Timeline for Model of a heating system in an industrial hall
Current License: CC BY-SA 3.0
10 events
when toggle format | what | by | license | comment | |
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Apr 4, 2016 at 14:42 | vote | accept | Matias | ||
Apr 3, 2016 at 20:15 | answer | added | Gert | timeline score: 0 | |
Apr 3, 2016 at 19:21 | comment | added | Gert | I'll formulate an answer. | |
Apr 3, 2016 at 19:18 | comment | added | Gert | For a thin-walled pipe transporting a liquid, with convective losses only, the approx. model is $T_{out}=T_{amb}+(T_{In}-T_{amb})e^{-\alpha L}$ with $\alpha=\frac{\pi Dh}{\dot{m}c_p}$ and $h$ the Heat Transfer Coefficient of the pipe. The higher the mass flow the more the system is invariant to heat withdrawn by the machine-tool. | |
Apr 3, 2016 at 18:38 | comment | added | Matias | @Gert thank you very much! What if I adopt the model: $ heatLossPipe = function(constant, Tin , Tambient)$ and I want to control eiher the massflow or the temperature (if the required heat of the machine change). | |
Apr 3, 2016 at 13:52 | comment | added | Gert | $\dot{Q_{mt}}=\dot{m}c_p(T_{in}-T_{out})$ is the power used by the machine tool. Since as there are no heat losses in the pipes, that power is what the heat plant needs to deliver to maintain all temperatures constant. Since as mass throughput $\dot{m}$ is constant, $\dot{Q_{hp}}=\dot{m}c_p(T_{out}-T_{in})$. There's nothing to model here, really. Heat out = Heat in. | |
Apr 3, 2016 at 13:33 | comment | added | Matias | @Gert Thank you for the comment: If the plant increases the temperature, it take some time till the temperature increases at the machine-tool (i call this "delay"). For simplificatoins, the pipes are adiabatic. I want to describe the system in a very simplified way. | |
Apr 3, 2016 at 13:17 | comment | added | Gert | What do you understand by 'the delay of the pipe'? W/o insulation both pipes will lose heat, of course. | |
Apr 3, 2016 at 10:59 | review | First posts | |||
Apr 3, 2016 at 16:51 | |||||
Apr 3, 2016 at 10:58 | history | asked | Matias | CC BY-SA 3.0 |