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.P 69
1. 解:P vac =P o -P 绝
即13.3⨯10-3Pa =98.7⨯10-3=P o -P 绝
⇒P 绝=85.4⨯10-3Pa
P a =P -P o
=85.4⨯10-3Pa -98.7⨯10-3Pa
=-13.3⨯10-3 Pa
2. 解:
πd 2π
de =4⨯1-d 2
2
πd d =(d 1-d 2) =70
1+π2
3. 解:对于稳定流态的流体通过圆形管道,有
u 2=d 2
1
1d 2 2
若将直径减小一半,即d 1d =2
2
⇒u 2=4
1
即液体的流速为原流速的4倍.
4. 解:
H L u 2
f =λ⋅d ⋅2g
L 2
H 1u 1
f 1=λ1⋅d ⋅
12g
H =λ⋅L 2u 2
2
f 22d ⋅
22g
1
L 2
H λ2u 22⋅⋅
f 2
H =d 22g
f 1L 2λ⋅1u 11d ⋅
12g
λ=64μ
=du ρ
u 2=4u 1, L 1=L 2, d 1=2d 2
64μL 2u 2
H f 2
H =d 2u 2ρ⋅2d ⋅
22g
f 164μL 21u
d ρ⋅d ⋅1
1u 112g
64μL 2
2u 2
H f 2
H =d u ⋅⋅
22ρd 22g
f 164μL (1u 2) 2
2d ⋅1
d ⋅u 22g
242ρ2
H f 2
H =1
=16
f 1⋅2⨯2⋅16
4
H f 2=16H f 1
即产生的能量损失为原来的16倍。
6. 解:1)雷诺数Re =ρud
μ
其中ρ=1000kg ⋅m -3,u =1.0m ⋅s -1
d =25mm =25⨯10-3m
μ=1cp =10-3Ps ⋅s 故Re =ρud
μ
2
1000⨯1.0⨯25⨯10-3
=10-3
=25000
故为湍流。
2)要使管中水层流,则Re ≤2000 1000⨯25⨯10-3m ⋅u ≤2000 即Re =-310
解得u ≤0.08m ⋅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
22p 1u 1p 2u 1z 1++=z 2+++H f ρg 2g ρg 2g
其中z 1=10m , z 2=2m ;
p 1=p 2; u 1=0; H f =∑h
g f 2u 216.15u 2
+则10=2+2⨯9.89.8
解得
1)A-A ′截面处流量u =u 2
u =2.17m ⋅s -1
2)q v =Au ρ 121πd =⨯3.14⨯(100⨯10-3) 2 44其中A =
=7.85⨯10-3m 2
u =2.17m ⋅s -1
q v =7.85⨯10-3⨯2.17⨯3600=61.32m
8. 解:对1-1′截面和2-2′截面,由伯努利方程得
22p 1u 1p 2u 1z 1++=z 2++ ρg 2g ρg 2g 3
其中z 1=z 2, p 1=1m H 2O =ρgh 1
3
u 1=0.5m ⋅s -1, p 1=ρgh 2
d 120.2u 2=2u 1=() 2⨯0.5=2.0m ⋅s -1 d 20.1
0.5222
∆h +==0.19m 2⨯9.82⨯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
22p 1u 1p 2u 1z 1+++H e =z 2+++H f ρg 2g ρg 2g
其中:z 1=2m , z 2=10m ; u 1=u 2=0
p 1=p vac =-100mmHg
=-13.6⨯103⨯9.8⨯0.1=-13332.2p a p 2=0
2+-13332.219.61000+H e =10+(+4⨯) ρg 9.8980 19.613332.2H e =12.08++=14.08+1.388=15.4689.8980⨯g
P =H e ⋅q V ⋅ρ=10215.468⨯2⨯π⨯(53⨯10-3) 2⨯980102=0.655kw
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
22p 1u 1p 2u 1z 1+++H e =z 2+++H f ρg 2g ρg 2g
其中:z 1=0,z 1=50m ; p 1-p 2=0
H f =H e =50+20=52.05 9.8
P =H e ⋅q V ⋅ρ52.05⨯36⨯1000==8.05kw 102η102⨯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
4
22p 1u 1p 2u 1z 1+++H e =z 2+++H f ρg 2g ρg 2g
其中,z 1=0,z 1=15m ;
p 1=0,
p 2=-200⨯10-3⨯13.6⨯103⨯9.8=-26656p a H f = 12026656H e =15+-=24.97 9.89.8⨯1200
H ⋅q ⋅ρ24.97⨯20⨯1200P =e V ==1.632kw 102102⨯3600
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
22p 1u 1p 2u 1z 1+++H e =z 2+++H f ρg 2g ρg 2g
其中,z 1=0,z 2=5.0m ;
p 1=0, p 2=2.5kgf . cm -2
=2.5⨯9.8=2.45⨯105p a -40.01
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 =u = 36
3600⨯(76⨯10-3) 2=2.2
2.45⨯1052.220.2H e =5.0+++1000⨯9.82⨯9.89.8
=5.0+25+0.25+0.02=30.27
P =
2)P H e ⋅q V ⋅ρ30.27⨯36⨯1000==3.0kw 102102⨯3600=H e ⋅q V ⋅ρ3.0==4.3kw 102η0.7
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
5