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Unformatted text preview: P BLEM 1 . M Air at 30° C and atmospheric pressure flows past a ﬂat plate at 15 m/s. The transition
Reynolds number is 0.3 x 105. EIHD: What is the thickness of the boundary layer and the wall shear stress at a point 0.6 m
from the leading edge of the plate? H AT] '
U = 15 m/s
“a
T0" = 30 C 5
I . x = 0.6 m . l
ASSUMPTIQNS:
ERQPERTIES; Air at 30°C from Table A8; v = 16.01 x 106 mzls
p = 1.164 kg/m3 A ALY I ' Rex = (Ux)lv = [(15.0)(0.60)]l(6.01 x 10'6) = 5.621 x 105 Since Rex > 3 x 105 the bomdarylayer is turbulent. From Table 61:
1 _l.
5}: 0.371{Rex]'; ; 2: 0.371{5.621 x 105] 5 = 2.627 x 1o2 8 = (0.60)(2.627 x 102) = 1.576 x 102 ¢=¢=¢=¢=¢=<=
Res = U5/v = (15.0)(l.576 x 102)/(16.01 x 106)= 1.477 x 104 I“, = 0.0225p'U2'[R65]1f4=(0.0225)~(1.1644)[l.477 x 10“)“4 Tw = 0.5347 Nt’l‘ﬂ2 ¢=¢=¢=¢=¢=¢= PRQBLEM 69
[(55 QWN; Air ﬂows pasta smooth ﬂat plate which is parallel to the ﬂow. El ND ; Determine the ratio of the friction drag over the front half of the plate. x = O to x = U2,
to the friction drag of the entire plate if:
(a) The flow is laminar over the entire plate. (b) The flow is turbulent over the entire plate 'and ReL < 107. H AT] '
U
# I
I
H———L———>l
ASSUMPTIQNS;
R PERTIE '
ANALYSIS;
(a) Flow is completely laminar: Leading onehalf; x = U2 ; Rex = UU(v2) ; A = bL/2
G = 1.32:3{Re,.]”2 From Table 61
U. L 1/2 _ _C'}oApU2 _ U 1/2 .ZU bL
Cf=l.:328‘[v—2] ,Df;2————2———l.328 'v—zJ' 2 ‘2 .
. _ PU2 ”11/2 L112
. Dm _ 1.328 2 (U) M2)
Full Length: x = L; Rex = UUv ; A = bL
_ _ "1&1“? . _E}ApU2 _ [U LII/2p. .U2
Cr— 1.328[ V 2 , Dm— % _ 1.328 v 2 b L
pU 1/2 _ 1I2 I2 _
Dr = 1328 2 '(ﬁ‘) 'b'Luz " ._f£r2=(L) (l) =?1f_ 0'70? ¢=¢=¢=¢=¢=<= (b) Flow is completely turbulent: Leading onehalf; x = U2; Rex = UlJ(v2) ; A = bU2 __ 1Is
Cr = 0.074[Red'1’5 From Table 61; Cf_ — 0. 074{%— 3‘] _ C_rApU2 U L 4’5 p_ U2 _____b L_ p U2 AL US L 415
D‘Q'T=Om4[v2 2 2 LOOM 2 'bi ] (— Full Length: x — L; Rex = UL/‘v; A: b L E
!
1 _ us [/5 2 . 2
Cf: m374[Uv——'I: mm: 0.074{U—'—VL] LE —bL — 0074'” E 0%]“5 (L IS
_' Dm_ = (Lr15(l)415_ l —O. 57 4
Dr 2 L 2 4/5 ¢=c=c=¢=¢=¢= LL________________ ERQBLEM 6815 . w A baseball pitcher is clocked throwing a fast ball at 90 mph when the air temperature is
60°F. ' HELL If the diameter of the baseball is 2.80 in.. calculate the drag force on the baseball
assuming the effects of surface roughness to be negligible, i.e. a new baseball. 512HEMATIQ; MP I ' (l) The baseball is traveling at a constant speed so the flow past it is
steadystate.
(2) The baseball is new and therefore its surface is assumed to be
smooth.
PRQPERTIES: Air @ 60 'F from Table 32 p = 0.07633 lbm/ft3
v = 0.5692 ft2 [hr = 1.58 x 10‘4ft21's
g = 32.17 {:152 LY I '
U = 90 mph (5280 fu’rnile)/(3600 s/hr) = 132 His Red = Udfv = l32(2.8l12)l(l.581x104): 1.948 x 105 C1) = 0.4 (Figure 615a for a smooth sphere)
D = C1) 1302 A12 = 0.4(0.07633l32.l7)(132)2 n.(2.3/12)2 14 Therefore. D = 0.3535 [bf <=<=<=¢=<=<= MMENT DI U I ° The drag on a new baseball is calculated assuming that the drag coefﬁcient of the baseball can be
represented by that for a smooth sphere, Figure 615a. At the Reynolds number of the ﬂow, Red = 1.948 x 105 , the flow has not naturally transitioned to turbulent. Hence the larger drag coefﬁcient is
used. If the baseball were assumed to be rough then the lower value of drag coefﬁcient would be justiﬁed. 1
i ERQBLEM QIBE A well streamlined automobile has a measured drag coefﬁcient of CD = 0.045. A
poorly streamlined automobile has a CD = 0.20. if each of these automobiles has an equivalent diameter of 6.0 ft, determine the power
required to overcome the air resistance at 60 mph at sea level through air at 60"F. a U.S. Standard Atmosphere, where PSL = 21 16.2 lbplft2 and p = 0.07633 lbmlftz. The speeds of the automobiles are constant with time (steadystate). Air @ 60 "C from Table B2 p = 0.07633 lbmlft3
P = 2116.21bp'ft2 T = 60 "F g = 32.17 fu's2 NALY I ' 60 mph = 88 Ms
Drag = C9 (pJ'(2g))U2(7td2 l4)
(Drag)1 = 0.0(115(0.07633116434)(88)2 7t(6.0)2 I4 = 11.69 lb1r (Drag)2 = 0.20100763376434)(88)2 7c(6.0)2 74 = 51.95 1h;
Horsepower = (drag)(velocity)l550 (Horsepower)1 = 11.69(88)l550 = 1.870 ¢==¢=¢=<=¢=
(Horsepowem = 51.95(88)!550 = 3312 <=¢=¢=¢=<=<= PR BLE 1 E KNQWN & FIND:
Repeat Prob. 6 18 for Standard Atmosphen’c cenditions at 5000 ft. (Denver, CO) where T = 41.2"F. PIPSL = 0.8321, and pipSL = 0.8616. Compare the power
required to overcome air resistance at the two altitudes. SSHEMAILQ:
W The speeds of the automobiles are constant with time (steadystate).
PRQPERTIES; Air @ 41.2 “F from Table 132 pSL = 0.07633 1bm/ft3
PSL = 2116.2 than? TSL = 60 "1:
g = 32.17 ft/s2
ANA LYSIS: 60 mph = 88 his
Drag = C9 (p1(2g))U2(zrc.d2 14) Drag = (Drag)SL~(P/PSL)
(Dragn = 11.69(0.8616) = 10.072
(Drag): = 51.95(O.8616) = 44.76 Horsepower = (drag)(velocity)1550 (Horsepowerh = 10.072(88)l550 = 1,611 <=<=¢=¢=¢=¢=
(Horsepowem = 44.76(88)l550 = 7.162 <=<=¢==¢=<=
E T DI v A 14% reduction of the drag on each automobile is experienced between sea level and Denver, CO
(5,000 ft). This is due to the change in air density between sea level and 5,000 ft. KW N; Water at a velocity of 5 mls flows over a horizontal ﬂat isothermal plate 20 cm long.
The temperature of the water is 30 C while that of the surface of the plate 13 60° C. FIND' Calculate the rate of heat transfer per unit width from the upward facing surface of the
plate.
HE ATIC:
Water @ 30°C
# b Wtdlh
U = 5 mls
“A“
A MPTI S: PR PERTIE : ANALYSIS; Use eq. 636 H921 2233x103 Use eq. 635 Nu: VIO
:1 c_
1 + 12,7.Et)”2.( pr2f3 _ 1) 1 + 12.7(2233 x 103)”2(5.42m  1) A =(AT) (20.6 x 103)(0.2)(6030) = 123.6 x 103 W/m = 123.6 kW/m ‘=¢=¢=<=¢=¢= ReL — (U L —p)ltt  ((5) (0. 2) (995. 7))/(797. 8 x 10 6)—  1.248 x 106 _6.?08 x103)(0 615):206x103 W/m2 c PRQBLEM ﬁZﬂ For water @ 30"C from Table A9; k = 0.615 WIm“C Ll. = 797.3 x 106 kglms p: 995. 7 kg/m3
Pr = 5.42 61,— _ 4.180 kJ/kg c (turbulent boundary layer) ‘c? = 0.074{Re._‘”5 = 0.074[1248 x 106]“: 4.466 x 103 = 6.708 x 103 (.02) EBQBLEM ﬁZﬁ
KNQWN; A fluid at 20° C ﬂows over a 50 cm long ﬂat plate with a velocity of 2 m/s. E IN D '. Detennine the average convection heat transfer coefﬁcient for the following fluids:
(:1) Air. (b) Water.
W
U = 2 mls I
~ “—255Mm—H A MPTI : Uniform surface temperature PBQPERTlEs; Air at 20' c, Table A8. v = 15.09 x 10*6 1112/3
' Pr = 0.713 k = 25.64 x 10'3 W!m°C
Water at 20' C. Table A9. v = 1.004 x 10'6 m2/s Pr = 6.99
k = 0.5996 WIm"C AMLXSIE:
(a) Air Re, =91 539.51% = 66270 a 66.27 x 103 (Laminar now)
V 15.09 x 1046175]
Use eq. 631.
__ 3
h = 0.664% Pr”3Re1”2 = 0,664.2_5§%_’;.1_0_ 0.713“3(66.27 ><103)”2 = 7.83 FE Ux 2 [11051.0 5 [m] (b) Water Re =
‘ V 1.004x 10061215] Use eq. 638. Turbulent Flow; m = 'V Nulzmiw + Nufurbulcm mam“, = 0.664~(Re)”2.(Pr)m = 0.6640 x 106 ”21699)“3 = 1270 eq. 630. —— 0.037(ReP'8(Pr)
NU b len =—————————————— = 6205 .6—37.
“" " ' 1 + 2.443(RerOl((Pr)2’3 — 1) eq
N1? = 6/0270? + (6205)2 = 6334
i; = FIEk = 6334(0.5996) = 7596. 111 .,._¢=¢=¢=¢=¢= L 0.5 m2'C =——————__— = 996 x 103 = l x 10‘5 (Turbulent ﬂow) PR BLEM 44 KNQWN: A window 0.5 m wide and 0.4 m high has an interior surface temperature of 10° C. A
small fan has been installed in order to minimize the danger of the window fogging up.
The fan creates a slight upward movement ofair over the window with its velocity
being of the order of 1 m/s. FIND ' Estimate the rate of heat loss through the window if the temperature of the air in the
room is 20" C.
S! :HEMATIQ:
= 10°C , 7,, = 20°C
3 [opposing flow] ﬁU = l mls
A MPTI N '
PRQPERTIEs: Air @ Tr: (20+10)/2= 15°C (233.15 K) from Table A8; Pr = 0.715
v = 14.64 x 10'6 m2/s k = 0.02526 W/m “C
13 = 171* = 3.47 x 103 UK p = 1.225 kglm3
ANALYSIS:
ReL = 955 = M = 27.32 x 103 (Laminar flow) V (14.64 x 106) .. .3. 3. _ . .
a=g914m PF=W=7MGHW V2 (14.64 x 105)2 The ratio of GIL [Re2 is 2 0.1 so combined natural and forced convection must be
considered in calculating the heat transfer. R The forced convection opposes the natural convection. Use eq. 6—30 to calculate the effect of forced convection.
mF = 0.664(ReL)”2(Pr)”3 = 0.664(27.32 x 103)“2(0.715)m = 93.17 25.6 eq. 652 to calculate the effect of natural convection.
NuN = 0.68 + 0.67‘[RaLW(Pr)]m MPr)=[+94—31)9”6]m=ﬂig’lﬁlmwsm [1 + 0.715
m” =0 68 + 0. 67 [(72 66 x 106M 0 343)]”“ = 43. 2 N1? = Nu]: — mm = 198.143 — 48.234”3 = 94.1
H —— h.(0 40) h“ “= 7* = 941 ; E = 5.942 wrmzrc
k T(.25 26 x 10 3)
Q= 11 A(A ).=(5 942)“). mm. 10)= 11 33 w <===¢=== MMENT DIS ION: The movement of this air upward by the fan decreases the heat transfer to the window. ...
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