(1)V
t
=2640 升,t=9.1℃,ρ
20
=0.8212g/cm
3
(用 f
和 0.0011)
查表求得 f=0.00086
V
20
=V
t
[1-f(t-20)]=2640X[1-0.00086(9.1-20)]=2665(升)
M=
V
20
(ρ
20
-0.0011)=2665X(0.8212-0.0011)=2186(kg)
(2)V
t
=140m
3
,t=11.3℃,ρ
20
=0.7384g/cm
3
(用K、
F计算)
查表求得
K=
1.0100+[(1.0100-1.0102)/(0.7400-0.7350)](0.7384-0.7400)+
[(1.0089-1.0100)/(12.0-11.0)](11.3-11)=1.0100+0.000064-
0.00033=1.00973
V
20
=K
×V
t
=1.00973×140=141.362(m
3
)
查表求得 F=0.9985
M=V
20
×ρ
20
×F=141.362X0.7384X0.9985=104.225(t)
5、某原油罐,已知标准体积为 3000 立方米,标准密度为
0.8796 克每立方厘米,原油含水率为 1.7%,求净油量(用
0.0011 法计算)(5分)
V
20
=3000X(1-1.7%)=
2949(
m
3
)
M=
V
20
(ρ
20
-0.0011)=2949X(0.8796-0.0011)=2590.697(t)
0.7350
0.7400
11.0
1.0102
1.0100
12.0
1.0090
1.0089