冶金传输原理(朱光俊版,第二章)

2-13

(1)q v =V.A V =q v A =10060∗60⁄π4×0.152=1.57m s ⁄

(2) q v =V.A V

=q v A =270060×60⁄0.3×0.4=6.25m s ⁄ A=q v V =270060×60⁄25

=0.03m 2 0.03/0.2=0.15 m 高度为150mm 2-15

q m =q v ×ρ ,ρ同 ,q m 也相同

V BC =q m ρ×A BC =q v 1×ρ1A BC ×ρ2=q v 1A BC =10×0.02520.052=2.5m s ⁄ V AB =q m 1ρ×A AB =q v 1×ρ1A AB ×ρ=q v 1A BA =10×0.02520.1×0.1=0.625m s ⁄

又因为q m =ρ×q v, ρ相同，q v 也相同

q v =q v 1=V CD ×A CD =10×0.0252×

π4=4.906×10−3m 3s ⁄ 2-17

ρgz 1+p 1+ρv 122=ρgz 2+p 2+ρv 222

1×10×5+p 1+12×32=1×10×0+p 2+12

×v 22 ,p 1=p 2,,v 2=√109m s ⁄ q v 1=q v 2,v 1A 1=v 2A 2

3×

π4×0.22=√109×π4×d 22,d 2=0.108m 1-18

q v 1=v 1×A ,0.2=π4

×d 2×V 1 , V 1=6.36m s ⁄ 用法，v 2=1.59m s ⁄,,,则可求 A,B 处的能量。 A 出能量，1×10×0+68.6×103+0.5×6.372=68620.28 J

B出能量，1×10×1+39.2×103+0.5×1.592=39211.27 J