By B B Laud
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Extra resources for Fundamentals Of Statistical Mechanics
20 × 602 whence Fair = 322 N T = φ(Re) or equivalent is readily proved. 7 ∴ Velocity ∝ (gl)1/2 Fr the same. 8 CD = φ(Re). 03 × 10−3 m3 · s−1 125/2 Equate Fr. 5 m · s−1 Skin friction = kAun . 9 N 252 29 30 Solutions manual ∴ (Resid. 19, where k = constant. E. E. 6 dh 2π rδr = (u + δu)2π(r + δr)h − u2π rh = 2π h(rδu + uδr) dt dh d ∴r = h (ur) dt dr r2 dh = hur + const. 7 12µu 6µr dh dp =− 2 =− 3 dr h h dt dp = −p = − F= a 0 Const. 194 × 10−3 m · s−1 Re = d d2g ud = ( ν ν 18µ dust − air ) Max. 1. 1 Difference of depths = 50 mm.
6392 For dynamic similarity, g 1/2 h3/2 /µ must be the same. Assume effect of γ negligible. 3 Equate ωD2 /µ. 4 rad · s−1 Equate P/ ω3 D5 . 4 ωd 2 d , ν D Q =φ ωD3 d/D is same in each case; so is ωd 2 /ν ∴ Systems are dynamically similar. 20 × 602 whence Fair = 322 N T = φ(Re) or equivalent is readily proved. 7 ∴ Velocity ∝ (gl)1/2 Fr the same. 8 CD = φ(Re). 03 × 10−3 m3 · s−1 125/2 Equate Fr. 5 m · s−1 Skin friction = kAun . 9 N 252 29 30 Solutions manual ∴ (Resid. 19, where k = constant. E. E.
Power for 1 pipe Pipe 1: hf + ∴ d2 = d1 /n Q 32f l(Q/n)2 hf = gQ n π 2 gd25 = ∴ 4(Q/n) π d2 ν × const −16/3 = 1/3 n−2 This is >1 u2 u21 [which is lost] = 1 2g 2g k d2 ∴ No advantage. 62 h = head at D above C. 2 m. 21 Let surface areas of tanks be A, B. 5 + h)1/2 An analytical solution is possible. 023 x 72 5 U but const. 332 × 108 ∴ Assume layer turbulent throughout. 6 Increase in θ Turb. lam. e. 9 p = const − 12 (2U sin θ)2 if boundary layer thin. 22◦ But separation occurs only where pressure gradient adverse.
Fundamentals Of Statistical Mechanics by B B Laud