Psychrometric Analysis of Complete Systems
Space Heating and Cooling Loads
Heating Load — the maximum probable net rate of heat loss from a conditioned space which would have to be made up by addition of heat from the heating system to maintain some desired temperature and humidity conditions in the space Cooling Load — for cooling
Cooling and heating load of a classroom at PSU with 10 occupants are estimated as follows:
|
Sensible |
Cooling (W) |
Heating (W) |
|
Walls |
1000 |
2000 |
|
Window (conduction) |
1000 |
2000 |
|
Window (radiation) |
1000 |
— |
|
People: 70 W/person |
700 |
— |
|
Lighting |
300 |
— |
|
Latent |
Cooling (W) |
Heating (W) |
|
People 30 W/person |
300 |
— |
|
Plants |
700 |
|
|
Outdoor Design Conditions |
||
|
31°C |
-14°C |
|
|
Twet |
23 °C |
— |
|
Indoor Design Conditions |
||
|
25°C |
22°C |
|
|
50% |
50% |
|
|
Minimum Outdoor Air |
||
|
Fresh air: 8 L/s person |
80 L/s |
80 L/s |
|
Air Supply Temperatures |
||
|
Maximum |
60°C |
|
|
Minimum Design the air-conditioning system. |
15°C |
|
|
|
|
|
Solution (Summer cooling conditions): |
Determine the enthalpy at all the status. We use the psych chart in sea level, p = 101325 Pa.
Outdoor (O): T0 = 31°C, T0,wet = 23°C
From Table A-lb, ps, wet = 2815 Pa Wo, wet — 0.622 ps, wet/(p — Ps, wet)
|
=0.0144 kgv/kga |
= 0.622 X 2815/(101325 — 2815) =0.0178 kgv/kga cp(T^,et — Ta) + W^Jfg 1.01(23 — 31) + 0.0178×2447
|
I —i O H |
2558-96
I0 = 1.01 T0 + Wo(2501 + 1.86T0)
= 1.01 kJ/(kga °C) x 31°C
+ 0.0144 kgv/kga(2501 kJ/kgv+ 1.86 kJ/(kgv °C) x 31°C) = 68.15 kj/kga
Room (R): TR = 25 °C, Q=50%
From Table A-1B, ps R = 3174 Pa WR = 0.622 (]) ps, R/(p — <]) Ps, r)
= 0.622×0.5×3174/(101325 -3174)
= 0.01 kgv/kga
IR = I. OITr + Wr(2501 + 1.86Tr)
= 1.01 kj/(kga °C) x 25°C
+ 0.01 kgv/kga(2501 kJ/kgv+ 1.86 kJ/(kgv °C) x 25°C)
= 50.72 kj/kga
O — h7 4000 W
Mixture (M): mn =——— se’m e— =—————- ————————— = 0396kg/ 5
V ° C^-Tj) 1.01×1000 J/kga(25°C — 15°C)
M0 = 80 L/s = 80xl0’3 m3/s x 1.2 kga/m3 = 0.096 kga/s hir = ma — m0 = 0.396 — 0.096 = 0.3 kga/s
TOC o "1-5" h z mi +mRiR 0.096×68.15 + 0.3×50.72 , ,,
= ^ =————————————————————————————————————— = 54 95kJ i k
M m +mR 0.096 + 0.3
Supply air at the inlet (I): Ti = 15 °C
AW = ma(WR — Wi) = Qiatent/ifg
Wl — Wr — Qlateiit/(ifg nia )
= 0.01kgv/kga — 1 kW /(2454 kJ/kgv x 0.396 kga/s ) = 0.009 kgv/kga ii= 1.01Ti + Wi(2501 + I.86T1)
= 1.01 x 15 + 0.009(2501 + 1.86 x 15)
= 37.91 kJ/kga Cooling coil (C): (]) = 90%
Wc = Wi = 0.009 kgv/kga Pc
|
P~Pc Pc |
Wc = 0.622-
0.009 = 0.622
101325 — pc pc= 1447.5 Pa
Ps, c = Pc / <> = 1447.5 / 0.9= 1608 Pa From Table Al-b, Tc = 14 °C ic = l. OlTc + Wc(2501 + 1.86TC)
= 1.01 x 14 + 0.009 (2501 + 1.86 x 14)
= 36.88 kJ/kga
Fan: V = ma v = 0.396 kg/s x 0.84 m3/kg = 0.332 m3/s = 1200 m3/hr
Cooling coil: Qcooiing = ma (iM — ic) = 0.396 kga/s (54.95 kJ/kga — 36.88 kJ/kga)
= 7.156 kW
Heating coil: Qheatmg = ma (ii — ic) = 0.396 kga/s (37.91 kJ/kga — 36.88 kJ/kga)
= 0.4 kW
The capacity of the heating coil will be larger in winter. Therefore, the final size of the equipment should be the greater of the summer and winter capacities. In many cases, economizers are used to recover energy. Then re-heat in the present design becomes unnecessary.
Posted in Fundamentals of Heating. Ventilating, and Air-Conditioning