psychromet prob - HVAG: Psychrometrics 9.. 500 ft3/min...

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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 t-b 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 one-third 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 (0-5 “13/ 3) Of 3131' 3-"- 50°F (10°C) (117- 70% relative humidity and discharges it at relative humidity are mixed with 1500 dry-bulb and 95% relative humidi ' tab and 95% ' ' ' 76°F (24°C) and 45% rel- fier uses tion cycle Operating between ' an 11-12 reirigera :ative humidity. What are the (a) dry-bulb 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) dry-bulb and 45°F (7°C) wet— _ - - bulb passes through an air washer with a humidifying ar- 00 Draw the temperature-entropy 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) dry-bulb 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 . ry-bulb 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. 31-2 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 = 0-011 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/kg-K, and cp for steam is approximately 1.675 kJ/ The gravimetric fraction of flow A is kg-K. 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 kg-K 3 TB — TA T - + (0.010) (1.675 i) C kg-K = 76°F — (0.413) (76°F — 50°F) = 1.0115 kJ/kg-K = 653“? SIGNAL PUBLICATIONS, INC. PROFES HVAc: PSYGHROMETRIOS 31-3 (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 0-70 = _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 — 0-70 = 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 = 0-412 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 10-3 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 wet-bulb 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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This note was uploaded on 09/11/2011 for the course ENV 4101 taught by Professor Wu during the Spring '08 term at University of Florida.

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psychromet prob - HVAG: Psychrometrics 9.. 500 ft3/min...

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