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

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

Psychrometric Analysis of Complete Systems

Psychrometric Analysis of Complete Systems

Solution (Summer cooling conditions):

подпись: 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

подпись: =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

подпись: 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

подпись: 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


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