Exam3_solution_SP11_Annotated

Exam3_solution_SP11_Annotated - <Problem 1>...

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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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