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Unformatted text preview: <Problem 1> (a) 2 pts Re D = 3 pts 0.05 kg/s
˙
4 ⋅ (0.5 kg / s)
4m
=
= 148, 051 > 2, 300 = Re D ,critcal π D µ π ⋅ (0.005 m) ⋅ (8.6 × 10 −4 N − s / m 2 ) ∴ Turbulent Flow ( Re D > Re D,critcal = 2, 300) (b) !'#(#')!*!!"# $%# $%#&#!$%# !"#
3 pts ˙
˙
˙
˙
E in − E out + E gen = E store ˙
˙
E gen = E store = 0
˙
˙
E in = m c p Tm + q"⋅P ⋅ dx ; P = π ⋅ D
˙
˙
E out = m c p (Tm + dTm )
3 pts ˙
˙
˙
˙
∴ E in − E out = m c p Tm + q"⋅P ⋅ dx − m c p (Tm + dTm ) = 0 3 pts ⎛π x⎞
˙
m c p dTm = q"⋅P ⋅ dx = q"max ⋅ sin⎜
⎟ ⋅ π ⋅ D ⋅ dx
⎝ 2 L⎠
x ∫ dT
0 3 pts m = q"max ⋅π ⋅ D x ⎛ π x ⎞
⋅ ∫ sin⎜
⎟ dx
0
˙
m cp
⎝ 2 L⎠ q" ⋅π ⋅ D ⎡ 2 L
⎛ π x⎞ ⎤
Tm ( x ) − Tin = max
⋅⎢− cos
⎝ 2L ⎠ ⎥
˙
m cp
⎣π
⎦
∴ Tm ( x ) = Tin + x =
0 2 ⋅ q"max ⋅L ⋅ D ⎡
⎛ π x⎞ ⎤
⋅ ⎢1 − cos
⎝ 2L ⎠ ⎥
˙
m cp
⎣
⎦ 2 ⋅ q"max ⋅L ⋅ D ⎡
⎛ π x⎞ ⎤
⋅ ⎢1 − cos
⎝ 2L ⎠ ⎥
˙
m cp
⎣
⎦ (c) 10 pts q" ( x ) = h ⋅ [Ts ( x ) − Tm ( x )] ⇒ Ts ( x ) = Tm ( x ) +
5 pts 2 pts 3 pts ∴ Ts ( x ) = Tin + q" ( x )
h ⎛π x⎞
2 ⋅ q"max ⋅L ⋅ D ⎡
⎛ π x ⎞ ⎤ q"max
⋅ ⎢1 − cos
+
⋅ sin⎜
⎟
⎝ 2L ⎠ ⎥
˙
m cp
h
⎝ 2 L ⎠ ⎣
⎦ (d) The surface temperature is max at x = L because temperature rise in the fluid is induced by heat addition, which is largest at the end of the tube. ∴ x max Ts = L <Problem 2> 5 pts 3 pts 5 pts (a) ˙
q = mcondensation ⋅ h fg = (2.73 kg / s) × (2.342 × 10 6 J / kg) = 6.39366 × 10 6 W ∴ q = 6.39366 × 10 W
6 (b) ˙
q = Cc (Tc,o − Tc ,i ) = mc ,total ⋅ c p ,c ⋅ (Tc,o − Tc ,i ) 5 pts ˙
mc,total 5 pts N= q
6.39366 × 10 6 W
=
=
= 101.9 kg / s c p ,c ⋅ (Tc ,o − Tc,i ) ( 4,181 J / kg − K ) ⋅ (15 °C ) ˙
mc,total 101.9 kg / s
=
= 720
˙
mc ,1
0.141 kg / s ∴ N = 720 3 pts (c) ˙
Ch = ∞, Cc = mc ⋅ c p ,c = 4.26 × 10 5 W / K ⇒ Cmin = Cc 4 pts ε= 4 pts NTU = − ln (1 − ε ) = 0.340326 =
1 pt q
qmax = Cc (Tc ,o − Tc,i ) (Tc,o − Tc ,i ) ( 303 − 288)K
=
=
= 0.288462
Cmin (Th ,i − Tc ,i ) (Th ,i − Tc ,i ) ( 340 − 288)K
U ⋅ Atotal U ⋅ ( N ⋅ 2 π ⋅ D ⋅ L) U ⋅ 720 ⋅ 2 π ⋅ (0.019 m) ⋅ (0.8 m)
=
=
Cmin
Cc
4.26 × 10 5 W / K ∴ U = 2,108.38 W / m − K 2 <Problem 3> (a) $%# $!#
!"# 4 pts q1→ 2 = I1 A1 cosθ1 dω 2 −1 θ1 = θ 2 = 0°
2 pts dω 2 −1 = A2 cos θ 2
r2 2 pts G2 = 2 pts = q1→ 2 I1 A1 cosθ1 dω 2 −1
=
A2
A2 I1 A1 cosθ1 A2 cosθ 2 I1 A1 cos θ1 cos θ 2
⋅
=
A2
r2
r2 (80,000 W / m
=
2 − sr) ⋅ (10 −4 m 2 ) ⋅ (1) ⋅ (1) (2 m) 2 = 2 W / m2 ∴ G2 = 2 W / m 2
(b) $%# $!#
%"# %"# *%#+#,'# *!#+#,'# &'()"# !"# !"# *&#+#&'# $&#
4 pts q1→ 3 = I1 A1 cosθ1 dω 3 −1 θ1 = θ 2 = 60°
θ 3 = 30° 2 pts dω 3 −1 = 2 pts G3 = 2 pts = = A3 cos θ 3
r2 q1→ 3 I1 A1 cosθ1 dω 3 −1
=
A3
A3 I1 A1 cosθ1 A3 cosθ 3 I1 A1 cos θ1 cos θ 3
⋅
=
A3
r2
r2 (80,000 W / m 2 − sr) ⋅ (10 −4 m 2 ) ⋅ (2 m ) 2 ⎛ 1⎞ ⎛ 3 ⎞
⋅
⎝ 2⎠ ⎜ 2 ⎟
⎝⎠ = 3
W / m 2 = 0.866 W / m 2
2 ∴ G3 = 3
W / m 2 = 0.866 W / m 2 2 (c) G3 0.866 W / m 2
=
= 0.275664 W / m 2 − sr
π
π 4 pts I3 = 4 pts q3→ 2 = I3 A3 cosθ 3 dω 2 − 3
dω 2 − 3 = A2 cos θ 2
r2
G2 =
2 pts = =
q3→ 2 I3 A3 cosθ 3 dω 2 − 3
=
A2
A2 I3 A3 cosθ 3 A2 cosθ 2 I3 A3 cos θ 3 cos θ 2
⋅
=
A2
r2
r2 (0.275664 W / m 2 ⎛ 3 ⎞ ⎛ 1⎞
− sr) ⋅ ( 3 × 10 −4 m 2 ) ⋅ ⎜ ⎟ ⋅
⎝ 2 ⎠ ⎝ 2⎠ (2 m ) 2 = 8.95245 × 10 −6 W / m 2 ∴ G2 = 8.95245 × 10 −6 W / m 2 ...
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 Fall '08
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