This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ChE 210 A Fall 2007 Name: M“ Exam #3 _ .w_ Show your Work for credit! Good luck! \ #1. (25 points) VLE Heaven Below is a Txy plot of ethanol and water at 1 atm using Raoult’s Law (and Antoine’s
equation for vapor pressure). W?” “WPOS . huh 0 on 0.4 0.6 as ‘1' 3°”th P“ 0+ PW‘ Exes
inethanol) ' From the Txy diagram:
4195. (a). What is the boiling point of pure ethanol? '38 5 °C l 0 #5 (b). A liquid solution at 85°C containing 0.25 mol fraction of ethanol and 0.75 mol
fraction of water is heated until the ﬁrst bubble point is formed. What is the bubble point temperature? What is the composition of the liquid at this point? What ' .
composition of the vapor at this point? pr’if QESOC .: 0,25
4m Va. = 0.34 [w 0 gangs] \ 8P3 (c). For the 25 mol% ethanol/75% water solution, instead of usingt e tagﬁm, w'ﬁie
down the equations and solution strategy that you must use to SOlVe for the bubble point
temperature * but do not solve” I! You do not have to look up Antoine’s eqn data or write Antoine's eqn out. 53; attached. The experimentally determined bubble point temperature for a 25 111.01% ethanol/75% water
solution is 82.5”C.
3‘05 (d). Why is the bubble point you found in (b) so different from this? Rmomﬁ's Law ”Sums Menu 8AA: \‘clml l.‘ w'cl mde'bm},
E’WH 3' H10 art MO'l' it“ CouponLads; 50 ll (deaf
urban Wcli‘hnh fails. 54195 1,05 Know , Ab‘mkeﬁs n Tit” L ”it OLASVQ/ I; 1mm
W / 3‘» “Q“  ‘F ’ “3
’N _\ P: 750 MM “3 ms, gut15‘.) a. 7159' Om u: be wH'hté. 80:? too; digg Ccalodaﬁa. pa" k4). Fab“
Calamioi’m' r’r hawcL skit 39 55"
49$ {'1' 23m 1:, CWMIA SMIL CW) If— MT , ﬂuxes: m Tb at awn, 'T' (In $321M {Wt125%) “F Giro'13 sun+4.3, «45¢ a» 0w; T of old“! Way esﬁM/‘tz.:{ em»—
mpMIa 50*” Mr Mk7) . ChE 210 A Fall 2007 Name: M Exam #3 #2. (40 points) Energy Balance Around a Boiler. Two streams of water are fed to a boiler. Process data are given here: welsh?" ‘ Silt?) i=1
Nanak ..
Feed stream 1: 120 kgfmin at 34"C ‘g‘ﬁgm 3 4"! Sig";
Feed stream 2: 175 kgfmin at 56°C 13 15 V "dd
Boiler Pressure: 17 bars (absolute) 1 55°C. _ ' " I a 5 55:77.3
H‘Cfﬁt “ﬁllHr: “1 The exiting steam emerges from the boiler through a 6cm diameter pipe. The emerging wet
steam has a quality of 0.90 at the boiler pressure. The problem is speciﬁed (no degree of
freedom analysis is necessary). apt.5 (3) Draw and label a ﬂow chart. 10 P5 (b) Write and simplify the appropriate energy balance for this boiler. State assumptions
when neglecting terms, and also state whether nonzero heat and works tenns‘are .
positive or negativePWAH « LE... +039 = i5 "' W; ﬁ bl: = Q Q > ¢: _ 3' bkh‘kO Wﬁm‘J'IWW‘D Elva“) 1” A‘ﬁ': d M L1 Ml like? “MW9
(c) Calculate the required heat input to the boiler in Isl/min. Neglect kinetic energy M
changes. ' A:
11°" Reference: ? 0 31.019) ml DM'L 3gb or H 150(6)“ new: $2. il'n'h'll p} ‘h‘l‘Fill Pf
Substance A
' I Ham H20 .9 in streaml —_—
H20 2) in stream 2 _l!——
Wet steam, quality 0.9 ———z l 12'.
kg ((1) Calculate the volumetric ﬂowrate of the emerging wet steam. . o ‘ 1* h C” 5:635?“ MFrluﬁ.”
L‘s Lowit‘d. AH = G. 51” ‘ “L as: hdﬁfihi: +6 .I 5.11" Jere'34 T: . N ‘ “ '1
3'5 2:153. . ﬁéatlicg‘K‘ﬂl T1 by}; Xtoﬂ)  [M gm H(H,b‘£3’34q_) "' 11C?%,_o®,55‘£)t
. WV“ .
A I L 01
Cd) i’rass'ura we fume calRd on H ol “LO (6'). 5: an. Telv“ 6:5 “"6 ° w? Sui'cl weir M .3e't z...r 55°C.
:2 0.“? M32. Table 3J5.
) A 3n; '
3"” I: G CHI—‘7’ L95 1.34%? it'd) = “2'4 ”IE3 H,(“zbl¢).5‘”‘.w'd‘ ): 2544 ”1‘11 1.1 A
a»: T4 (“in ﬁa'ﬂJ'V ., rm c...) = 2743.4 ”jig arul'nglw, we “in we”) = 8W8 “it:
 is. ‘  2,}; ii 5 0.40 {3“ + w ill = Miami1+3 +°~‘(“‘3)= 2901.24 UM,
3 pm, 5164 glam wFHx X 245%.; (nolJA %) — 120 a; 142.4 353") — ”'53 . ($554$733
'3 Lilia
: 100126 he: {13“ :2 2;} MIN A
“cad V3 ﬁ. \fi LWJu 4:3, Briana») = 00“le walks '\
24315 V3 ( WSW, x30ﬁ3 3': th CDJIBL» *— DJ £5.03nb33
.= 0.1050510 ”Elkj 30m h; hurv ChE 210 A Fall 2007 Name:
Exam #3 #3. (35 points) Energy Balance Around a Condensor. A stream of pure cyclOpentane vapor ﬂowing at a rate of 44.66 mols/s at 150°C and 1
atm enters a cooler in which 55% of the feed is condensed at constant pressure. The
problem is Speciﬁed (no degree offreedorn analysis is necessary). CsHloU) 9P5 (a) What is the temperature at the condenser outlet? Explain how you know (a single
sentence should sufﬁce). {If you cannot determine (a), pick a Tout less than 150 and continue with(b)}. Tka “hm“ boilt' P1" {5 m a. m P1» 5,1 ”13 Fri
a} wkﬁ'ok 'V andl (LIn Mum's? M in ._ '1:me “Ta lc ﬂat) W
waywat bulimic, p1” EN cydopﬁnﬁme u: 49.3°L (b) Fill in an inletoutlet enthalpy table and calculate the required cooling rate in kW.
Choose as the reference state (351110(3) at the temp. in (a). Reference: (331110(6) at temp. in (a) 401.30]; Substance C5H10 E
Cme V 3pc (1.3.39.th = 0.55“ (44.51.) = 24.51,
ﬁlers, ﬁgswm m oww sinum = aaua  24. 5'1,» 29.“; Loan»? 1°an dam 99" ngopm‘ﬁtM (”melts GALA 62‘) K Luv "' 21.30 ELIwal ””9“ CP “"33 h 3"” Cw c n. £0 = 73.391 “0'3 + 3918 Mo'E‘T  2554x5313 bemaﬂs
”Incl ”CD1.13936. '5 O T‘
n A
= C ( clT‘
lli"b“*1n:WXT3 A ”V + Salaam 1”)
A ‘\ ‘YO'C. __
q?“ HCSHIO (v3 ( '57an : AsHV + all” gogifi‘g‘t: ' 38' 3" hr)”,
A N. “.99,
#5 3pm 14:94.14”) (491‘3‘C) 1 AN  31.3 Ky’mD‘
t I , . n. ..
stats Q 2. AH : i no"; EM "' 2%.. H“;  SlouchL3) q MmﬂBaa)
: ”Ml. 11039" _—. ".tt. X103 tau}
5 L—__.. ...... .4 ...
View
Full Document
 Fall '02
 Shanks

Click to edit the document details