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:  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... ...
View
Full Document
 '08
 Rhee,Jinny
 Fluid Dynamics, Thermodynamics, Energy, tube, engine oil

Click to edit the document details