无法在OpenModelica中评估直管的传热系数
我正在尝试适应 Modelica.Fluid.Dissipation
。我想使用函数 kc_overall
评估直管的传热系数。我尝试遵循UsersGuide
示例。我不确定我是否理解如何编写它以及我必须使用哪些输入。这是我的代码:
model Heat_tranfer_calcul
Modelica.SIunits.MassFlowRate M_FLOW=0.3 "input mass flow rate";
Modelica.Fluid.Dissipation.HeatTransfer.StraightPipe.kc_overall_IN_con IN_con(d_hyd=13e-3,L=15);
Modelica.Fluid.Dissipation.HeatTransfer.StraightPipe.kc_overall_IN_var IN_var(cp=4184,eta=8.9e-4,lambda=0.6,rho=1000) ;
equation
Modelica.Fluid.Dissipation.HeatTransfer.StraightPipe.kc_overall(IN_con,IN_var,M_FLOW)
end Heat_tranfer_calcul;
我还尝试直接从 Modelica.Media.Water.StandardWater 获取流体属性。 如果有人可以帮助我了解该功能的工作原理,那将非常有帮助。
马克西姆
I'm trying to get used to the Modelica.Fluid.Dissipation
. I want to evaluate the heat transfer coefficient of a straight pipe using the function kc_overall
. I tried to followthe UsersGuide
example. I'm not sure I understood how to write it and which inputs I have to use. Here's my code:
model Heat_tranfer_calcul
Modelica.SIunits.MassFlowRate M_FLOW=0.3 "input mass flow rate";
Modelica.Fluid.Dissipation.HeatTransfer.StraightPipe.kc_overall_IN_con IN_con(d_hyd=13e-3,L=15);
Modelica.Fluid.Dissipation.HeatTransfer.StraightPipe.kc_overall_IN_var IN_var(cp=4184,eta=8.9e-4,lambda=0.6,rho=1000) ;
equation
Modelica.Fluid.Dissipation.HeatTransfer.StraightPipe.kc_overall(IN_con,IN_var,M_FLOW)
end Heat_tranfer_calcul;
I also tried to get the fluid properties directly from the Modelica.Media.Water.StandardWater
.
If someone could help me to uderstand how the function works that would be really helpful.
Maxime
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函数
kc_overall
仅接受两个参数——记录IN_con
和IN_var
。质量流量必须通过IN_var
指定,因此您的代码应该是:顺便说一句,使用小写字母作为变量名称是很好的 Modelica 编码习惯,即
m_flow
而不是M_FLOW
。The function
kc_overall
only takes two arguments — the recordsIN_con
andIN_var
. The mass flow rate must be specified throughIN_var
so your code should be:By the way, it is good Modelica coding practice to use lowercase letters for variable names, that is
m_flow
instead ofM_FLOW
.