安全阀计算实例讨论

安全阀计算实例讨论

设计数据

（1）容器数据：

设计压力：1.0 MPa G 设计温度：80 C

外壁不保温安全阀定压：1.0 MPa G

直径：2000mm 切线长度：6500mm

椭圆型封头

（2）介质

10％NaOH溶液（物性按照水计算）

（3）安全阀计算工况：火灾；按有合适的消防设施和良好的下水系统计算；设备允许超压按10%计

计算

1 按照《石油化工设计手册》第四卷P421页公式

(1) 选用计算公式

按照《石油化工设计手册》第四卷P421页公式

G=155400FA^0.82/L

其中

G——火灾工况时安全阀所需的排放量，kg/h

F——容器外壁校正系数，此处取1

A——容器湿表面积，m2

L——容器在泄压工况时的气化前热，kJ/kg

（2）所需泄放量的计算

A=π*2.0*6.5+2.61*2.0^2

=51.26 m2

安全阀泄放时的入口压力1.0*1.1=1.1 MPa G；对应水的气化潜热L=1987.7kj/kg G=155400FA^0.82/L

=155400*1*51.26^0.82/1987.7

=1973.7kg/h

2 用chemCAD算

Device type = Relief valve

Valve type = Balanced valve

Vent model = HEM (Homogeneous Equilibrium Model)

Vessel model = Bubbly model

Design model = API-520/521

Design, Pressure vessels.

Short cut method used for design case.

API 520-521, Adequate firefighting and drainage facilities exist. Horizontal vessel

Head type = Ellipsoidal

Head K factor (dpth / R)=0.5

Vapor Z factor=0.91599

Cp/Cv=1.4177

Vapor MW=18.015

Liquid heat capacity kcal/kmol-C =19.43

Latent heat kJ/kg =1966.3

Relief device analysis:

Set pressureMPa =1.2

Back pressureMPa =0.1

% Overpressure=10

TemperatureC =192.03

Discharge coefficient=0.953

C0 radial distribution parameter=1.2

Kb Backpressure correction factor =1

Exposed aream2 =49.245

Environmental factor=1

Heat ratekJ/h =3.7971e+006

Calculated nozzle aream2 =0.0038775 (For heat model 1)

The following calculation is base on vent area 0.0038775 m2. Calculated vent ratekg/h =1.0208e+005

Calc criticalrate kg/h =1.0208e+005

Calc critical press MPa =1.1477

Nozzle inlet vapor mass fraction= 7.6261e-008

Device inlet densitykg/m3 = 873.54

（3）从上面的计算结果可以看出手算的结果是“1973.7kg/h”，而软件计算的结果是“102080 kg/h ”。问题出在哪里量，哪一个准确，或者应该怎么算。

附件是我所用的CHEMCAD计算文件和报告。

问题可能在这里：火灾模型选用的是API520/521，API520/521认为泄放时排出单纯为气相；而平衡流量模型选择了HEM，HEM是用于计算两相流的。因此平衡流量模型应该选用single phase vapor。我按照楼主的条件计算了下，如果物性只按水计算，结果：

Relief Device Sizing for Stream 1

Device type = Relief valve

Valve type = Balanced valve

Vent model = All vapor

Design model = API-520/521

Design, Pressure vessels.

API 520-521, Adequate firefighting and drainage facilities exist. Horizontal vessel

Head type = Ellipsoidal

Head K factor (dpth / R) = 0.5

Vessel dimensions:

Diameter m = 2

Length (T to T) m = 6.5

Vessel volume m3 = 22.515

Liquid level m = 2

Initial vapor volume fraction = 1e-005

Height above ground m = 0

Fluid properties:

Vapor mass kg = 0.0013803

Liquid mass kg = 19767

Vapor density kg/m3 = 6.1307

Liquid density kg/m3 = 877.97

Surface tension N/m = 0.040495

Liquid viscosity N-s/m2 = 0.00014156

Vapor Z factor = 0.92065

Cp/Cv = 1.4111

Vapor MW = 18.015

Liquid heat capacity kJ/kg-K = 4.4947

Latent heat kJ/kg = 1981.2

Relief device analysis:

Set pressure MPa-G = 1

Back pressure MPa-G = 0.1

[wiki]%[/wiki] Overpressure = 10

Temperature C = 188.08

Discharge coefficient = 0.975

Kb Backpressure corr. factor = 1

Exposed area m2 = 49.245

Environmental factor = 1

Heat rate kJ/h = 3.7971e+006

Calculated nozzle area m2 = 0.00029276 (For heat model 1)

CHEMCAD 5.6.4 Page 2

Job Name: relief valve Date: 03/21/2008 Time: 11:48:55

Selected valve type: 1.5G2.5

Actual nozzle area m2 = 0.00032452

The following calculation is base on vent area 0.00032452 m2.

Calculated vent rate kg/h = 2122.4

Calc critical rate kg/h = 2122.4

Calc critical press MPa-G = 0.53108

Nozzle inlet vap. mass fraction = 1

Device inlet density kg/m3 = 6.1307

Nozzle inlet vap. vol. fraction = 1