kW preheat

ma=99.9g /s@0.0999 kg/ s

Inlet Outlet

T dry=26.8℃ T dry=31.1℃

T wet=22.2℃ T wet=24.4℃

∅ i=67 % ∅ o=59 %

w i=15 x10−3kgH 2O/kgair wo=16.75 x10−3 kgH2O /kgair

Temperature rises at exit

∆T dry :31.1−26.8=4.3℃

∆T wet :24.4−22.2=2.2℃

1.5kW preheat

ma=99.9g /s@0.0999 kg/ s

Inlet Outlet

T dry=27.3℃ T dry=38.9℃

T wet=22.6℃ T wet=26.2℃

∅ i=68 % ∅ o=36 %

w i=15.25 x10−3 kgH 2O/kgair wo=15.75 x10−3 kgH2O /kgair

Temperature rises at exit

∆T dry :38.9−27.3=11.6℃

∆T wet :26.2−22.6=3.6℃

Test 3 : Steam Humidification

ma=100.7 g /s@0.1007 kg/ s

Inlet Outlet

T dry=27.9℃ T dry=30.9℃

T wet=23.4℃ T wet=30.7℃

∅ i=68% ∅ o=100%

w i=16.25 x10−3 kgH 2O/kgair wo=29 x10−3 kgH 2O /kgair

Temperature rises at exit

∆T dry :30.9−27.9=3℃

∆T wet :30.7−23.4=7.3℃

Dry air mass balance

ma1=ma2

=ma

∆ w=w2−w1

¿ (29 x10−3 )−(16.25 x10−3 )

¿12.75 x10−3kgH 2O/kgair

Water mass balance

ma2w2=ma1

w1+mw

mw=ma2w2−ma1

w1

¿ ma(w2−w1)

¿0.1007(12.75x 10−3)

¿1.2839 x10−3 kg /s

The relative humidity changes

∆∅=∅ o−∅ i

¿100 %−68 %

¿32%@0.32

Test 4 : Cooling and Dehumidification

ma=100.8 g/ s@0.1008 kg /s

Inlet Outlet

T dry=29.1℃ T dry=18.8℃

T wet=23.5℃ T wet=18.2℃

∅ i=63 % ∅ o=100%

w i=16 x10−3 kgH2O /kgair wo=13.75 x10−3 kgH2O /kgair

hi=70 kJ /kgair ho=54kJ /kgair

Temperature drops at exit

∆T dry :29.1−18.8=10.3℃

∆T wet :23.5−18.2=5.3℃

∆ w=w1−w2

¿ (16 x10−3 )−(13.75 x 10−3 )

¿2.25 x10−3kgH 2O/kgair

Water mass balance

ma1w1=ma2

w2+mw

mw=ma1w1−ma2

w2

¿ ma(w2−w1)

¿0.1008(2.25 x 10−3)

¿0.2268 x10−3kg /s

The relative humidity changes

∆∅=∅ o−∅ i

¿100 %−63 %

¿37%@0.37

Assumption:

T w=18.8℃

T h

15 62.982

18.8 hw

20 83.915

18.8−1520−15

=hw−62.982

83.815−62.982

hw=78.81508 kJ /kg

Energy balance

E¿=Eout

ma1h1=mw hw+ma2

h2+ Qout

Qout=ma (h1−h2 )−mwhw

¿0.1008 (70−54 )−0.0002(78.81508)

¿1.629kW

ma=103.7 g /s@0.1037 kg/ s

Inlet Outlet

T dry=29.5℃ T dry=18.6℃

T wet=23.4℃ T wet=18.0℃

∅ i=60% ∅ o=95%

w i=15.75 x10−3 kgH 2O/kgair wo=12.75 x10−3 kgH2O /kgair

hi=70kJ /kgair ho=51kJ /kgair

Temperature drops at exit

∆T dry :29.5−18.6=10.9℃

∆T wet :23.4−18.0=5.4℃

∆ w=w1−w2

¿ (15.75 x10−3 )−(12.75 x10−3 )

¿3 x10−3kgH 2O/kgair

Water mass balance

ma1w1=ma2

w2+mw

mw=ma1w1−ma2

w2

¿ ma(w2−w1)

¿0.1037(3x 10−3)

¿0.3111x 10−3 kg /s

The relative humidity changes

∆∅=∅ o−∅ i

¿95%−60 %

¿35 %@0.35

Assumption:

T w=18.6℃

T h

15 62.982

18.6 hw

20 83.915

18.6−1520−15

=hw−62.982

83.815−62.982

hw=77.98176 kJ /kg

Energy balance

E¿=Eout

ma1h1=mw hw+ma2

h2+ Qout

Qout=ma (h1−h2 )−mwhw

¿0.1037 (70−51 )−0.0003(77.98176)

¿1.947 kW