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Methods of handling Supply air in HVAC

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Design and Education

Design and Education

Published in: Design, Business, Technology

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  • 1. a. All Outside Air supplied b. Outside air supplied with External Bypass system
  • 2. c. All Outside air supplied with Re – Heater d. Outside and Recirculated air supplied
  • 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. 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. 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. 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. 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