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Unformatted text preview:  auvuuuLVJ.  II V" ‘7 U 301W: (CV15 M5114: Far/f 2009 KNOWN: Temperature 3 moving through a. tube and mass ﬂow rate of various liquid
of prescribed diameter. Mercury, wafer; 0: engine oil ‘ 7
11': = 0. 03kg/s Tm :27; a" ASSUMPTIONS: Constant properties. PROPERTIES: (T = 300K) «De 7r(0.025m),u —— p: “UL RED = 4d: = 4 x 0.03kg/s '_ 1.53kg/s'111 '
0. Q)
 .12, 1 m JC Hence, even for water (,0 = 0.855x10'3 N's/m2), Rep < 2300 and the ﬂow is laminar.
(7‘!j % From qu. 8.3 and 8.23 it follows that '
.( A —.3 . p.
‘ —3
Hence:
Liquid “m0” /9} 31rd,“ 7”) mtfm) <}
Oil 0.069 0.0030 25.2
Mercury 0.0045 1.257 0.031
Water 0.061 2.234 13.02 COMNTS: Note the eﬁ‘e '
effect of Pr on the thermal entry length. PROBLEM neg KNOWN: Flow rate and inlet temperature of engine oil in a tube of prescribed length,
diameter. and surface temperature. Total heat transfer and oil outlet temperature with and without the assumption of fully developed ﬂow. SCHEMATIC: I vi: = 05k 9/5 Tmi=25°C/V K,’ i} "1 '§;%ai.=5aétat£;;;ais }'.'Er . ASSUMPTIONS: (1) Steadystate conditions, (2) Negligible kinetic energy, potential energy
and ﬂow work changes, (3) Constant properties.
@301: $152.! £0135
PROPERTIES: Table 21.5, Engine oil, (Tm = 340K): p=_860‘kg/m3, (razors J/kgK,
u: samtc'z’Eg/sm, k: ouaewlmK, Pr=393T
575‘ 5:59 5.72. {at (1751. 13
ANALYSIS: From Eqs. 3:43!» and 89?, “521.96; .1} ratingat; .sa‘. '( 1'i}E21‘5p’1‘trtE{11ﬁrewaiw'giétiiigétéliﬁ ﬁlm" an _ ._ _ 41:21:,— = a _ a (u 1:0.025mx10m a)
“rm—T, m Tm,i)exp( mp 11) 100 C 75 Cexp ___H~.~0.5kglsx2076mg_l{ h P D
‘ — :2 r
Tm = 100°C  75°C exp(— momma) L4: 0 105—qu _
q=n'lc (T  )=05kg/s>c2g(a}'iﬁ?fkgK(T 45°C) I ‘5 ﬁligm r
1: mm 111.!  _ mun . , Una With Rep == 4 Iii/qu = 4(O.5 kg/s)/rt(0.025 m)0.0531 kg/sm = Wthe flow is laminar. ﬂow
Considering the thermal entry (re) region, it follows from Eq. ﬁthat 5% [3mm] cw [awn2.61m] =mwm. 1+0.04[(nn.)a= mm 0.025 m 14034940)”
' D a; i I a
Tmm, = 100°C — 75°Cexp(0.000757x92.5) = 100°C — 693°C =3§19C Q
.3 “ ii
‘5'! 51 Q “'5’” t
qt“) =1033wm<30.125)°c=529w‘t Q
If fully developed (fd) conditions are assumed for the entire tube,
— 1: 0.139 W/mK 2
a _. _ = _......— . = 20. 
h D366 0.025!“ 366 3W/m K
med, = 100°C — 739°C = 25.1°C Q
qm = 1038W/K(26.1—25)°C = 1190w . Q CONIMENTS: The assumption of fully developed fonditions throughout the tube leads to a large error in the calculation of h and hence q. Note that
Liairm = 0.05 D ReD Pr = 0.05(0.025 m)480(793) = 476 m, which is much_1arger than the tube
length. The calculations should be repeated with properties evaluated atTm = 300K. PROBLEM 1:63 KNOWN: Tubing with glycerin Welded to transformer lateral surface to remove dissipated power.
Maximum allowable temperature rise of coolant is 6°C. FIND: (a) Required coolant rate m, tube length L and lateral spacing S between turns, and (b) Effect of
flowrate on outlet temperature and maximum power. SCHEMATICL ' .
Transformer. 1000 W (:alycerint T J: 24 0c,
. T T Ts = 47 °C ‘
H = 500 mm '5
i l
D = 20 mm k——>l—D =aoo mm ASSUMPTIONS: (l) Steadystate conditions, (2) All heat dissipated by transformer transferred to
glycerin, (3) Fullydeveloped flow (part a), (4) Negligible kinetic and potential energy changes, (5)
Negligible tube wall thermal resistance. PROPERTIES: Table A5. Glycerin (Tm = 300 K): p = 1259.9 kgrmi. cp = 2427 MtgK, a = 79.9 x 10'1
Ns/ml. k = 286 x 10'3 W/rnK. Pr = 6730. ANALYSIS: (a) From an overall energy balance assuming the maximum temperature rise of the
glycerin coolant is 6°C, ﬁnd the flow rate as .q=ch(Tm'n— mJ) rh=q/cp(Tm‘n— m)=1000W/2427I/kgK(6K)=6.87x10'2kg/s < From Eqrﬂsﬁ', the length of tubing can be determined, ﬂy : eitp(—PLh/r'ncp) Ts "Tami
where P = TED. For the tube flow, ﬁnd
RED_ 4:11 = 4x6.87x10 'kg/s =54? ' itDu it x 0.020m x799 x 10'2 N  s/m1 which implies laminar flow, and if fully developed, _ 3
Nunzﬂgziﬁﬁ E=3.66x286><10 W/m Kzsmw/mLK
k 0.020m ' %‘%:_¥% = exp[—('J't(0.020m) x 52.3W/m3  K xL)/(6.87 x10"2 kg/s x 2427J/kg  K)] L = 15.3 m. < The number of turns of the tubing. N, is N # L/(er) = (15.3 m)!rt(0.3 in) = 16.2 and hence the spacing S
will be s =HIN=500 min/16.2=30.8 mm. ‘ < continued... PROBLEM 8.26 (Cont)
 053. Ema i, 1355?: W“ Cad, ¢_7LC__
(b) Parametric calculations Were‘performed using theHHLGarrehrrizﬁooipad—basedoa—Eq: 8 26 (a then‘nal entry length condition), and the following results were obtained. 35  E
y ' u g :5,  a1 S E
E E E 29
ﬂ — = .
E 7 Z “III... gt a,
25 . . .
0.05 0.09 0.13 0.17 0.21 0.25 0.05 . ' . 0.1T . 0.25 Mass llowrata. mdaltkgls) Mass flowrata. mdollkgis) With T, maintained at 47°C, the maximum allowable transforrne power (heat rate) and glycerin outlet
temperature increase and decrease. respectchly, with increasin m. The increase in q is due to an increase in 'ﬁﬁn (and hence h) with increasing Reg. The va a of NED increased from 5.3 to 9.4 with
increasing In from 0.05 to 0.25 kg/s. CONﬂVIENTS: Since G2},l = (L/DVReD Pr = (15.3 .02 m)l(5.47 x 6780) = 0.0206 < 0.05. entrance
length effects are significant, and Eq. 8.56 should be sed. to determine my. Nw aim 1/040WWQR60’
“43.04 EDI!) Edyaﬁ PROBLEM KNOWN: Flow rate, inlet temperature and desired outlet temperature of water passing through a tube of
prescribed diameter'and surface temperature. FIND: (a) Required tube length. L, for prescribed conditions, (b) Required length using tube diameters
ovor the range 30 S D s 50 mm with flow rates In = 1. 2 and 3 kg/s; represcnt this design information graphically. and (c) Pressure gradient as a function of tube diameter for the three flow rates assuming the '
tube wall is smooth. SCHEMATIC: m = 2 kg/s /. Tm'i = 00/7 .
ASSUMPTIONS: (l) Steadystate conditions, (2) Negligible potential energy, kinetic energy and flow
' work changes, (3) Constant preperties. PROPERTIES: Table A6, Water (Tm = 323 K): on = 4181 J/kgK. it = 547 x 10'6 Ns/mz, 1: = 0.643
W/mK. P: = 3.56. ANALYSIS: (a) From Eq. 8.6. the Reynolds number is R6D=iri=_‘l§3k_gii___=msx105. ' (1) Hence the flow is turbulent, and assuming fully developed conditions throughout the tube, it follows
from the DittusBoelter correlation, Eq‘ 8.60. i  k 0.643W/m K ___ 4,5 M_ 5 415 0.4” 2_
11.130.023RsD Pr .. 004m 0.023(1.16x10 ) (3.56) _6919w/m K (2) From Eq. 3.4221, we then obtain .. —' 2 AT AT 2k‘s41811kKfn25T75‘C 
L: mcp n( _D/ 0:“ g/( /E ) ( q / )=10I6m' <
non 1t(0.04m)6919W/m* K Tube 15 glh. L (m) equations, the required length  .
ﬂow rate is computed and pi Tuba diameter. D (mm) —8— Flow rate. rndot=1 kgls
— mdot = 2 kgls
—¢— mdct = 3 kgfs Continued... ...
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This note was uploaded on 09/08/2010 for the course ME 114 at San Jose State University .
 '08
 Rhee,Jinny

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