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Energy calculation through inlet and outlet face

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Old   November 29, 2017, 14:31
Default Energy calculation through inlet and outlet face
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I have an Inlet face where fluid at 500C at the rate 1 Kg/s is given as input. Outlet boundary condition is 0 Pa relative pressure. Inlet and outlet face area are 2 m^2.

How do you calculate the Energy fed at Inlet face and Energy exited at oulet face in Joules? may be for each time step?

I am doing transient simulation with time step 20 sec. Total simulation time for 5 hr.
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Old   November 29, 2017, 14:52
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You have to define what kind of energy you are accounting for.

If you want to account for total energy (internal + flow work + kinetic energy + etc), I would use Total Enthalpy, then

Energy at inlet = Time Integral (from 0 -> 5hr) of massFlowInt(Total Enthalpy)@Inlet

Energy at outlet = Time Integral (from 0 -> 5hr) of massFlowInt(Total Enthalpy)@Outlet

Not sure how valuable those integrals are for open systems since I usually work with energy rates; therefore, no need for the time integral.
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Old   November 29, 2017, 15:07
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Quote:
Originally Posted by Opaque View Post
You have to define what kind of energy you are accounting for.

If you want to account for total energy (internal + flow work + kinetic energy + etc), I would use Total Enthalpy, then

Energy at inlet = Time Integral (from 0 -> 5hr) of massFlowInt(Total Enthalpy)@Inlet

Energy at outlet = Time Integral (from 0 -> 5hr) of massFlowInt(Total Enthalpy)@Outlet

Not sure how valuable those integrals are for open systems since I usually work with energy rates; therefore, no need for the time integral.
Thanks for the quick response. I want to account total energy. I do not understand the Time integral part(0 --> 5hr). Till now I was using 20[s]*(massFlowInt(totalEnthalpy)@Inlet for 1 time step and for 5 hours I just times up by 900(total iterations) for Total Input energy for the whole simulation time of 5 hour(18000s). Is it correct?
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Old   December 4, 2017, 17:23
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Difficult to answer w/o every detail about your boundary conditions.

To be able to convert the time integral to a "number of time steps" (not iterations) times "Delta Energy Flow", you must be certain that every variable involved in the calculation of the Delta Energy Flow is constant in time. Are you certain of this ?

My first initial response would be No until a careful review of your setup is done.
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