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Unformatted text preview: HVAG: Psychrometrics 9.. 500 ft3/min (0.25 1113/3) of air at 80°F (27°C) dry ' bulb and 70% relative humidity are removed from a room. 150 its/min (0.075 1113/5) pass through an air
d at 50°F (10°C). The ~  ' t 80°F 27°C d b b
3193 a“ a ( ) ry ul and conditioner and leave saturate
3/5) bypass the air con A room cont .' "'o 0 tb 1b. Th t t 1 ' . 57 F (19 C) we “ e 0 a pressure 15 1 at!“ remaining 350 ft3/min (0.175 in
th the conditioned air at 1 atm. 5 h h 'd't , b th 1 , d 'ﬁ ,
Egg? aret e (a) um1 1 y ( )en apy an (c)spec1 c ditioner and mm M
' What is the mixture’s (a) temperature, (b) humidity
ratio, and (c) relative humidity? (d) What is the heat of cooling coils bypasses onethird of the load (in tons) of the 3:“. conditioner? 2; If one row
air passing through it, what is the theoretical bypass
. . . . . . 7
four rows of identical cooling c0113 in series. 10 (Time “mi? one h
ft3 /min (2.36 m°/s) of air {ta/min (05 “13/ 3) Of 3131' 3" 50°F (10°C) (117 70% relative humidity and discharges it at
relative humidity are mixed with 1500 drybulb and 95% relative humidi ' tab and 95%
' ' ' 76°F (24°C) and 45% rel ﬁer uses tion cycle Operating between ' an 1112 reirigera
:ative humidity. What are the (a) drybulb temperature. 100%“ (saturated) (38°C) and 50°F (10°C). (a) Locate ' idity, and (c) dew point of the mixture? ints on the psychromet—
ric chart. (b) Find the quant our) A dehumidiﬁer takes 5000
at 95°F (35°C) dry—bulb and
60°F (16°C) Ventilation 1ty of water removed from 4. Air at 60°F (16°C) drybulb and 45°F (7°C) wet— _   bulb passes through an air washer with a humidifying ar 00 Draw the temperatureentropy and enthalpy ' entropy diagrams for the refrigeration cycle. (e) Find I
enthalpy, entropy, and spe '; l . What are the (a) effective bypass
the temperature, pressure,
ndpoint of the refrigeration cycle. .I d (b) drybulb temperature of
ciﬁc volume for each e 11. (Time limit: one hour) 1500 its/min (0.71 m3/s) C d —b lb, 75°F 24°C t— lb ' .
) Ty u ( ) we bu an of saturated 25 psia (170 kPa) air are heated from 200 ﬁsses through a cooling tower and leaves at 85°F (29° C) o o .
rybulb and 90% relative humidity. What are the to ‘l‘oo F (93 ‘30 204 C) m a constant pFessure’ Constant
morsture drying process. (a) What IS the ﬁnal rela
is the ﬁnal speciﬁc humidity? tent per cubic foot (meter) of air? _ .
. (c) How much heat is required p
(d) What is the ﬁnal dew point? ives 1800 ft3/min (0.85 m3 /s) of
) 410 lbm/hr (0.052 kg/s) ‘ ‘3 An air washer race
40% relative humidity and dis— 12. (Time limit: one hour at 70°F (21°C) and
larges the air at 75% relative humidity. A recirculat— o 
_ . o of dry 800°]? (42'? C) au’ pass through a scrubber to
mg water Spray Wlth a conStam temperature 0f 50 F reduce particulate emissions. To protect the elastomeric
the air temperature is reduced to "£7100   _
(Y C) IS used. (a) What mass of makeup water is re seals in the scrubber,
ing the air through a spray of the spray chamber illlired per minute? (b) What will be the condition of 350%. (17700) by pass
. . .,
I e discharged alr ' 80°F (27°C) water. The pressure in is 20 psia (140 kPa). (a) How much water is evaporated
(b) What will be the relative humidity of the aturated steam at atmospheric per hour?
spray chamber? Repeat Prob. 6 using s
' 0°F (10° C) water spray. air leaving the pressure in place of the 5
evaporative counter 3 During performances, a theater experiences a sensi 13. (Time limit: one hour) An
3’18 heat load of~ 500,000 Btu/ hr (150 kW) and a mois— ﬂow air cooling towar removes 1 x 10° Btu {hr (290 kW)
tare load of 175 lbm/h'r (80 kg/h). Air enters the from a water ﬂow. The temperature of the water is re
?ater at 65°F (18°C) and 55% relative humidity and duced from 120 to 110°F (49 to 43°C). Air enters the
.Temoved when it reaches 75°F (24° C) or 60% relative Cooling tower at 91°F (33°C) and 60% relative humid 100°F (38°C) and 82% relative Ilifnidity, whichever comes ﬁrst. (a) What is the venti— ity, and air leaves at
lion rate in mass 0 humidity. Calculate the (a f air per hour? (b)'What are the
' the theater? tity of makeup water. ‘.I
«u
i" —— . , rr_r,g.:1;Iz.,___g_—'l‘I '1
I. 312 N'I'AL ENGINEERING p: “u;
BFn rows = (BFl row)” 5.919119”? ..... .. . 1. Customary U.S. Solution (a) and (b) Locate the intersection of 80°F dry bulb and 67°F wet bulb on the psychrometric chart
(App. 31A). Read the value of humidity and enthalpy. 3. w = 0.0112 ibrn moisture/lbm dry air
0 = 31.5 Btu/lbm dry air trically weighted. op for air is 0.240 Btu/ (c) cp is gravime
lbm—°F, and GP for steam is approximately 0.40 Btu / lbm— °F.
1 50°F [10°C] 76°F (24°C)
G ~ = = .
a" 1 + 0.0112 0 989
0.0112
Gsteam = 1 + O 0112 = 0011 Customary U.S. Solution
6 . # G I c _ + G (a) Locate the two points on the psychrometric p’mmure — a" “I” Steamcp’swam and draw a line between them. . Reading from the chart (speciﬁc volumes),
71A = 13.0 113 /1bm
0,3 = 13.7 ft3/ibm Btu
= . .2
(0 989) (0 40 1bm_OF> Btu
. .4
+ (0 011) (0 0 1bm_0F>
The density at each point is = 0.242 Btu/1bm—°F 1 1
pA = —— = 3
’U
A 13.0 3— lbm
3 = 0.0730 lbm/ft3 = 0.0769 lbm/ft3 SI Solution (51) and (b) 1 1 Locate the intersection of 27°C dry bulb and 19°C wet bulb on the psycin'ometric chart p3 = =
(App. 31.B). Read the vaiue of humidity and enthalpy. “B 13 7 i
' lbm The mass ﬂow at each point is _ . lbm 113
mA _ pAVA _ (0.0769 E37) (1000 = 76.9 lbm/min , . lbm 113
m3 _ pBVB _ (0.0730 ET) (1500 = 109.51bm/min _ g 1 kg w _ (105 kg dry air) (1000 g)
= 0.0105 kg/kg dry air h = 53.9 kJ/kg dry air (c) cp is gravimetrically weighted. 017 for air is 1.0048
kJ/kgK, and cp for steam is approximately 1.675 kJ/ The gravimetric fraction of flow A is kgK.
1
G i 2:. I = . 9
M 1 +0.0105 0 9 0 76.9 E
0.0105 A = 0.413 I
= d. = _ 1b lb 
Gsteam 1 + 0 + 
. _ G . I + G . mm mm I
Commune — were“ alrcpxsteam Since the scales are all linear,
kJ T —T
= (0.990) (1.0048 4—) 0.41 = L—g
kgK 3 TB — TA T 
+ (0.010) (1.675 i) C
kgK = 76°F — (0.413) (76°F — 50°F) = 1.0115 kJ/kgK = 653“?
SIGNAL PUBLICATIONS, INC. PROFES HVAc: PSYGHROMETRIOS 313 (b) w = 0.0082 lbm moisture/lbm dry air 4' Customary U'S‘ Sclmio"
(a) From Eq. 31.24, the bypass factor is ( ) p BF = 1 _ nsat
= 1 — 0.70 = (b) From Eq. 31.34, the dry—bulb temperature of air
leaving the washer can be determined. SI Solution (a) Locate the two points on the psychrometric chart
and draw a line between them. Reading from the chart (speciﬁc volumes), Tdb‘ # Tab L
‘3" ‘01]
715m =
T in — Tw 'n
M = 0.813 m3/kg dry air db. b4
60°F _ Tdbput U]; = 0.856 m3/kg dry air 070 = _600F __ 450}; The density at each point is Tdbiaut = 49.5°F — i — ——‘1 — 1 23 k / 3
PA — UA — m3 —— . g m
0'813 kg SI Solution
p 1 1 1 17 kg/m3 (a) From Eq. 31.24, the bypass factor is
B = — = 3 = . :
’03 m
0.856 —— .9
kg = 1 —' nsat E
*3
The mass ﬂow at each point is = 1 — 070 = g   kg m3 b Fr E 3134 h d bib ‘
m = V = 123 __ 0.5 __ = 0615 k ( ) om q. . , t e ry— u temperature of an
V A M A < m3) < s ) g/S leaving the washer can be determined.
_ . kg m3
mB = pBVB = —3‘ '— = le’ in — Tdb out
In S ﬁnal. = I ‘
Team F wa.in 15°C '— Tdbput
16°C — 7°C kg ——' T = 9.700
' S = 0412 db’m gravimetric fraction of ﬂow A is
0.70 = k
0.615 —g + 0.878 155
s s
ince the scales are linear, 5. Customary U.S. Solution
(a) Refer to the psychrometric chart (App. 31A).
0 412 — __——TB _ TC
' — TB __ T A At point 1, properties of air at Tdb = 95°F and wa =
To = TB — (0.412)(TB  TA) 75 F are
= 24°C — (0412x2400 _ 10°C) col = 0.0141 lbm moisture/lbm air
= 182°C h1 = 38.41bm air
= _ 3 '
w_<80 g )(1kg) v1 143ft/lbma1r
kg dry at 1000 g At point 2, properties of air at Tdb = 85°F and 90%
= (1008 kg/kg dry air relatwe humldity are
T 10 600 Log = 0.0237 lbm moisture/lbm air
d” * h2 = 46.6 Btu/lbm air PROFESSIONAL PUBLICATIONS, INS. Btu Btu
4 . — 38.4
h? _ hl 6 ,6 lbm air lbm air
U1 14 3 E3
' lbm air = 0.573 Btu/ft3 air (b) The moisture added is lbm moisture lbm air lbm moisture
#0. 141 ————————
“’2 ’ “’1 = 0 lbm air
3 ________——
Ur 14.3 0.0237 lbm air = 6.71 x 10—4 lbm/ft3 air
SI Solution (a) Refer to the psychrometric chart (App. 31.B). At point 1, properties of air at Tdb = 35°C and wa = 24°C are
wl = 14.3 g/kg air h1 = 71.8 kJ/kg air
01 = 0.8893 m3/kg air At point 2, properties of air at Tdb = 29°C and 90%
relative humidity are 012 = 23.1 g/kg air
h2 = 88 kJ/kg air The enthalpy change is id id
88 — 1.
h: r hi _ kg air 7 8 kg air
— 3
“1 0.8893 m _
kg air = 18.2 kJ/m3 air (b) The moisture added is g kg
23.1 — 4.
( kg air 1 3 kg air) 1 kg
X (1000 g) ____________‘_______#_————
F 3 0.8893 kg air = 9.90 x 103 kg/m3 air 70°F (21°C) Customary U.S. Solution I
(a) Refer to the psychrometric chart (App. 31.A).
At point 1, properties of air at Tdb = 70°F and 05 = 40% h1 = 23.6 Btu/lbm air
0.21 = 0.00623 lbm moisture/lbm air vl = 13.48 ft3/1bm air The mass flow rate of incoming air is _ ft3
V1 1800 ——,—
ma 1 = — = mm = 133.53 lbm air/min
, 3
U1 ft
13.48 _
lbm a1r Locate point 1 on the psychrometric chart. Notice that the temperature of the recirculating water
is constant but not equal to the air’s entering wetbulb
temperature. Therefore, this is not an adiabatic pro cess . Locate point 3 as 50°F saturated condition (water being
sprayed) on the psychrometric chart. Draw a line from point 1 to point 3. The intersection
of this line with 75% relative humidity deﬁnes point 2 a5 h2 = 21.4 Btu/lbm air
W2 =0.0072 lbm moisture/lbm air TM = 56°F wa 2 = 51.8°F (b) The moisture (water) added is mm = ma,1(w2 — 001)
b  1 .
= (133.531 m a") (0mm ME
:11 1'1 lbm air 4 .062 ———e——
U 0 5 lbmair = 0.127 lbm/min lbm moisture) ...
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 Spring '08
 Wu
 Psychrometrics, Air Cooling

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