Success

Summer

Conservation of mass:

For air:

ma1=ma3=ma

For the water:

mv1=mw2'+mv3

mw2'=mv1ma-mv3mama

mw2'=ω1-ω3ma

For state 1, since the dry-bulb is 80℉ and the wet-bulb temperature is 100℉ ,

For state 3, since the dry-bulb is 75℉ and the wet-bulb temperature is 68℉ ,

Thus,

ν1=14.5ft3lbm ω1=125grainswaterlbmair ω2=ω3=92grainswaterlbmair

h1*=44.2 Btulbmair h2*= 29.9Btulbmair h3*=32.7Btulbmair

ma=∀ν1=212000ft3lbm14.5ft3lbm=14620.69lbmairmin

mw2'=ω1-ω3ma=125-92grainswaterlbmair*14620.69lbmairmin*1 lbm7000 grains=68.926lbm

Cooling process 1-2:

dEdt=-Qcool-W+imeh+V22+gzi-eme(h+V22+gz)e

The assumptions for this process: 1, 2, 3

Qcool=mv1hv1+maha1-mw2'hw2'-mv2hv2+maha2

Qcool=mamv1mahv1+ha1-mw2'mahw2'-mv2mahv2+ha2

Qcool=maω1hv1+ha1-ω1-ω2hw2'-ω2hv2+ha2

Qcool=mah1*-h2*-ω1-ω2hf@68℉

=14620.69lbmairmin44.2-29.9Btulbmair-125-92grainswaterlbmair*1Btu7000grainswater*1 min60sec

=16845.44Btusec

First law for part 1:

dEdt=-Q-W+imeh+V22+gzi-eme(h+V22+gz)e

The assumptions for this process: 1, 2, 3

Q=mv1hv1+maha1-mw2'hw2'-mv3hv3+maha3

Q=mamv1mahv1+ha1-mw2'mahw2'-mv3mahv3+ha3

Q=maω1hv1+ha1-ω1-ω3hw2'-ω3hv3+ha3

Q=mah1*-h3*-ω1-ω3hf@68℉

=14620.69lbmairmin44.2-32.7Btulbmair-125-92grainswaterlbmair*1Btu7000grainswater*1 min60sec

=16163.14Btusec

Winter

Conservation of mass:

For air:

mA1=mA3=mA

For the water:

mV1=mW2'+mV3

mW2'=W3-W1mA

For state 1, since the dry-bulb is 0℉ and the relative humidity is 1.0 ,

For state 3, since the dry-bulb is 70℉ and the wet-bulb temperature is 62℉ ,

Thus,

ν1= ft3lbm W1= 0.9 grainswaterlbmair W3= 70.5 grainswaterlbmair

mA=∀ν1=212000ft3lbmft3lbm= lbmairmin

mW2'=W1-W3ma=70.5-6.5grainswaterlbmair*lbmairmin*1 lbm7000 grains=lbm

First law for part 1:

dEdt=Q-W+imeh+V22+gzi-eme(h+V22+gz)e

The assumptions for this process: 1, 2, 3

Q=mV3hV3+mAhA3-mV1hV1+mAhA1-mW2'hW2'

Q=mAh3*-h1*-W3-W1hg@62℉...