Heat Transfer: Space Conditioning a) Determine the enthalpy of the outside air,

Question

Heat Transfer:  Space Conditioning

a) Determine the enthalpy of the outside air, mixed air, supply air, room air, and return air (in BTU/lbA).

b) Determine the rate (in cubic feet per minute) that the fan must supply air into the space.

c) Determine the load on the cooling coil (in BTU/hr and W)

d) There is a fault in the mechanical system and the outside air damper becomes stuck open. As a result, the outside air is now entering at a rate of 1200 cfm. What is the load on the cooling coil now. How much additional energy am I using as result of this fault?

A given space is to be maintained at 78 degrees F DB and 65 degrees WB. The total heat gain in the space is 60,000 BTU/hr. The occupants of the space require 500cfm of outside air. The outdoor air is 90 degrees DB and 55% RH. Assume that both the drybulb and wetbulb temperatures are some 20 degrees below that of the space. Assume that the return air is the same as the room (space) air Return a Space Cooling Cou outside mixed air air 3 supply air an

Explanation / Answer

Enthalpy of moist air=

h = (0.240 Btu/lboF) t + x [(0.444 Btu/lboF) t + (1061 Btu/lb)]

outside air condition=

DB=90F,RH=55%,w=0.01643lb/lb of air,WB=76.7F

h=40btu/lb

Room Air=

DB=78F,WB=65F,w=0.0101lb/lb of air

h=30Btu/lba

as return air and room air is same so enthalpy will be same.

Supply air ccondition=

DB=58F,WB=45F,w=0.00336lb/lb of air,

h=17.58Btu/lba

Room total load=60000Btu/hr

=Room sensible load+Room Latent Load

=(1.085*Cfm*temp change+.7*Cfm*change in wet bulb temp) b/w point 3 and 4

60000= 1.085*Cfm*20+.7*Cfm*(76.7-65)

Cfm=2000

this is supply air quantity.

Mixed air enthaply =

(houtside air* Cfmoutside air+hreturn air*Cfmreturn air)/Total Cfm

=32.5Btu/lba

mixed air dry bulb temp

=(toutside air* Cfmoutside air+treturn air*Cfmreturn air)/Total Cfm

=81F

mixed air specific humidity

=(woutside air* Cfmoutside air+wreturn air*Cfmreturn air)/Total Cfm

=.0.0116825lb/lba

Load on cooling coil

in btu/r

=(1.085*Cfm*temp change+.7*Cfm*change in wet bulb temp)

=1.085*2000*(81-58)+.7*2000*(68.4-45)=82670btu/hr

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