化学工程基础习题答案(武汉大学__第二版)

化学工程基础习题 第二章． P69 1.解： Pvac = Po − P绝
即13.3 ×10 −3 Pa = 98.7 ×10 −3 = Po − P绝 ⇒ P绝 = 85.4 × 10−3 Pa
Pa = P − Po
= 85.4 × 10−3 Pa − 98.7 × 10−3 Pa = −13.3 ×10−3 Pa
( ρ − ρ H 2 O ) × R + ρ H2 O × 6 d × ( Hg − 6) Lλ ρH2O
u = 2g×
= 2g ×
0.05 × 0.504 = 2.03m ⋅ s −1 6 × 0.02
7
3） P =
25 × 2.03 ×
π × 0.05 2 ×1000 4 = 0.976 kw 102
4）根据静力学基本方程式：
p B + ρ H 2 O g (6 + H ) = ρ Hg gR '+ p0
⇒ p B = ρ Hg gR '+ p0 − ρ H 2 O g(6 + H )
p A + ρ H 2 O gh = p B + ρ H 2 Og 6 + ρ H 2 Og (h − R ) + ρ Hg gR
2.5 × 9.8 = 2.45 × 105 pa 0.01−4
=
忽略出水管路水泵至压力表之间的阻力损失， 则：衡算系统的阻力损失主要为吸入管路的阻力损失：
H f = 0.2 9.8 u=
36 3600 × π 4 (76 ×10−3 )2 = 2.2
H e = 5.0 +
2.45 × 105 2.22 0.2 + + 1000 × 9.8 2 × 9.8 9.8
z1 +
2 p 1 u1 p u2 + = z2 + 2 + 1 + H f ρ g 2g ρ g 2g
其中 z1 = 10m, z 2 = 2m ;
p1 = p2 ; u1 = 0; H f =
∑h
g
f
2 2 则 10 = 2 + u 2 + 16.15u
2 × 9.8
9.8
解得 1）A-A′截面处流量 u = u2
其中： z1 = 0, z1 = 50 m; p1 − p2 = 0
H f = 20 9.8 H e = 50 + P=
20 = 52.05 9.8
H e ⋅ qV ⋅ ρ 52.05 × 36 × 1000 = Fra Baidu bibliotek 8.05kw 102η 102 × 0.6 × 3600
19.解：取贮槽液面为 1-1′截面， 蒸发器内管路出口为 2-2′截面， 由伯努利方程得
0.5 2 22 ∆h + = = 0.19 m 2 × 9.8 2 × 9.8
15.解：选取贮槽液面为 1-1′截面，高位槽液面为 2-2′截面， 由伯努利方程得
z1 +
2 p1 u1 p u2 + + H e = z2 + 2 + 1 + H f ρ g 2g ρ g 2g
其中： z1 = 2 m, z 2 = 10m;u1 = u 2 = 0
H e = 15 +
20.解：1）取贮水池液面为 1-1′截面， 出口管路压力表所在液面为 2-2′截面， 由伯努利方程得
z1 +
2 p 1 u1 p u2 + + He = z2 + 2 + 1 + H f ρ g 2g ρ g 2g
其中， z1 = 0, z 2 = 5.0 m;
5
p1 = 0, p2 = 2.5kgf .cm −2
φ 2π (260 − 35) = L 1 l 25 + 1 l 55 + 1 l 85 n n n 16 20 0.2 25 0.07 55
= 2 × 3.14 × 225 1413 = 0.223 0.788 1.435 0.0139 + 3.94 + 6.214 + + 16 0.2 0.007 1413 = 138.96W ⋅ m−1 10.168 220 + 180 = 200 °C 2
q `= 300 =
R 0 = 3.73 − 0.901 = 2.83°C ⋅ m 2 ⋅ W −1
ex2 解：
φ=
(T1 − T4 ) (T1 − T4 ) = δ δi ∑ i ∑ λi A 2π rm L λi
=
2π L(T1 − T4 ) 2π L( T1 − T4 ) = ( ri +1 − ri ) 1 1 r ∑ ∑ l n i +1 λi ( ri +1 − ri ) λi ri ri +1 ln ri
8.解：对 1-1′截面和 2-2′截面，由伯努利方程得
z1 +
2 p 1 u1 p u2 + = z2 + 2 + 1 ρ g 2g ρ g 2g
其中 z1 = z 2 , p1 = 1mH O = ρ gh1
2
u1 = 0.5m ⋅ s −1 , p1 = ρ gh2 u2 = d12 0.2 u = ( ) 2 × 0.5 = 2.0m ⋅ s −1 2 1 d2 0.1
π −3 2 H e ⋅ qV ⋅ ρ 15.468 × 2 × 4 × (53 ×10 ) ×980 P= = = 0.655kw 102 102
4
17.解：取水池液面为 1-1′截面，高位截面为 2-2′截面， 由伯努利方程得
2 2 p1 u1 p 2 u1 z1 + + + H e = z2 + + +Hf ρ g 2g ρ g 2g
z1 +
2 p1 u1 p u2 + = z2 + 2 + 1 + H f ρ g 2g ρ g 2g
其中： z1 = 0, z 2 = 4.8 m;
p1 = 0, p2 = ?
忽略进水管路水泵中真空表至水泵之间的阻力损失， 则：衡算系统的阻力损失为吸入管路的阻力损失：
H f = 0.2 9.8
6
p2 = − (4.8 +
Hf = λ ⋅ L u2 ⋅ d 2g L 1 u12 ⋅ d1 2 g L 2 u 22 ⋅ d 2 2g
H f 1 = λ1 ⋅
H f 2 = λ2 ⋅
H f2 H f1
L 2 u2 2 ⋅ d2 2g = L u2 λ1 ⋅ 1 ⋅ 1 d1 2 g λ2 ⋅
64µ du ρ
λ = 64 Re =
2.2 2 0.2 + )× 1000× 9.8 = − 49600pa 2 × 9.8 9.8
得真空表的读数为 P vac = 49600 Pa 23.解：1）取低位槽液面为 1-1′截面 高位槽液面为 2-2′截面 由伯努利方程得
z1 +
2 p 1 u1 p u2 + + He = z2 + 2 + 1 + H f ρ g 2g ρ g 2g
u 2 = 2g ×
d p p × ( A − B − zB + z A ) Lλ ρ g ρ g
p A − pB = ( ρHg − ρH 2 O ) × gR + ρH 2 O × g × 6 pA − pB ( ρ Hg − ρ H 2 O ) × R + ρ H 2 O × 6 = ρ H 2O g ρ H 2O
2.解：
π 2 π 2 d1 − d 2 4 de = 4 × 4 = (d1 − d 2 ) = 70 π d1 + π d 2
3.解：对于稳定流态的流体通过圆形管道，有
u2 u1 =
2 d1
d
2 2
若将直径减小一半，即 d 1
⇒ u2 u = 4
1
d
=2
2
即液体的流速为原流速的 4 倍.
1
4.解：
H f2 H f1
H f2 H f1
H f2 H f1
=
= 16
H f 2 = 16H f 1
即产生的能量损失为原来的 16 倍。
2
6.解：1）雷诺数 Re =
ρud µ
其中 ρ = 1000 kg ⋅ m−3 ， u = 1.0m ⋅ s −1
d = 25mm = 25 ×10−3 m
µ = 1cp = 10 −3 Ps ⋅ s
其中， z1 = 0, z 2 = 20 m; p1 = 0, p2 = 0
H f = 5, H e = 20 + 5 = 25
w e = 25 × 9.8 = 245J/kg
2）在管路 A、B 截面间列伯努利方程得：
p A u2 p B u2 A zA + + = zB + + B +Hf ρg 2g ρg 2g pA pB L u2 − = zB − z A + λ ρg ρg d 2g
⇒ p A = p B + ρ H 2 O g 6 − ρ H 2 O gR + ρ Hg gR ⇒ p A = ρHg gR '+ p0 − ρ H 2 O g(6 + H ) + ρ H 2 O g 6 − ρ H 2 O gR + ρ Hg gR ⇒ p A − p0 = ρ Hg gR '− ρ H 2 O gH + ( ρ Hg − ρ H 2 O ) gR = [13.6 × 1.2 − 1× 1+ (13.6 − 1) × 0.04] × 9.8 × 1000 = 1.55 × 105 pa
u2 = 4u1 , L 1 = L2 , d 1 = 2d 2
64 µ L 2 u2 2 ⋅ ⋅ d 2 u 2 ρ d 2 2g = 64µ L 1 u1 2 ⋅ ⋅ d1u 1 ρ d1 2 g 64µ L 2 u2 2 ⋅ ⋅ d2 u 2 ρ d 2 2 g = 1 2 64 µ L1 ( 4 u 2 ) ⋅ ⋅ 1 2 d 2g 2 2 d2 u 2 ρ 4 1 1 1 ⋅ ⋅ 1 2 16 2× 4 1
p1 = p vac = −100mmHg
= − 13.6× 103 × 9.8× 0.1 = − 13332.2p a p2 = 0 2+ −13332.2 19.6 1000 + He = 10 + ( +4× ) ρg 9.8 980 19.6 13332.2 H e = 12.08 + + = 14.08 + 1.388 = 15.468 9.8 980 × g
= 5.0 + 25 + 0.25 + 0.02 = 30.27
P=
H e ⋅ qV ⋅ ρ 30.27 × 36 ×1000 = = 3.0kw 102 102 × 3600
2） P =
H e ⋅ qV ⋅ ρ 3.0 = = 4.3kw 102η 0.7
3）取贮槽液面为 1-1′截面， 水泵吸入管路上真空表处液面为 2-2′截面， 由伯努利方程得
=
Ex4 解：空气的定性温度 T =
200℃时空气的物性参数为： ρ = 0.746 Kg / m3
9
λ = 3.91× 10−2W ⋅ m−1 ⋅ K −1
µ = 2.6 ×10−5 Pa ⋅ s Cp = 1.034 KJ ⋅ Kg−1 ⋅°C
u = 2.17m ⋅ s−1
3
2） q v = Au ρ 其中 A = π d 2 = × 3.14 × (100 ×10− 3 ) 2
= 7.85× 10−3 m 2
u = 2.17 m ⋅ s −1
q v = 7.85 ×10−3 × 2.17 × 3600 = 61.32 m h
3
1 4
1 4
z1 +
2 p 1 u1 p u2 + + He = z2 + 2 + 1 + H f ρ g 2g ρ g 2g
其中， z1 = 0, z1 = 15m;
p1 = 0, p2 = −200 × 10−3 × 13.6 × 103 × 9.8 = −26656pa H f = 120 9.8
120 26656 − = 24.97 9.8 9.8 ×1200 H ⋅ q ⋅ ρ 24.97 × 20 ×1200 P= e V = = 1.632kw 102 102 × 3600
8
第三章 传热过程 p105 ex1 解：
q=
∆t t1 − t 4 1150 − 30 = = δ δ δ 0 . 2 0.1 0.006 ∑R 1 + 2 + 3 + + λ1 λ2 λ3 1.07 0.14 45
=
1120 1120 = = 1243W m2 0.187 − 0.714 − 0.0001 0.901 1120 ⇒ ∑ R + R0 = 3.73°C ⋅ m2 ⋅ W −1 ∑( R + R 0 )
故 Re =
ρud µ
1000 ×1.0 × 25 × 10−3 = 10 −3 = 25000
故为湍流。 2）要使管中水层流，则 Re ≤ 2000 即 Re =
1000 × 25 × 10− 3 m ⋅ u ≤ 2000 10−3
解得 u ≤ 0.08 m ⋅ s −1 7.解：取高位水槽液面为 1-1′， A-A′截面为 2-2′截面，由伯努利方程