螺旋板式换热器2013年5月

已知数据物性参数传热量M1= 2000 kg/h = 0.556 kg/s

t 1’ = 140 C t 1’’ = 40 C

t m1 = 40 + 0.3×(140-40) = 70 C

参考[2]：表7-8 表7-10 表7-11

υ 1 = 778.525×10^- 6 kg/(m×s)

λ 1 = 0.1085 W /(m×K)

C p1 = 2.23kJ/(kg×K) ρ 1 = 788 kg/m3

Q = M1×Cp1×( t 1’- t 1’’) = 123.988kJ

用试算法确定t2’’

t 2‘ =25 C t m2 = (t2‘+t2‘)/2

(1)假定水的出口温度

t 2‘’ = 35 C 则 t m2 = 30 C

ρ 1 = 995.7 kg/m3 C p2 =4.18kJ/(kg×K)

t 2‘’ = 123.988/( 995.7×4.18×15/3600 )

= 32.14 C 相差太大，重新取值。

(2)假定水的出口温度

t2‘’ = 33 C t m2 = 29 C

ρ 1 = 995.98 kg/m3 C p2 =4.18kJ/(kg×K)

t 2‘’ = 123.988/( 995.98×4.18×15/3600 )

= 32.15 C 相差太大，重新取值。

Q =

123.99kJ

出口温度物性参数选型当量直径(3)假定水的出口温度

t2‘’ = 32.15 C t m2 = 28.6 C

ρ 1 = 996.09 kg/m3 C p2 =4.18kJ/(kg×K)

t2‘’= 123.988/( 996.09×4.18×15/3600 )

= 32.147 C 相差0.003 C，合适。

t m2= ( t1’+t2’’)/2 = ( 25+32.15 )/2 = 28.6 C

参考[2]：表2-1

υ 2 = 794.69×10^-6 kg/(m×s) λ 2 = 0.609 W/(m×

K)

C p2 =4.18kJ/(kg×K) ρ 2 = 996.09 kg/m3

换热器为液—液热交换器，煤油和水相对清洁，选用

Ⅰ型

参考[1]附录F.1

冷却水 w2 = 0.5 m/s 煤油w1 = 0.4 m/s

A2 = m2 /( 3600×w2×ρ 2 ) = 15/(3600×0.5 )

=0.0083 m2

A1 = M1/( 3600×w1×ρ 1 ) = 2000/(3600×0.4×788 )

=0.0018 m2

参考[1]表3.1

H = 0.6m H e = H-2×б = 0.6-2×0.01 = 0.58 m

b2 = A2/H e = 0.0083/0.58 = 0.014 m

t2‘’=

32.15 C

b1 = A1/H e =0.0018/0.58 = 0.003 m

设计内容计算过程结果

参考[1]表3.1

查看产品样本：取b2 = 15 mm b1 = 5 mm

de2 = 2×He×b2/( He+b2 )= 2×0.58×

0.015/(0.58+0.015 )

=0.0292 m

de1 = 2×He×b1/( He+b1 ) = 2×0.58×0.005/(0.58 +0.005 )

=0.0099 m

ω2 = m2 / ( 3600×A2×ρ2 ) = 15 / ( 3600×

0.0083 )

= 0.50 m / s

Re2 = ω2×de2×ρ 2 /υ2 = 0.5×0.0292×996.09 / ( 794.69×10 ^ -6 ) = 18300

Pr2 =υ2×Cp2 / λ2 = 794.69×10^ -6×4.18×10^3 =5.45

ω1 = M1 / ( 3600×A1×ρ1 ) = 2000 /

( 3600×0.0018×788 ) = 0.39 m / s

Re1 = ω1×de1×ρ 1 /υ1 = 0.39×0.0099×788 / ( 778.525×10 ^ -6 ) = 3908 de2 =

=0.0292 m

de1 =

=0.0099 m

设计内容计算过程结果

换热系数传热系数平均温差Pr1 =υ 1 × Cp1 /λ1=778.525×10^ -6×2.23×10^3

=16

α 2 = 0.0397×( λ2 /de2 )×Re2^0.784×Pr2^0.4

= 0.0397×( 0.609 /0.0292 )×18300^0.784×

5.45^0.4

= 3583 W/( m2× C)

α1 = 0.0397×( λ1 /de1 )×Re1^0.784×Pr1^0.4 =

0.0397×( 0.1085 /0.0099 )×3908^0.784×

16^0.4

= 655 W/( m2× C)

查询网络资料，介质为水和煤油，选用Q235卷筒

钢板，厚度4mm，导热系数为λ= 46.5 W/（ m× C )

参考[1]附录C

取r1,s = r2,s = 0.00017 m2× C /W

1 /K = 1 / α1 + 1/α

2 + б/λ + r1,s + r2,s =

=1 /3583 + 1 /655 + 0.004 /46.5+0.00017×2

=2.23×10^ -3

K = 448 W / ( m2 / C )

附图1在题后。

Δt m= [( t1’ - t2’’ ) - ( t1’’ - t2’)] /[ ln ( t1’

α 2 =

3583W

/(m2× C)

α1 =

655W

/(m2× C)

K = 448 W

/( m2/ C )

设计内容计算过程结果