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# Methods of handling Supply air in HVAC

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### Methods of handling Supply air in HVAC

1. 1. a. All Outside Air supplied b. Outside air supplied with External Bypass system
2. 2. c. All Outside air supplied with Re – Heater d. Outside and Recirculated air supplied
3. 3. 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
4. 4. 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 26C and 50% RH. Outside air is at 32C DB and 24C 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 14C and 100% RH then it is reheated to 19C and is supplied to building. Determine a. The fan capacity in m 3/sec to the building b. The tonnage capacity of the AC unit (Assume tw = 14C ) c. The heat required by the re-heater Figure Qs = 36 KW Q L = 10.2 KW t 4 = 26C ; RH4 = 50% t 0 = 32C DB ; tw0 = 24C t 2 = 14C ; RH2 = 100% 0.45m = m 0 0.55m = m R t 2 = 14C ; t3 = 19C Other Data hw = 58.679 KJ/kg h 0 = 71.944 KJ/kgda ; W0 = 0.015538 kg m/kgda h 4 = 52.943 KJ/kgda ; W 4 = 0.010518 kgm /kgda h 2 =39.293 KJ/kgda ; W2 = W 3 = 0.009981 kg m/kgda h f g at 19C=2456.49 KJ/kg Processes: 014 – Adiabatic Mixing 1 to 2 – Cooling & De-humidifying 2 to 3 – Sensible Heating 3 to 4 – Heating &Humidifying
5. 5. Qs  m(1.02)(t 4 36  m(1.02)(26  t3) - 19) m  5.042 kg/sec Q L  m(W 4 - W 3 )hfg where : h fg  2500 KJ/kg (For air conditioni ng applicatio n) 10.2  5.042(0.01 W 3  0.0097 0518 - W 3 )2456.49 kgm kgda 36  5.042(52.9 43  h 3 ) KJ h 3  45.803 kgda From Psychromet ric chart and formulas at t 3  19  C and h 3  45.803 KJ/kgda m υ 3  0.8421 3 kgda A) Fan Capacity  5.042(0.84 21)  4.245 mo m  0.45  h1  h4 ho  h 4  h 1  52.943 0.45  71.944  52.943 KJ h 1  61.49 kgda W 1  0.0128 m 3 sec W1  W4 Wo  W 4  W 1  0.010518 0.015538  0.010518 kgm kgda m 0  0.45(5.042 ) m o  2.27 kg/sec m R  m  m 0  5.042  2.27 m R  3.15 kg/sec Q  m(h 1 - h 2 ) - m(W 1 - W 2 )h w Q  5.042(61.4 9  39.293)  5.042(0.01 28  0.009981)5 Q  111.92  0.834 Q  111.086 Tonnage 211 KJ min KW  6665.16 KJ min Capacity  1 Ton of Refrigerat ion Q  31.58 TONS Heat required by the re - heater Q R  m(h 3 - h 2 )  5.042(45.8 03 - 39.293) Q R  32.82 KW 8.679
6. 6. Example: Dehumidifier – Heater Moist air at 32C 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 10C and is re-heated to 16C before entering to the conditioned space. Condensate leaves the de-humidifier at 10C . 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: Psychrometric Chart h1 60% 1 W1 To conditioned space h3 h2 2 10°C S Processes: 1 to 2 – Cooling & Dehumifying 2 to 3 – Sensible Heating Using Fundamental Formulas or Psychrometric chart kgm W 1  0.01807 kgda KJ h 1  78 . 429 kgda W 2  0 . 007632 h 2  29 . 278 kgm kgda KJ kgda W 3  W 2  0 . 007632 kgda KJ h 3  35 . 393 h w  41 . 92 kgm kgda KJ kg a. Condensate removed m w  m(W 1  W 2 ) m w  1 . 5( 0 . 01807  0 . 007632 ) m w  0 . 016 kg sec b. Capacity of AC-unit in TR Q  m ( h 1  h 2 )  mwhw Q  m ( h 1  h 2 )  m ( W 1  W 2 )h w Q  1 . 5( 78 . 429  29 . 278 )  0 . 016 ( 41 . 92 ) Q  73 . 15 KW  4389 KJ min   KJ  1 TR     20 . 8 Tons of Refrigerat ion Q  4389 min  211 KJ    min   n atio atur 3 16°C W2 = W3 ve Cur 32°C
7. 7. Example: Outside air and re-circulated air supplied An auditorium is to be maintained at 27C 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 29C DB and 22C WB is cooled to 17C DB and 15C 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 = 34C 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 Re-circulated Air Exhaust air 3 mR h0 50% RH Exhaust Fan 3 h1 Q m m0 m Fan 2 1 0 1 22°C Qs = 145 KW QL = 95 KW Supply air W0 h3 Auditorium m 0 Psychrometric Chart 3 4 W3 3 h2 W1 AC Unit 15°C Processes: 013 – Adiabatic mixing 1 to 2 – Cooling & dehumidifying 2 to 3 – Heating & humidifying From Chart or Fundamenta kgm W 0  0.016813 h 0  77.281 kgda KJ m0 m m0 14.2  kgm KJ m  14 . 2 kgda From mR KJ m kgda mR kgm 14.2 kgda KJ  sec h0  h1 h0  h3  77 . 281  64 . 298 77 . 281  55 . 626 kg sec 64 . 298  55 . 626 77 . 281  55 . 626 kg sec  mass flow rate of outside air m w  m ( W 1  W 2 )  14 . 2 ( 0 . 013759  0 . 009831 ) m w  0 . 056 kg sec  mass flow rate of condensate leaving AC - unit Q  m(h 1 - h 2 ) - m w h w  14 . 2 ( 64 . 298  41 . 984 )  0 . 056 ( 121 . 63 ) Q  316.8588  0.38143168 Q  317 . 24 ( 60 ) 211  317 . 24 KW  90 . 21 TR  Capacity of AC - unit  mass flow rate of supply air theory of mixing m R  8 . 51 kgda kg h0  h3 m 0  5 .7 34°C 145  m ( 1 . 02 )( 27  17 ) h1  h3  22°C 27°C 29°C Qs  m ( 1 . 02 )( t 3  t 2 ) kgda W 3  0.01117 kgda W 2  0 . 009831 h 2  41.984 15°C 17°C kgm W 1  0.013759 h 1  64.298 To conditioned space l Formulas h 3  55.626 kgda W2 2  mass flow rate of re - circulated air