7.21 On a hot September day, a room has a sensi- ble cooling load of 20,300 BTUh

Question

7.21 On a hot September day, a room has a sensi- ble cooling load of 20,300 BTUhr from oc- cupants, lights, walls, windows, and so on. The latent cooling load for the room is 9000 BTU/hr. The room design conditions are 76 FDB and 50% RH. Supply air will enter the room at 58 FDB. A. Sketch the equipment and duct arrange- ment, showing known information. B. Determine the required air flow rate into the room in CFM. C. For the above supply air, find the wet bulb temperature, enthalpy, relative humidity, and moisture content in grllb and lb/lb. D. t is known that 260CFM of outside air is required for ventilation in this room. The outside air is at 94 FDB and 76 F WB. It is mixed with return air from the room be- fore it enters the air conditioning unit. For the mixed air, determine the dry bulb tem- perature, wet bulb temperature, enthalpy, and moisture content in grlib and lb/lb en- tering the unit. E. Determine the required size of the refrig- eration equipment required to condition this room, in BTUhr and tons. (Includes ventilation cooling load.) F Determine the savings in equipment ca- pacity if the outside ventilation air re- quirement is reduced to 130 CFM. Give the answer in BTUhr and percent.

Explanation / Answer

solution:

1)here initially supply air is supply at To=58 F or 14.44 c and Tw=17 C and room have condition to be maintained at Ti=76 F or 24.44 c with sensible heat load and latent heat load,as onlys upply air is present hence all air present in room is exhausted and hence to get air at 76 F we have to heat air in first case with heater,at that moment equation are as follows

BTu/hr=3798 kw

OASH=.0204*Q*(Ti-To)

OASH=20300 BTu/hr or 5.949 kw and latent OALH=9000 BTu/hr or 2.637 kw

as outside air has to take all load then supply air flow rate through heater is

OASH=.0204*Q*(Ti-To)

Q=29.16 m3/min or 1029.72 CFM

here from latent heat we can get supply air condition of humidity as follows

OALH=50*Q*(Wi-Wo)

Wi=.009 kg/kg of dry air

so we get Wo=.01080 kg/kg of dry air

from psychometrics table we gt that

Ho=enthalphy=42 kj/kg

relative humidity=100%

3)when external supply air is mixed with return air from room then we have

Qsupply=7.3623 m3/min

Qreturn=Qroom-Qsupply

Qroom=Qsupply=29.16 m3/min

we get

Qreturn=21.7971 m3/min

supply air To=94 F or 34.44 C

so mixed air dry bulb temperature is given by

Tmix=To*Qsupply+Ti*Qreturn/Qroom

Tmix=26.96 C

wet bulb temperature

Tow=76 F or 24.44 C

Tmix=To*Qsupply+Ti*Qreturn/Qroom

Tmix=18.87 C

from table we get enthalphy and moisture content as

ho=74 kj/kg

wo=.0146

hmix=50.07 kj/kg

wmix=.01041 kg/kg of dry air

total load=OASH+OALH+RSH+RLH

OASH=.0204*Qsupply*(To-Ti)=1.5019 kw

OALH=50*Q*(Wo-Wi)=2.0614 kw

total load=12.1493 kw

BTu/hr=3798 kw

refrigertion system capacity=12.1493/3.517=3.454 ton of refrigeration or TR 13118.292 BTU/ hr

if flow rate decresed by 50% the OASH and OALH is also decreases by 50%

OASH1=OASH/2=.75095 kw

OALH1=OALH/2=1.0307 kw

now total load=OASH1+OALH1+RSH+RLH=10.3676 kw

hence reduction is by 14.66% in refrigeration capacity but fresh air is also important ofr human comfort

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