生物反应工程教学课件-Assignment-II-Solutions.docVIP

P6.1 (a) Re = Ni Di2 ρ/μ≥104 μ ≤ Ni Di2 ρ/104 = (800/60)×(5×10-2)2 ×1000/104 = 3.33×10-3 Pa·s (b) πNi Di = πNi’ Di’ Ni’ = Ni Di/ Di’ = (800/60)×(1/15) = 0.889 s-1 μ≤ Ni’ Di’2 ρ/104 = 0.889×(15×5×10-2)2 ×1000/104 = 0.05 Pa·s P6.2 (a) (i) Re = Ni Di2 ρ/μ = (90/60)×(3/3)2×1000/(1.005×10-3) = 1.49×106 From Figure 6.13, flow at this Re is fully turbulent, and from Table 6.1, Np’ = 5 Therefore, P = Np’ρNi3Di5 = 5×1000×(90/60)3×(3/3)5 = 1.69×104 W (ii) Re = Ni Di2 ρ/μ = (90/60)×(3/3)2×1000/(100×1.005×10-3) = 1.49×104≥104 Flow at this Re is fully turbulent and from Table 6.1: Np’=5 Therefore, P= Np’ ρNi3Di5=5×1000×(90/60)3×(3/3)5=1.69×104 W (iii) Re = Ni Di2 ρ/μ = (90/60)×(3/3)2×1000/(104×1.005×10-3) = 149 Flow at this Re is laminar and from Table 6.1: k1 = 70 Therefore, P= k1 ρμNi2 Di3 = 70×1000×(104×1.005×10-3)×(90/60)2×(3/3)3 =1.65×106 W (b) Re = Ni Di2 ρ/μ≥104 Ni ≥ 104×μ/(Di2 ρ) = 104×(1000×1.005×10-3)/(1×1000) = 10.05 s-1 Flow at this Re is fully turbulent and from Table 6.1: Np’=5 P= Np’ ρNi3Di5 = 5×1000×(10.05)3×(3/3)5 = 5.08×106 W = 5.08×103 KW P6.3 Re = Ni Di2 ρ/μ = (900/60)×(7/100)2×1000/(1.005×10-3) = 7.3×104 ≥ 104 Flow at this Re is fully turbulent and from Table 6.1: Np’=5 P = Np’ρNi3Di5 = 5×1000×(900/60)3×(7/100)5 = 28.36 W Friction in the stirrer motor gearbox and seals reduces the energy transmitted to the fluid; therefore the electrical power consumed by stirrer motors is always greater than the mixing power by an amount depending on the efficiency of the drive. P6.4 (a) Assumption: flow is turbulent, then Np’ = 5 P = Np’ ρNi3Di5 ≤ 1.5×103×(πR2×H) Ni3 ≤ 1.5×103×(πR2×H)/( Np’ ρDi5) = 1.5×103×(π×12×2)/(5×1000×(2/3)5) = 14.314 s-3 Then, Ni ≤ 2.428 s-1 Re = Ni Di2 ρ/μ = 2.428×(2/3)2×1000/(1.005×10-3) = 1.074×106≥104 The assumption is validated. Ni tm = 1.54V/Di3 tm = 1.54V/(Di3 Ni)=1.54×πR2×H/( Di3 Ni) = 1.54×π×12×2/( (2/3)3×2.428) = 13.45 s (b) Assumption: flow is turbulent, then Np’=5 P = Np’ ρNi3Di5 ≤ 0.5×1.5×103×(πR2×H) Ni3 ≤ 0.5×1.5×103×(πR2×H)/( Np’ ρDi5