# Psychrometric Analysis of Complete Systems

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

Example:

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

O r

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.