Methods of handling Supply air in HVAC

7
a. All Outside Air supplied b. Outside air supplied with External Bypass system

description

Design and Education

Transcript of Methods of handling Supply air in HVAC

Page 1: Methods of handling Supply air in HVAC

a. All Outside Air supplied

b. Outside air supplied with External Bypass system

Page 2: Methods of handling Supply air in HVAC

c. All Outside air supplied with Re – Heater

d. Outside and Recirculated air supplied

Page 3: Methods of handling Supply air in HVAC

e. Outside air and Re-circulated air supplied with Bypass (Re-circulated air)

f. Outside and Re-circulated air supplied with Re-heater

By: Engr. Yuri G. Melliza

Page 4: Methods of handling Supply air in HVAC

Example: Outside and Re-circulated air supplied with Re-heater A commercial building to be air conditioned has a sensible heat load of 36 KW and a latent heat load of 10.2 KW. The building is to be maintained at 26C and 50% RH. Outside air is at 32C DB and 24C WB. Forty five percent of the supply air is fresh air and the rest is re-circulated air. Conditioned air leaves the AC unit at 14C and 100% RH then it is reheated to 19C and is supplied to building. Determine

a. The fan capacity in m3/sec to the building b. The tonnage capacity of the AC unit (Assume tw = 14C ) c. The heat required by the re-heater

Figure

Qs = 36 KW QL = 10.2 KW t4 = 26C ; RH4 = 50% t0 = 32C DB ; tw0 = 24C t2 = 14C ; RH2 = 100% 0.45m = m0 0.55m = mR t2 = 14C ; t3 = 19C Other Data hw = 58.679 KJ/kg h0 = 71.944 KJ/kgda ; W0 = 0.015538 kgm/kgda h4 = 52.943 KJ/kgda ; W4 = 0.010518 kgm/kgda h2 =39.293 KJ/kgda ; W2 = W3 = 0.009981 kgm/kgda hfg at 19C=2456.49 KJ/kg Processes: 014 – Adiabatic Mixing 1 to 2 – Cooling & De-humidifying 2 to 3 – Sensible Heating 3 to 4 – Heating &Humidifying

Page 5: Methods of handling Supply air in HVAC

KW 32.82Q

39.293)-035.042(45.8)h-m(hQ

heater-re the by requiredHeat

TONS 31.58Q

ionRefrigerat of Ton 1min

KJ 211

Capacity Tonnage

min

KJ 6665.16KW 111.086Q

0.834111.92Q

8.6790.009981)5285.042(0.0139.293)95.042(61.4Q

)hW-m(W-)h-m(hQ

kg/sec 3.15m

2.275.042mmm

kg/sec 2.27m

)0.45(5.042m

kgda

kgm0.0128W

kgda

KJ 61.49h

0.0105180.015538

0.010518W

52.94371.944

52.943h0.45

WWo

WW

hho

hh0.45

m

m

sec

m 4.24521)5.042(0.84Capacity Fan A)

kgda

m 0.8421

KJ/kgda 45.803 h and C19 tat

formulas andchart ricPsychromet From

kgda

KJ 45.803h

)h435.042(52.936

kgda

kgm 0.0097 W

)2456.49W-05185.042(0.0110.2

n)applicatio ngconditioniair (For KJ/kg 2500 h :where

)hfgW-m(WQ

kg/sec 5.042m

19)-m(1.02)(2636

)tm(1.02)(tQs

R

23R

w2121

R

0R

o

0

1

1

11

4

41

4

41o

3

3

33

3

3

3

3

fg

34L

34

3

υ

Page 6: Methods of handling Supply air in HVAC

60%

1

2

To conditioned

space

Psychrometric

Chart

3

Saturation Curve

32°C10°C 16°C

W2 = W3

W1

h1

h3

h2

Example: Dehumidifier – Heater Moist air at 32C and 60% RH enters the refrigeration coil of a De-humidifier with a flow rate of 1.5 kg/sec. The air leaves saturated at 10C and is re-heated to 16C before entering to the conditioned space. Condensate leaves the de-humidifier at 10C . Determine

a. The condensate removed in kg/sec b. The Capacity of the de-humidifier unit (AC – unit) in Tons of refrigeration c. The heat required by the Re-heater

Figure:

Processes: 1 to 2 – Cooling & Dehumifying 2 to 3 – Sensible Heating Using Fundamental Formulas or Psychrometric chart

kg

KJ 92.41h

kgda

KJ 393.35h

kgda

kgm 007632.02WW

kgda

KJ 278.29h

kgda

kgm 007632.0W

kgda

KJ 429.78h

kgda

kgm 0.01807W

w

3

3

2

2

1

1

a. Condensate removed

sec

kg 016.0m

)007632.001807.0(5.1m

)WW(mm

w

w

21w

b. Capacity of AC-unit in TR

ionRefrigerat of Tons 8.20

min

KJ 211

TR 1

min

KJ 4389Q

min

KJ 4389KW 15.73Q

)92.41(016.0)278.29429.78(5.1Q

h)WW(m)hh(mQ

mwhw)hh(mQ

w2121

21

Page 7: Methods of handling Supply air in HVAC

0 1

Exhaust air

Supply

air

AuditoriumQ

2

Re-circulated Air

AC Unit

3

3

Fanm m m

3mR

m0

Exhaust Fan

Qs = 145 KW

QL = 95 KW

0

1

2

To conditioned space

Psychrometric Chart

43

34°C27°C 29°C

22°C

22°C17°C

15°C

15°C

50% RH

W0

W1

W3

W2

h0

h1

h3

h2

kgda

KJ 41.984h

kgda

kgm 009831.0W

kgda

KJ 64.298h

kgda

kgm 0.013759W

kgda

KJ 55.626 h

kgda

KJ 77.281h

kgda

kgm 0.01117W

kgda

kgm 0.016813W

Formulas lFundamentaor Chart From

2

2

1

1

30

30

air circulated-re of rateflow mass sec

kg 51.8m

626.55281.77

298.64281.77

14.2

m

hh

hh

m

m

mixing of theory From

air supply of rateflow mass sec

kg 2.14m

)1727)(02.1(m145

)tt)(02.1(mQs

R

R

30

10R

23

unit-AC of Capacity TR 21.90211

)60(24.317Q

KW 24.3170.38143168 316.8588Q

)63.121(056.0)984.41298.64(2.14hm-)h-m(hQ

unit - AC leaving condensate of rateflow mass sec

kg 056.0m

)009831.0013759.0(2.14)WW(mm

air outside of rateflow mass sec

kg 7.5m

626.55281.77

626.55298.64

14.2

m

hh

hh

m

m

ww21

w

21w

0

0

30

310

Example: Outside air and re-circulated air supplied An auditorium is to be maintained at 27C and 50% RH. The calculated sensible heat load in the space Qs = 145 KW and latent heat load QL = 95 KW (hfg = 2437.4 KJ/kg). The air mixture at 29C DB and 22C WB is cooled to 17C DB and 15C WBin a Chilled-water AC unit and delivered as supply-air to the space, calculate

a. The kg/sec of supply air b. The kg/sec of re-circulated air c. The kg/sec of outside air if t0 = 34C and 50% RH d. The kg/sec of condensate from AC-unit (assume hw = 121.63 KJ/kg) e. The refrigeration capacity of the AC-unit in TR

Processes: 013 – Adiabatic mixing 1 to 2 – Cooling & dehumidifying 2 to 3 – Heating & humidifying